Instructors’ Solution Manual
Introduction to Quantum Mechanics, 3rd ed.
David Griffiths, Darrell Schroeter
Reed College
August 3, 2018
2
Contents
1 The Wave Function
4
2 The Time-Independent Schrödinger Equation
16
3 Formalism
78
4 Quantum Mechanics in Three Dimensions
109
5 Identical Particles
168
6 Symmetries and Conservation Laws
197
7 Time-Independent Perturbation Theory
235
8 The Variational Principle
301
9 The WKB Approximation
333
10 Scattering
354
11 Quantum Dynamics
372
12 Afterword
420
A Linear Algebra
427
3
Preface
These are our own solutions to the problems in Introduction to Quantum Mechanics, 3rd ed. We have
made every effort to insure that they are clear and correct, but errors are bound to occur, and for this
we apologize in advance. We would like to thank the many people who pointed out mistakes in the solution manual for the second edition, and encourage anyone who finds defects in this one to alert us (griffith@reed.edu or schroetd@reed.edu). We especially thank Kenny Scott, Alain Thys, and Sergei Walter,
who found many errors in the 2nd edition solution manual. We maintain a list of errata on the web page
(http://academic.reed.edu/physics/faculty/griffiths.html), and incorporate corrections in the manual itself from
time to time. We also thank our students for many useful suggestions, and Neelaksh Sadhoo, who did much of
the typesetting for the second edition.
David Griffiths and Darrell Schroeter
4
CHAPTER 1. THE WAVE FUNCTION
Chapter 1
The Wave Function
Problem 1.1
(a)
hji2 = 212 = 441.
hj 2 i =
=
1 X 2
1 2
j N (j) =
(14 ) + (152 ) + 3(162 ) + 2(222 ) + 2(242 ) + 5(252 )
N
14
1
6434
(196 + 225 + 768 + 968 + 1152 + 3125) =
= 459.571.
14
14
j
14
15
16
22
24
25
(b)
σ2 =
=
σ=
∆j = j − hji
14 − 21 = −7
15 − 21 = −6
16 − 21 = −5
22 − 21 = 1
24 − 21 = 3
25 − 21 = 4
1 X
1 (∆j)2 N (j) =
(−7)2 + (−6)2 + (−5)2 · 3 + (1)2 · 2 + (3)2 · 2 + (4)2 · 5
N
14
260
1
(49 + 36 + 75 + 2 + 18 + 80) =
= 18.571.
14
14
√
18.571 = 4.309.
(c)
hj 2 i − hji2 = 459.571 − 441 = 18.571.
[Agrees with (b).]
CHAPTER 1. THE WAVE FUNCTION
5
Problem 1.2
(a)
Z h
2
1
1
x √ dx = √
2 hx
2 h
2
hx i =
0
h2
−
σ = hx i − hxi =
5
2
2
2
2 5/2
x
5
h
=
0
h2
.
5
2
h
4 2
2h
=
h ⇒ σ = √ = 0.2981h.
3
45
3 5
(b)
√ x+
1 √
1
1
√ √ dx = 1 − √ (2 x)
=1− √
x+ − x− .
2 hx
2 h
h
x−
Z x+
P =1−
x−
x+ ≡ hxi + σ = 0.3333h + 0.2981h = 0.6315h;
P =1−
√
0.6315 +
√
x− ≡ hxi − σ = 0.3333h − 0.2981h = 0.0352h.
0.0352 = 0.393.
Problem 1.3
(a)
Z ∞
1=
2
Ae−λ(x−a) dx.
Let u ≡ x − a, du = dx, u : −∞ → ∞.
−∞
Z ∞
1=A
e
r
−λu2
du = A
−∞
r
π
λ
⇒ A=
λ
.
π
(b)
Z ∞
hxi = A
2
xe−λ(x−a) dx = A
Z ∞
−∞
−∞
Z ∞
=A
−λu
ue
2
Z ∞
du + a
−∞
hx2 i = A
2
(u + a)e−λu du
Z ∞
e
−λu2
−∞
r π
= a.
du = A 0 + a
λ
2
x2 e−λ(x−a) dx
−∞
Z ∞
=A
2
u2 e−λu du + 2a
Z ∞
−∞
1
=A
2λ
r
2
ue−λu du + a2
−∞
π
+ 0 + a2
λ
σ 2 = hx2 i − hxi2 = a2 +
r
π
λ
Z ∞
−∞
1
= a2 +
.
2λ
1
1
− a2 =
;
2λ
2λ
1
σ=√ .
2λ
2
e−λu du
6
CHAPTER 1. THE WAVE FUNCTION
(c)
l(x)
A
x
a
Problem 1.4
(a)
|A|2
1= 2
a
Z a
Z b
|A|2
2
(
2
2
(b − x) dx = |A|
2
(b − a) a
r
b−a
3
2 a
2b
= |A|
+
⇒ A=
.
= |A|
3
3
3
b
x dx +
0
1
a2
3 a
b)
x
1
(b − x)3
+
−
3 0 (b − a)2
3
a
(b)
^
A
a
b
x
(c) At x = a.
(d)
Z a
|A|2
|Ψ| dx = 2
P =
a
0
2
Z a
a
x dx = |A| = .
3
b
0
2
2a
P = 1 if b = a, X
P = 1/2 if b = 2a. X
(e)
Z a
Z b
1
1
2
x3 dx +
x(b
−
x)
dx
x|Ψ|2 dx = |A|2 2
a 0
(b − a)2 a
(
4 a
b)
2
x
1
x3
x4
3 1
2x
+
b
− 2b +
=
b a2
4 0 (b − a)2
2
3
4 a
3
=
a2 (b − a)2 + 2b4 − 8b4 /3 + b4 − 2a2 b2 + 8a3 b/3 − a4
2
4b(b − a)
4
2 3
1
3
b
2a + b
2 2
−a b + a b =
(b3 − 3a2 b + 2a3 ) =
.
=
2
2
4b(b − a)
3
3
4(b − a)
4
Z
hxi =
CHAPTER 1. THE WAVE FUNCTION
7
Problem 1.5
(a)
Z
1=
2
|Ψ| dx = 2|A|
2
Z ∞
e
−2λx
0
−2λx ∞
e
|A|2
dx = 2|A|
=
;
−2λ 0
λ
2
A=
√
λ.
(b)
Z
hxi =
2
x|Ψ| dx = |A|
2
Z ∞
xe−2λ|x| dx = 0.
[Odd integrand.]
−∞
2
hx i = 2|A|
2
Z ∞
2 −2λx
x e
0
2
1
=
.
dx = 2λ
(2λ)3
2λ2
(c)
σ 2 = hx2 i − hxi2 =
1
;
2λ2
√
1
σ=√ .
2λ
√
|Ψ(±σ)|2 = |A|2 e−2λσ = λe−2λ/ 2λ = λe− 2 = 0.2431λ.
|^| 2
h
.24h
x
<m
+m
e−2λx dx = 2λ
−2λx ∞
√
e
= e−2λσ = e− 2 = 0.2431.
−2λ σ
Probability outside:
Z ∞
2
|Ψ|2 dx = 2|A|2
σ
Z ∞
σ
Problem 1.6
For integration by parts, the differentiation has to be with respect to the integration variable – in this case the
differentiation is with respect to t, but the integration variable is x. It’s true that
∂x 2
∂
∂
∂
(x|Ψ|2 ) =
|Ψ| + x |Ψ|2 = x |Ψ|2 ,
∂t
∂t
∂t
∂t
but this does not allow us to perform the integration:
Z b
∂
x |Ψ|2 dx =
∂t
a
Z b
∂
b
(x|Ψ|2 )dx 6= (x|Ψ|2 ) a .
∂t
a
8
CHAPTER 1. THE WAVE FUNCTION
Problem 1.7
From Eq. 1.33, dhpi
dt = −i~
∂
∂t
Ψ
∗ ∂Ψ
∂x
R ∂
∂t
∂2Ψ
∂2Ψ
Ψ∗ ∂Ψ
∂x dx. But, noting that ∂x∂t = ∂t∂x and using Eqs. 1.23-1.24:
i
i
∂Ψ∗ ∂Ψ
∂Ψ
i~ ∂ 2 Ψ∗
i~ ∂ 2 Ψ
∗ ∂
∗ ∂Ψ
∗ ∂
+ VΨ
− VΨ
=
+Ψ
= −
+Ψ
∂t ∂x
∂x ∂t
2m ∂x2
~
∂x
∂x 2m ∂x2
~
2 ∗
3
∂
Ψ
∂Ψ
i~
i
∂
Ψ
∂Ψ
∂
=
Ψ∗ 3 −
+
V Ψ∗
− Ψ∗ (V Ψ)
2m
∂x
∂x2 ∂x
~
∂x
∂x
The first term integrates to zero, using integration by parts twice, and the second term can be simplified to
∗ ∂Ψ
∗ ∂V
2 ∂V
V Ψ∗ ∂Ψ
∂x − Ψ V ∂x − Ψ ∂x Ψ = −|Ψ| ∂x . So
dhpi
= −i~
dt
Z
i
∂V
∂V
−|Ψ|2
dx = h−
i.
~
∂x
∂x
QED
Problem 1.8
2
2
~ ∂ Ψ
Suppose Ψ satisfies the Schrödinger equation without V0 : i~ ∂Ψ
∂t = − 2m ∂x2 + V Ψ. We want to find the solution
2
2
∂Ψ0
∂
Ψ
~
0
Ψ0 with V0 : i~ ∂t = − 2m
∂x2 + (V + V0 )Ψ0 .
Claim: Ψ0 = Ψe−iV0 t/~ .
i
−iV t/~ h ~2 ∂ 2 Ψ
iV0
∂Ψ −iV0 t/~
0
0
Proof: i~ ∂Ψ
=
i~
e
+
i~Ψ
−
e
=
−
+
V
Ψ
e−iV0 t/~ + V0 Ψe−iV0 t/~
2
∂t
∂t
~
2m ∂x
2
2
~ ∂ Ψ0
= − 2m
∂x2 + (V + V0 )Ψ0 .
QED
This has no effect on the expectation value of a dynamical variable, since the extra phase factor, being independent of x, cancels out in Eq. 1.36.
Problem 1.9
(a)
2
Z ∞
1 = 2|A|
−2amx2 /~
e
21
dx = 2|A|
0
2
r
π
= |A|2
(2am/~)
r
π~
;
2am
A=
2am
π~
1/4
.
(b)
∂Ψ
= −iaΨ;
∂t
∂Ψ
2amx
=−
Ψ;
∂x
~
∂2Ψ
2am
=−
∂x2
~
2
2
∂Ψ
Ψ+x
∂x
2am
=−
~
~ ∂ Ψ
Plug these into the Schrödinger equation, i~ ∂Ψ
∂t = − 2m ∂x2 + V Ψ:
~2
2am
2amx2
V Ψ = i~(−ia)Ψ +
−
1−
Ψ
2m
~
~
2amx2
Ψ = 2a2 mx2 Ψ, so
= ~a − ~a 1 −
~
V (x) = 2ma2 x2 .
2amx2
1−
~
Ψ.
CHAPTER 1. THE WAVE FUNCTION
9
(c)
Z ∞
hxi =
x|Ψ|2 dx = 0.
[Odd integrand.]
Z ∞
1
dx = 2|A| 2
2 (2am/~)
−∞
2
hx i = 2|A|
2
2 −2amx2 /~
x e
0
hpi = m
r
2
π~
~
=
.
2am
4am
dhxi
= 0.
dt
2
Z
~ ∂
∂2Ψ
hp i = Ψ
Ψdx = −~2 Ψ∗ 2 dx
i ∂x
∂x
Z
Z
Z
2
2am
2amx
2am
2
2
2
2
∗
|Ψ| dx −
x |Ψ| dx
= −~
Ψ −
1−
Ψ dx = 2am~
~
~
~
2am 2
2am ~
1
= 2am~ 1 −
hx i = 2am~ 1 −
= 2am~
= am~.
~
~ 4am
2
2
Z
∗
(d)
σx2 = hx2 i − hxi2 =
σx σp =
q
~
4am
√
~
=⇒ σx =
4am
r
~
;
4am
σp2 = hp2 i − hpi2 = am~ =⇒ σp =
From Math Tables: π = 3.141592653589793238462643 · · ·
P (0) = 0
P (5) = 3/25
P (1) = 2/25
P (6) = 3/25
P (2) = 3/25
P (7) = 1/25
P (3) = 5/25
P (8) = 2/25
P (4) = 3/25
P (9) = 3/25
In general, P (j) = NN(j) .
(b) Most probable: 3.
am~.
am~ = ~2 . This is (just barely) consistent with the uncertainty principle.
Problem 1.10
(a)
√
Median: 13 are ≤ 4, 12 are ≥ 5, so median is 4.
1
Average: hji = 25
[0 · 0 + 1 · 2 + 2 · 3 + 3 · 5 + 4 · 3 + 5 · 3 + 6 · 3 + 7 · 1 + 8 · 2 + 9 · 3]
1
= 25
[0 + 2 + 6 + 15 + 12 + 15 + 18 + 7 + 16 + 27] = 118
25 = 4.72.
1
(c) hj 2 i = 25
[0 + 12 · 2 + 22 · 3 + 32 · 5 + 42 · 3 + 52 · 3 + 62 · 3 + 72 · 1 + 82 · 2 + 92 · 3]
1
= 25
[0 + 2 + 12 + 45 + 48 + 75 + 108 + 49 + 128 + 243] = 710
25 = 28.4.
√
σ 2 = hj 2 i − hji2 = 28.4 − 4.722 = 28.4 − 22.2784 = 6.1216; σ = 6.1216 = 2.474.
10
CHAPTER 1. THE WAVE FUNCTION
Problem 1.11
(a)
1
mv 2 + V = E
2
r
→
v(x) =
2
(E − V (x)) .
m
(b)
Z b
T =
a
r
1
q
2
m
E − 12 kx2
dx =
m
k
Z b
1
p
dx.
(2E/k) − x2
a
p
Turning points: v = 0 ⇒ E = V = 12 kb2 ⇒ b = 2E/k; a = −b.
r
r Z b
r
b
1
m
m
m
−1 x
√
=2
T =2
dx = 2
sin
sin−1 (1)
2
2
k 0
k
b
k
0
b −x
r r
m π
m
=2
=π
.
k 2
k
1
ρ(x) =
π
pmq 2
k
Z b
ρ(x) dx =
a
m
2
π
Z b
E − 12 kx2 1
√
0
1
= π √ b2 − x 2 .
ρ(x)
1
2 π
= 1. X
dx =
π 2
b2 − x 2
-b
b
x
(c) hxi = 0.
Z
x2
x2
2 b
√
dx =
dx
π 0
b2 − x2
b2 − x2
−b
x b
2
xp 2
b2
b2
b2 π
b2
E
=
−
b − x2 +
sin−1
=
sin−1 (1) =
=
=
.
π
2
2
b
π
π 2
2
k
0
hx2 i =
1
π
Z b
√
p
p
b
σx = hx2 i − hxi2 = hx2 i = √ =
2
r
E
.
k
Problem 1.12
(a)
dt
|dt/dp| dp
=
T
T
where dt is now the time it spends with momentum in the range dp (dt is intrinsically positive, but
dp/dt = F = −kx runs negative—hence the absolute value). Now
s p2
1 2
2
p2
+ kx = E ⇒ x = ±
E−
,
2m 2
k
2m
ρ(p) dp =
CHAPTER 1. THE WAVE FUNCTION
11
so
ρ(p) =
π
pm
k
k
1
r 2
k
p2
E − 2m
=
1
π
p
2mE − p2
=
1
π
p
c2 − p2
,
√
where c ≡ 2mE. This is the same as ρ(x) (Problem 1.11(b)), with c in place of b (and, of course, p in
place of x).
(b) From Problem 1.11(c), hpi = 0, hp2 i =
E√
mE =
k
uncertainty principle!
r
r
(c) σx σp =
√
c2
c
, σp = √ = mE .
2
2
m
E
E = . If E ≥ 12 ~ω, then σx σp ≥ 12 ~, which is precisely the Heisenberg
k
ω
x@t_D := Cos@tD
snapshots = Table@x@ p RandomReal@jDD, 8j, 10 000<D
Problem 1.13
8- 0.977661, 0.371147, - 0.0389023, - 0.0257086, - 0.968682, - 0.681951, - 0.874704,
x@t_D :=
Cos@tD 0.584236, - 0.974667, 0.10798, 0.996311, - 0.767106, 0.926913<
Ü9987á,
snapshots = Table@x@ p RandomReal@jDD, 8j, 10 000<D
Histogram@snapshots, 100, "PDF", PlotRange Æ 80, 2<D
8- 0.977661, 0.371147, - 0.0389023, - 0.0257086, - 0.968682, - 0.681951, - 0.874704,
2.0
Ü9987á,
0.584236, - 0.974667, 0.10798, 0.996311, - 0.767106, 0.926913<
1.5
Histogram@snapshots, 100, "PDF", PlotRange Æ 80, 2<D
2.0
1.0
1.5
0.5
1.0
0.5
-0.5
0.0
0.5
1.0
1
r@x_D :=
p
1 - x2
Plot@ r@xD, 8x, - 1, 1<, PlotRange -> 80, 2<D
-0.5
0.0 2.0
0.5
1
r@x_D :=
p
1 - x2
1.5
Plot@ r@xD, 8x, - 1, 1<, PlotRange -> 80, 2<D
1.0
1.5
1.0
0.5
12
CHAPTER 1. THE WAVE FUNCTION
-0.5
0.0
0.5
1.0
1
r@x_D :=
1 - x2
p
Plot@ r@xD, 8x, - 1, 1<, PlotRange -> 80, 2<D
2.0
1.5
1.0
0.5
2
1p13.nb
-1.0
-0.5
0.5
1.0
Show@Histogram@snapshots, 100, "PDF", PlotRange Æ 80, 2<D,
Plot@ r@xD, 8x, - 1, 1<, PlotRange -> 80, 2<DD
2.0
1.5
1.0
0.5
-0.5
0.0
0.5
|Ψ(x, t)|2 dx,
so dPdtab =
1.0
Problem 1.14
(a) Pab (t) =
Rb
a
Rb ∂
a ∂t
|Ψ|2 dx. But (Eq. 1.25):
∂|Ψ|2
∂
i~
∂Ψ∗
∂
∗ ∂Ψ
=
−
Ψ
= − J(x, t).
Ψ
∂t
∂x 2m
∂x
∂x
∂x
dPab
∴
=−
dt
Z b
∂
b
J(x, t)dx = − [J(x, t)]|a = J(a, t) − J(b, t).
∂x
a
QED
Probability is dimensionless, so J has the dimensions 1/time, and units seconds−1 .
2
∗
df
−iat df iat
(b) Here Ψ(x, t) = f (x)e−iat , where f (x) ≡ Ae−amx /~ , so Ψ ∂Ψ
= f dx
,
∂x = f e
dx e
df
and Ψ∗ ∂Ψ
∂x = f dx too, so J(x, t) = 0.
CHAPTER 1. THE WAVE FUNCTION
13
Problem 1.15
Use Eqs. [1.23] and [1.24], and integration by parts:
d
dt
Z ∞
Z ∞ ∗
∂
∂Ψ1
∂Ψ2
(Ψ∗1 Ψ2 ) dx =
Ψ2 + Ψ∗1
dx
∂t
∂t
−∞ ∂t
−∞
Z ∞ i
i
−i~ ∂ 2 Ψ∗1
i~ ∂ 2 Ψ2
∗
∗
+
−
=
V
Ψ
Ψ
+
Ψ
V
Ψ
dx
2
2
1
1
2m ∂x2
~
2m ∂x2
~
−∞
Z ∞ 2 ∗
2
∂ Ψ1
i~
∗ ∂ Ψ2
Ψ2 − Ψ1
dx
= −
2m −∞ ∂x2
∂x2
"
#
Z ∞
Z ∞
∞
∞
i~ ∂Ψ∗1
∂Ψ∗1 ∂Ψ2
∂Ψ∗1 ∂Ψ2
∗ ∂Ψ2
= −
−
+
Ψ2
dx − Ψ1
dx = 0. QED
2m ∂x
∂x −∞
−∞ ∂x ∂x
−∞ ∂x ∂x
−∞
Ψ∗1 Ψ2 dx =
−∞
Z ∞
Problem 1.16
(a)
a
3
x5
4
2x
1 = |A|
+
a − x dx = 2|A|
a − 2a x + x dx = 2|A| a x − 2a
3
5 0
−a
0
r
2 1
16 5 2
15
= 2|A|2 a5 1 − +
=
a |A| , so A =
.
3 5
15
16a5
2
Z a
2
2 2
2
Z a
4
2 2
4
2
(b)
Z a
hxi =
x|Ψ|2 dx = 0.
(Odd integrand.)
−a
(c)
hpi =
~ 2
A
i
Z a
a2 − x2
−a
d 2
a − x2 dx = 0.
|dx {z
}
(Odd integrand.)
−2x
Since we only know hxi at t = 0 we cannot calculate dhxi/dt directly.
(d)
hx2 i = A2
Z a
−a
2
x2 a2 − x2 dx = 2A2
Z a
a4 x2 − 2a2 x4 + x6 dx
0
a
3
5
15
x7
15 7 1 2 1
2x
4x
=2
a
−
2a
+
=
a
−
+
16a5
3
5
7 0
8a5
3 5 7
2 35 − 42 + 15
15a
a2 8
a2
=
=
· =
.
8
8 7
7
3
·5
·7
14
CHAPTER 1. THE WAVE FUNCTION
(e)
hp2 i = −A2 ~2
Z a
a 2 − x2
−a
d2 2
a − x2 dx = 2A2 ~2 2
2
|dx {z
}
Z a
a2 − x2 dx
0
−2
15 2
=4·
~
16a5
a
x3
15~2 3 a3
15~2 2
5 ~2
2
a x−
=
a
−
·
.
=
=
3 0
4a5
3
4a2 3
2 a2
(f )
r
σx =
p
σp =
p
hx2 i − hxi2 =
1 2
a
a = √ .
7
7
(g)
r
hp2 i − hpi2 =
5 ~2
=
2 a2
r
5~
.
2a
(h)
a
σx σp = √ ·
7
r
5~
=
2a
r
5
~=
14
r
~
10 ~
> .X
7 2
2
Problem 1.17
∗
2
∗
i~ ∂ Ψ
i ∗ ∗
(a) Eq. 1.24 now reads ∂Ψ
∂t = − 2m ∂x2 + ~ V Ψ , and Eq. 1.25 picks up an extra term:
∂
i
i
2Γ 2
|Ψ|2 = · · · + |Ψ|2 (V ∗ − V ) = · · · + |Ψ|2 (V0 + iΓ − V0 + iΓ) = · · · −
|Ψ| ,
∂t
~
~
~
R
2Γ ∞
2Γ
2
and Eq. 1.27 becomes dP
QED
dt = − ~ −∞ |Ψ| dx = − ~ P .
(b)
dP
2Γ
2Γ
~
= − dt =⇒ ln P = − t + constant =⇒ P (t) = P (0)e−2Γt/~ , so τ =
.
P
~
~
2Γ
Problem 1.18
√
h
h2
>d ⇒ T <
.
3mkB d2
3mkB T
(a) Electrons (m = 9.1 × 10−31 kg):
T <
(6.6 × 10−34 )2
= 1.3 × 105 K.
3(9.1 × 10−31 )(1.4 × 10−23 )(3 × 10−10 )2
Silicon nuclei (m = 28mp = 28(1.7 × 10−27 ) = 4.8 × 10−26 kg):
T <
(6.6 × 10−34 )2
= 2.4 K.
3(4.8 × 10−26 )(1.4 × 10−23 )(3 × 10−10 )2
CHAPTER 1. THE WAVE FUNCTION
15
(b) P V = N kB T ; volume occupied by one molecule (N = 1, V = d3 ) ⇒ d = (kB T /P )1/3 .
h2
T <
3mkB
P
kB T
2/3
⇒ T
5/3
h2 P 2/3
1
<
⇒T <
5/3
3m k
kB
B
h2
3m
3/5
For helium (m = 4mp = 6.8 × 10−27 kg) at 1 atm = 1.0 × 105 N/m2 :
1
T <
(1.4 × 10−23 )
(6.6 × 10−34 )2
3(6.8 × 10−27 )
3/5
(1.0 × 105 )2/5 = 2.8 K.
For atomic hydrogen (m = mp = 1.7 × 10−27 kg) with d = 0.01 m:
T <
(6.6 × 10−34 )2
= 6.2 × 10−14 K.
3(1.7 × 10−27 )(1.4 × 10−23 )(10−2 )2
At 3 K it is definitely in the classical regime.
P 2/5 .
16
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
Chapter 2
The Time-Independent Schrödinger
Equation
Problem 2.1
(a)
Ψ(x, t) = ψ(x)e−i(E0 +iΓ)t/~ = ψ(x)eΓt/~ e−iE0 t/~ =⇒ |Ψ|2 = |ψ|2 e2Γt/~ .
Z ∞
Z ∞
2
2Γt/~
|Ψ(x, t)| dx = e
|ψ|2 dx.
−∞
−∞
The second term is independent of t, so if the product is to be 1 for all time, the first term (e2Γt/~ ) must
also be constant, and hence Γ = 0. QED
2
2
~ d ψ
(b) If ψ satisfies Eq. 2.5, − 2m
dx2 + V ψ = Eψ, then (taking the complex conjugate and noting that V and
2
2
∗
~ d ψ
∗
∗
∗
E are real): − 2m
dx2 + V ψ = Eψ , so ψ also satisfies Eq. 2.5. Now, if ψ1 and ψ2 satisfy Eq. 2.5, so
too does any linear combination of them (ψ3 ≡ c1 ψ1 + c2 ψ2 ):
~2
d2 ψ1
d2 ψ2
~2 d2 ψ3
+ V ψ3 = −
c1
+ c2
−
+ V (c1 ψ1 + c2 ψ2 )
2m dx2
2m
dx2
dx2
~2 d2 ψ1
~2 d2 ψ2
= c1 −
+ V ψ1 + c2 −
+ V ψ2
2m dx2
2m dx2
= c1 (Eψ1 ) + c2 (Eψ2 ) = E(c1 ψ1 + c2 ψ2 ) = Eψ3 .
Thus, (ψ + ψ ∗ ) and i(ψ − ψ ∗ ) – both of which are real – satisfy Eq. 2.5. Conclusion: From any complex
solution, we can always construct two real solutions (of course, if ψ is already real, the second one will be
zero). In particular, since ψ = 21 [(ψ + ψ ∗ ) − i(i(ψ − ψ ∗ ))], ψ can be expressed as a linear combination of
two real solutions. QED
(c) If ψ(x) satisfies Eq. 2.5, then, changing variables x → −x and noting that d2 /d(−x)2 = d2 /dx2 ,
−
~2 d2 ψ(−x)
+ V (−x)ψ(−x) = Eψ(−x);
2m dx2
so if V (−x) = V (x) then ψ(−x) also satisfies Eq. 2.5. It follows that ψ+ (x) ≡ ψ(x) + ψ(−x) (which is
even: ψ+ (−x) = ψ+ (x)) and ψ− (x) ≡ ψ(x) − ψ(−x) (which is odd: ψ− (−x) = −ψ− (x)) both satisfy Eq.
2.5. But ψ(x) = 21 (ψ+ (x) + ψ− (x)), so any solution can be expressed as a linear combination of even and
odd solutions. QED
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
17
Problem 2.2
2
00
Given ddxψ2 = 2m
~2 [V (x) − E]ψ, if E < Vmin , then ψ and ψ always have the same sign: If ψ is positive(negative),
00
then ψ is also positive(negative). This means that ψ always curves away from the axis (see Figure). However,
it has got to go to zero as x → −∞ (else it would not be normalizable). At some point it’s got to depart from
zero (if it doesn’t, it’s going to be identically zero everywhere), in (say) the positive direction. At this point its
slope is positive, and increasing, so ψ gets bigger and bigger as x increases. It can’t ever “turn over” and head
back toward the axis, because that would require a negative second derivative—it always has to bend away from
the axis. By the same token, if it starts out heading negative, it just runs more and more negative. In neither
case is there any way for it to come back to zero, as it must (at x → ∞) in order to be normalizable. QED
s
x
Problem 2.3
2
2
2
Equation 2.23 says ddxψ2 = − 2mE
= A + Bx;
~2 ψ; Eq. 2.26 says ψ(0) = ψ(a) = 0. If E = 0, d ψ/dx = 0, so ψ(x) √
2
2
ψ(0) = A = 0 ⇒ ψ = Bx; ψ(a) = Ba = 0 ⇒ B = 0, so ψ = 0. If E < 0, d ψ/dx = κ2 ψ, with κ ≡ −2mE/~
real, so ψ(x) = Aeκx + Be−κx . This time ψ(0) = A + B = 0 ⇒ B = −A, so ψ = A(eκx − e−κx ), while
ψ(a) = A (eκa − e−κa ) = 0 ⇒ either A = 0, so ψ = 0, or else eκa = e−κa , so e2κa = 1, so 2κa = ln(1) = 0,
so κ = 0, and again ψ = 0. In all cases, then, the boundary conditions force ψ = 0, which is unacceptable
(non-normalizable).
Problem 2.4
Z
nπ 2 a
nπ
a
hxi = x|ψ| dx =
x dx.
Let y ≡
x, so dx =
dy;
x sin2
a 0
a
a
nπ
Z
nπ
2 a 2 nπ
y2
y sin 2y cos 2y
2a
=
−
−
y sin2 y dy = 2 2
a nπ
n π
4
4
8
0
0
2 2
2a n π
cos 2nπ 1
a
= 2 2
−
+
= . (Independent of n.)
n π
4
8
8
2
Z
2
Z
Z
2 a 2 2 nπ 2 a 3 nπ 2 2
hx i =
x sin
x dx =
y sin y dy
a 0
a
a nπ
0
nπ
2a2 y 3
y2
1
y cos 2y
=
−
−
sin
2y
−
(nπ)3 6
4
8
4
0
2
3
2a
(nπ)
nπ cos(2nπ)
1
2 1
=
−
= a
−
.
(nπ)3
6
4
3 2(nπ)2
2
y : 0 → nπ.
18
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
hpi = m
dhxi
= 0.
dt
(Note : Eq. 1.33 is much faster than Eq. 1.35.)
2 2
Z
~ d
d ψn
2
∗
ψn
hp i =
ψn dx = −~
dx
i dx
dx2
Z
2
2mEn
nπ~
ψn∗ ψn dx = 2mEn =
= (−~2 ) − 2
.
~
a
2
Z
ψn∗
σx2 = hx2 i − hxi2 = a2
1
1
1
−
−
3 2(nπ)2
4
2
a2
=
4
1
2
−
3 (nπ)2
;
a
σx =
2
s
2
1
.
−
3 (nπ)2
r
(nπ)2
− 2.
3
q
2
The product σx σp is smallest for n = 1; in that case, σx σp = ~2 π3 − 2 = (1.136)~/2 > ~/2. X
σp2 = hp2 i − hpi2 =
nπ~
a
;
nπ~
σp =
.
a
~
∴ σx σp =
2
Problem 2.5
(a)
|Ψ|2 = Ψ∗ Ψ = |A|2 (ψ1∗ + ψ2∗ )(ψ1 + ψ2 ) = |A|2 [ψ1∗ ψ1 + ψ1∗ ψ2 + ψ2∗ ψ1 + ψ2∗ ψ2 ].
Z
Z
√
1 = |Ψ|2 dx = |A|2 [|ψ1 |2 + ψ1∗ ψ2 + ψ2∗ ψ1 + |ψ2 |2 ]dx = 2|A|2 ⇒ A = 1/ 2.
(b)
i
1 h
Ψ(x, t) = √ ψ1 e−iE1 t/~ + ψ2 e−iE2 t/~
2
1
=√
2
(but
En
= n2 ω)
~
r 2
2π
1 −iωt
2π
π
π
−iωt
−i4ωt
−3iωt
x e
+ sin
x e
= √ e
sin
x + sin
x e
.
sin
a
a
a
a
a
a
π π 1
2π
2π
sin2
x + sin
x sin
x e−3iωt + e3iωt + sin2
x
a
a
a
a
a
π π 1
2π
2π
=
sin2
x + sin2
x + 2 sin
x sin
x cos(3ωt) .
a
a
a
a
a
|Ψ(x, t)|2 =
(c)
Z
x|Ψ(x, t)|2 dx
Z
π π 1 a
2π
2π
=
x sin2
x + sin2
x + 2 sin
x sin
x cos(3ωt) dx
a 0
a
a
a
a
hxi =
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
Z a
π x sin
x dx =
a
0
"
2
19
# a
Z a
x sin 2π
cos 2π
a2
x2
2π
2
a x
a x
=
−
−
=
x
sin
x
dx.
4
4π/a
8(π/a)2
4
a
0
0
Z a
Z
π π 2π
1 a
3π
x sin
x cos
x sin
x dx =
x − cos
x dx
a
a
2 0
a
a
0
a
π ax
π 3π
1 a2
a2
ax
3π
cos
=
x +
sin
x − 2 cos
x −
sin
x
2 π2
a
π
a
9π
a
3π
a
0
2
1
1 a
a2
a2
8a2
1
−
.
=
cos(π)
−
cos(0)
−
cos(3π)
−
cos(0)
=
−
=
−
2 π2
9π 2
π2
9
9π 2
1 a2
a2
16a2
a
32
cos(3ωt)
=
cos(3ωt)
.
+
−
1
−
a 4
4
9π 2
2
9π 2
∴ hxi =
32 a = 0.3603(a/2);
9π 2 2
Amplitude:
angular frequency: 3ω =
3π 2 ~
.
2ma2
(d)
hpi = m
a 32 8~
dhxi
=m
− 2 (−3ω) sin(3ωt) =
sin(3ωt).
dt
2
9π
3a
(e) You could get either E1 = π 2 ~2 /2ma2 or E2 = 2π 2 ~2 /ma2 , with equal probability P1 = P2 = 1/2.
So hHi =
1
5π 2 ~2
(E1 + E2 ) =
;
2
4ma2
it’s the average of E1 and E2 .
Problem 2.6
From Problem 2.5, we see that
Ψ(x, t) =
√1 e−iωt
a
|Ψ(x, t)|2 =
1
a
π
ax
sin
2
sin
π
ax
+ sin
+ sin2
2π
a x
2π
a x
−3iωt iφ e
e ;
+ 2 sin
π
ax
sin
2π
a x
cos(3ωt − φ) ;
32
and hence hxi = a2 1 − 9π
This amounts physically to starting the clock at a different time
2 cos(3ωt − φ) .
(i.e., shifting the t = 0 point).
If φ =
π
a
, so Ψ(x, 0) = A[ψ1 (x) + iψ2 (x)], then cos(3ωt − φ) = sin(3ωt); hxi starts at .
2
2
a
32
If φ = π, so Ψ(x, 0) = A[ψ1 (x) − ψ2 (x)], then cos(3ωt − φ) = − cos(3ωt); hxi starts at
1+ 2 .
2
9π
20
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
Problem 2.7
^(x,0)
Aa/2
a
a/2
(a)
Z a/2
1 = A2
x2 dx + A2
Z a
(a − x)2 dx = A2
a/2
0
x
a/2
x3
3 0
a
−
(a − x)3
3
a/2
√
A a3
2 3
a3
A2 a3
=
+
=
⇒ A= √ .
3
8
8
12
a3
2
(b)
√ Z a/2
Z a
22 3
nπ
nπ
√
x dx +
(a − x) sin
x dx
x sin
aa a 0
a
a
a/2
√ 2
a/2
2 6
nπ
xa
nπ
a
= 2
sin
x −
cos
x
a
nπ
a
nπ
a
0
2
a
a a
a
nπ
ax
nπ
nπ
−
+a −
cos
x
x −
x
sin
cos
nπ
a
nπ
a
nπ
a
a/2
a/2
√ 2
2
2
2
2 6
a
nπ
nπ
a
a
nπ − a cos
= 2
sin
− cos
nπ +
cos
a
nπ
2
2nπ
2
nπ
nπ 2
2
a
nπ
a2 a2 nπ
+
cos
sin
+ cos nπ − nπ
2
nπ
2nπ
2
√
√
(
2
0,
n even,
2 6 a
nπ
4 6
nπ
√
=
2
sin
=
sin
=
(n−1)/2 4 6
2
2
2
(−1)
,
(nπ)
2
(nπ)
2
a
(nπ)2 n odd.
r
cn =
√ r
4 6 2 X
1
nπ
n2 π 2 ~2
So Ψ(x, t) = 2
(−1)(n−1)/2 2 sin
x e−iEn t/~ , where En =
.
π
a n=1,3,5,...
n
a
2ma2
(c)
P1 = |c1 |2 =
16 · 6
= 0.9855.
π4
(d)
hHi =
X
|cn |2 En =
96 π 2 ~2
π 4 2ma2
1
1
1
1
+ 2 + 2 + 2 + ···
1
3
5
7
|
{z
}
π 2 /8
=
48~2 π 2
6~2
=
.
2
2
π ma 8
ma2
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
21
Problem 2.8
A
2
r
Z a/2
2
dx = A (a/2) = 1 ⇒ A =
0
2
.
a
From Eq. 2.37,
r Z a/2
π π
i
π i a/2
2
2
2h a
2h
sin
− cos
− cos 0 = .
c1 = A
x dx =
x
= − cos
a 0
a
a
π
a
π
2
π
0
P1 = |c1 |2 = (2/π)2 = 0.4053.
Problem 2.9
ĤΨ(x, 0) = −
Z
~2 ∂ 2
~2 ∂
~2
[Ax(a
−
x)]
=
−A
(a
−
2x)
=
A
.
2m ∂x2
2m ∂x
m
2
a
x
x3
a −
Ψ(x, 0) ĤΨ(x, 0) dx = A
x(a − x) dx = A
m 0
m
2
3
0
3
2
3
2 3
2
a
~
a
30 ~ a
5~
= A2
−
= 5
=
m 2
3
a m 6
ma2
∗
2~
2 Z a
2~
2
(same as Example 2.3).
Problem 2.10
(a) Using Eqs. 2.48 and 2.60,
mω 1/4 − mω x2
1
d
+ mωx
e 2~
−~
dx
π~
2~mω
i mω 2
mω 1/4 h mω mω 1/4
mω 2
1
1
=√
−~ −
2x + mωx e− 2~ x = √
2mωxe− 2~ x .
2~
2~mω π~
2~mω π~
mω 1/4
mω 2
d
1
2mω −~
+ mωx xe− 2~ x
(a+ )2 ψ0 =
2~mω π~
dx
i mω 2 mω 1/4 2mω
mω 2
1 mω 1/4 h mω 2
− 2~ x
2
−~ 1 − x
2x + mωx e
x − 1 e− 2~ x .
=
=
~ π~
2~
π~
~
a+ ψ0 = √
Therefore, from Eq. 2.68,
1 mω 1/4
1
ψ2 = √ (a+ )2 ψ0 = √
2
2 π~
mω 2
2mω 2
x − 1 e− 2~ x .
~
22
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
(b)
s0
s1
(c) Since ψ0 and ψ2Rare even, whereas ψ1 is odd,
we need to check is ψ2∗ ψ0 dx:
Z
R
s2
ψ0∗ ψ1 dx and
R
ψ2∗ ψ1 dx vanish automatically. The only one
Z
mω 2
mω ∞ 2mω 2
x − 1 e− ~ x dx
~
2 π~ −∞
r
Z ∞
Z
mω 2
mω
2mω ∞ 2 − mω x2
e− ~ x dx −
x e ~ dx
=−
2π~
~
−∞
−∞
r
r
r
2mω ~
mω
π~
π~
−
= 0. X
=−
2π~
mω
~ 2mω mω
1
ψ2∗ ψ0 dx = √
r
Problem 2.11
R
(a) Note that ψ0 is even, and ψ1 is odd. In either case |ψ|2 is even, so hxi = x|ψ|2 dx = 0. Therefore
hpi = mdhxi/dt = 0. (These results hold for any stationary state of the harmonic oscillator.)
√
2
2
From Eqs. 2.60 and 2.63, ψ0 = αe−ξ /2 , ψ1 = 2αξe−ξ /2 . So
n = 0:
hx2 i = α2
Z ∞
2
x2 e−ξ dx = α2
−∞
2
~
mω
3/2 Z ∞
2
1
ξ 2 e−ξ dξ = √
π
−∞
Z ∞
~
mω
√
π
~
=
.
2
2mω
d2 −ξ2 /2
e
e
dξ
hp i = ψ0
ψ0 dx = −~ α
dξ 2
−∞
√
−ξ2
m~ω
m~ω
π √
m~ω
2
√
√
=−
ξ − 1 e dξ = −
− π =
.
2
2
π −∞
π
2
Z
~ d
i dx
Z ∞
2 2
r
mω
~
−ξ 2 /2
n = 1:
hx2 i = 2α2
Z ∞
2
x2 ξ 2 e−ξ dx = 2α2
−∞
Z ∞
~
mω
3/2 Z ∞
−∞
2
ξ 4 e−ξ dξ = √
√
2~ 3 π
3~
=
.
2mω
πmω 4
d2
−ξ 2 /2
hp i = −~ 2α
ξe
ξe
dξ
dξ 2
−∞
√ Z
2mω~ 3 √
π
3m~ω
2mω~ ∞ 4
2 −ξ 2
√
√
=−
ξ − 3ξ e dξ = −
π−3
=
.
4
2
2
π −∞
π
2
2
2
r
mω
~
−ξ 2 /2
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
(b) n = 0:
r
σx =
p
hx2 i − hxi2 =
r
σx σp =
~
2mω
r
3~
;
2mω
r
p
~
; σp = hp2 i − hpi2 =
2mω
mω~
~
= .
2
2
r
m~ω
;
2
(Right at the uncertainty limit.)X
n = 1:
σx =
r
σp =
3m~ω
;
2
σx σp = 3
~
~
> .X
2
2
(c)
1

 4 ~ω (n = 0) 
1 2
;
hp i =
hT i =
3

2m
~ω
(n
=
1)
4
hT i + hV i = hHi =
1

 4 ~ω (n = 0) 
1
.
hV i = mω 2 hx2 i =
3

2
~ω
(n
=
1)
4
1

 2 ~ω (n = 0) = E0 
3
2 ~ω (n = 1) = E1
, as expected.

Problem 2.12
From Eq. 2.70,
r
~
~mω
(a+ + a− ), p = i
(a+ − a− ),
2mω
2
r
Z
~
hxi =
ψn∗ (a+ + a− )ψn dx.
2mω
r
x=
so
But (Eq. 2.67)
a+ ψn =
So
√
n + 1ψn+1 ,
a− ψn =
√
nψn−1 .
r
Z
Z
√
√
~
∗
∗
hxi =
n + 1 ψn ψn+1 dx + n ψn ψn−1 dx = 0 (by orthogonality).
2mω
dhxi
~
~
hpi = m
= 0. x2 =
(a+ + a− )2 =
a2+ + a+ a− + a− a+ + a2− .
dt
2mω
2mω
Z
~
ψn∗ a2+ + a+ a− + a− a+ + a2− ψn . But
hx2 i =
2mω
p
√
 2
√
√
a
ψ
=
a
n
+
1ψ
=
n
+
1
n
+
2ψ
=
(n + 1)(n + 2)ψn+2 .

n+1
n+2
n
+
+

√ √
 a a ψ = a √nψ
=
n
nψ
=
nψ
+ − n
+
n−1
n
n.
p
√
√
a
a
ψ
=
a
n
+
1ψ
=
n
+
1)
n
+
1ψ
=
(n
+

−
+
n
−
n+1
n

p 1)ψn .
√ √
√
 2
a− ψn
= a− nψn−1
= n n − 1ψn−2
= (n − 1)nψn−2 .
So
Z
Z
~
~
1 ~
2
2
hx i =
0 + n |ψn | dx + (n + 1) |ψn | dx + 0 =
(2n + 1) = n +
.
2mω
2mω
2 mω
2
23
24
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
~mω
~mω 2
(a+ − a− )2 = −
a+ − a+ a− − a− a+ + a2− ⇒
2
2
~mω
~mω
1
hp2 i = −
[0 − n − (n + 1) + 0] =
(2n + 1) = n +
m~ω.
2
2
2
p2 = −
1
1
hT i = hp /2mi =
n+
~ω .
2
2
r
r
p
1
~
2
2
σx = hx i − hxi = n +
;
2 mω
2
r
p
σp = hp2 i − hpi2 =
n+
1√
m~ω;
2
σx σp =
n+
1
~
~≥ .X
2
2
Problem 2.13
(a)
Z
1=
|Ψ(x, 0)|2 dx = |A|2
Z
9|ψ0 |2 + 12ψ0∗ ψ1 + 12ψ1∗ ψ0 + 16|ψ1 |2 dx
= |A|2 (9 + 0 + 0 + 16) = 25|A|2 ⇒ A = 1/5.
(b)
Ψ(x, t) =
i
i
1h
1h
3ψ0 (x)e−iE0 t/~ + 4ψ1 (x)e−iE1 t/~ =
3ψ0 (x)e−iωt/2 + 4ψ1 (x)e−3iωt/2 .
5
5
(Here ψ0 and ψ1 are given by Eqs. 2.60 and 2.63; E0 and E1 by Eq. 2.62.)
i
1 h 2
9ψ0 + 12ψ0 ψ1 eiωt/2 e−3iωt/2 + 12ψ0 ψ1 e−iωt/2 e3iωt/2 + 16ψ12
25
1 2
=
9ψ0 + 16ψ12 + 24ψ0 ψ1 cos(ωt) .
25
|Ψ(x, t)|2 =
(With ψ2 in place of ψ1 the frequency would be (E2 − E0 )/~ = [(5/2)~ω − (1/2)~ω]/~ = 2ω.)
(c)
hxi =
But
R
xψ02 dx =
R
Z
Z
Z
1
9 xψ02 dx + 16 xψ12 dx + 24 cos(ωt) xψ0 ψ1 dx .
25
xψ12 dx = 0 (see Problem 2.11 or 2.12), while
r
Z
xψ0 ψ1 dx =
r
=
mω
π~
r
2mω
~
Z
xe
2 mω √
2 π2
π
~
2
− mω
2~ x
1
2
r
xe
~
mω
2
− mω
2~ x
!3
r
dx =
r
=
2 mω π
~
Z ∞
mω
−∞
~
.
2mω
So
24
hxi =
25
r
~
cos(ωt);
2mω
d
24
hpi = m hxi = −
dt
25
r
2
x2 e− ~ x dx
mω~
sin(ωt).
2
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
25
Ehrenfest’s theorem says dhpi/dt = −h∂V /∂xi. Here
r
dhpi
24 mω~
1
∂V
=−
ω cos(ωt), V = mω 2 x2 ⇒
= mω 2 x,
dt
25
2
2
∂x
so
r
r
~
∂V
24 ~mω
2
2 24
−
= −mω hxi = −mω
cos(ωt) = −
ω cos(ωt),
∂x
25 2mω
25
2
so Ehrenfest’s theorem holds.
(d) You could get E0 = 12 ~ω, with probability |c0 |2 = 9/25, or E1 = 32 ~ω, with probability |c1 |2 = 16/25.
Problem 2.14
r
r
r
Z
Z ∞
2
mω ∞ −ξ2
mω
~
e dx = 2
e−ξ dξ.
ψ0 =
e
, so P = 2
π~
π~ x0
π~ mω ξ0
q
~
Classically allowed region extends out to: 12 mω 2 x20 = E0 = 12 ~ω, or x0 = mω
, so ξ0 = 1.
mω 1/4
2
P =√
π
Z ∞
−ξ 2 /2
√
2
e−ξ dξ = 2(1 − F ( 2)) (in notation of CRC Table) = 0.157.
1
Problem 2.15
−2(5−1)
−2(5−3)
4
n = 5: j = 1 ⇒ a3 = (1+1)(1+2)
a1 = − 43 a1 ; j = 3 ⇒ a5 = (3+1)(3+2)
a3 = − 51 a3 = 15
a1 ; j = 5 ⇒ a7 = 0. So
a1
4
4
3
5
3
5
H5 (ξ) = a1 ξ − 3 a1 ξ + 15 a1 ξ = 15 (15ξ − 20ξ + 4ξ ). By convention the coefficient of ξ 5 is 25 , so a1 = 15 · 8,
and H5 (ξ) = 120ξ − 160ξ 3 + 32ξ 5 (which agrees with Table 2.1).
−2(6−0)
−2(6−2)
n = 6: j = 0 ⇒ a2 = (0+1)(0+2)
a0 = −6a0 ; j = 2 ⇒ a4 = (2+1)(2+2)
a2 = − 23 a2 = 4a0 ; j = 4 ⇒ a6 =
−2(6−4)
2
8
8 6
2
4
6
(4+1)(4+2) a4 = − 15 a4 = − 15 a0 ; j = 6 ⇒ a8 = 0. So H6 (ξ) = a0 − 6a0 ξ + 4a0 ξ − 15 ξ a0 . The coefficient of ξ
8
a0 ⇒ a0 = −15 · 8 = −120. H6 (ξ) = −120 + 720ξ 2 − 480ξ 4 + 64ξ 6 .
is 26 , so 26 = − 15
Problem 2.16
(a)
2
2
2
2
d
d
d −ξ2
−ξ 2
(e ) = −2ξe ;
e−ξ =
(−2ξe−ξ ) = (−2 + 4ξ 2 )e−ξ ;
dξ
dξ
dξ
3
2
2
d
d
−ξ 2
2 −ξ 2
2
e
=
(−2 + 4ξ )e
= 8ξ + (−2 + 4ξ )(−2ξ) e−ξ = (12ξ − 8ξ 3 )e−ξ ;
dξ
dξ
4
2
2
d
d
−ξ 2
3 −ξ 2
2
3
e
=
(12ξ − 8ξ )e
= 12 − 24ξ + (12ξ − 8ξ )(−2ξ) e−ξ = (12 − 48ξ 2 + 16ξ 4 )e−ξ .
dξ
dξ
3
4
2
d
d
ξ2
−ξ 2
3
ξ2
H3 (ξ) = −e
e
= −12ξ + 8ξ ; H4 (ξ) = e
e−ξ = 12 − 48ξ 2 + 16ξ 4 .
dξ
dξ
26
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
(b)
H5 = 2ξH4 − 8H3 = 2ξ(12 − 48ξ 2 + 16ξ 4 ) − 8(−12ξ + 8ξ 3 ) = 120ξ − 160ξ 3 + 32ξ 5 .
H6 = 2ξH5 − 10H4 = 2ξ(120ξ − 160ξ 3 + 32ξ 5 ) − 10(12 − 48ξ 2 + 16ξ 4 ) = −120 + 720ξ 2 − 480ξ 4 + 64ξ 6 .
(c)
dH5
= 120 − 480ξ 2 + 160ξ 4 = 10(12 − 48ξ 2 + 16ξ 4 ) = (2)(5)H4 . X
dξ
dH6
= 1440ξ − 1920ξ 3 + 384ξ 5 = 12(120ξ − 160ξ 3 + 32ξ 5 ) = (2)(6)H5 . X
dξ
(d)
2
d −z2 +2zξ
(e
) = (−2z + 2ξ)e−z +2zξ ; setting z = 0, H1 (ξ) = 2ξ.
dz
d
dz
2
d
dz
3
(e
(e
−z 2 +2zξ
−z 2 +2zξ
d
−z 2 +2zξ
)=
(−2z + 2ξ)e
dz
2 −z 2 +2zξ
= − 2 + (−2z + 2ξ) e
; setting z = 0, H2 (ξ) = −2 + 4ξ 2 .
d
2 −z 2 +2zξ
)=
− 2 + (−2z + 2ξ) e
dz
2
2
= 2(−2z + 2ξ)(−2) + − 2 + (−2z + 2ξ) (−2z + 2ξ) e−z +2zξ ;
setting z = 0, H3 (ξ) = −8ξ + (−2 + 4ξ 2 )(2ξ) = −12ξ + 8ξ 3 .
Problem 2.17
Aeikx + Be−ikx = A(cos kx + i sin kx) + B(cos kx − i sin kx) = (A + B) cos kx + i(A − B) sin kx
= C cos kx + D sin kx, with C = A + B; D = i(A − B).
ikx
ikx
e + e−ikx
e − e−ikx
1
1
C cos kx + D sin kx = C
+D
= (C − iD)eikx + (C + iD)e−ikx
2
2i
2
2
= Aeikx + Be−ikx , with A =
1
1
(C − iD); B = (C + iD).
2
2
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
Problem 2.18
~k2
Equation 2.95 says Ψ = Aei(kx− 2m t) , so
i
h
~k2
~k2
~k2
~k2
i~
∂Ψ∗
i~
∗ ∂Ψ
J =
Ψ
−Ψ
=
|A|2 ei(kx− 2m t) (−ik)e−i(kx− 2m t) − e−i(kx− 2m t) (ik)ei(kx− 2m t)
2m
∂x
∂x
2m
=
i~
~k 2
|A|2 (−2ik) =
|A| .
2m
m
It flows in the positive (x) direction (as you would expect).
Problem 2.19
(a)
∞
X
bn inπx/a
einπx/a − e−inπx/a +
e
+ e−inπx/a
2i
2
n=1
n=1
∞
∞
X
X
an
bn
an
bn
inπx/a
= b0 +
+
e
+
−
+
e−inπx/a .
2i
2
2i
2
n=1
n=1
f (x) = b0 +
∞
X
an Let
c0 ≡ b0 ; cn = 12 (−ian + bn ) , for n = 1, 2, 3, . . . ; cn ≡ 12 (ia−n + b−n ) , for n = −1, −2, −3, . . . .
Then f (x) =
∞
X
cn einπx/a .
QED
n=−∞
(b)
Z a
f (x)e
−imπx/a
dx =
−a
Z a
∞
X
n=−∞
Z a
ei(n−m)πx/a dx. But for n 6= m,
cn
−a
a
ei(n−m)πx/a dx =
−a
ei(n−m)πx/a
ei(n−m)π − e−i(n−m)π
(−1)n−m − (−1)n−m
=
=
= 0,
i(n − m)π/a −a
i(n − m)π/a
i(n − m)π/a
whereas for n = m,
Z a
Z a
ei(n−m)πx/a dx =
dx = 2a.
−a
−a
So all terms except n = m are zero, and
Z a
Z a
1
f (x)e−inπx/a dx.
f (x)e−imπx/a = 2acm , so cn =
2a −a
−a
(c)
f (x) =
∞
X
n=−∞
r
π1
1 X
F (k)eikx = √
F (k)eikx ∆k,
2a
2π
QED
27
28
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
where ∆k ≡
π
is the increment in k from n to (n + 1).
a
r
F (k) =
2 1
a
π 2a
Z a
f (x)e
−ikx
−a
1
dx = √
2π
Z a
f (x)e−ikx dx.
−a
(d) As a → ∞, k becomes a continuous variable,
1
f (x) = √
2π
Z ∞
1
F (k)eikx dk; F (k) = √
2π
−∞
Z ∞
f (x)e−ikx dx.
−∞
Problem 2.20
(a)
Z ∞
2
2
Z ∞
|Ψ(x, 0)| dx = 2|A|
1=
−∞
∞
e−2ax dx = 2|A|2
0
√
|A|2
e−2ax
=
⇒ A = a.
−2a 0
a
(b)
A
φ(k) = √
2π
Z ∞
A
e−a|x| e−ikx dx = √
2π
−∞
Z ∞
e−a|x| (cos kx − i sin kx)dx.
−∞
The cosine integrand is even, and the sine is odd, so the latter vanishes and
Z ∞
Z ∞
A
A
−ax
φ(k) = 2 √
e
cos kx dx = √
e−ax eikx + e−ikx dx
2π 0
2π 0
(ik−a)x
∞
Z ∞
e
A
A
e−(ik+a)x
= √
e(ik−a)x + e−(ik+a)x dx = √
+
−(ik + a) 0
2π 0
2π ik − a
r
A
−1
1
A −ik − a + ik − a
a
2a
= √
+
=√
=
.
−k 2 − a2
2π k 2 + a2
2π ik − a ik + a
2π
(c)
r
Z
Z
2
~k2
1
a3 ∞
1
a3/2 ∞
1
i(kx− ~k
2m t) dk =
Ψ(x, t) = √ 2
e
ei(kx− 2m t) dk.
2
2
2
2
2π −∞ k + a
π −∞ k + a
2π
p
(d) For large a, Ψ(x, 0) is a sharp narrow spike whereas φ(k) ∼
= 2/πa is broad and flat; position
p is welldefined but momentum is ill-defined. For small a, Ψ(x, 0) is a broad and flat whereas φ(k) ∼
= ( 2a3 /π)/k 2
is a sharp narrow spike; position is ill-defined but momentum is well-defined.
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
29
Problem 2.21
(a)
1 = |A|
2
Z ∞
e
−2ax2
2
r
dx = |A|
−∞
π
;
2a
A=
2a
π
1/4
.
(b)
Z ∞
e
−(ax2 +bx)
Z ∞
dx =
e
−∞
−y 2 +(b2 /4a)
−∞
1
φ(k) = √ A
2π
Z ∞
−ax2 −ikx
e
e
−∞
1
1
Ψ(x, t) = √
2π (2πa)1/4
1 2
1
√ dy = √ eb /4a
a
a
1
dx = √
2π
Z ∞
2a
π
2
1/4 r
Z ∞
e
−y 2
r
dy =
−∞
π b2 /4a
e
.
a
2
π −k2 /4a
1
e
=
e−k /4a .
1/4
a
(2πa)
2
t/2m)
e−k /4a ei(kx−~k
|
{z
} dk
−∞
1
2
e−[( 4a +i~t/2m)k −ixk]
=√
1/4 −ax2 /(1+2i~at/m)
√
2
1
π
e
1
2a
q
p
e−x /4( 4a +i~t/2m) =
.
1/4
π
1
2π(2πa)
1 + 2i~at/m
+ i~t/2m
4a
(c)
r
2
Let θ ≡ 2~at/m. Then |Ψ| =
2
2
2a e−ax /(1+iθ) e−ax /(1−iθ)
p
. The exponent is
π
(1 + iθ)(1 − iθ)
r
2
2
ax2
ax2
−2ax2
2a e−2ax /(1+θ )
2 (1 − iθ + 1 + iθ)
2
√
−
−
= −ax
=
; |Ψ| =
.
(1 + iθ) (1 − iθ)
(1 + iθ)(1 − iθ)
1 + θ2
π
1 + θ2
r
r
a
2 −2w2 x2
2
Or, with w ≡
, |Ψ| =
we
. As t increases, the graph of |Ψ|2 flattens out and broadens.
2
1+θ
π
|^| 2
|^|2
x
t=0
t>0
x
(d)
Z ∞
hxi =
x|Ψ|2 dx = 0 (odd integrand); hpi = m
−∞
r
2
hx i =
2
w
π
Z ∞
2 −2w2 x2
x e
−∞
r
dx =
2
1
w
π 4w2
r
dhxi
= 0.
dt
π
1
=
.
2w2
4w2
2
hp i = −~
2
Z ∞
−∞
Ψ∗
∂2Ψ
dx.
∂x2
30
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
2
Write Ψ = Be−bx , where B ≡
2a
π
1/4
√
1
a
and b ≡
.
1 + iθ
1 + iθ
2
∂2Ψ
∂ −bx2
=
B
−2bxe
= −2bB(1 − 2bx2 )e−bx .
2
∂x
∂x
∗
2
∂2Ψ
a
a
2a
= −2b|B|2 (1 − 2bx2 )e−(b+b )x ; b + b∗ =
= 2w2 .
+
=
2
d∂x
1 + iθ 1 − iθ
1 + θ2
r
r
r
2
2 2
2a
2
2
1
2
∗∂ Ψ
√
=
|B| =
= −2b
w. So Ψ
w(1 − 2bx2 )e−2w x .
2
2
π 1+θ
π
∂x
π
Ψ∗
r
2
hp i = 2b~
2
r
= 2b~
2
2
w
π
Z ∞
2
w
π
r
2
−∞
b
But 1 −
=1−
2w2
hp2 i = 2b~2
2
(1 − 2bx2 )e−2w x dx
π
1
− 2b 2
2
2w
4w
a
1 + iθ
a
= ~2 a.
2b
r
π
2w2
1 + θ2
2a
σx =
1
;
2w
2
= 2b~
=1−
b
1−
2w2
√
σp = ~ a.
~p
~p
~
1 √
1 + θ2 =
1 + (2~at/m)2 ≥ . X
~ a=
2w
2
2
2
Closest at t = 0, at which time it is right at the uncertainty limit.
Problem 2.22
(a)
(−2)3 − 3(−2)2 + 2(−2) − 1 = −8 − 12 − 4 − 1 = −25.
(b)
cos(3π) + 2 = −1 + 2 = 1.
(c)
0 (x = 2 is outside the domain of integration).
.
1 + iθ
a
(1 − iθ)
=
= , so
2
2
2b
(e)
σx σp =
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
31
Problem 2.23
If c > 0, y : −∞ → ∞.
If c < 0, y : ∞ → −∞.
 1 R∞
1
Z ∞
 c −∞ f (y/c)δ(y)dy = c f (0) (c > 0); or
f (x)δ(cx)dx =
R
 1 R −∞
1 ∞
1
−∞
c ∞ f (y/c)δ(y)dy = − c −∞ f (y/c)δ(y)dy = − c f (0) (c < 0).
Z ∞
Z ∞
1
1
1
In either case,
f (x)δ(cx)dx =
f (0) =
f (x) δ(x)dx. So δ(cx) =
δ(x). X
|c|
|c|
|c|
−∞
−∞
1
(a) Let y ≡ cx, so dx = dy.
c
(b)
Z ∞
f (x)
−∞
Z ∞
∞
df
dθ
−
dx = f θ
θdx (integration by parts)
dx
−∞ dx
−∞
Z ∞
Z ∞
df
= f (∞) −
dx = f (∞) − f (∞) + f (0) = f (0) =
f (x)δ(x)dx.
dx
0
−∞
So dθ/dx = δ(x). X [Makes sense: The θ function is constant (so derivative is zero) except at x = 0, where
the derivative is infinite.]
Problem 2.24
√
ψ(x) =
mα −mα|x|/~2
e
=
~
√
mα
~
(
2
e−mαx/~ , (x ≥ 0),
2
emαx/~ ,
(x ≤ 0).
hxi = 0 (odd integrand).
2 3
2mα
~
~2
mα
~4
2 −2mαx/~2
2
2
2
√
x e
dx = 2 2
;
σ
=
.
hx i =
x |ψ| dx = 2 2
=
x
~
~
2mα
2m2 α2
2mα
0
−∞
 mα −mαx/~2

3
√
− ~2 e
, (x ≥ 0)  √

i
2
2
mα
mα h
dψ
=
−θ(x)e−mαx/~ + θ(−x)emαx/~ .
=

dx
~  mα mαx/~2
~
,
(x ≤ 0)
~2 e
Z ∞
Z ∞
d2 ψ
=
dx2
√
mα
~
3 h
i
2
2
2
mα
mα
θ(x)e−mαx/~ − δ(−x)emαx/~ + 2 θ(−x)emαx/~
2
~
~
√
3 h
i
2
mα
mα
−2δ(x) + 2 e−mα|x|/~ .
=
~
~
2
−δ(x)e−mαx/~ +
In the last step I used the fact that δ(−x) = δ(x) (Eq. 2.145), f (x)δ(x) = f (0)δ(x) (Eq. 2.115), and θ(−x) +
θ(x) = 1 (Eq. 2.146). Since dψ/dx is an odd function, hpi = 0.
√
3 Z ∞
√
Z ∞
h
i
2
2
d2 ψ
mα
mα
2
2
2 mα
hp i = −~
ψ 2 dx = −~
e−mα|x|/~ −2δ(x) + 2 e−mα|x|/~ dx
dx
~
~
~
−∞
−∞
Z
mα 2 mα 2 mα ∞ −2mαx/~2
mα ~2
mα 2
=
2−2 2
e
dx = 2
1− 2
=
.
~
~
~
~ 2mα
~
0
Evidently
σp =
mα
,
~
so
σx σp = √
~2 mα √ ~
~
= 2 > . X
2
2
2mα ~
32
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
Problem 2.25
hψbound |ψscattering i
Z 0
√
Z ∞
2
2
mα
emαx/~ Aeikx + Be−ikx dx +
=
e−mαx/~ F eikx + Ge−ikx dx
~
−∞
0
Z 0
√
Z 0
Z ∞
Z ∞
mα
mα
mα
− mα
− mα
(
(
(
(
2 +ik )x
2 −ik )x
2 +ik )x
2 −ik )x
~
~
~
~
=
A
e
dx + B
e
dx + F
e
dx + G
e
dx
~
−∞
−∞
0
0
#
"
mα
mα
mα
mα
√
0
0
∞
∞
e( ~2 −ik)x
e(− ~2 +ik)x
e(− ~2 −ik)x
mα
e( ~2 +ik)x
+ B mα
+F
+G
=
A mα
mα
mα
~
− ~2 + ik 0
− ~2 − ik 0
~2 + ik −∞
~2 − ik −∞
√
√
mα
A
B
F
G
mα A + G
B+F
=
+
−
−
=
+
mα
mα
mα
mα
~
− mα
− mα
~
~2 + ik
~2 − ik
~2 + ik
~2 − ik
~2 + ik
~2 − ik
"
#
"
#
√
√
mα
mα
mα
mα
mα ~2 (A + G + B + F ) + ik(B + F − A − G)
~2 − ik (A + G) + ~2 + ik (B + F )
=
=
mα 2
mα 2
~
~
+ k2
+ k2
2
2
~
~
But Equation 2.136 says (A + G + B + F ) = 2(A + B), and Equation 2.137 says ik(B + F − A − G) =
−(2mα/~2 )(A + B), so
"
#
√
2mα
mα mα
~2 2(A + B) − ~2 (A + B)
hψbound |ψscattering =i
= 0. X
mα 2
~
+ k2
2
~
Problem 2.26
Z ∞
1
1
Put f (x) = δ(x) into Eq. 2.103: F (k) = √
δ(x)e−ikx dx = √ .
2π −∞
2π
Z ∞
Z ∞
1
1
1
√ eikx dk =
∴ f (x) = δ(x) = √
eikx dk. QED
2π
2π −∞ 2π
−∞
Problem 2.27
V(x)
(a)
-a
a
x
(b) From Problem 2.1(c) the solutions are even or odd. Look first for even solutions:
 −κx
(x > a),
 Ae
ψ(x) = B(eκx + e−κx ) (−a < x < a),
 κx
Ae
(x < −a).
Continuity at a : Ae−κa = B(eκa + e−κa ), or A = B(e2κa + 1).
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
Discontinuous derivative at a, ∆
33
dψ
2mα
= − 2 ψ(a) :
dx
~
2mα −κa
2mα
Ae
⇒ A + B(e2κa − 1) = 2 A; or
~2
~ κ
2mα
2mα
2mα
2mα
2κa
2κa
2κa
2κa
B(e
− 1 = B(e
−1 ⇒e
− 1 + 2 − 1.
− 1) = A
+ 1)
−1=e
2
2
2
~ κ
~ κ
~ κ
~ κ
−κAe−κa − B(κeκa − κe−κa ) = −
1=
2mα
2mα
~2 κ
~2 κ
− 1 + 2 e−2κa ;
= 1 + e−2κa , or e−2κa =
− 1.
2
~ κ
~ κ
mα
mα
2
~
This is a transcendental equation for κ (and hence for E). I’ll solve it graphically: Let z ≡ 2κa, c ≡ 2amα
,
−z
so e = cz − 1. Plot both sides and look for intersections:
1
cz-1
e -z
z
1/c
From the graph, noting that c and z are both positive, we see that there is one (and only one) solution
~2
z2
(for even ψ). If α = 2ma
, so c = 1, the calculator gives z = 1.278, so κ2 = − 2mE
= (2a)
2
2 ⇒ E =
~
2 2 (1.278)2
~
~
− 8
ma2 = −0.204 ma2 .
Now look for odd solutions:
 −κx
(x > a),
 Ae
ψ(x) = B(eκx − e−κx ) (−a < x < a),

−Aeκx
(x < −a).
Continuity at a : Ae−κa = B(eκa − e−κa ), or A = B(e2κa − 1).
2mα
Discontinuity in ψ : −κAe
− B(κe + κe
) = − 2 Ae−κa ⇒ B(e2κa + 1) = A
~
2mα
2mα
2mα
2κa
2κa
2κa
e
+ 1 = (e
− 1)
−1 =e
− 1 − 2 + 1,
2
2
~ κ
~ κ
~ κ
0
1=
−κa
−κa
κa
2mα
−1 ,
~2 κ
2mα −2κa ~2 κ
~2 κ
2mα
−2κa
−2κa
−
1
−
e
;
=
1
−
e
,
e
=
1
−
, or e−z = 1 − cz.
~2 κ
~2 κ
mα
mα
1
1/c
1/c
z
34
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
This time there may or may not be a solution. Both graphs have their y-intercepts at 1, but if c is too
large (α too small), there may be no intersection (solid line), whereas if c is smaller (dashed line) there
will be. (Note that z = 0 ⇒ κ = 0 is not a solution, since ψ is then non-normalizable.) The slope of e−z
(at z = 0) is −1; the slope of (1 − cz) is −c. So there is an odd solution ⇔ c < 1, or α > ~2 /2ma.
Conclusion:
One bound state if α ≤ ~2 /2ma; two if α > ~2 /2ma.
s
s
-a
-a
a
a
x
Even
α=
1
~2
⇒c= .
ma
2
x
Odd
Even: e−z = 12 z − 1 ⇒ z = 2.21772,
Odd: e−z = 1 − 12 z ⇒ z = 1.59362.
E = −0.615(~2 /ma2 ); E = −0.317(~2 /ma2 ).
α=
~2
⇒ c = 2. Only even: e−z = 2z − 1 ⇒ z = 0.738835;
4ma
E = −0.0682(~2 /ma2 ).
(c) (i) There is one bound state (even); c is huge, so z is small, so e−z ≈ 1 = cz − 1, which means z = 2/c,
or 2κa = 2(2amα/~2 ) ⇒ κ = (2mα/~2 ).
E=−
~2 κ2
2mα2
= − 2 .
2m
~
This makes sense: the two delta-functions coincide, so there is really just one delta-function, with
“strength” 2α. Putting this into Equation 2.130 we recover the answer in the box.
(ii) There are two bound states, one even and one odd; c is small, so z is huge, and e−z ≈ 0. For the
even case, 0 = cz − 1 ⇒ z = 1/c ⇒ κ = (mα/~2 ). For the odd case, 0 = 1 − cz, which leads to the
mα2
. Any linear combination of the two
2~2
will be an eigenstate (with the same energy); the sum (properly normalized) would represent a particle
in the delta-function out at large positive x, and the difference would be a particle in the delta-function
at large negative x (the other—distant—delta-function becomes irrelevant), so it makes sense that we get
two states, each with the energy of a particle in a single delta-function well.
same result: the two states are degenerate, each with energy −
Problem 2.28
 ikx

 Ae + Be−ikx (x < −a)

ψ = Ceikx + De−ikx (−a < x < a) . Impose boundary conditions:
 ikx

Fe
(x > a)
(1) Continuity at −a : Ae−ika + Beika = Ce−ika + Deika ⇒ βA + B = βC + D, where β ≡ e−2ika .
(2) Continuity at +a : Ceika + De−ika = F eika ⇒ F = C + βD.
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
35
−ika
(3) Discontinuity in ψ 0 at −a : ik(Ce−ika − Deika ) − ik(Ae−ika − Beika ) = − 2mα
+ Beika )
~2 (Ae
⇒ βC − D = β(γ + 1)A + B(γ − 1), where γ ≡ i2mα/~2 k.
ika
(4) Discontinuity in ψ 0 at +a : ikF eika − ik(Ceika − De−ika ) = − 2mα
)
~2 (F e
⇒ C − βD = (1 − γ)F.
To solve for C and D,
add (2) and (4) :
2C = F + (1 − γ)F ⇒ 2C = (2 − γ)F.
subtract (2) and (4) : 2βD = F − (1 − γ)F ⇒ 2D = (γ/β)F.
add (1) and (3) :
2βC = βA + B + β(γ + 1)A + B(γ − 1) ⇒ 2C = (γ + 2)A + (γ/β)B.
subtract (1) and (3) : 2D = βA + B − β(γ + 1)A − B(γ − 1) ⇒ 2D = −γβA + (2 − γ)B.
Equate the two expressions for 2C : (2 − γ)F = (γ + 2)A + (γ/β)B.
Equate the two expressions for 2D : (γ/β)F = −γβA + (2 − γ)B.
Solve these for F and B, in terms of A. Multiply the first by β(2 − γ), the second by γ, and subtract:
2
β(2 − γ)2 F = β(4 − γ 2 )A + γ(2 − γ)B ;
(γ /β)F = −βγ 2 A + γ(2 − γ)B .
4
F
=
.
⇒ β(2 − γ)2 − γ 2 /β F = β 4 − γ 2 + γ 2 A = 4βA ⇒
A
(2 − γ)2 − γ 2 /β 2
Let g ≡ i/γ =
~2 k
i
F
4g 2
; φ ≡ 4ka, so γ = , β 2 = e−iφ . Then:
=
.
2mα
g
A
(2g − i)2 + eiφ
Denominator: 4g 2 − 4ig − 1 + cos φ + i sin φ = (4g 2 − 1 + cos φ) + i(sin φ − 4g).
|Denominator|2 = (4g 2 − 1 + cos φ)2 + (sin φ − 4g)2
= 16g 4 + 1 + cos2 φ − 8g 2 − 2 cos φ + 8g 2 cos φ + sin2 φ − 8g sin φ + 16g 2
= 16g 4 + 8g 2 + 2 + (8g 2 − 2) cos φ − 8g sin φ.
T =
F
A
2
=
8g 4
(8g 4 + 4g 2 + 1) + (4g 2 − 1) cos φ − 4g sin φ
, where g ≡
~2 k
and φ ≡ 4ka.
2mα
Problem 2.29
 −κx

(x > a)
 Fe

In place of Eq. 2.154, we have: ψ(x) = D sin(lx) (0 < x < a) .


−ψ(−x) (x < 0)
Continuity of ψ : F e−κa = D sin(la); continuity of ψ 0 : −F κe−κa = Dl cos(la).
q
p
Divide: − κ = l cot(la), or − κa = la cot(la) ⇒ z02 − z 2 = −z cot z, or − cot z = (z0 /z)2 − 1.
Wide, deep well: Intersections are at π, 2π, 3π, etc. Same as Eq. 2.160, but now for n even. This fills in the
rest of the states for the infinite square well.
Shallow, narrow well: If z0 < π/2, there is no odd bound state. The corresponding condition on V0 is
V0 <
π 2 ~2
⇒ no odd bound state.
8ma2
36
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
/
z0
2/
z
Problem 2.30
Z ∞
2
2
Z a
2
2
Z ∞
−2κx
cos lx dx + |F |
e
|ψ| dx = 2 |D|
dx
0
a
a
∞
1
x
1
a sin 2la
e−2κa
= 2 |D|2
+ |F |2 − e−2κx
+ sin 2lx
= 2 |D|2
+
+ |F |2
.
2 4l
2κ
2
4l
2κ
0
a
1=2
0
sin(2la) cos2 (la)
+
.
But F = Deκa cos la (Eq. 2.152), so 1 = |D|2 a +
2l
κ
Furthermore κ = l tan(la) (Eq. 2.157), so
cos la
2 sin la cos la cos3 la
1 = |D|2 a +
+
= |D|2 a +
(sin2 la + cos2 la)
2l
l sin la
l sin la
1
1
eκa cos la
1
= |D|2 a +
D= p
F =p
= |D|2 a +
.
,
.
l tan la
κ
a + 1/κ
a + 1/κ
Problem 2.31
√
Equation 2.158 ⇒ z0 = ~a 2mV0 . We want α = area of potential = 2aV0 held constant as a → 0. Therefore
p
√
α
α
V0 = 2a
; z0 = ~a 2m 2a
= ~1 mαa → 0. So z0 is small, and the intersection in Fig. 2.17 occurs at very small
z. Solve Eq. 2.159 for very small z, by expanding tan z:
q
p
2
(z
/z)
−
1
=
(1/z)
z02 − z 2 .
tan z ∼
z
=
=
0
Now (from Eqs. 2.149, 2.151
and 2.158) z02 −z 2 = κ2 a2 , so z 2 = κa. But z02 −z 2 = z 4 1 ⇒ z ∼
= z0 , so κa ∼
= z02 .
1
mα
1√
∼
But we found that z0 = ~ mαa here, so κa = ~2 mαa, or κ = ~2 . (At this point the a’s have canceled, and
we can go to the limit a → 0.)
√
−2mE
mα
m2 α2
mα2
= 2 ⇒ −2mE =
.
E
=
−
(which agrees with Eq. 2.132).
~
~
~2
2~2
√
V02
α
2mV0 . But V0 = 2a
In Eq. 2.172, V0 E ⇒ T −1 ∼
sin2 2a
, so the argument of the sine is small,
= 1+ 4EV
~
0
2
V0 2a
mα2
−1 ∼
2 m
and we can replace sin by : T = 1 + 4E ~ 2mV0 = 1 + (2aV0 ) 2~2 E . But 2aV0 = α, so T −1 = 1 + 2~
2E ,
in agreement with Eq. 2.144.
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
37
Problem 2.32
Multiply Eq. 2.168 by sin la, Eq. 2.169 by 1l cos la, and add:
ik
C sin2 la + D sin la cos la = F eika sin la
ika
C
=
F
e
sin
la
+
cos
la
.
C cos2 la − D sin la cos la = ikl F eika cos la
l
Multiply Eq. 2.168 by cos la, Eq. 2.169 by 1l sin la, and subtract:
ik
C sin la cos la + D cos2 la = F eika cos la
ika
sin
la
.
D
=
F
e
cos
la
−
C sin la cos la − D sin2 la = ikl F eika sin la
l
Put these into Eq. 2.166:
(1) Ae
−ika
+ Be
ika
ik
ik
ika
= −F e
sin la +
cos la sin la + F e
sin la cos la
cos la −
l
l
ik
ik
2
ika
2
= Fe
cos la −
sin la cos la − sin la −
sin la cos la
l
l
ik
= F eika cos(2la) −
sin(2la) .
l
ika
Likewise, from Eq. 2.167:
ik
ik
il ika
sin la +
cos la cos la + cos la −
sin la sin la
(2) Ae
− Be = − F e
k
l
l
il ika
ik
ik
2
2
= − Fe
sin la cos la +
cos la + sin la cos la −
sin la
k
l
l
il ika
ik
il
ika
= − Fe
sin(2la) +
cos(2la) = F e
cos(2la) − sin(2la) .
k
l
k
k
l
Add (1) and (2): 2Ae−ika = F eika 2 cos(2la) − i
+
sin(2la) , or:
l
k
−ika
ika
e−2ika A
F =
(confirming Eq. 2.171). Now subtract (2) from (1):
2
2
cos(2la) − i sin(2la)
2kl (k + l )
l
k
sin(2la) 2
2Beika = F eika i
−
sin(2la) ⇒ B = i
(l − k 2 )F (confirming Eq. 2.170).
k
l
2kl
T −1 =
A
F
2
= cos(2la) − i
sin(2la) 2
(k + l2 )
2kl
But cos2 (2la) = 1 − sin2 (2la), so
2
−1
T = 1 + sin (2la)
√
2
= cos2 (2la) +
(k 2 + l2 )2
−1
(2lk)2
|
{z
}
sin2 (2la) 2
(k + l2 )2 .
(2lk)2
=1+
(k 2 − l2 )2
sin2 (2la).
(2kl)2
(k2 −l2 )2
1
1
4
2 2
4
[k4 +2k2 l2 +l4 −4k2 l2 ]= (2kl)
2 [k −2k l +l ]= (2kl)2 .
(2kl)2
p
2m(E + V0 )
2a p
2mV0
; so (2la) =
2m(E + V0 ); k 2 − l2 = − 2 , and
~
~
~
2m 2 2
2
V
(k 2 − l2 )2
V
0
0
~2
.
=
=
2
(2kl)2
4E(E
+ V0 )
4 2m
E(E
+
V
)
0
~2
V02
2a p
∴ T −1 = 1 +
sin2
2m(E + V0 ) , confirming Eq. 2.172.
4E(E + V0 )
~
But k =
2mE
, l=
~
38
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
Problem 2.33
 ikx

 Ae + Be−ikx (x < −a)

ψ = Ceκx + De−κx (−a < x < a)
 ikx

Fe
(x > a)
E < V0 .
√
k=
2mE
; κ=
~
p
2m(V0 − E)
.
~
(1) Continuity of ψ at −a: Ae−ika + Beika = Ce−κa + Deκa .
(2) Continuity of ψ 0 at −a: ik(Ae−ika − Beika ) = κ(Ce−κa − Deκa ).
κ κa
κ −κa Ce
+ 1+i
De .
⇒ 2Ae−ika = 1 − i
k
k
(3) Continuity of ψ at +a: Ceκa + De−κa = F eika .
(4) Continuity of ψ 0 at +a: κ(Ceκa − De−κa ) = ikF eika .
ik
ik
κa
ika
−κa
⇒ 2Ce = 1 +
F e ; 2De
= 1−
F eika .
κ
κ
2Ae
−ika
−2κa
ik
iκ
ik
e2κa
iκ
ika e
1+
Fe
+ 1+
1−
F eika
= 1−
k
κ
2
k
κ
2
k
κ
κ
k
F eika
−
−
=
1+i
+ 1 e−2κa + 1 + i
+ 1 e2κa
2
κ k
k
κ
ika
2
2
Fe
(κ − k ) 2κa
=
2 e−2κa + e2κa + i
e
− e−2κa .
2
kκ
x
−x
ex + e−x
e −e
, cosh x ≡
, so
But sinh x ≡
2
2
F eika
(κ2 − k 2 )
=
2 sinh(2κa)
4 cosh(2κa) + i
2
kκ
(κ2 − k 2 )
ika
= 2F e
cosh(2κa) + i
sinh(2κa) .
2kκ
2
(κ2 − k 2 )2
sinh2 (2κa). But cosh2 = 1 + sinh2 , so
(2κk)2
V02
(κ2 − k 2 )2
2a p
2
2
1
+
sinh
2m(V
−
E)
,
T −1 = 1 + 1 +
sinh
(2κa)
=
0
(2κk)2
4E(V0 − E)
~
|
{z
}
T −1 =
A
F
= cosh2 (2κa) +
F
2 2
where F =
4
4
2 2
2
2 2
4κ k + k + κ − 2κ k
(κ + k )
=
=
(2κk)2
(2κk)2
2m(V0 −E)
2mE
~2 +
~2
2
2m(V0 −E)
4 2mE
~2
~2
=
V02
.
4E(V0 − E)
(You can also get this from Eq. 2.172 by switching the sign of V0 and using sin(iθ) = i sinh θ.)
 ikx


 Ae + Be−ikx (x < −a)
(−a < x < a)
E = V0 .
ψ = C + Dx

 ikx
Fe
(x > a)
(In central region −
~2 d 2 ψ
d2 ψ
+ V0 ψ = Eψ ⇒
= 0, so ψ = C + Dx.)
2
2m dx
dx2
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
39
(1) Continuous ψ at −a : Ae−ika + Beika = C − Da.
(2) Continuous ψ at +a : F eika = C + Da.
⇒ (2.5) 2Da = F eika − Ae−ika − Beika .
(3) Continuous ψ 0 at −a : ik Ae−ika − Beika = D.
(4) Continuous ψ 0 at +a : ikF eika = D.
⇒ (4.5) Ae−2ika − B = F.
Use (4) to eliminate D in (2.5): Ae−2ika + B = F − 2aikF = (1 − 2iak)F , and add to (4.5):
2Ae−2ika = 2F (1 − ika), so T −1 =
2
A
F
= 1 + (ka)2 = 1 +
2mE 2
a .
~2
(You can also get this from Eq. 2.172 by changing the sign of V0 and taking the limit E → V0 , using sin ∼
= .)
This case is identical to the one in the book, only with V0 → −V0 . So
E > V0 .
T
−1
V02
=1+
sin2
4E(E − V0 )
2a p
2m(E − V0 ) .
~
Problem 2.34
(a)
ψ=
Aeikx + Be−ikx (x < 0)
F e−κx
(x > 0)
√
where k =
p
2mE
; κ=
~
2m(V0 − E)
.
~
(1) Continuity of ψ : A + B = F.
(2) Continuity of ψ 0 : ik(A − B) = −κF.
ik
ik
ik
⇒ A + B = − (A − B) ⇒ A 1 +
= −B 1 −
.
κ
κ
κ
R=
B
A
2
=
|(1 + ik/κ)|2
1 + (k/κ)2
=
= 1.
2
|(1 − ik/κ)|
1 + (k/κ)2
Although the wave function penetrates into the barrier, it is eventually all reflected.
(b)
ψ=
Aeikx + Be−ikx (x < 0)
F eilx
(x > 0)
√
where k =
(1) Continuity of ψ : A + B = F.
(2) Continuity of ψ 0 : ik(A − B) = ilF.
2mE
; l=
~
p
2m(E − V0 )
.
~
40
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
k
k
k
⇒ A + B = (A − B); A 1 −
= −B 1 +
.
l
l
l
2
(1 − k/l)2
(k − l)2
(k − l)4
=
=
.
(1 + k/l)2
(k + l)2
(k 2 − l2 )2
√
p
2m
2m √
2m
2
2
Now k − l = 2 (E − E + V0 ) =
V
;
k
−
l
=
[
E
−
E − V0 ], so
0
~
~2
~
√
√
( E − E − V 0 )4
R=
.
V02
R=
B
A
=
(c)
vi dt
vi
vt dt
vt
From the diagram, T = Pt /Pi = |F |2 vt /|A|2 vi , where Pi is the probability of finding the incident particle in the
box corresponding to the time interval dt, and Pt is the probability of finding the transmitted particle in the
associated box to the right of the barrier.
r
√
2
vt
E − V0
E − V0 F
√
But
=
(from Eq. 2.98). So T =
. Alternatively, from Problem 2.18:
vi
E
A
E
r
2
2
~k 2
~l 2
Jt
F
E − V0
F l
Ji =
|A| ; Jt = |F | ; T =
=
.
=
m
m
Ji
A k
A
E
For E < V0 , of course, T = 0.
(d)
k
2k/l
2k
l −1
=A k
=
A.
k
k+l
l +1
l +1
For E > V0 , F = A + B = A + A
T =
F
A
2
T +R=
l
=
k
2k
k+l
2
√ √
√
√
l
4kl
4kl(k − l)2
4 E E − V0 ( E − E − V0 )2
=
=
=
.
k
(k + l)2
(k 2 − l2 )2
V02
4kl
(k − l)2
4kl + k 2 − 2kl + l2
k 2 + 2kl + l2
(k + l)2
+
=
=
=
= 1. X
(k + l)2
(k + l)2
(k + l)2
(k + l)2
(k + l)2
Problem 2.35
(a)
ψ(x) =
Aeikx + Be−ikx (x < 0)
F eilx
(x > 0)
√
where k ≡
2mE
, l≡
~
Continuity of ψ ⇒ A + B = F
=⇒
Continuity of ψ 0 ⇒ ik(A − B) = ilF
k
k
k
= −B 1 +
;
A + B = (A − B); A 1 −
l
l
l
B
=−
A
p
2m(E + V0 )
.
~
1 − k/l
1 + k/l
.
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
41
√ !2
√
2
2
B
l−k
E + V0 − E
√
=
R=
= √
A
l+k
E + V0 + E
!2 √
p
2 2
1 + V0 /E − 1
1+3−1
2−1
1
= p
= .
=
= √
2
+
1
9
1+3+1
1 + V0 /E + 1
(b) The cliff is two-dimensional, and even if we pretend the car drops straight down, the potential as a function
of distance along the (crooked, but now one-dimensional) path is −mgx (with x the vertical coordinate),
as shown.
V(x)
x
-V0
(c) Here V0 /E = 12/4 = 3, the same as in part (a), so R = 1/9, and hence T = 8/9 = 0.8889.
Problem 2.36
Start with Eq. 2.25: ψ(x) = A sin kx + B cos kx. This time the boundary conditions are ψ(a) = ψ(−a) = 0:
A sin ka + B cos ka = 0; −A sin ka + B cos ka = 0.
(
Subtract : A sin ka = 0 ⇒ ka = jπ or A = 0,
Add :
B cos ka = 0 ⇒ ka = (j − 21 )π or B = 0,
(where j = 1, 2, 3, . . .).
If B = 0 (so A 6= 0), k = jπ/a. In Rthis case let n ≡ 2j (so n is an even√integer); then k = nπ/2a,
a
ψ = A sin(nπx/2a). Normalizing: 1 = |A|2 −a sin2 (nπx/2a) dx = |A|2 a ⇒ A = 1/ a.
If A = 0 (so B 6= 0), k = (j − 21 )π/a. In
R athis case let n ≡ 2j − 1 (n is an odd√integer); again k = nπ/2a,
ψ = B cos(nπx/2a). Normalizing: 1 = |B|2 −a cos2 (nπx/2a)dx = |B|2 a ⇒ B = 1/ a.
2 2
2
2 2
k
n π ~
In either case Eq. 2.24 yields E = ~2m
= 2m(2a)
2 (in agreement with Eq. 2.30 for a well of width 2a).
The substitution x → (x + a)/2 takes Eq. 2.31 to
q

2
nπx
n/2

(−1)
sin
(n even),
r
r

a
2a

nπx nπ 2
nπ (x + a)
2
sin
=
sin
+
=
q

a
a
2
a
2a
2

(−1)(n−1)/2 2 cos nπx (n odd).
a
2a
So (apart from normalization) we recover the results above. The graphs are the same as Figure 2.2, except that
some are upside down (different normalization).
cos(/x/2a)
sin(2/x/2a)
cos(3/x/2a)
42
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
Problem 2.37
Use the trig identity sin 3θ = 3 sin θ − 4 sin3 θ to write
3
sin
πx
a
3
= sin
4
πx
a
Normalize using Eq. 2.20: |A|2
1
− sin
4
r a 3
3πx
1
. So (Eq. 2.31): Ψ(x, 0) = A
ψ1 (x) − ψ3 (x) .
a
2 4
4
a 9
1
5
4
+
=
a|A|2 = 1 ⇒ A = √ .
2 16 16
16
5a
So Ψ(x, 0) = √110 [3ψ1 (x) − ψ3 (x)] , and hence (Eq. 2.17)
i
1 h
Ψ(x, t) = √
3ψ1 (x)e−iE1 t/~ − ψ3 (x)e−iE3 t/~ .
10
1
E3 − E1
9ψ12 + ψ32 − 6ψ1 ψ3 cos
t ; so
10
~
Z a
Z a
E3 − E1
1
3
9
2
hxi1 + hxi3 − cos
t
hxi =
xψ1 (x)ψ3 (x)dx,
x|Ψ(x, t)| dx =
10
10
5
~
0
0
|Ψ(x, t)|2 =
where hxin = a/2 is the expectation value of x in the nth stationary state. The remaining integral is
Z
Z
πx
3πx
2πx
4πx
2 a
1 a
x sin
sin
dx =
x cos
− cos
dx
a 0
a
a
a 0
a
a
2
2
a
a
2πx
xa
2πx
a
4πx
xa
4πx
1
= 0.
cos
+
sin
−
cos
−
sin
=
a
2π
a
2π
a
4π
a
4π
a
0
Evidently then,
9 a
1 a
a
hxi =
+
= .
10 2
10 2
2
Using Eq. 2.21,
2
2
hHi = |c1 | E1 + |c3 | E3 =
9
10
π 2 ~2
+
2ma2
1
10
9π 2 ~2
9π 2 ~2
=
.
2ma2
10ma2
Problem 2.38
(a) According to Eq. 2.39, the most general solution to the time-dependent Schrödinger equation for the
infinite square well is
∞
X
2 2
2
Ψ(x, t) =
cn ψn (x)e−i(n π ~/2ma )t .
n=1
2 2
2
2 2
2
2
n2 π 2 ~
n2 π 2 ~ 4ma2
T =
= 2πn2 , so e−i(n π ~/2ma )(t+T ) = e−i(n π ~/2ma )t e−i2πn , and since n2 is
2
2 π~
2ma
2ma
2
an integer, e−i2πn = 1. Therefore Ψ(x, t + T ) = Ψ(x, t). QED
Now
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
43
(b) The classical revival time is the time it takes the particle to go down and back: Tc = 2a/v, with the
velocity given by
r
r
1
2E
2m
2
⇒ Tc = a
.
E = mv ⇒ v =
2
m
E
(c) The two revival times are equal if
r
2m
4ma2
=a
,
π~
E
E=
or
π 2 ~2
E1
=
.
2
8ma
4
Problem 2.39
(a)
√ √ dΨ
2 3
2 3
a
1, (0 < x < a/2)
= √ ·
= √ 1 − 2θ x −
.
−1, (a/2 < x < a)
dx
2
a a
a a
(b)
√ √ d2 Ψ
4 3
2 3
a
a
= √ − 2δ x −
= − √ δ x−
.
dx2
2
2
a a
a a
(c)
~2
hHi = −
2m
√ Z
√
4 3
2 3~2 ∗ a
2 · 3 · ~2
a
6~2
∗
√ Ψ
− √
dx =
=
=
. X
Ψ δ x−
2
2
m·a·a
ma2
a a
ma a
| {z }
√
3/a
Problem 2.40
(a) In the standard notation ξ ≡
p
mω/~ x, α ≡ (mω/π~)1/4 ,
2
2
Ψ(x, 0) = A(1 − 2ξ)2 e−ξ /2 = A(1 − 4ξ + 4ξ 2 )e−ξ /2 .
It can be expressed as a linear combination of the first three stationary states (Eq. 2.60 and 2.63, and
Problem 2.10):
2
ψ0 (x) = αe−ξ /2 ,
ψ1 (x) =
So Ψ(x, 0) = c0 ψ0 + c1 ψ1 + c2 ψ2 = α(c0 +
 √

α√2c2 = 4A
α 2c1 = −4A

√

α(c0 − c2 / 2) = A
√
√
2
2 αξe−ξ /2 ,
2ξc1 +
√
2
α
ψ2 (x) = √ (2ξ 2 − 1)e−ξ /2 .
2
2
2ξ 2 c2 − √12 c2 )e−ξ /2 with (equating like powers)
√
⇒ c2 = 2 2A/α,
√
⇒ c1 = −2 2A/α,
√
⇒ c0 = (A/α) + c2 / 2 = (1 + 2)A/α = 3A/α.
44
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
Normalizing: 1 = |c0 |2 + |c1 |2 + |c2 |2 = (8 + 8 + 9)(A/α)2 = 25(A/α)2 ⇒ A = α/5 =
c0 =
(b) You could get
3
,
5
c1 = −
√
2 2
,
5
c2 =
1 mω 1/4
.
5 π~
√
2 2
.
5
1
9 3
8 5
8
~ω, probability
; ~ω, probability
; ~ω, probability
.
2
25 2
25 2
25
9
hHi =
25
1
8 3
8 5
~ω
73
~ω +
~ω +
~ω =
(9 + 24 + 40) =
~ω.
2
25 2
25 2
50
50
(c)
"
#
√
√
√
√
3
2 2
2 2
−iωt/2 2 2
−3iωt/2 2 2
−5iωt/2
−iωt/2 3
−iωt
−2iωt
Ψ(x, t) = ψ0 e
−
ψ1 e
+
ψ2 e
=e
ψ0 −
ψ1 e
+
ψ2 e
.
5
5
5
5
5
5
To change the sign of the middle term we need e−iωT = −1 (then e−2iωT = 1); evidently ωT = π, or
T = π/ω.
Problem 2.41
Everything in Section 2.3.2 still applies, except that there is an additional boundary condition: ψ(0) = 0. This
eliminates all the even solutions (n = 0, 2, 4, . . .), leaving only the odd solutions. So
En =
n+
1
2
~ω, n = 1, 3, 5, . . . .
Problem 2.42
(a) Normalization is the same as before: A =
2a 1/4
.
π
(b) Equation 2.104 says
1
φ(k) = √
2π
2a
π
1/4 Z ∞
2
e−ax eilx e−ikx dx
[same as before, only k → k − l] =
−∞
1
1
Ψ(x, t) = √
2π (2πa)1/4
Z ∞
2
2
e−(k−l) /4a ei(kx−~k t/2m) dk
−∞
Let u ≡ k − l, so k = u + l and dk = du:
Z ∞
2
2
2
1
1
Ψ(x, t) = √
e−u /4a ei[ux+lx−(~t/2m)(u +2ul+l )] du
2π (2πa)1/4 −∞
Z ∞
2
1
~t
~lt
~lt
1
1
il(x− 2m
)
=√
e
e−u ( 4a +i 2m )+iu(x− m ) du.
2π (2πa)1/4
−∞
2
1
e−(k−l) /4a .
(2πa)1/4
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
45
Using the hint in Problem 2.21, the integral becomes
√
Z ∞
2
2√
2
1
~lt 2
~t
2 a −a(x− ~lt
1
m ) /γ
q
e(x− m ) /4( 4a +i 2m )
e−y dy =
π,
e
γ
~t
1
−∞
+
i
4a
2m
so
Ψ(x, t) =
2a
π
1/4
2
2
~lt
1 −a(x− ~lt
m ) /γ eil(x− 2m ) .
e
γ
The gaussian envelope (the first exponential) travels at speed ~l/m; the sinusoidal wave (the second
exponential) travels at speed ~l/2m .
(c)
r
2
|Ψ(x, t)| =
h
i
2
1
2a 1 a(x− ~lt
+ (γ ∗1 )2
m )
γ2
e
.
2
π |γ|
The term in square brackets simplifies:
1
1
1
1
2i~t
2i~t
2
∗ 2
2
+
=
[(γ
)
+
γ
]
=
1
−
+
1
+
=
.
γ2
(γ ∗ )2
|γ|4
|γ|4
m
m
|γ|4
s
p
2ia~t
2ia~t
2
1+
|γ| =
1−
= 1 + θ2 ,
m
m
where (as before) θ ≡ 2~at/m. So
r
r
2
2
2a
2 −2w2 (x− ~lt
1
/(1+θ 2 )
2a(x− ~lt
2
)
m
m ) .
√
|Ψ(x, t)| =
=
e
we
π 1 + θ2
π
p
~l
where (as before) w ≡ a/(1 + θ2 ). The result is the same as in Problem 2.21, except that x → x − m
t ,
so |Ψ|2 has the same (flattening gaussian) shape – only this time the center moves at constant speed
v = ~l/m.
(d)
Z ∞
hxi =
x|Ψ(x, t)|2 dx.
Let y ≡ x − vt, so x = y + vt.
−∞
Z ∞
=
r
(y + vt)
−∞
2 −2w2 y2
~l
we
dy = vt =
t.
π
m
(The first integral is trivially zero; the second is 1 by normalization.)
hpi = m
2
dhxi
= ~l.
dt
Z ∞
hx i =
r
2
(y + vt)
−∞
2 −2w2 y2
1
1
we
dy =
+ 0 + (vt)2 =
+
2
π
4w
4w2
~lt
m
(The first integral is same as in Problem 2.21).
Z ∞
2
∂Ψ
2a
~lt
2
2
∗∂ Ψ
= − 2 x−
+ il Ψ,
hp i = −~
Ψ
dx;
∂x2
∂x
γ
m
−∞
2
∂2Ψ
2a
2a
~lt
=
−
Ψ
+
−
x
−
+
il
Ψ = Ax2 + Bx + C Ψ,
2
2
2
∂x
γ
γ
m
2
.
46
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
where
A≡
2a
C≡− 2 +
γ
2
2a
γ2
2a
γ2
2
B≡−
,
2 ~lt
m
2
+
2a
γ2
2
4ial
γ2
2~lt 4ial
4ial
− 2 =− 4 ,
m
γ
γ
~lt
m
1
− l2 = − 4 (2aγ 2 + l2 ).
γ
Z ∞
Ψ∗ [Ax2 + Bx + C]Ψ dx = −~2 [Ahx2 i + Bhxi + C]
"
#
2 !
~2
1
~lt
~lt
= − 4 4a2
+
− 4ial
− (2aγ 2 + l2 )
γ
4w2
m
m
"
("
#
2
2 #)
2~at
4ia2 ~t
4ia~t
~at
~2
2
a+a
− 2a −
+ l −1 −
=− 4
+4
γ
m
m
m
m
hp i = −~
2
−∞
~2
= − 4 (−aγ 4 − l2 γ 4 ) = ~2 (a + l2 ).
γ
σx2 = hx2 i − hxi2 =
1
+
4w2
~lt
m
2
−
~lt
m
2
=
1
1
⇒ σx =
;
4w2
2w
√
σp2 = hp2 i − hpi2 = ~2 a + ~2 l2 − ~2 l2 = ~2 a, so σp = ~ a.
(e) σx and σp are same as before, so the uncertainty principle still holds.
Problem 2.43
√
Equation 2.25 ⇒ ψ(x) = A sin kx + B cos kx, 0 ≤ x ≤ a, with k = 2mE/~2 .
Even solutions: ψ(x) = ψ(−x) = A sin(−kx) + B cos(−kx) = −A sin kx + B cos kx (−a ≤ x ≤ 0).

 ψ continuous at 0 : B = B (no new condition).
Boundary
~2 k
ψ 0 discontinuous (Eq. 2.128 with sign of α switched): Ak + Ak = 2mα
2 B ⇒ B = mα A.
~
conditions 
2
2
ψ → 0 at x = a : A sin(ka) + ~mαk A cos(ka) = 0 ⇒ tan(ka) = − ~mαk .
~2 k
ψ(x) = A sin kx +
cos kx (0 ≤ x ≤ a); ψ(−x) = ψ(x).
mα
/
tan(ka)
2/
3/
ka
-h2k
m_
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
47
From the graph, the allowed energies are slightly above
ka =
nπ
(n = 1, 3, 5, . . .) so
2
En &
n 2 π 2 ~2
(n = 1, 3, 5, . . .).
2m(2a)2
These energies are somewhat higher than the corresponding energies for the infinite square well (Eq. 2.30, with
a → 2a). As α → 0, the straight line (−~2 k/mα) gets steeper and steeper, and the intersections get closer to
nπ/2; the energies then reduce to those of the ordinary infinite well. As α → ∞, the straight line approaches
2 2 2
π ~
horizontal, and the intersections are at nπ (n = 1, 2, 3, . . .), so En → n2ma
– these are the allowed energies for
2
the infinite square well of width a. At this point the barrier is impenetrable, and we have two isolated infinite
square wells.
Odd solutions: ψ(x) = −ψ(−x) = −A sin(−kx) − B cos(−kx) = A sin(kx) − B cos(kx) (−a ≤ x ≤ 0).

 ψ continuous at 0 : B = −B ⇒ B = 0.
ψ 0 discontinuous: Ak − Ak = 2mα
Boundary conditions
~2 (0) (no new condition).

ψ(a) = 0 ⇒ A sin(ka) = 0 ⇒ ka = nπ
2 (n = 2, 4, 6, . . .).
ψ(x) = A sin(kx), (−a < x < a);
En =
n 2 π 2 ~2
(n = 2, 4, 6, . . .).
2m(2a)2
These are the exact (even n) energies (and wave functions) for the infinite square well (of width 2a). The point
is that the odd solutions (even n) are zero at the origin, so they never “feel” the delta function at all.
Problem 2.44
−

~2
d2 ψ1
~2 d2 ψ1


+
V
ψ
=
Eψ
⇒
−
ψ
+
V
ψ
ψ
=
Eψ
ψ
1
1
2
1 2
1 2

2m dx2
2m
dx2
⇒−
~2
d2 ψ1
d2 ψ2
ψ2
−
ψ
= 0.
1
2m
dx2
dx2


~2
~2 d2 ψ2
d2 ψ2


+
V
ψ
=
Eψ
⇒
−
ψ
+
V
ψ
ψ
=
Eψ
ψ
2
2
1
1
2
1
2
2
2
2m dx
2m
dx
d
dψ2 dψ1
dψ1 dψ2
dψ1
dψ2
d2 ψ1
d2 ψ2
d2 ψ1
d2 ψ2
But
ψ2
− ψ1
=
+ ψ2
−
− ψ1
= ψ2
− ψ1
. Since this is
2
2
2
dx
dx
dx
dx dx
dx
dx dx
dx
dx
dx2
dψ1
dψ2
zero, it follows that ψ2
− ψ1
= K (a constant). But ψ → 0 at ∞ so the constant must be zero. Thus
dx
dx
dψ1
dψ2
1 dψ1
1 dψ2
ψ2
= ψ1
, or
=
, so ln ψ1 = ln ψ2 + constant, or ψ1 = (constant)ψ2 . QED
dx
dx
ψ1 dx
ψ2 dx
−
Problem 2.45
(a)
−
~2 d2 ψn
d2 ψn
2m
+
V
ψ
=
E
ψ
⇒
= − 2 (En − V )ψn ;
n
n
n
2m dx2
dx2
~
d2 ψm
2m
= − 2 (Em − V )ψm .
dx2
~
dψm
dψn
d2 ψm
dψm dψ
dψm dψ
d2 ψn
n
n
ψn − ψm
=
ψn + − − ψm
2
dx
dx
dx
dx2
dx dx
dx dx
2m
2m
= − 2 [(Em − V )ψm ψn − ψm (En − V )ψn ] = − 2 (Em − En )ψm ψn . X
~
~
d
dx
48
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
(b) Integrate both sides:
Z x2
d dψm
dψn
)ψ 0 (x ) − ψ 0 (x )ψ (x ) + ψ (x
0
0
ψn − ψm
dx = [ψm
(x2 )ψn (x2 ) − ψm(x
2
n 1
m 1 )ψn (x1 )]
n 2
m 1
dx
dx
x1 dx
Z x2
2m
0
0
ψm ψn dx. [F]
= ψm (x2 )ψn (x2 ) − ψm (x1 )ψn (x1 ) = 2 (En − Em )
~
x1
(c) Because x1 and x2 are adjacent nodes of ψm , ψm (x) must either be posiitive or negative throughout the
interval. We might as well make it positive (if it’s not, multiply ψm by −1). Then
ψm (x) ≥ 0 for x1 ≤ x ≤ x2 ;
ψ 0 (x1 ) ≥ 0 and ψ 0 (x2 ) ≤ 0.
If ψn (x) has no nodes between x1 and x2 then it too must have the same sign throughout the interval (and we
may as well choose it to be positive):
ψn (x) ≥ 0 for x1 ≤ x ≤ x2 .
Rx
0
0
In that case ψm
(x2 )ψn (x2 ) − ψm
(x1 )ψn (x1 ) ≤ 0, but (En − Em ) x12 ψm ψn dx > 0, in contradiction to Equation
[F]. Conclusion: ψn (x) must have at least one node between x1 and x2 .
Problem 2.46
−
~2 d2 ψ
d2 ψ
= Eψ (where x is measured around the circumference), or
= −k 2 ψ, with k ≡
2
2m dx
dx2
√
2mE
, so
~
ψ(x) = Aeikx + Be−ikx .
But ψ(x + L) = ψ(x), since x + L is the same point as x, so
Aeikx eikL + Be−ikx e−ikL = Aeikx + Be−ikx ,
and this is true for all x. In particular, for x = 0 :
(1) AeikL + Be−ikL = A + B.
And for x =
π
:
2k
Aeiπ/2 eikL + Be−iπ/2 e−ikL = Aeiπ/2 + Be−iπ/2 , or iAeikL − iBe−ikL = iA − iB, so
(2) AeikL − Be−ikL = A − B.
Add (1) and (2): 2AeikL = 2A.
Either A = 0, or else eikL = 1, in which case kL = 2nπ (n = 0, ±1, ±2, . . .). But if A = 0, then Be−ikL = B,
leading to the same conclusion. So for every positive n there are two solutions: ψn+ (x) = Aei(2nπx/L) and
RL
ψn− (x) = Be−i(2nπx/L) (n = 0 is ok too, but in that case there is just one solution). Normalizing: 0 |ψ± |2 dx =
√
1 ⇒ A = B = 1/ L. Any other solution (with the same energy) is a linear combination of these.
1
2n2 π 2 ~2
ψn± (x) = √ e±i(2nπx/L) ; En =
(n = 0, 1, 2, 3, . . .).
mL2
L
The theorem fails because here ψ does not go to zero at ∞; x is restricted to a finite range, and we are unable
to determine the constant K (in Problem 2.44).
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
49
Problem 2.47
(a) (i) b = 0 ⇒ ordinary finite square well. Exponential decay outside; sinusoidal inside (cos for ψ1 , sin for
ψ2 ). No nodes for ψ1 , one node for ψ2 .
s1
s2
-a
-a
a
a
x
x
(ii) Ground state is even. Exponential decay outside, sinusoidal inside the wells, hyperbolic cosine in
barrier. First excited state is odd – hyperbolic sine in barrier. No nodes for ψ1 , one node for ψ2 .
s1
s2
-(b/2+a) -b/2
-(b/2+a) -b/2
b/2 b/2+a
x
b/2 b/2+a
x
(iii) For b a, same as (ii), but wave function very small in barrier region. Essentially two isolated finite
square wells; ψ1 and ψ2 are degenerate (in energy); they are even and odd linear combinations of the
ground states of the two separate wells.
s1
s
2
b/2
-b/2
-(b/2+a)
b/2+a x
-(b/2+a)
-b/2
(b) From Eq. 2.160 we know that for b = 0 the energies fall slightly below
2 2
π ~
h
E1 + V0 ≈ 2m(2a)
2 = 4
4π 2 ~2
E2 + V0 ≈ 2m(2a)
2 = h
)
where h ≡
π 2 ~2
.
2ma2
For b a, the width of each (isolated) well is a, so
E1 + V0 ≈ E2 + V0 ≈
π 2 ~2
= h (again, slightly below this).
2ma2
b/2
b/2+a x
50
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
E+V0
h
E 2 +V0
E 1+V0
h/4
b
2
[Within each well, ddxψ2 = − 2m
~2 (V0 + E)ψ, so the more curved the wave function, the higher the energy.]
(c) In the (even) ground state the energy is lowest in configuration (i), with b → 0, so the electron tends to
draw the nuclei together, promoting bonding of the atoms. In the (odd) first excited state, by contrast,
the electron drives the nuclei apart.
Problem 2.48
(a) Let V0 ≡ 32~2 /ma2 . This is just like the odd bound states for the finite square well, since they are the
ones that go to zero at the origin. Referring to the solution to Problem 2.29, the wave function is
(
p
D sin lx,
l ≡ 2m(E + V0 )/~ (0 < x < a),
√
ψ(x) =
F e−κx ,
κ ≡ −2mE/~
(x > a),
and the boundary conditions at x = a yield
− cot z =
with
p
(z0 /z)2 − 1
√
p
2m(32~2 /ma2 )
2mV0
a=
a = 8.
~
~
Referring to the figure (Problem 2.29), and noting that (5/2)π = 7.85 < z0 < 3π = 9.42, we see that there
are three bound states.
z0 =
(b) Let
a
x
1
1
2 a
= |D|
I1 ≡
|ψ| dx = |D|
sin lx dx = |D|
− sin lx cos lx
− sin la cos la ;
2 2l
2 2l
0
0
0
−2κx ∞
Z ∞
Z ∞
e
e−2κa
I2 ≡
|ψ|2 dx = |F |2
e−2κx dx = |F |2 −
= |F |2
.
2κ
2κ
a
a
a
Z a
2
2
Z a
2
2
2
But continuity at x = a ⇒ F e−κa = D sin la, so I2 = |D|2 sin2κla .
Normalizing:
h
i
a
1
sin2 la
1
κ
1 = I1 + I2 = |D|
− sin la cos la +
=
|D|2 κa − sin la cos la + sin2 la
2 2l
2κ
2κ
l
But (referring again to Problem 2.29) κ/l = − cot la, so
1
(1 + κa)
=
|D|2 κa + cot la sin la cos la + sin2 la = |D|2
.
2κ
2κ
2
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
51
So |D|2 = 2κ/(1 + κa), and the probability of finding the particle outside the well is
P = I2 =
2κ sin2 la
sin2 la
=
.
1 + κa 2κ
1 + κa
We can express this in terms of z ≡ la and z0 : κa =
sin2 la = sin2 z =
p
z02 − z 2 (page 80),
1
1
=
=
1 + (z0 /z)2 − 1
1 + cot2 z
z
z0
2
⇒P =
z02 (1 +
z2
p
.
z02 − z 2 )
So far, this ispcorrect for any bound state. In the present case z0 = 8 and z is the third solution
to − cot z = (8/z)2 − 1, which occurs somewhere in the interval 7.85 < z < 8. Mathematica gives
z = 7.9573 and P = 0.54204.
Problem 2.49
(a)
i~
∂Ψ mω x20
−2iωt
−iωt
= −
−2iωe
+
− 2x0 x(−iω)e
Ψ, so
∂t
2~
2
m
∂Ψ
1
1
2 2 −2iωt
2 −iωt
i~
= − mx0 ω e
+ ~ω + mx0 xω e
Ψ.
∂t
2
2
i
∂Ψ h mω mω
= −
2x − 2x0 e−iωt Ψ = −
x − x0 e−iωt Ψ;
∂x
2~
~
∂2Ψ
mω
mω
mω mω 2
−iωt ∂Ψ
−iωt 2
=−
Ψ−
= −
+
Ψ.
x − x0 e
x − x0 e
∂x2
~
~
∂x
~
~
52
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
1
~2 ∂ 2 Ψ 1
~2
mω mω 2
2 2
−iωt 2
+ mω x Ψ = −
x − x0 e
Ψ + mω 2 x2 Ψ
−
−
+
2m ∂x2
2
2m
~
~
2
1
1
1
2
2
−iωt
2 −2iωt
2 2
=
~ω − mω x − 2x0 xe
+ x0 e
+ mω x Ψ
2
2
2
1
1
~ω + mx0 xω 2 e−iωt − mω 2 x20 e−2iωt Ψ
=
2
2
∂Ψ
= i~
(comparing second line above). X
∂t
(b)
r
x2
x2
iωt
−iωt
x2 + 20 (1+e2iωt )− i~t
+ x2 + 20 (1+e−2iωt )+ i~t
mω − mω
2~
m −2x0 xe
m −2x0 xe
|Ψ|2 =
e
π~
r
2
2
2
mω − mω
=
e 2~ [2x +x0 +x0 cos(2ωt)−4x0 x cos(ωt)] . But x20 [1 + cos(2ωt)] = 2x20 cos2 ωt, so
π~
r
r
mω − mω
mω − mω (x−x0 cos ωt)2
x2 −2x0 x cos(ωt)+x20 cos2 (ωt)]
[
~
=
e
e ~
.
=
π~
π~
The wave packet is a Gaussian of fixed shape, whose center oscillates back and forth sinusoidally, with
amplitude x0 and angular frequency ω.
(c) Note that this wave function is correctly normalized (compare Eq. 2.60). Let y ≡ x − x0 cos ωt :
Z
Z
Z
hxi = x|Ψ|2 dx = (y + x0 cos ωt)|Ψ|2 dy = 0 + x0 a cos ωt |Ψ|2 dy = x0 cos ωt.
hpi = m
h−
dhxi
= −mx0 ω sin ωt.
dt
dhpi
= −mx0 ω 2 cos ωt.
dt
V =
1
dV
mω 2 x2 =⇒
= mω 2 x.
2
dx
dV
dhpi
i = −mω 2 hxi = −mω 2 x0 cos ωt =
, so Ehrenfest’s theorem is satisfied.
dx
dt
Problem 2.50
(a)
(E + 12 mv 2 )
∂Ψ
mα ∂
= − 2
|x − vt| − i
Ψ;
∂t
~ ∂t
~
∂
|x − vt| =
∂t
−v, if x − vt > 0
v, if x − vt < 0
We can write this in terms of the θ-function (Eq. 2.146):
2θ(z) − 1 =
∂Ψ
i~
=
∂t
1, if z > 0
−1, if z < 0
, so
∂
|x − vt| = −v[2θ(x − vt) − 1].
∂t
mαv
1
2
i
[2θ(x − vt) − 1] + E + mv Ψ.
~
2
[F]
.
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
53
∂Ψ
mα ∂
imv
= − 2
|x − vt| +
Ψ
∂x
~ ∂x
~
∂
|x − vt| = {1, if x > vt; −1, if x < vt} = 2θ(x − vt) − 1.
∂x
imv
mα
= − 2 [2θ(x − vt) − 1] +
Ψ.
~
~
∂2Ψ
=
∂x2
−
mα
imv
[2θ(x − vt) − 1] +
~2
~
2
Ψ−
2mα ∂
θ(x
−
vt)
Ψ.
~2
∂x
∂
But (from Problem 2.23(b)) ∂x
θ(x − vt) = δ(x − vt), so
−
~2 ∂ 2 Ψ
− αδ(x − vt)Ψ
2m ∂x2
!
2
imv
mα
~2
+ αδ(x − vt) − αδ(x − vt) Ψ
= −
− 2 [2θ(x − vt) − 1] +
2m
~
~
~2 m 2 α 2
m2 v 2
mv mα
2
=−
[2θ(x
−
vt)
−
1]
−
−
2i
[2θ(x
−
vt)
−
1]
Ψ
{z
}
2m
~4 |
~2
~ ~2
1
2
mα
1
mvα
∂Ψ
= − 2 + mv 2 + i
[2θ(x − vt) − 1] Ψ = i~
(compare [F]). X
2~
2
~
∂t
(b)
mα −2mα|y|/~2
e
(y ≡ x − vt).
~2
Z
2mα ~2
mα ∞ −2mαy/~2
Check normalization: 2 2
e
dy = 2
= 1.
~
~ 2mα
0
|Ψ|2 =
Z ∞
X
∂Ψ
, which we calculated above [F].
∂t
−∞
Z imαv
1
1
2
=
[2θ(y) − 1] + E + mv |Ψ|2 dy = E + mv 2 .
~
2
2
hHi =
Ψ∗ HΨdx.
But HΨ = i~
(Note that [2θ(y) − 1] is an odd function of y.) Interpretation: The wave packet is dragged along (at speed
v) with the delta-function. The total energy is the energy it would have in a stationary delta-function
(E), plus kinetic energy due to the motion ( 21 mv 2 ).
Problem 2.51
2a
π
1/4
p
1 −a(x+ 1 gt2 )2 /γ 2
2
e
; γ = 1 + 2ia~t/m.
γ
∂Ψ
∂Ψ0
img
1 2
mgt
1 2
=
+ Ψ0 −
x + gt
exp −i
x + gt
,
∂t
∂t
~
2
~
6
Ψ0 =
54
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
"
1/4 (
2 #)
2 2
2
1
1 2
1
2a
1 2
2 dγ
1 dγ
+
− 2 x + gt gt − a x + gt
− 3
e−a(x+ 2 gt ) /γ
− 2
γ
dt
γ
γ
2
2
γ dt
#
"
2
1
1
2agt
2a
1 dγ
dγ
x + gt2 + 3 x + gt2
Ψ0 .
− 2
= −
γ dt
γ
2
γ
2
dt
∂Ψ0
=
∂t
But
2a
π
dγ
1 2ia~
ia~
=
=
, so
dt
2γ
m
γm
"
2 #
1 2
∂Ψ0
−ia~ 2agt
2ia2 ~
1 2
x + gt + 4
=
− 2
x + gt
Ψ0 , and hence
∂t
γ2m
γ
2
γ m
2
"
2
#
1 2
2agt
1 2
img
1 2
∂Ψ
ia~
2ia2 ~
x + gt + 4
= − 2 − 2
x + gt
−
x + gt
Ψ.
∂t
γ m
γ
2
γ m
2
~
2
[F]
Meanwhile
∂Ψ0
imgt
1 2
imgt
∂Ψ
=
− Ψ0
exp −
x + gt
∂x
∂x
~
~
6
∂Ψ0
1 2
= −2a x + gt /γ 2 Ψ0 , so
∂x
2
∂Ψ
1 2
imgt
2
= −2a x + gt /γ −
Ψ, and hence
∂x
2
~
(
2 )
∂2Ψ
2a
1 2
imgt ∂Ψ
1 2
imgt
2a
2
2
= − 2 Ψ + −2a x + gt /γ −
= − 2 + −2a x + gt /γ −
Ψ.
∂x2
γ
2
~
∂x
γ
2
~
~2 ∂ 2 Ψ
+VΨ=
−
2m ∂x2
(
)
2
~2
1 2
imgt
~2 a
2
−
−2a x + gt /γ −
+ mgx Ψ.
mγ 2
2m
2
~
[F F]
So the time-dependent Schrödinger equation is satisfied if i~ times the square bracket in Equation [F] is equal
to the curly bracket in Equation [F F]:
a~2
2ia~gt
−
2
γ m
γ2
1
x + gt2
2
2
1 2
1 2
x + gt
+ mg x
+ 2 gt
2
2
2
~2
1 2
imgt
? ~ a
2
=
−
−2a
x
+
gt
/γ
−
+
mgx.
mγ 2
2m
2
~
2a2 ~2
− 4
γ m
I have cancelled the first terms on either side, and also the mgx terms. This leaves
2ia~gt
−
γ2
1
x + gt2
2
2
1 2
mg 2 t2
x + gt
+
2
2
"
#
2
2
2
~
4a
4iamgt
m2 g 2 t2
1 2
1 2
?
=−
x + gt
+
x + gt −
.
2m γ 4
2
~γ 2
2
~2
2a2 ~2
− 4
γ m
The terms quadratic in x + 12 gt2 cancel, as do the linear terms, and so do those of zeroth order. This confirms
that Ψ satisfies the Schrödinger equation.
To calculate the expectation value of x, first note that
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
r
2
2
|Ψ| = Ψ0 | =
r
hxi =
r
=
2a 1 −(x+ 12 gt2 )2 γ12 + (γ ∗1 )2
e
. But
π |γ|2
2a 1
π |γ|2
Z ∞
2a 1
π |γ|2
r
2
xe
−2a(x+ 21 gt2 ) /|γ|4
dx =
−∞
1
− gt2
2
Z ∞
2
55
1
1
(γ ∗ ) + γ 2
2
+
=
=
, so (letting y ≡ x+ 12 gt2 )
2
∗
2
4
γ
(γ )
|γ|
|γ|4
2a 1
π |γ|2
Z ∞
−∞
1 2 −2ay2 /|γ|4
y − gt e
dy
2
4
e−2ay /|γ| dy.
−∞
1
π/2a |γ|2 , so hxi = − gt2 . This is precisely the classical motion under free fall—as we should
2
have anticipated from Ehrenfest’s theorem.
The integral is
p
Problem 2.52
V(x)
(a)
x
(b)
dψ0
= −Aa sech(ax) tanh(ax);
dx
d2 ψ0
= −Aa2 − sech(ax) tanh2 (ax) + sech(ax) sech2 (ax) .
2
dx
~2 d 2 ψ 0
~2 a2
−
sech2 (ax)ψ0
2
2m dx
m
~2 a2
~2
=
Aa2 − sech(ax) tanh2 (ax) + sech3 (ax) −
A sech3 (ax)
2m
m
~2 a2 A =
− sech(ax) tanh2 (ax) + sech3 (ax) − 2 sech3 (ax)
2m
~2 a2
=−
A sech(ax) tanh2 (ax) + sech2 (ax) .
2m
sinh2 θ
1
sinh2 θ + 1
But (tanh2 θ + sech2 θ) =
+
=
= 1, so
2
2
cosh θ cosh θ
cosh2 θ
Hψ0 = −
=−
1 = |A|
2
~2 a2
ψ0 . QED
2m
Z ∞
Evidently E = −
~2 a2
.
2m
∞
2
= |A|2 =⇒ A =
sech (ax)dx = |A|
tanh(ax)
a
a
−∞
−∞
2
2 1
r
a
.
2
56
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
s(x)
x
(c)
dψk
A =
(ik − a tanh ax)ik − a2 sech2 ax eikx .
dx
ik + a
A d2 ψk
=
ik (ik − a tanh ax)ik − a2 sech2 ax − a2 ik sech2 ax + 2a3 sech2 ax tanh ax eikx .
dx2
ik + a
2
~2 a2
A
−~ ik 2
~2 d2 ψk
+
V
ψ
=
−k − iak tanh ax − a2 sech2 ax +
ik sech2 ax
−
k
2
2m dx
ik + a
2m
2m
~2 a3
~2 a2
−
sech2 ax tanh ax −
sech2 ax(ik − a tanh ax) eikx
m
m
=
Aeikx ~2
ik 3 − ak 2 tanh ax + ia2 k sech2 ax + ia2 k sech2 ax
ik + a 2m
−2a3 sech2 ax tanh ax − 2ia2 k sech2 ax + 2a3 sech2 ax tanh ax
~2 k 2
Aeikx ~2 2
k (ik − a tanh ax) =
ψk = Eψk . QED
ik + a 2m
2m
ik − a ikx
As x → +∞, tanh ax → +1, so ψk (x) → A
e , which represents a transmitted wave.
ik + a
=
R = 0.
ik − a
T =
ik + a
2
=
−ik − a
−ik + a
ik − a
ik + a
= 1.
Problem 2.53
(a) (1) From Eq. 2.136: F + G = A + B.
(2) From Eq. 2.138: F − G = (1 + 2iβ)A − (1 − 2iβ)B, where β = mα/~2 k.
1
Subtract: 2G = −2iβA + 2(1 − iβ)B ⇒ B =
(iβA + G). Multiply (1) by (1 − 2iβ) and add:
1 − iβ
1
1
iβ 1
2(1 − iβ)F − 2iβG = 2A ⇒ F =
(A + iβG). S =
.
1 − iβ
1 − iβ 1 iβ
(b) For an even potential, V (−x) = V (x), scattering from the right is the same as scattering from the left, with
x ↔ −x, A ↔ G, B ↔ F (see Fig. 2.21): F = S11 G + S12 A, B = S21 G + S22 A. So S11 = S22 , S21 = S12 .
(Note that the delta-well S matrix in (a) has this property.) In the case of the finite square well, Eqs. 2.170
and 2.171 give
S21 =
2
e−2ika
2
2
+l )
sin 2la
cos 2la − i (k 2kl
;
S11 =
2
−k )
i (l 2kl
sin 2la e−2ika
2
2
+l )
sin 2la
cos 2la − i (k 2kl
. So
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
S=
e−2ika
2
2
+l )
sin 2la
cos 2la − i (k 2kl
57
!
2
−k2 )
i (l 2kl
sin 2la
1
.
2
−k2 )
1
i (l 2kl
sin 2la
Problem 2.54
(a)
B = S11 A + S12 G ⇒ G =
1
S11
1
(B − S11 A) = M21 A + M22 B ⇒ M21 = −
, M22 =
.
S12
S12
S12
S22
(S11 S22 − S12 S21 )
S22
(B − S11 A) = −
A+
B = M11 A + M12 B.
S12
S12
S12
1
det S
S22
− det(S) S22
⇒ M11 = −
, M12 =
. M=
. Conversely:
−S11
1
S12
S12
S12
F = S21 A + S22 G = S21 A +
G = M21 A + M22 B ⇒ B =
1
M21
1
(G − M21 A) = S11 A + S12 G ⇒ S11 = −
; S12 =
.
M22
M22
M22
(M11 M22 − M12 M21 )
M12
M12
(G − M21 A) =
A+
G = S21 A + S22 G.
M22
M22
M22
1
det M
M12
−M21 1
⇒ S21 =
; S22 =
. S=
.
M22
M22
M22 det(M) M12
F = M11 A + M12 B = M11 A +
[It happens that the time-reversal invariance of the Schrödinger equation, plus conservation of probability,
∗
∗
requires M22 = M11
, M21 = M12
, and det(M) = 1, but I won’t use this here. See Merzbacher’s Quantum
Mechanics. Similarly, for even potentials S11 = S22 , S12 = S21 (Problem 2.53).]
Rl = |S11 |2 =
M21
M22
2
, Tl = |S21 |2 =
det(M)
M22
2
, Rr = |S22 |2 =
M12
M22
2
, Tr = |S12 |2 =
1
.
|M22 |2
(b)
A
C
F
B
D
G
M2
M1
x
F
C
C
A
F
A
A
= M2
,
= M1
, so
= M2 M1
=M
, with M = M2 M1 . QED
G
D
D
B
G
B
B
(c)
ψ(x) =
Aeikx + Be−ikx (x < a)
F eikx + Ge−ikx (x > a)
.
Continuity of ψ :
Aeika + Be−ika = F eika + Ge−ika
2mα
0
Discontinuity of ψ : ik F eika − Ge−ika − ik Aeika − Be−ika = − 2mα
Aeika + Be−ika .
~2 ψ(a) = − ~2
58
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
(1) F e2ika + G = Ae2ika + B.
2ika
(2) F e2ika − G = Ae2ika − B + i 2mα
+B .
~2 k Ae
Add (1) and (2):
2F e2ika = 2Ae2ika + i
mα
2mα
mα 2ika
A + i 2 e−2ika B = M11 A + M12 B.
Ae
+
B
⇒
F
=
1
+
i
~2 k
~2 k
~ k
So M11 = (1 + iβ); M12 = iβe−2ika ; β ≡
mα
.
~2 k
Subtract (2) from (1):
2G = 2B − 2iβe2ika A − 2iβB ⇒ G = (1 − iβ)B − iβe2ika A = M21 A + M22 B.
2ika
So M21 = −iβe
(1 + iβ) iβe−2ika
.
−iβe2ika (1 − iβ)
; M22 = (1 − iβ).
M=
(d)
M2 =
(1 + iβ) iβe−2ika
(1 + iβ) iβe2ika
;
to
get
M
,
just
switch
the
sign
of
a:
M
=
.
1
1
−iβe2ika (1 − iβ)
−iβe−2ika (1 − iβ)
M = M2 M1 =
[1 + 2iβ + β 2 (e−4ika − 1)] 2iβ[cos 2ka − β sin 2ka]
.
−2iβ[cos 2ka − β sin 2ka] [1 − 2iβ + β 2 (e4ika − 1)]
T = Tl = Tr =
1
⇒
|M22 |2
T −1 = [1 + 2iβ + β 2 (e−4ika − 1)][1 − 2iβ + β 2 (e4ika − 1)]
= 1 − 2iβ + β 2 e4ika − β 2 + 2iβ + 4β 2 + 2iβ 3 e4ika − 2iβ 3 + β 2 e−4ika
− β 2 − 2iβ 3 e−4ika + 2iβ 3 + β 4 (1 − e−4ika − e4ika + 1)
= 1 + 2β 2 + β 2 (e−4ika + e4ika ) − 2iβ 3 (e−4ika − e4ika ) + 2β 4 − β 4 (e−4ika + e4ika )
= 1 + 2β 2 + 2β 2 cos 4ka + 2iβ 3 2i sin 4ka + 2β 4 − 2β 4 cos 4ka
= 1 + 2β 2 (1 + cos 4ka) − 4β 3 sin 4ka + 2β 4 (1 − cos 4ka)
= 1 + 4β 2 cos2 2ka − 8β 3 sin 2ka cos 2ka − 4β 4 sin2 2ka
T =
1
1 + 4β 2 (cos 2ka − β sin 2ka)2
Problem 2.55
I’ll just show the first two graphs, and the last two. Evidently K lies between 0.9999 and 1.0001
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
59
Problem 2.56
The correct values (in Eq. 2.73) are K = 2n + 1 (corresponding to En = (n + 12 )~ω). I’ll start by “guessing”
2.9, 4.9, and 6.9, and tweaking the number until I’ve got 5 reliable significant digits. The results (see below)
are 3.0000, 5.0000, 7.0000. (The actual energies are these numbers multiplied by 12 ~ω.)
60
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
61
Problem 2.57
2
~
ψ 00 = Eψ, or, with the correct energies (Eq. 2.30) and a = 1, ψ 00 + (nπ)2 ψ =
The Schrödinger equation says − 2m
0. I’ll start with a “guess” using 9 in place of π 2 (that is, I’ll use 9 for the ground state, 36 for the first excited
state, 81 for the next, and finally 144). Then I’ll tweak the parameter until the graph crosses the axis right
at x = 1. The results (see below) are, to five significant digits: 9.8696, 39.478, 88.826, 157.91. (The actual
energies are these numbers multiplied by ~2 /2ma2 .)
62
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
Problem 2.58
(a) The total energy is simply N times the ground state energy of the infinite square well:
Ea = N
π 2 ~2
.
2ma2
(b) Filling the lowest N energy levels of the infinite square well (with width N a) gives
Eb =
X
N
n=1
N
n2 π 2 ~2
π 2 ~2 X 2
=
n .
2m(N A)2
2mN 2 a2 n=1
The sum is N (N + 1)(2N + 1)/6, so
Eb =
N
1
1
+ +
3
2 6N
π 2 ~2
.
2ma2
π 2 ~2
π 2 ~2
=
.
2ma2
3ma2
(c)
Ea − Eb
∆E
=
≈
N
N
N − (N/3)
N
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
63
(d) The binding energy of hydrogen (13.6 eV) is ~2 /2ma2B , where aB = 0.529 Å is the Bohr radius, so
2
∆E
2 2 aB 2
2 0.529 π
= π
Ebinding =
(13.6) eV = 1.6 eV.
N
3
a
3
4
Problem 2.59
(a)
−
~2 d2 ψ
+ mgx ψ = E ψ
2m dx2
dψ
dψ dz
dψ
=
=a ;
dx
dz dx
dz
−
(x ≥ 0;
ψ = 0 for x ≤ 0).
2
d2 ψ dz
d2 ψ
2d ψ
=
a
.
=
a
dx2
dz 2 dx
dz 2
√
z
z
~2 2 d2 ψ
~2 2 √ 00
z√
2m
2m
+
mg
a
ψ
=
Eψ
⇒
−
a a y + mg
a y = E a y ⇒ −y 00 + 2 2 mg y = 2 2 E y.
2m dz 2
a
2m
a
~ a
a
~ a
Let
2m
1
mg = 1, or a ≡
~2 a2
a
2m2 g
~2
1/3
and ≡
2m
2m
E= 2
~2 a2
~
~2
2m2 g
2/3
E=
Then −y 00 + zy = y. X
DSolve@- y ''@xD + x y@xD ä s y@xD, y@xD, xD
Out[1]=
88y@xD Æ AiryAi@- s + xD C@1D + AiryBi@- s + xD C@2D<<
In[2]:=
Plot@AiryAi@xD, 8x, - 15, 5<D
In[1]:=
0.4
0.2
Out[2]=
-15
-10
-5
5
-0.2
-0.4
In[3]:=
Plot@AiryBi@xD, 8x, - 15, 5<D
5
4
3
Out[3]=
2
1
-15
-10
-5
FindRoot@AiryAi@xD == 0, 8x, - 2<D
Out[4]=
8x Æ - 2.33811<
In[5]:=
FindRoot@AiryAi@xD == 0, 8x, - 12.8<D
In[4]:=
5
Out[5]=
8x Æ - 12.8288<
In[6]:=
NIntegrate@HAiryAi@xDL ^ 2, 8x, - 2.338107410459767`, •<D
Out[6]=
0.491697
In[7]:=
NIntegrate@HAiryAi@xDL ^ 2, 8x, - 12.828776752865757`, •<D
Out[7]=
1.14018
2
m~2 g 2
1/3
E.
64
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
FindRoot@AiryAi@xD == 0, 8x, - 2<D
Out[12]=
8x Æ - 2.33811<
In[13]:=
FindRoot@AiryAi@xD == 0, 8x, - 12.8<D
In[12]:=
Out[13]=
8x Æ - 12.8288<
In[14]:=
NIntegrate@HAiryAi@xDL ^ 2, 8x, - 2.338107410459767`, •<D
Out[14]=
0.491697
In[15]:=
NIntegrate@HAiryAi@xDL ^ 2, 8x, - 12.828776752865757`, •<D
Out[15]=
1.14018
In[16]:=
Pone@x_D := H0.4916966179009774`L ^ H- 1 ê 2L * AiryAi@x - 2.338107410459767`D
In[17]:=
Plot@Pone@xD, 8x, 0, 16<D
0.7
0.6
0.5
Out[17]=
0.4
0.3
0.2
0.1
5
10
15
10
15
In[18]:=
Pten@x_D := H1.1401837256117164`L ^ H- 1 ê 2L * AiryAi@x - 12.828776752865757`D
In[19]:=
Plot@Pten@xD, 8x, 0, 16<D
0.4
0.2
Out[19]=
5
-0.2
-0.4
0.1
5
10
15
Pten@x_D := H1.1401837256117164`L ^ H- 1 ê 2L * AiryAi@x - 12.828776752865757`D
Plot@Pten@xD, 8x, 0, 16<D
0.4
2
Problem 2.59 copy.nb
CHAPTER
2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
0.2
In[20]:=
65
NIntegrate@Pone@xD * Pten@xD, 8x, 0, •<D
5 :
NIntegrate::ncvb
NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in x near 8x< = 81.72352<. NIntegrate
15
obtained -9.19403 ¥ 10-17 and 1.5861121045064646`*^-16 for the integral and error estimates. á
-0.2
Out[20]=
10
- 9.19403 ¥ 10-17
-0.4
(b)
NIntegrate@Pone@xD * Pten@xD, 8x, 0, •<D
For ψ :
σ = 0.697089,
σ = 0.882819 ~,
σ σ = 0.615403 ~ > 0.5~ X.
NIntegrate::ncvb
: x
1
p
x p
NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in x near 8x< = 81.72352<. NIntegrate
For obtained
ψ10 : -9.19403
σx =¥3.8248,
σp = 2.06791 ~, σ
σ = 7.90935 ~ > 0.5~ X.
10-17 and 1.5861121045064646`*^-16
forxthep integral and error estimates. á
- 9.19403 ¥ 10-17
(See print-out.)
NIntegrate@x HPone@xDL ^ 2, 8x, 0, •<D
1.55874
NIntegrate@x ^ 2 HPone@xDL ^ 2, 8x, 0, •<D
2.9156
2.9155980068599967` - H1.558738273638599`L ^ 2
Problem 2.59 copy.nb
0.697089
NIntegrate@x HPten@xDL ^ 2, 8x, 0, •<D
8.55252
NIntegrate@x ^ 2 HPten@xDL ^ 2, 8x, 0, •<D
87.7747
87.77467357595424` - H8.552517834822023`L ^ 2
3.8248
NIntegrate@- ‰ Pone@xD HPone '@xDL, 8x, 0, •<D
NIntegrate::ncvb :
NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in x near 8x< = 81.2858<. NIntegrate
obtained 0. + 6.07153 ¥ 10-18 ‰ and 7.766131095614155`*^-17 for the integral and error estimates. á
0. + 6.07153 ¥ 10-18 ‰
NIntegrate@- Pone@xD HPone ''@xDL, 8x, 0, •<D
0.779369
NIntegrate@- Pten@xD HPten ''@xDL, 8x, 0, •<D
4.27626
0.7793691368188985`
0.882819
4.276258918045443`
2.06791
0.6970889478066314` * 0.8828188584409027`
0.615403
3.8248022512288737` * 2.067911728784728`
7.90935
3
66
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
1
(Equation 1.43). Here
T v(x)
r
2
1
2
E = mv + mgx ⇒ v =
(E − mgx) and
2
m
(c) ρC (x) =
1 2
E
gT = h =
⇒T =
2
mg
s
2E
,
mg 2
so
mg
1
mg
(2m2 g/~2 a2 )
1
a
1
ρC (x) = p
= p
= q
= p
→ p
.
2 g/~2 a2 )
2
2
~2 a2
~2 a2
2 E(E − mgx)
(
−
(2m
(
−
a)
2
(
− 1)
2
− mgx
4
2m
Problem 2.59 copy.nb
2m
For ψ10 , = 12.82877 (Out[13] on page 64). The graphs are
Plot@H4 * 12.828776752865757` H12.828776752865757` - xLL ^ H- 1 ê 2L,
8x, 0, 12.5<, PlotRange Æ 80, .26<D
0.25
0.20
0.15
0.10
0.05
2
4
6
8
10
12
2
4
6
8
10
12
4
6
8
10
12
Plot@HPten@xDL ^ 2, 8x, 0, 12.5<, PlotRange Æ 80, .26<D
0.25
0.20
0.15
0.10
0.05
Show@%74, %75D
0.25
0.20
0.15
0.10
0.05
2
Comment: Well, they agree, in a kind of averaged sense.
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
67
Problem 2.60
(a)
q
n
kg m2
kg m4
kg m2
p
n+p+q
2 n+2q
−(n+2q)
E0 = (~) (m) (α) =
(kg)
=
(kg)
(m
)
(s)
=
.
s
s2
s2
So n + p + q = 1, n + 2q = 1, n + 2q = 2. The last two are incompatible. Evidently there is, on purely
dimensional grounds, no possible formula for E0 .
n
p
q
(b) Let ψλ (x) ≡ ψ(y), where y ≡ λx. Then
2
2
d2 ψλ (x)
d2 ψ(y) 2 dy
β
β
2 d ψ(y)
2
2
=
=λ
= λ − 2 ψ(y) + κ ψ(y) = −λ2 2 2 ψλ (x) + λ2 κ2 ψλ (x),
y
dx2
d
dx
dy 2
y
λ x
β
d2 ψλ (x)
+ 2 ψλ (x) = (λκ)2 ψλ (x) = (κ0 )2 ψλ (x). So ψλ (x) is a solution to Equation 2.190, with κ0 ≡ λκ.
dx2
x
−2mE
−2mE 0
0 2
2 2
2
=
(κ
)
=
λ
κ
=
λ
⇒ E 0 = λ2 E.
The corresponding energy is given by
~2
~2
(c)
or
f@x_D := A
b - H1 ê 4L , k xF
x BesselKB‰
FullSimplifyBf ''@xD +
b
x^2
f@xD - k ^ 2 f@xD ä 0F
True
h@x_D :=
x BesselK@4 ‰, xD
Plot@h@xD, 8x, 0, 5<D
0.004
0.003
0.002
0.001
1
2
3
4
5
0.0006
0.0008
0.0010
-0.001
-0.002
Plot@h@xD, 8x, 0, .001<D
0.00006
0.00004
0.00002
0.0002
0.0004
-0.00002
-0.00004
‡ Hf@xDL ^ 2 ‚ x
•
0
ConditionalExpressionB
A2
- 1 + 4 b p CschB
4 k2
1
- 4 + b pF
, Re@kD > 0 && - 2 < ImA - 1 + 4 b E < 2F
68
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
s
So
A=κ
2 sinh(πg)
.
πg
(d) From the first plot on page 67 we see that the highest zero crossing occurs at κx ≈ 1.7; to find the exact
value, use FindRoot (see print-out below). We want this to occur at x = = 1, so κ = 1.69541.
h@x_D :=
x BesselK@4 ‰, xD
FindRoot@h@xD ä 0, 8x, 1.7<D
8x Æ 1.69541 + 0. ‰<
j@x_D := If@x >= 1, h@1.6954066933505485` xD, 0D
Plot@j@xD, 8x, 0, 6<D
0.004
0.003
0.002
0.001
1
2
3
4
5
6
The parameter β ≡ 2mα/~2 is dimensionless, so we may as well
eliminate ~ (in favor of β, m, and α) in
4
q
2
(m)r = (kg)p+q (m)4q+r (s)−2q = kgsm
, so
the dimensional analysis. This leaves E0 = mp αq r = (kg)p kgsm
2
2
p + q = 1, 4q + r = 2, q = 1, ⇒ p = 0, r = −2. On dimensional grounds, therefore, the expression for E0 has
to be of the form E0 = −α/2 times some function of β. As → 0 (for fixed values of m and α), E0 → −∞,
indicating once again that there is no ground state for this system.
Problem 2.61
(a) From Equation 2.197:
N = 1 : j = 1 : −λψ2 + (2λ)ψ1 − λψ0 = Eψ1
(
j = 1 : −λψ2 + (2λ)ψ1 − λψ0 = Eψ1
N =2:
j = 2 : −λψ3 + (2λ)ψ2 − λψ1 = Eψ2


j = 1 : −λψ2 + (2λ)ψ1 − λψ0 = Eψ1
N = 3 : j = 2 : −λψ3 + (2λ)ψ2 − λψ1 = Eψ2


j = 3 : −λψ4 + (2λ)ψ3 − λψ2 = Eψ2
(b) Denote the eigenvalues by Ẽn :
⇒
H = (2λ).
⇒
H=
2λ −λ
.
−λ 2λ

⇒

2λ −λ 0
H = −λ 2λ −λ .
0 −λ 2λ
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
69
8~2
n 2 π 2 ~2
π 2 ~2
2~2
=
.
The
exact
energies
are
E
=
,
so
E
=
; the agreement
n
1
2m(a/2)2
2ma2
2ma2
2ma2
2
is not too bad: 8 ≈ π = 9.87.
N = 1 : Ẽ1 = 2λ =
N =2:
2λ − Ẽ −λ
det
−λ 2λ − Ẽ
= 0 ⇒ (2λ − Ẽ)2 − λ2 = 0 ⇒ 2λ − Ẽ = ±λ.
9~2
~2
=
. This is better: 9 is closer to π 2 than 8 was.
2m(a/3)2
2ma2
3~2
27~2
Ẽ2 = 2λ + λ = 3λ =
=
. The exact answer has 4π 2 = 39.5 instead of 27.
2m(a/3)2
2ma2
Ẽ1 = 2λ − λ = λ =
N =3:


2λ − Ẽ −λ
0
√
det  −λ 2λ − Ẽ −λ  = 0 ⇒ (2λ − Ẽ)3 − 2λ2 (2λ − Ẽ) = 0 ⇒ 2λ − Ẽ = 0 or 2λ − Ẽ = ± 2 λ.
0
−λ 2λ − Ẽ
√
√
√
√
√
16(2 − 2)~2
(2 − 2)~2
=
. This is better yet: 16(2 − 2) = 9.37.
Ẽ1 = 2λ − 2 λ = λ(2 − 2) =
2m(a/4)2
2ma2
2~2
32~2
=
. Improving: the exact answer is 4π 2 = 39.5 instead of 32.
Ẽ2 = 2λ =
2m(a/4)2
2ma2
√
√
√
√
√
16(2 + 2)~2
(2 + 2)~2
Ẽ3 = 2λ + 2 λ = λ(2 + 2) =
=
; 16(2 + 2) = 54.6 ≈ 9π 2 = 88.8.
2m(a/4)2
2ma2
(c)
h = Table@If@i == j, 2 l, 0D, 8i, 10<, 8j, 10<D
k = Table@If@i ä j + 1, - l, 0D, 8i, 10<, 8j, 10<D
m = Table@If@i ä j - 1, - l, 0D, 8i, 10<, 8j, 10<D
p = Table@h@@i, jDD + k@@i, jDD + m@@i, jDD, 8i, 10<, 8j, 10<D
P = MatrixForm@%D
2 l -l 0
0
0
0
0
0
0
0
-l 2 l -l 0
0
0
0
0
0
0
0 -l 2 l -l 0
0
0
0
0
0
0
0 -l 2 l -l 0
0
0
0
0
0
0
0 -l 2 l -l 0
0
0
0
0
0
0
0 -l 2 l -l 0
0
0
0
0
0
0
0 -l 2 l -l 0
0
0
0
0
0
0
0 -l 2 l -l 0
0
0
0
0
0
0
0 -l 2 l -l
0
0
0
0
0
0
0
0 -l 2 l
l=1
1
EIG = Eigenvalues@N@pDD
83.91899, 3.68251, 3.30972, 2.83083, 2.28463,
1.71537, 1.16917, 0.690279, 0.317493, 0.0810141<
70
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
2
Ten.nb
~2
=
To get the eigenvalues, multiply by λ =
2m(a/11)2
In[16]:=
Out[16]=
121 * EIG ê Hp ^ 2L
121
π2
E1 :
848.0462, 45.147, 40.5767, 34.7056,
28.0092, 21.0302, 14.3339, 8.46272, 3.89242, 0.993221<
EVE
Thus theIn[17]:=
lowest
five= Eigenvectors@N@pDD
eigenvalues, in units of E1 , are 0.9932, 3.8924, 8.4627, 14.3339, 21.0302, as compared to
the exactOut[17]=
values
1,
4,
9,
16, 25. -Doing
the same
for N =-100:
880.120131,
0.23053,
0.322253,
0.387868, 0.422061, - 0.422061,
- 0.120131<,
8- 0.23053, 0.387868, - 0.422061,
h = 0.387868,
Table@If@i- 0.322253,
== j, 2 l, 0.23053,
0D, 8i, 100<,
8j, 100<D
0.322253, - 0.120131, - 0.120131, 0.322253, - 0.422061, 0.387868, - 0.23053<,
8- 0.322253, 0.422061, - 0.23053, - 0.120131, 0.387868, - 0.387868, 0.120131,
In[23]:=
- 0.422061,
- 0.322253, - 0.120131,
k =0.23053,
Table@If@i
ä j + 1, 0.322253<,
- l, 0D, 8i,80.387868,
100<, 8j, 100<D
0.422061, - 0.23053, - 0.23053, 0.422061, - 0.120131, - 0.322253, 0.387868<,
In[24]:= m = Table@If@i ä j - 1, - l, 0D, 8i, 100<, 8j, 100<D
80.422061, - 0.120131, - 0.387868, 0.23053, 0.322253, - 0.322253, - 0.23053,
0.387868, 0.120131, - 0.422061<, 80.422061, 0.120131, - 0.387868,
In[25]:= p = Table@h@@i, jDD + k@@i, jDD + m@@i, jDD, 8i, 100<, 8j, 100<D
- 0.23053, 0.322253, 0.322253, - 0.23053, - 0.387868, 0.120131, 0.422061<,
80.387868, - 0.322253, 0.120131, 0.422061, 0.23053, - 0.23053, - 0.422061,
In[26]:= l = 1
- 0.120131, 0.322253, 0.387868<, 80.322253, 0.422061, 0.23053, - 0.120131,
Out[26]= 1
- 0.387868, - 0.387868, - 0.120131, 0.23053, 0.422061, 0.322253<,
8- 0.23053, - 0.387868, - 0.422061, - 0.322253, - 0.120131, 0.120131,
= Eigenvalues@N@pDD
In[27]:= EIG0.322253,
0.422061, 0.387868, 0.23053<, 80.120131, 0.23053, 0.322253,
0.387868, 0.422061, 0.422061, 0.387868, 0.322253, 0.23053, 0.120131<<
In[28]:= 10 201 * EIG ê Hp ^ 2L
In[22]:=
ListLinePlot@EVE@@1DD,
PlotRange
Æ 80.5, 0.5<D
84133.31,
4130.31, 4125.32,
4118.33,
4109.36,
4098.41, 4085.5, 4070.64, 4053.84,
4035.11, 4014.49, 3991.97, 3967.6, 3941.39, 3913.36, 3883.55, 3851.98,
3818.69,
3783.7, 3747.04, 3708.77, 3668.9, 3627.49, 3584.57, 3540.18, 3494.36,
0.4
3447.16, 3398.63, 3348.81, 3297.75, 3245.5, 3192.11, 3137.63, 3082.12,
3025.62, 2968.2, 2909.9, 2850.79, 2790.92, 2730.35, 2669.14, 2607.35, 2545.03,
0.2
2482.25,
2419.07, 2355.55, 2291.76, 2227.74, 2163.57, 2099.3, 2035.01,
1970.74, 1906.57, 1842.55, 1778.75, 1715.24, 1652.06, 1589.28, 1526.96,
1465.17, 1403.96, 1343.39, 1283.52, 1224.41, 1166.11, 1108.69, 1052.19,
Out[18]=
996.679, 942.201,
888.81,
836.559,
785.499,
735.679,
687.147, 639.95, 594.133,
2
4
6
8
10
549.742, 506.819, 465.405, 425.541, 387.265, 350.614, 315.624, 282.328,
250.76, 220.948, 192.922, 166.71, 142.336, 119.824, 99.1963, 80.4724,
-0.2
63.6704, 48.8067, 35.8956, 24.9496, 15.9794, 8.99347, 3.99871, 0.999919<
In[18]:=
Out[28]=
-0.4
This time the lowest five eigenvalues are 0.9999, 3.9987, 8.9934, 15.9794, 24.9496.
(d) Always,
ψ0 = ψN +1 = 0; might as wellPlotRange
set λ = 1 for
this 0.5<D
part.
Æ 80,
In[19]:= ListLinePlot@EVE@@10DD,
0.5
N = 1 : 2ψ1 = Ẽψ1 , Ẽ1 = 2. Up to normalization:
0.4
0.3
Out[19]=
0.2
0.1
2
4
6
8
10
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
N =2:
2 −1
−1 2
ψ1
ψ1
= Ẽ
⇒ 2ψ1 − ψ2 = Ẽψ1 and − ψ1 + 2ψ2 = Ẽψ2 .
ψ2
ψ2
n = 1 : Ẽ1 = 1 ⇒ 2ψ1 − ψ2 = ψ1 ⇒ ψ2 = ψ1
n = 2 : Ẽ2 = 3 ⇒ 2ψ1 − ψ2 = 3ψ1 ⇒ ψ2 = −ψ1

 
 
2 −1 0
ψ1
ψ1
N = 3 : −1 2 −1 ψ2  = Ẽ ψ2  ⇒ 2ψ1 − ψ2 = Ẽψ1 , −ψ1 + 2ψ2 − ψ3 = Ẽψ2 , −ψ2 + 2ψ3 = Ẽψ3 .
0 −1 2
ψ3
ψ3
√
√
√
n = 1 : Ẽ1 = 2 − 2 ⇒ 2ψ1 − ψ2 = (2 − 2)ψ1 ⇒ ψ2 = 2ψ1 ;
√
√
−ψ1 + 2ψ2 − ψ3 = (2 − 2)ψ2 ⇒ ψ1 + ψ3 = 2ψ2 = 2ψ1 ⇒ ψ3 = ψ1
n = 2 : Ẽ2 = 2 ⇒ 2ψ1 − ψ2 = 2ψ1 ⇒ ψ2 = 0; −ψ1 + 2ψ2 − ψ3 = 2ψ2 ⇒ ψ3 = −ψ1
√
√
√
n = 3 : Ẽ3 = 2 + 2 ⇒ 2ψ1 − ψ2 = (2 + 2)ψ1 ⇒ ψ2 = − 2ψ1 ;
√
√
−ψ1 + 2ψ2 − ψ3 = (2 + 2)ψ2 ⇒ ψ1 + ψ3 = − 2ψ2 = 2ψ1 ⇒ ψ3 = ψ1
71
72
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
For N = 10 we get
EVE = Eigenvectors@N@pDD
ListLinePlot@EVE@@10DD, PlotRange Ø 80, 0.5<D
0.5
0.4
0.3
0.2
0.1
2
4
6
8
10
ListLinePlot@EVE@@9DD, PlotRange Ø 8- 0.5, 0.5<D
0.4
0.2
2
4
6
8
10
-0.2
-0.4
ListLinePlot@EVE@@8DD, PlotRange Ø 8- 0.5, 0.5<D
0.4
0.2
2
-0.2
-0.4
4
6
8
10
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
and for N = 100
EVE = Eigenvectors@N@pDD
ListLinePlot@EVE@@100DD, PlotRange -> 80, 0.2<D
0.20
0.15
0.10
0.05
20
40
60
80
100
20
40
60
80
100
ListLinePlot@EVE@@99DD, PlotRange -> 8- 0.2, 0.2<D
0.2
0.1
-0.1
-0.2
ListLinePlot@EVE@@98DD, PlotRange -> 8- 0.2, 0.2<D
0.2
0.1
20
-0.1
-0.2
40
60
80
100
73
74
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
Problem 2.62
~2
(N + 1)2 = (N + 1)2 V0 (here N = 100);
2ma2
πx πj∆x
πj
j
Vj = bV0 sin
= bV0 sin
= bV0 sin
a
a
N +1
Vj
πj
b
vj =
sin
=
.
λ
(N + 1)2
N +1
λ=
2
(here b = 500);
2p62.nb
Factoring out λ, the diagonal elements of H are
In[8]:=
ListLinePlot@EVE@@98DD, PlotRange Æ 8.1, 0.2<D
b
2 + vj = 2 +
sin
(N + 1)2
0.20
πj
N +1
500
=2+
sin
10201
πj
101
.
0.15
In[1]:=
Out[8]=
In[2]:=
In[3]:=
h = Table@If@i == j, 2 + H500 ê 10 201L Sin@p j ê 101D, 0D, 8i, 100<, 8j, 100<D
0.10
k = Table@If@i ä j + 1, - 1, 0D, 8i, 100<, 8j, 100<D
0.05
20
40 - 1, 0D,
60 8i, 100<,
80 8j, 100<D
100
m = Table@If@i
ä j - 1,
p = Table@h@@i, jDD + k@@i, jDD + m@@i, jDD, 8i, 100<, 8j, 100<D
-0.05
In[4]:=
-0.10
In[5]:=
In[9]:=
EVE = Eigenvectors@N@pDD
EIG = Eigenvalues@N@pDD
10 201 * EIG
ListLinePlot@EVE@@100DD, PlotRange Æ 80, 0.2<D
Out[10]= 841 255., 41 158.1, 41 063.4, 40 968.8, 40 869.4, 40 758.8, 40 632., 40 486.7, 40 322.1,
0.20
40 138.4, 39 935.6, 39 714.1, 39 474., 39 215.7, 38 939.5, 38 645.5, 38 334.2,
38 005.7, 37 660.6, 37 299., 36 921.3, 36 528., 36 119.3, 35 695.8, 35 257.7,
34 805.6, 34 339.9, 33 860.9, 33 369.3, 32 865.4, 32 349.7, 31 822.8, 31 285.1,
0.15
30 737.3, 30 179.7, 29 613., 29 037.7, 28 454.3, 27 863.4, 27 265.6, 26 661.5,
26 051.7, 25 436.7, 24 817.1, 24 193.6, 23 566.7, 22 937., 22 305.2, 21 671.9,
21 037.6, 20 403.1, 19 768.8, 19 135.5, 18 503.7, 17 874.1, 17 247.2, 16 623.6,
16 004., 15 389., 14 779.2, 14 175.1, 13 577.3, 12 986.5, 12 403.1, 11 827.8,
Out[6]= 0.10
11 261.1, 10 703.5, 10 155.6, 9617.98, 9091.08, 8575.44, 8071.55, 7579.91,
7100.98, 6635.24, 6183.13, 5745.1, 5321.57, 4912.95, 4519.64, 4142.02, 3780.47,
3435.34, 3106.97, 2795.68, 2501.8, 2225.62, 1967.43, 1727.51, 1506.15, 1303.63,
0.05
1120.24, 956.36, 812.457, 689.191, 590.237, 499.854, 476.163, 304.8, 304.66<
In[10]:=
In[6]:=
So the lowest three energies are 304.66 V0 , 304.8 V0 , and 476.163 V0 . Notice that the ground state is almost
20
40 two separated
60
100 barrier in between them, and the particle can
degenerate—essentially
we have
wells80with a huge
be either in the left one or in the right one (or the even and odd linear combinations thereof).
In[7]:=
ListLinePlot@EVE@@99DD, PlotRange Æ 8- .2, 0.2<D
0.2
0.1
Out[7]=
20
40
60
80
100
In[1]:=
In[2]:=
h = Table@If@i == j, 2 + H500 ê 10 201L Sin@p j ê 101D, 0D, 8i, 100<, 8j, 100<D
k = Table@If@i ä j + 1, - 1, 0D, 8i, 100<, 8j, 100<D
In[3]:=
m = Table@If@i ä j - 1, - 1, 0D, 8i, 100<, 8j, 100<D
In[4]:=
p = Table@h@@i, jDD + k@@i, jDD + m@@i, jDD, 8i, 100<, 8j, 100<D
In[5]:=
EVE = Eigenvectors@N@pDD
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
In[6]:=
ListLinePlot@EVE@@100DD, PlotRange Æ 80, 0.2<D
0.20
0.15
Out[6]= 0.10
0.05
In[7]:=
20
40
60
80
100
20
40
60
80
100
ListLinePlot@EVE@@99DD, PlotRange Æ 8- .2, 0.2<D
0.2
0.1
Out[7]=
-0.1
2
2p62.nb
-0.2
In[8]:=
ListLinePlot@EVE@@98DD, PlotRange Æ 8- .1, 0.2<D
0.20
0.15
0.10
Out[8]=
0.05
20
40
60
80
100
-0.05
-0.10
Notice
that
the central barrier pushes the wave function out to the wings.
= Eigenvalues@N@pDD
In[9]:= EIG
In[10]:=
Out[10]=
10 201 * EIG
841 255., 41 158.1, 41 063.4, 40 968.8, 40 869.4, 40 758.8, 40 632., 40 486.7, 40 322.1,
40 138.4, 39 935.6, 39 714.1, 39 474., 39 215.7, 38 939.5, 38 645.5, 38 334.2,
38 005.7, 37 660.6, 37 299., 36 921.3, 36 528., 36 119.3, 35 695.8, 35 257.7,
34 805.6, 34 339.9, 33 860.9, 33 369.3, 32 865.4, 32 349.7, 31 822.8, 31 285.1,
30 737.3, 30 179.7, 29 613., 29 037.7, 28 454.3, 27 863.4, 27 265.6, 26 661.5,
26 051.7, 25 436.7, 24 817.1, 24 193.6, 23 566.7, 22 937., 22 305.2, 21 671.9,
21 037.6, 20 403.1, 19 768.8, 19 135.5, 18 503.7, 17 874.1, 17 247.2, 16 623.6,
16 004., 15 389., 14 779.2, 14 175.1, 13 577.3, 12 986.5, 12 403.1, 11 827.8,
11 261.1, 10 703.5, 10 155.6, 9617.98, 9091.08, 8575.44, 8071.55, 7579.91,
7100.98, 6635.24, 6183.13, 5745.1, 5321.57, 4912.95, 4519.64, 4142.02, 3780.47,
75
76
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
Problem 2.63
X
∂
1 ∂Z
1 X
ln(Z) = −
=−
(−En )e−βEn =
En P (n). X
∂β
Z ∂β
Z n
n
1
(b) Geometric series: 1 + x + x2 + x3 + · · · =
. Here x = e−β~ω :
1−x
(a) −
Z=
∞
X
1
e−β(n+ 2 )~ω = e−β~ω/2
∞
X
e(−β~ω)n = e−β~ω/2
n=0
n=0
1
. X
1 − e−β~ω
(c)
β~ω
ln Z = ln e−β~ω/2) − ln 1 − e−β~ω = −
− ln 1 − e−β~ω ,
2
~ω 1 − e−β~ω + 2e−β~ω
~ω 1 + e−β~ω
∂
~ω
~ωe−β~ω
=
−
⇒
Ē
=
. X
ln Z = −
−
−β~ω
−β~ω
∂β
2
1−e
2
1−e
2 1 − e−β~ω
(d)
∂ Ē
∂ Ē dβ
1 ∂ Ē
=
=−
.
∂T
∂β dT
kB T 2 ∂β
1 − e−β~ω −~ωe−β~ω − 1 + e−β~ω ~ωe−β~ω
~ω
∂ Ē
~ω −2~ωe−β~ω
(~ω)2 eβ~ω
=
=
=
−
2
2.
∂β
2
2 (1 − e−β~ω )2
(1 − e−β~ω )
(eβ~ω − 1)
1
eβ~ω
C=3
(~ω)2
2.
2
kB T
(eβ~ω − 1)
Or, using β = 1/kB T and ~ω = kB θE ,
C=3
(e)
In[12]:=
1
kB T 2
2
(kB θE )
eθE /T
2 = 3kB
eθE /T − 1
θE
T
2
eθE /T
eθE /T − 1
2 . X
Plot@3 * Hx ^ H- 2LL * Exp@1 ê xD ê HExp@1 ê xD - 1L ^ 2, 8x, 0, .7<, PlotRange Æ 80, 2.6<D
2.5
2.0
1.5
Out[12]=
1.0
0.5
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Incidentally, comparing the graphs suggests that x = 0.7 corresponds to T = 1000 K, so θE = 1000/0.7 K
= 1400 K. Then ~ω = (1400kB )K = (1400)(8.6 × 10−5 ) eV = 0.12 eV.
CHAPTER 2. THE TIME-INDEPENDENT SCHRÖDINGER EQUATION
77
Problem 2.64
(a) Plugging into the differential equation we have
∞
X
an n (n − 1) xn−2 −
n=2
∞
X
an n (n − 1) xn − 2
n=2
∞
X
an n xn + ` (` + 1)
n=1
∞
X
an xn = 0 .
n=0
Reindexing the sums so that all the powers of x match we have
∞
X
ap+2 (p + 2) (p + 1) xp −
p=0
∞
X
ap p (p − 1) xp − 2
p=2
∞
X
ap p xp + ` (` + 1)
p=1
∞
X
a p xp = 0 .
p=0
We can then combine the sums (extending the second and third sums to begin at p = 0) to get
∞
X
{(p + 2) (p + 1) ap+2 − [p (p − 1) + 2 p − ` (` + 1)] ap } xp = 0 .
p=0
From this we read off the recursion relation
p (p + 1) − ` (` + 1)
ap .
(p + 2) (p + 1)
ap+2 =
(b) For large values of p,
ap+2 ≈
p
ap .
p+2
with the (approximate) solution
ap ≈
C
.
p
and this gives the behavior
f (x) ≈ C
X1
p
xp ≈ log
1
1−x
which diverges at x = 1. (There will be finite corrections coming from the low values of p, but these cannot fix
the divergence.)
(c) For ` = 0 we need the even function (a1 = 0) and the recursion relation gives a2 = 0 so that
P0 (x) = a0 .
For ` = 1 we need the odd function (a0 = 0) and the recursion relation gives a3 = 0 so that
P1 (x) = a1 x .
For ` = 2 we need the even function again and the recursion relation gives a2 = −3 a0 and a4 = 0 so that
P2 (x) = a0 1 − 3 x2 .
For ` = 3 we need the odd function again and the recursion relation gives a3 = −5/3 a1 and a5 = 0 so that
5 3
P3 (x) = a1 x − x
.
3
78
CHAPTER 3. FORMALISM
Chapter 3
Formalism
Problem 3.1
(a) All conditions are trivial except Eq. A.1: we need to show that the sum of two square-integrable functions
is itself square-integrable. Let h(x) = f (x) + g(x), so that |h|2 = (f + g)∗ (f + g) = |f |2 + |g|2 + f ∗ g + g ∗ f
and hence
Z
∗
Z
Z
Z
Z
2
2
2
∗
∗
|h| dx = |f | dx + |g| dx + f g dx +
f g dx .
If f (x) and g(x)
R are square-integrable, then the first two terms are finite, and (by Eq. 3.7) so too are the
last two. So |h|2 dx is finite. QED
The set of all normalized functions is certainly not a vector space: it doesn’t include 0, and the sum of
two normalized functions is not (in general) normalized—in fact, if f (x) is normalized, then the square
integral of 2f (x) is 4.
(b) Equation A.19 is trivial:
Z b
hg|f i =
∗
Z b
g(x) f (x) dx =
a
!∗
∗
f (x) g(x) dx
= hf |gi∗ .
a
Equation A.20 holds (see Eq. 3.9) subject to the understanding in footnote 6. As for Eq. A.21, this is
pretty obvious:
Z
Z
Z
hf |(b|gi + c|hi) = f (x)∗ (bg(x) + ch(x)) dx = b f ∗ g dx + c f ∗ h dx = bhf |gi + chf |hi.
Problem 3.2
(a)
Z 1
hf |f i =
x2ν dx =
0
1
1
1
x2ν+1 =
1 − 02ν+1 .
2ν + 1
2ν + 1
0
Now 02ν+1 is finite (in fact, zero) provided (2ν + 1) > 0, which is to say, ν > − 21 . If (2ν + 1) < 0 the
integral definitely blows up. As for the critical case ν = − 21 , this must be handled separately:
Z 1
1
hf |f i =
x−1 dx = ln x 0 = ln 1 − ln 0 = 0 + ∞.
0
CHAPTER 3. FORMALISM
79
So f (x) is in Hilbert space only for ν strictly greater than -1/2.
(b) For ν = 1/2, we know from (a) that f (x) is in Hilbert space: yes.
Since xf = x3/2 , we know from (a) that it is in Hilbert space: yes.
For df /dx = 21 x−1/2 , we know from (a) that it is not in Hilbert space: no.
[Moral: Simple operations, such as differentiating (or multiplying by 1/x), can carry a function out of
Hilbert space.]
Problem 3.3
Suppose hh|Q̂hi = hQ̂h|hi for all functions h(x). Let h(x) = f (x) + cg(x) for some arbitrary constant c. Then
hh|Q̂hi = h(f + cg)|Q̂(f + cg)i = hf |Q̂f i + chf |Q̂gi + c∗ hg|Q̂f i + |c|2 hg|Q̂gi;
hQ̂h|hi = hQ̂(f + cg)|(f + cg)i = hQ̂f |f i + chQ̂f |gi + c∗ hQ̂g|f i + |c|2 hQ̂g|gi.
Equating the two and noting that hf |Q̂f i = hQ̂f |f i and hg|Q̂gi = hQ̂g|gi leaves
chf |Q̂gi + c∗ hg|Q̂f i = chQ̂f |gi + c∗ hQ̂g|f i.
In particular, choosing c = 1:
hf |Q̂gi + hg|Q̂f i = hQ̂f |gi + hQ̂g|f i,
whereas if c = i:
hf |Q̂gi − hg|Q̂f i = hQ̂f |gi − hQ̂g|f i.
Adding the last two equations:
hf |Q̂gi = hQ̂f |gi. QED
Problem 3.4
(a) hf |(Ĥ + K̂)gi = hf |Ĥgi + hf |K̂gi = hĤf |gi + hK̂f |gi = h(Ĥ + K̂)f |gi. X
(b) hf |αQ̂gi = αhf |Q̂gi; hαQ̂f |gi = α∗ hQ̂f |gi. Hermitian ⇔ α is real.
(c) hf |Ĥ K̂gi = hĤf |K̂gi = hK̂ Ĥf |gi, so Ĥ K̂ is hermitian ⇔ Ĥ K̂ = K̂ Ĥ, or [Ĥ, K̂] = 0.
R
R
(d) hf |x̂gi = f ∗ (xg) dx = (xf )∗ g dx = hx̂f |gi. X
Z
Z
Z
2
~2 d 2
~2
∗d g
∗
hf |Ĥgi = f −
+
V
g
dx
=
−
f
dx
+
f ∗ V g dx.
2m dx2
2m
dx2
Integrating by parts (twice):
Z ∞
Z ∞ ∗
Z ∞ 2 ∗
∞
∞
∞
2
df dg
df ∗
d f
∗d g
∗ dg
∗ dg
f
dx = f
−
dx = f
−
g
+
g dx.
2
2
dx
dx
dx
dx
dx
dx
−∞
−∞
−∞ dx
−∞
−∞
−∞
But for functions f (x) and g(x) in Hilbert space the boundary terms vanish, so
Z ∞
Z ∞ 2 ∗
2
d f
∗d g
f
dx =
g dx, and hence (assuming that V (x) is real):
2
2
dx
−∞
−∞ dx
∗
Z ∞
~2 d 2 f
hf |Ĥgi =
−
+
V
f
g dx = hĤf |gi. X
2m dx2
−∞
80
CHAPTER 3. FORMALISM
Problem 3.5
R
f ∗ (xg) dx = (xf )∗ g dx = hxf |gi, so x† = x.
R
R
hf |igi = f ∗ (ig) dx = (−if )∗ g dx = h−if |gi, so i† = −i.
R
(a) hf |xgi =
dg
hf | i =
dx
Z ∞
f
−∞
∗ dg
dx
∞
∗
Z ∞
−
dx = f g
−∞
−∞
df
dx
∗
df
g dx = −h |gi, so
dx
d
dx
†
=−
d
.
dx
1
1
(−ip + mωx). But p and x are hermitian, and i† = −i, so (a+ )† = √
(ip + mωx), or
2~mω
2~mω
(a+ )† = (a− ).
(b) a+ = √
(c) hf |(Q̂R̂)gi = hQ̂† f |R̂gi = hR̂† Q̂† f |gi = h(Q̂R̂)† f |gi, so (Q̂R̂)† = R̂† Q̂† . X
Problem 3.6
Z 2π 2 ∗
Z 2π ∗
2π
2π
2π
df ∗
d f
df dg
d2 g
∗ dg
∗ dg
+
−
dφ
=
f
−
g
dφ
=
f
g dφ.
2
dφ
dφ
dφ
dφ
dφ
dφ
dφ2
0
0
0
0
0
0
As in Example 3.1, for periodic functions (Eq. 3.26) the boundary terms vanish, and we conclude that hf |Q̂gi =
hQ̂f |gi, so Q̂ is hermitian: yes.
Z 2π
hf |Q̂gi =
f∗
√
d2 f
= qf ⇒ f± (φ) = Ae± qφ .
2
dφ
√
√
The periodicity condition (Eq. 3.26) requires that q(2π) = 2nπi, or q = in, so the eigenvalues are
Q̂f = qf ⇒
q = −n2 , (n = 0, 1, 2, . . .). The spectrum is doubly degenerate; for a given n there are two eigenfunctions
(the plus sign or the minus sign, in the exponent), except for the special case n = 0, which is not degenerate.
Problem 3.7
(a) Suppose Q̂f = qf and Q̂g = qg. Let h(x) = af (x) + bg(x), for arbitrary constants a and b. Then
Q̂h = Q̂(af + bg) = a(Q̂f ) + b(Q̂g) = a(qf ) + b(qg) = q(af + bg) = qh. X
(b)
d2 f
d2
d x
d2 g
d2
d
= 2 (ex ) =
(e ) = ex = f,
= 2 e−x =
−e−x = e−x = g.
2
2
dx
dx
dx
dx
dx
dx
So both of them are eigenfunctions, with the same eigenvalue 1. The simplest orthogonal linear combinations are
sinh x =
1
1 x
e − e−x = (f − g)
2
2
and
cosh x =
1
1 x
e + e−x = (f + g).
2
2
(They are clearly orthogonal, since sinh x is odd while cosh x is even.)
CHAPTER 3. FORMALISM
81
Problem 3.8
(a) The eigenvalues (Eq. 3.29) are 0, ±1, ±2, . . ., which are obviously real. X For any two eigenfunctions,
0
f = Aq e−iqφ and g = Aq0 e−iq φ (Eq. 3.28), we have
2π
Z 2π
0
i
A∗q Aq0 h i(q−q0 )2π
ei(q−q )φ
∗
iqφ −iq 0 φ
∗
=
e
−
1
.
hf |gi = Aq Aq0
e e
dφ = Aq Aq0
i(q − q 0 )
i(q − q 0 )
0
0
0
But q and q are integers, so e
nonzero). X
i(q−q 0 )2π
= 1, and hence hf |gi = 0 (provided q 6= q 0 , so the denominator is
(b) In Problem 3.6 the eigenvalues are q = −n2 , with n = 0, 1, 2, . . ., which are obviously real. X
0
two eigenfunctions, f = Aq e±inφ and g = Aq0 e±in φ , we have
2π
Z 2π
0
i
A∗q Aq0 h ±i(n0 −n)2π
e±i(n −n)φ
∗
∓inφ ±in0 φ
∗
=
e
e
hf |gi = Aq Aq0
e
dφ = Aq Aq0
−
1
=0
±i(n0 − n)
±i(n0 − n)
0
For any
0
(provided n 6= n0 ).
Problem 3.9
(a) Infinite square well (Eq. 2.22).
(b) Delta-function barrier (Fig. 2.15), or the finite rectangular barrier (Prob. 2.33).
(c) Delta-function well (Eq. 2.114), or the finite square well (Eq. 2.148) or the sech2 potential (Prob. 2.52).
Problem 3.10
From Eq. 2.31, with n = 1:
~ d
p̂ ψ1 (x) =
i dx
r
π ~
2
sin
x =
a
a
i
r
π π~
π 2π
cos
x = −i
cot
x ψ1 (x).
aa
a
a
a
Since p̂ ψ1 is not a (constant) multiple of ψ1 , ψ1 is not an eigenfunction of p̂: no.
Problem 3.11
Ψ0 (x, t) =
mω
π~
1/4
e
2
− mω
−iωt/2
2~ x
e
;
1/4
Z ∞
mω 2
1
mω
Φ(p, t) = √
e−iωt/2
e−ipx/~ e− 2~ x dx.
2π~ π~
−∞
From Problem 2.21(b):
r
1/4
2
1
mω
2π~ −p2 /2mω~
1
−iωt/2
Φ(p, t) = √
e
e
=
e−p /2mω~ e−iωt/2 .
1/4
mω
(πmω~)
2π~ π~
82
CHAPTER 3. FORMALISM
√
2
1
p2
1
e−p /mω~ . Maximum classical momentum:
= E = ~ω =⇒ p = mω~.
2m
2
πmω~
So the probability it’s outside classical range is:
|Φ(p, t)|2 = √
Z −√mω~
P =
|Φ|2 dp +
−∞
Z √mω~
Z ∞
√
Z √mω~
|Φ|2 dp = 1 − 2
|Φ|2 dp.
Now
0
mω~
r
r
Z √mω~
1
2
mω~
−p2 /mω~
|Φ| dp = √
e
dp. Let z ≡
p, so dp =
dz.
mω~
2
πmω~ 0
0
Z √2
√
2
1
1
=√
e−z /2 dz = F ( 2) − , in CRC Table notation.
2
2π 0
h
√
√ i
√
1
P = 1 − 2 (F ( 2) −
= 1 − 2F ( 2) + 1 = 2 1 − F ( 2) = 0.157.
2
2
To two digits: 0.16 (compare Prob. 2.14).
Problem 3.12
Let k ≡ p/~ in Equation 2.101:
1
Ψ(x, t) = √
2π
Z ∞
2
φ(p/~)ei(px/~) e−ip t/2m
−∞
dp
.
~
Comparing Equation 3.55, read off
Φ(p, t) =
1 −ip2 t/2m
e
φ(p/~).
~
1
|φ(p/~)|2 is independent of t.
~
Alternatively, plug 2.101 into 3.54, and reverse the order of integration:
Z ∞
Z ∞
1
1
−ipx/~
i(kx−~2 k2 t/2m)
√
Φ(p, t) = √
e
φ(k)e
dk dx
2π −∞
2π~ −∞
Z ∞
Z ∞
2 2
1
1
=√
φ(k)e−i~ k t/2m
ei(k−p/~)x dx dk
2π −∞
~ −∞
Z ∞
2 2
2
1
1
=√
φ(k)e−i~ k t/2m δ (k − p/~) dk = √ e−ip t/2m φ(p/~)
~ −∞
~
Note that |Φ(p, t)|2 =
(I used Equation 2.147 in the third line).
Problem 3.13
Z ∞
1
eipx/~ Φ(p, t)dp.
2π~ −∞
Z
Z Z
Z
1
1
∗
−ip0 x/~ ∗ 0
0
+ipx/~
√
hxi = Ψ xΨdx =
e
Φ (p , t)dp x √
e
Φ(p, t)dp dx.
2π~
2π~
From Eq. 3.55: Ψ(x, t) = √
CHAPTER 3. FORMALISM
83
d ipx/~ e
, so (integrating by parts):
dp
Z
Z
Z
~ d ipx/~
~ ∂
x eipx/~ Φ dp =
e
)Φ dp = eipx/~ −
Φ(p, t) dp.
i dp
i ∂p
ZZZ ~ ∂
1
−ip0 x/~ ∗ 0
ipx/~
e
Φ (p , t)e
−
Φ(p, t) dp0 dp dx.
So hxi =
2π~
i ∂p
But xeipx/~ = −i~
Do the x integral first, letting y ≡ x/~:
Z
Z
0
1
1
−ip0 x/~ ipx/~
e
e
dx =
ei(p−p )y dy = δ(p − p0 ), (Eq. 2.144), so
2π~
2π
ZZ
Z
~ ∂
~ ∂
hxi =
Φ∗ (p0 , t)δ(p − p0 ) −
Φ(p, t) dp0 dp = Φ∗ (p, t) −
Φ(p, t) dp. QED
i ∂p
i ∂p
Problem 3.14
(a)
[Â + B̂, Ĉ] = (Â + B̂)Ĉ − Ĉ(Â + B̂) = ÂĈ + B̂ Ĉ − Ĉ Â − Ĉ B̂ = [Â, Ĉ] + B̂, Ĉ] X
[ÂB̂, Ĉ] = ÂB̂ Ĉ − Ĉ ÂB̂ = ÂB̂ Ĉ − ÂĈ B̂ + ÂĈ B̂ − Ĉ ÂB̂ = Â[B̂, Ĉ] + [Â, Ĉ]B̂.
X
(b) Introducing a test function g(x), as in Eq. 2.51:
~ d n
~
n
n ~ dg
n ~ dg
n−1
n dg
[x , p̂]g = x
−
(x g) = x
−
nx
g+x
= i~nxn−1 g.
i dx
i dx
i dx
i
dx
So, dropping the test function, [xn , p̂] = i~nxn−1 . X
~ dg
~ d
~ dg
~ df
dg
df
df
(c) [f, p̂]g = f
−
(f g) = f
−
g+f
= i~ g ⇒ [f, p̂] = i~ . X
i dx
i dx
i dx
i dx
dx
dx
dx
(d)
n
o
] = ~ω {â [â , â ] + [â , â ]â } .
[Ĥ, â± ] = [~ω(â− â+ − 21 ), â± ] = ~ω [â− â+ , â± ] − [ 21 , â±
− + ±
− ± +
So, using [â− , â+ ] = 1 (Equation 2.56),
] + [â , â ]â } = ~ωâ , and
, [Ĥ, â+ ] = ~ω {â−
[â+
â+
− + +
+
, â− ]â+ } = ~ω(−â− ), or, combining them
[â−
[Ĥ, â− ] = ~ω {â− [â+ , â− ] + [Ĥ, â± ] = ±~ωâ± . X
Problem 3.15
p2
1 x,
+V =
x, p2 + [x, V ];
2m
2m
Using Eq. 2.52: x, p2 = i~p + pi~ = 2i~p.
x, p2 = xp2 − p2 x = xp2 − pxp + pxp − p2 x = [x, p]p + p[x, p].
And [x, V ] = 0,
p2
1
i~p
+V =
2i~p =
.
so x,
2m
2m
m
84
CHAPTER 3. FORMALISM
The generalized uncertainty principle (Eq. 3.62) says, in this case,
2
σx2 σH
≥
2 2
1 i~
~
~
hpi =
hpi ⇒ σx σH ≥
|hpi|. QED
2i m
2m
2m
For stationary states σH = 0 and hpi = 0, so it just says 0 ≥ 0.
Problem 3.16
Suppose P̂ fn = λn fn and Q̂fn = µn fn (that is: fn (x) is an eigenfunction both of P̂ and of Q̂), and the set
P {fn }
is complete, so that any function f (x) (in Hilbert space) can be expressed as a linear combination: f = cn fn .
Then
X
X
X
X
X
[P̂ , Q̂]f = (P̂ Q̂ − Q̂P̂ )
cn fn = P̂
cn µn fn − Q̂
cn λn fn =
cn µn λn fn −
cn λn µn fn = 0.
Since this is true for any function f , it follows that [P̂ , Q̂] = 0.
Problem 3.17
dΨ
i
a
i
= (iax − iahxi + hpi)Ψ =
−x + hxi + hpi Ψ.
dx
~
~
a
dΨ
a
ihpi
a
x2
ihpi
=
− x + hxi +
dx ⇒ ln Ψ =
−
+ hxix +
x + constant.
Ψ
~
a
~
2
a
Let constant = −
hxi2 a
a
ihpi
+ B (B a new constant). Then ln Ψ = − (x − hxi)2 +
x + B.
2~
2~
~
2
a
2
Ψ = e− 2~ (x−hxi) eihpix/~ eB = Ae−a(x−hxi) /2~ eihpix/~ , where A ≡ eB .
Problem 3.18
(a) 1 commutes with everything, so
d
hΨ|Ψi = 0 (this is the conservation of normalization, which we origidt
nally proved in Eq. 1.27).
d
hHi = 0 (assuming H has no explicit time
dt
dependence); this is conservation of energy, in the sense of the comment following Eq. 2.21.
i~p
dhxi
i
i~hpi
hpi
(c) [H, x] = −
(see Problem 3.15). So
=
−
=
(Eq. 1.33).
m
dt
~
m
m
(b) Anything commutes with itself, so [H, H] = 0, and hence
(d) [H, p] =
p2
∂V
dhpi
i
∂V
∂V
+ V, p = [V, p] = i~
(Problem 3.14(c)). So
=
i~
= −
.
2m
∂x
dt
~
∂x
∂x
This is Ehrenfest’s theorem (Eq. 1.38).
CHAPTER 3. FORMALISM
85
Problem 3.19
dhpi
∂V
(a) Problem 3.18(d) says
= −
= 0, so hpi = p0 , a constant (independent of t). Then Problem
dt
∂x
dhxi
hpi
p0
p0
3.18(c) says
=
=
, and hence hxi =
t + x0 (where x0 is another constant). So hxi moves at
dt
m
m
m
constant velocity (v0 = p0 /m). QED
(b) In the same way,
dhpi
∂V
=−
= −hmω 2 xi = −mω 2 hxi.
dt
∂x
1 dhpi
dhxi
hpi
d2 hxi
1
=
=
⇒
= (−mω 2 )hxi = −ω 2 hxi.
dt
m
dt2
m dt
m
So hxi satisfies the classical harmonic oscillator equation, and the general solution is
hxi = A sin(ωt) + B cos(ωt);
it oscillates at the classical frequency. QED
Problem 3.20
1
Ψ(x, t) = √ ψ1 e−iE1 t/~ + ψ2 e−E2 t/~ .
2
1 H 2 Ψ = √ (H 2 ψ1 )e−E1 t/~ + (H 2 ψ2 )e−iE2 t/~ .
2
Hψ1 = E1 ψ1 ⇒ H 2 ψ1 = E1 Hψ1 = E12 ψ1 , and H 2 ψ2 = E22 ψ2 , so
1
h ψ1 e−iE1 t/~ + ψ2 e−iE2 t/~ | E12 ψ1 e−iE1 t/~ + E22 ψ2 e−iE2 t/~ i
2
1
= hψ1 |ψ1 ieiE1 t/~ E12 e−iE1 t/~ + hψ1 |ψ2 ieiE1 t/~ E22 e−iE2 t/~
2
1
+ hψ2 |ψ1 ieiE2 t/~ E12 e−iE1 t/~ + hψ2 |ψ2 ieiE2 t/~ E22 e−iE2 t/~ = E12 + E22 .
2
hH 2 i =
Similarly, hHi = 12 (E1 + E2 )
(Problem 2.5(e)).
1
1 2
1
E1 + E22 − (E1 + E2 )2 = 2E12 + 2E22 − E22 − E12 − 2E1 E2
2
4
4
1
1
1 2
= E1 − 2E1 E2 + E22 = (E2 − E1 )2 . σH = (E2 − E1 ).
4
4
2
2
σH
= hH 2 i − hHi2 =
1
hψ1 |x2 |ψ1 i + hψ2 |x2 |ψ2 i + hψ1 |x2 |ψ2 iei(E1 −E2 )t/~ + hψ2 |x2 |ψ1 iei(E2 −E1 )t/~ .
2
Z
Z
2 a 2
nπ
mπ
1 a 2
n−m
n+m
2
hψn |x |ψm i =
x sin
x sin
x dx =
x cos
πx − cos
πx dx.
a 0
a
a
a 0
a
a
hx2 i =
Z a
Now
2
x cos
0
2
3 2
a
k
2a x
k
a
kπx
k
πx dx =
cos
πx +
− 2 sin
πx
a
k2 π2
a
kπ
a
a
0
2a3
2a3
= 2 2 cos(kπ) = 2 2 (−1)k
k π
k π
(for k = nonzero integer).
86
CHAPTER 3. FORMALISM
(−1)n+m
2a2
2a2 (−1)n−m
4nm
−
=
(−1)n+m 2
.
∴ hψn |x |ψm i = 2
π (n − m)2
(n + m)2
π2
(n − m2 )2
2
16a2
1
1
So hψ1 |x2 |ψ2 i = hψ2 |x2 |ψ1 i = − 2 . Meanwhile, from Problem 2.4, hψn |x2 |ψn i = a2
.
−
9π
3 2(nπ)2
1 2 1
1
16a2 i(E2 −E1 )t/~
1
1
−i(E2 −E1 )t/~
Thus hx2 i =
e
+
e
.
a
− 2 + a2
− 2 −
{z
}
2
3 2π
3 8π
9π 2 |
E2 −E1
2 cos(
t)
~
E2 − E1
(4 − 1)π 2 ~2
3π 2 ~
= 3ω [in the notation of Problem 2.5(b)].
=
=
~
2ma2 ~
2ma2 a2 2
32
a
5
32
hx2 i =
− 2 − 2 cos(3ωt) . From Problem 2.5(c), hxi =
1 − 2 cos(3ωt) .
2 3 8π
9π
2
9π
2
2
a 4
64
64
32
5
2
2
2
2
cos (3ωt) .
So σx = hx i − hxi =
− 2 cos(3ωt) − 1 + 2 cos(3ωt) −
−
4 3 4π 2
9π
9π
9π 2
2
5
32
dhxi
8~
a2 1
2
2
cos (3ωt) . And, from Problem 2.5(d):
σx =
−
−
=
sin(3ωt).
4 3 4π 2
9π 2
dt
3ma
2
~2 dhxi
2 2
Meanwhile, the energy-time uncertainty principle (Eq. 3.74) says σH σx ≥
. Here
4
dt
"
#
2
2
2
2
1
a
5
32
5
32
1
3
1
2 2
σH
σx = (3~ω)2
−
−
−
−
cos2 (3ωt) = (~ωa)2
cos2 (3ωt) .
4
4 3 4π 2
9π 2
4
3 4π 2
9π 2
2 2
2
~2 dhxi
~ 8~
8
2aω
~
2
=
·
sin (3ωt) =
= 2 .
(~ωa)2 sin2 (3ωt), since
4
dt
2 3ma
3π 2
ma
π
So the uncertainty principle holds if
2 2
2
5
32
3
1
8
2
−
−
cos (3ωt) ≥
sin2 (3ωt),
4
3 4π 2
9π 2
3π 2
which is to say, if
2
2
2
1
5
32
4 8
32
2
2
− 2 ≥
cos
(3ωt)
+
sin
(3ωt)
=
.
3 4π
9π 2
3 3π 2
9π 2
1
5
32 2
Evaluating both sides: − 2 = 0.20668;
= 0.12978. So it holds. (Whew!)
3 4π
9π 2
Problem 3.21
From Problem 2.42, we have:
hxi =
~l
dhxi
~l
1
1 + θ2
2~at
1 2
1 2
t, so
= , σx2 =
=
, where θ =
; hHi =
hp i =
~ (a + l2 ).
2
m
dt
m
4w
4a
m
2m
2m
p2
We need hH 2 i (to get σH ). Now, H =
, so
2m
Z ∞
Z ∞
1
1
1
2
4
4
2
hH i =
hp i =
p |Φ(p, t)| dp, where (Eq. 3.54): Φ(p, t) = √
e−ipx/~ Ψ(x, t) dx.
4m2
4m2 −∞
2π~ −∞
CHAPTER 3. FORMALISM
87
From Problem 2.42: Ψ(x, t) =
2a
π
1/4
√
2
l
l2
1
e− 4a ea(ix+ 2a ) /(1+iθ) .
1 + iθ
1/4
Z ∞
2
2
l
1
2a
il
1
√
e−ipx/~ ea(ix+ 2a ) /(1+iθ) dx. Let y ≡ x − .
e−l /4a
π
2a
1
+
iθ
2π~
−∞
1/4
Z ∞
2
2
1
2a
1
√
=√
e−l /4a epl/2a~
e−ipy/~ e−ay /(1+iθ) dy.
1 + iθ
2π~ π
−∞
[See Prob. 2.22(b) for the integral.]
r
1/4
1
2a
π(1 + iθ) − p2 (1+iθ)
1
−l2 /4a pl/2a~
√
e
e
=√
e 4a~2
a
1 + iθ
2π~ π
1/4
p2 (1+iθ)
pl
l2
1
1
e− 4a e 2a~ e− 4a~2 .
=√
~ 2aπ
So Φ(p, t) = √
p2
2
1 1 −l2 /2a pl/a~ −p2 /2a~2
1
1
− 1 l2 − 2pl
~ + ~2
e
e
e
= √
= √
e 2a
e−(l−p/~) /2a .
2aπ ~
~ 2aπ
~ 2aπ
2
|Φ(p, t)| = √
Z ∞
2
1
p
hp4 i = √
p4 e−(l−p/~) /2a dp. Let − l ≡ z, so p = ~(z + l).
~
~ 2aπ −∞
Z ∞
2
1
= √
~5
(z + l)4 e−z /2a dz. Only even powers of z survive:
~ 2aπ
−∞
Z ∞
√
2
~4
3(2a)2 √
(2a) √
~4
z 4 + 6z 2 l2 + l4 e−z /2a dz = √
2aπ + 6l2
2aπ + l4 2aπ
=√
4
2
2aπ −∞
2aπ
4
~
= ~4 3a2 + 6al2 + l4 . ∴ hH 2 i =
3a2 + 6al2 + l4 .
4m2
2
σH
= hH 2 i − hHi2 =
~4 a
~4
~4
2
2
4
2
2
4
2
2
3a
+
6al
+
l
−
a
−
2al
−
l
=
2a
+
4al
=
a + 2l2 .
4m2
4m2
2m2
2 2 ~4 a
2~at
~4 l 2
a
2~at
2 1
a + 2l
1+
=
1+ 2 1+
2m2
4a
m
4m2
2l
m
2
2
~4 l 2
~2 ~l
~2 dhxi
≥
=
=
, so it works.
2
4m
4 m
4
dt
2 2
σH
σx =
Problem 3.22
~
~ dhxi
dhxi
, so σx σH ≥
|hpi|, which is the Griffiths
. But hpi = m
2 dt
dt
2m
uncertainty principle of Problem 3.15.
For Q = x, Eq. 3.74 says σH σx ≥
Problem 3.23
P 2 |βi = P (P |βi) = P (hα|βi|αi) = hα|βi(P |αi) = hα|βi hα|αi |αi = hα|βi|αi = P |βi.
| {z }
1
88
CHAPTER 3. FORMALISM
Since P 2 |βi = P |βi for any vector |βi, P 2 = P . QED [Note: To say two operators are equal means that
they have the same effect on all vectors.]
If |γi is an eigenvector of P̂ with eigenvalue λ, then P̂ |γi = λ|γi, and it follows that P̂ 2 |γi = λP̂ |γi = λ2 |γi.
But P̂ 2 = P̂ , and |γi =
6 0, so λ2 = λ, and hence the eigenvalues of P̂ are 0 and 1. Any (complex) multiple of
|αi is an eigenvector of P̂ , with eigenvalue 1; any vector orthogonal to |αi is an eigenvector of P̂ , with eigenvalue
0.
Problem 3.24
In the usual casual notation, write Q̂en as the name of the vector Q̂|en i. Then
Qmn ≡ hem |Q̂|en i = hem |Q̂en i = hQ̂em |en i = hen |Q̂em i∗ = hen |Q̂|em i∗ = Q∗nm . QED
Problem 3.25
Write the eigenvector as |ψi = c1 |1i + c2 |2i, and call the eigenvalue E. The eigenvalue equation is
Ĥ|ψi = (|1ih1| − |2ih2| + |1ih2| + |2ih1|) (c1 |1i + c2 |2i) = (c1 |1i + c1 |2i − c2 |2i + c2 |1i)
= [(c1 + c2 )|1i + (c1 − c2 )|2i] = E|ψi = E(c1 |1i + c2 |2i).
E
E
(c1 + c2 ) = Ec1 ⇒ c2 =
− 1 c1 ; (c1 − c2 ) = Ec2 ⇒ c1 =
+ 1 c2 .
2
√
E
E
E
c2 =
−1
+ 1 c2 ⇒
− 1 = 1 ⇒ E = ± 2 .
i
h
√
√
The eigenvectors are: c2 = (± 2 − 1)c1 ⇒ |ψ± i = c1 |1i + (± 2 − 1)|2i .
1 1
.
The Hamiltonian matrix is
H=
1 −1
Problem 3.26
(a) hα| = −ih1| − 2h2| + ih3|;
hβ| = −ih1| + 2h3|.
(b) hα|βi = (−ih1| − 2h2| + ih3|) (i|1i + 2|3i) = (−i)(i)h1|1i + (i)(2)h3|3i = 1 + 2i.
hβ|αi = (−ih1| + 2h3|) (i|1i − 2|2i − i|3i) = (−i)(i)h1|1i + (2)(−i)h3|3i = 1 − 2i = hα|βi∗ . X
(c)
A11 = h1|αihβ|1i = (i)(−i) = 1; A12 = h1|αihβ|2i = (i)(0) = 0; A13 = h1|αihβ|3i = (i)(2) = 2i;
A21 = h2|αihβ|1i = (−2)(−i) = 2i; A22 = h2|αihβ|2i = (−2)(0) = 0; A23 = h2|αihβ|3i = (−2)(2) = −4;
A31 = h3|αihβ|1i = (−i)(−i) = −1; A32 = h3|αihβ|2i = (−i)(0) = 0; A33 = h3|αihβ|3i = (−i)(2) = −2i.


1 0 2i
A =  2i 0 −4  . No, it’s not hermitian.
−1 0 −2i
CHAPTER 3. FORMALISM
89
Problem 3.27
(a)
!
|αi =
X
cn |en i ⇒ Q̂|αi =
n
X
X
cn Q̂|en i =
hen |αiqn |en i =
n
n
X
qn |en ihen | |αi ⇒ Q̂ =
n
X
qn |en ihen |. X
n
(b) Noting that Q̂|en i = qn |en i ⇒ Q̂2 |en i = qn Q̂|en i = qn2 |en i · · · ⇒ Q̂j |en i = qnj |en i, we have


!
!
∞
X
X
X
X
X
1 j
1
Q̂
j
q

e |αi =
Q̂
|en ihen | |αi =
q  |en ihen | |αi =
e n |en ihen | |αi,
j!
j! n
n
n
n
j=0
j
so
!
Q̂
e =
X
qn
e |en ihen | . QED
n
Problem 3.28
1 3
1 5
(a) The Taylor expansion of sin z is z − 3!
z + 5!
z − · · · , so
3
5
d
1 d
1 d
1
1
(sin D̂)x5 =
−
+
− · · · x5 = 5x4 − 5 · 4 · 3x2 + 5 · 4 · 3 · 2 · 1 x0
3
5
dx 3! dx
5! dx
3!
5!
= 5x4 − 10x2 + 1.
(b)
1
= 1 + z + z 2 + z 3 + · · · , so
1−z
!
1 d
1 d2
1 d3
1 d4
1
cos x = 1 +
+
+
+
+
·
·
·
cos x
2 dx 4 dx2
8 dx3
16 dx4
1 − D̂/2
1
1
1
1
= cos x − sin x − cos x + sin x +
cos x + · · ·
4
2
8
16
1
1
1
1
1
= cos x 1 − +
+ · · · − sin x 1 − +
+ ···
4 16
2
4 16
1
1
= (cos x − sin x)
= 25 (2 cos x − sin x).
2
1 + 1/4
Problem 3.29
(a) It’s trivial for n = 1. Assume it is true for (n − 1) (i.e. [Ân−1 , B̂] = (n − 1)Ân−2 Ĉ), and prove it then holds
for n:
[Ân , B̂] = [Â(Ân−1 ), B̂] = Â[Ân−1 , B̂] + [Â, B̂]Ân−1 = Â(n − 1)Ân−2 Ĉ + Ĉ Ân−1 = nÂn−1 Ĉ.
(b)
h
e
λÂ
i
, B̂ =
"∞
X 1
=λ
n=0
∞
X
n!
#
n
n
λ Â , B̂ =
∞
X
1
∞
h
i X
1 n n−1
λn Ân , B̂ =
λ nÂ
Ĉ
n!
n!
n=0
n=0
∞
X
1
1 m m
λn−1 Ân−1 Ĉ = λ
λ  Ĉ = λeλ Ĉ. QED
(n
−
1)!
m!
n=1
m=0
90
CHAPTER 3. FORMALISM
d λQ̂ d
(c) Note first that
=
e
dλ
dλ
1
1 + λQ̂ + λ2 Q̂2 + · · ·
2
1
= Q̂ + λQ̂2 + λ2 Q̂3 + · · · = Q̂eλQ̂ , so
2
dfˆ
= (Â + B̂)eλ(Â+B̂) = (Â + B̂)fˆ.
dλ
2
2
2
dĝ
= Âeλ eλB̂ e−λ Ĉ/2 + eλ B̂eλB̂ e−λ Ĉ/2 + eλ eλB̂ (−λĈ)e−λ Ĉ/2 .
dλ
In the second term on the right, we can move B̂ to the left, using (b): eλ B̂ = B̂eλ + λeλ Ĉ. (Of course, Ĉ
moves freely through the other terms, because it commutes with  and B̂). Thus
2
dĝ
= ( + B̂ + λĈ − λĈ)eλ eλB̂ e−λ Ĉ/2 = ( + B̂)ĝ.
dλ
Since fˆ(λ) and ĝ(λ) are equal for all λ, they are equal in particular for λ = 1:
eÂ+B̂ = e eB̂ e−Ĉ/2 .
This generalizes the old rule for numbers (as opposed to operators): ex+y = ex ey .
Problem 3.30
Let |ni be the nth eigenstate of the Hamiltonian.
Z
Z
cn (t) = hn|S(t)i = hn|
dx |xihx| |S(t)i = dx hn|xi hx|S(t)i.
But hx|S(t)i = Ψ(x, t), and hx|ni is the nth energy eigenstate in position space, ψn (x). So
cn (t) =
R
ψn (x)∗ Ψ(x, t) dx.
Note that cn (t) here includes the wiggle factor, e−iEn t/~ .
Problem 3.31
Z 1
|e1 i = 1; he1 |e1 i =
1
1 dx = 2. So |e01 i = √ .
2
−1
Z 1
Z 1
1
x3
2
|e2 i = x;
x dx = 0; he2 |e2 i =
x dx =
= . So |e02 i =
3
3
2 −1
−1
−1
r Z 1
Z 1
1
1 2
2
|e3 i = x2 ; he01 |e3 i = √
x2 dx = √ ; he02 |e3 i =
x3 dx = 0.
3 −1
2 −1
23
1
he01 |e2 i = √
2
1 2
1
So (Problem A.4): |e003 i = |e3 i − √ |e01 i = x2 − .
3
23
2
5
Z 1
1
1
x
2 x3
x
2 4 2
8
he003 |e003 i =
x2 −
dx =
− ·
+
= − + =
. So
3
5
3
3
9
5
9
9
45
−1
−1
r
3
x.
2
CHAPTER 3. FORMALISM
r
|e03 i =
91
r 45 2 1
5 3 2 1
x −
=
x −
.
8
3
2 2
2
r Z 1
r
Z 1
1
3
3 2
|e4 i = x3 . he01 |e4 i = √
x3 dx = 0; he02 |e4 i =
x4 dx =
· ;
2 −1
2 5
2 −1
r
r Z 1
r
5
3 5 1 3
3 2 3
3
0
00
0
0
3
he3 |e4 i =
x − x dx = 0. |e4 i = |e4 i − he2 |e4 i|e2 i = x −
x = x3 − x.
2 −1 2
2
2 5 2
5
2
7
1
Z 1
2 12 18
3
x
2 · 3 x5
9 x3
8
= −
x3 − x dx =
−
+
+
=
.
he004 |e004 i =
5
7
5 5
25 3 −1
7 25 75
7 · 25
−1
r r 7 3 3
7 5 3 3
x − x =
x − x .
2 2
5
2 2
2
5
|e04 i =
Problem 3.32
(a) hQi = hψ|Q̂ψi = hQ̂† ψ|ψi = −hQ̂ψ|ψi = −(hψ|Q̂ψi)∗ = −hQi∗ , so hQi is imaginary. X
(b) Q̂|ψi = λ|ψi ⇒ hψ|Q̂ψi = λhψ|ψi = −hQ̂ψ|ψi = −λ∗ hψ|ψi. Therefore λ∗ = −λ, and hence λ is imaginary.
(c) If Q̂|ψi = λ|ψi and Q̂|φi = µ|φi, then hφ|Q̂ψi = λhφ|ψi = −hQ̂φ|ψi = −µ∗ hφ|ψi = µhφ|ψi. Therefore
(λ − µ)hφ|ψi = 0, and if λ 6= µ the eigenvectors must be orthogonal.
(d) From Problem 3.5(b) we know that (P̂ Q̂)† = Q̂† P̂ † , so if P̂ = P̂ † and Q̂ = Q̂† then
[P̂ , Q̂]† = (P̂ Q̂ − Q̂P̂ )† = Q̂† P̂ † − P̂ † Q̂† = Q̂P̂ − P̂ Q̂ = −[P̂ , Q̂]. X
If P̂ = −P̂ † and Q̂ = −Q̂† , then [P̂ , Q̂]† = Q̂† P̂ † − P̂ † Q̂† = (−Q̂)(−P̂ ) − (−P̂ )(−Q̂) = −[P̂ , Q̂].
So in either case the commutator is antihermitian.
(e) Let  ≡ 12 (Q̂ + Q̂† ) and B̂ ≡ 21 (Q̂ − Q̂† ). Clearly Q̂ =  + B̂. But † = 12 (Q̂† + Q̂) = Â, so  is hermitian,
while B̂ † = 12 (Q̂† − Q̂) = −B̂, so B̂ is antihermitian
Problem 3.33
(a) ψ1 .
(b) b1 (with probability 9/25) or b2 (with probability 16/25).
(c) Right after the measurement of B:
• With probability 9/25 the particle is in state φ1 = (3ψ1 + 4ψ2 )/5; in that case the probability of
getting a1 is 9/25.
• With probability 16/25 the particle is in state φ2 = (4ψ1 − 3ψ2 )/5; in that case the probability of
getting a1 is 16/25.
92
CHAPTER 3. FORMALISM
So the total probability of getting a1 is
16 16
9 9
337
·
+
·
=
= 0.5392.
25 25 25 25
625
[Note: The measurment of B (even if we don’t know the outcome of that measurement) collapses the wave
function, and thereby alters the probabilities for the second measurment of A. If the graduate student
inadvertantly neglected to measure B, the second measurement of A would be certain to reproduce the
result a1 .]
Problem 3.34
r
n2 π 2 ~2
.
2ma2
r
Z ∞
Z
nπ 1
1
2 −iEn t/~ a −ipx/~
Φn (p, t) = √
e−ipx/~ Ψn (x, t) dx = √
e
sin
e
x dx
a
2π~ −∞
2π~ a
0
Z
i
1 a h i(nπ/a−p/~)x
1
e−iEn t/~
e
− ei(−nπ/a−p/~)x dx
=√
2i 0
π~a
i(nπ/a−p/~)x
a
1
e
ei(−nπ/a−p/~)x
−iEn t/~ 1
=√
e
−
2i i(nπ/a − p/~) i(−nπ/a − p/~) 0
π~a
−1 −iEn t/~ ei(nπ−pa/~) − 1 e−i(nπ+pa/~) − 1
= √
e
+
(nπ/a − p/~)
(nπ/a + p/~)
2 π~a
n −ipa/~
−1
−1 −iEn t/~ (−1) e
(−1)n e−ipa/~ − 1
= √
e
a+
a
(nπ − ap/~)
(nπ + ap/~)
2 π~a
r
h
i
1
a −iEn t/~
2nπ
n −ipa/~
=−
e
(−1)
e
−
1
2 π~
(nπ)2 − (ap/~)2
r
aπ
ne−iEn t/~
=
1 − (−1)n e−ipa/~ .
~ (nπ)2 − (ap/~)2
(a) Ψn (x, t) =
nπ 2
sin
x e−iEn t/~ ,
a
a
with En =
(b) Noting that
(
n −ipa/~
1 − (−1) e
−ipa/2~
=e
ipa/2~
e
− (−1)n e−ipa/2~ = 2e−ipa/2~
cos(pa/2~) (n odd),
i sin(pa/2~) (n even),
we have
4πa
n2
×
|Φn (p, t)| =
~ [(nπ)2 − (ap/~)2 ]2
2
(
cos2 (ap/2~) n odd,
sin2 (ap/2~) n even.
From the formula it looks as though the maxima should occur at p = ±(nπ~/a) (though since the numerator
also vanishes there, the function itself is perfectly finite). Physically I would expect the most probable values
of p to be given by En = (n2 π 2 ~2 )/(2ma2 ) = p2 /2m, or (again) p = ±(nπ~/a). For the graphs I used a = 1
and ~ = 1, so the maxima should be at nπ = (3.1, 6.3, 15.7, 30.1); actually (by eyeball) they are at (0, 5, 15,
30). As n increases they get closer to the “expected” value, but the interesting thing is that there is a range
of possible momenta, not just two discrete values, as one might naively have guessed (but in Problem 3.10 we
already noted that the stationary states are not eigenstates of p̂).
CHAPTER 3. FORMALISM
In[1]:=
f@x_D :=
In[2]:=
g@x_D :=
In[3]:=
n=1
In[4]:=
93
n2
HCos@x ê 2DL ^ 2
n2
HSin@x ê 2DL ^ 2
HHn pL ^ 2 - x ^ 2L ^ 2
HHn pL ^ 2 - x ^ 2L ^ 2
Plot@f@xD, 8x, - 15, 15<, PlotRange Æ 80, 0.011<D
0.010
0.008
0.006
Out[4]=
0.004
0.002
-15
In[5]:=
In[6]:=
-10
-5
5
10
15
n=2
Plot@g@xD, 8x, - 20, 20<, PlotRange Æ 80, 0.008<D
0.008
0.006
0.004
Out[6]=
0.002
-20
In[7]:=
n=5
-10
10
20
94
2
CHAPTER 3. FORMALISM
3Problem34.nb
In[8]:=
Plot@f@xD, 8x, - 36, 36<, PlotRange Æ 80, 0.008<D
0.008
0.006
0.004
Out[8]=
0.002
-30
In[9]:=
In[10]:=
-20
-10
10
20
30
n = 10
Plot@g@xD, 8x, - 50, 50<, PlotRange Æ 80, 0.007<D
0.007
0.006
0.005
0.004
Out[10]=
0.003
0.002
0.001
-40
-20
20
40
(c)
hp2 i =
Z ∞
p2 |Φn (p, t)|2 dp =
−∞
4n~2
=
a2
4n2 πa
~
Z ∞
p2
−∞
2
[(nπ)2 − (ap/~)2 ]
cos2 (pa/2~)
sin2 (pa/2~)
dp
[let x ≡
ap
]
nπ~
Z ∞
x2
4n~2
T
(x)
dx
=
In ,
n
2 2
a2
−∞ (1 − x )
where
Tn (x) ≡
cos2 (nπx/2),
sin2 (nπx/2),
if n is odd,
if n is even.
The integral can be evaluated by partial fractions:
x2
1
1
1
1
1
=
+
+
−
⇒
(x2 − 1)2
4 (x − 1)2
(x + 1)2
(x − 1) (x + 1)
Z ∞
Z ∞
Z ∞
Z ∞
1
1
1
1
1
In =
T
(x)
dx
+
T
(x)
dx
+
T
(x)
dx
−
T
(x)
dx
.
n
n
n
2 n
4 −∞ (x − 1)2
−∞ (x + 1)
−∞ (x − 1)
−∞ (x + 1)
For odd n:
Z ∞
1
2 nπx
cos
dx =
k
2
−∞ (x ± 1)
Z ∞
h
i
1
2 nπ
cos
(y
∓
1)
dy =
k
2
−∞ y
Z ∞
1
2 nπy
sin
dy.
k
2
−∞ y
CHAPTER 3. FORMALISM
95
For even n:
Z ∞
nπx 1
sin2
dx =
k
2
−∞ (x ± 1)
In either case, then,
In =
1
2
Z ∞
h nπ
i
1
sin2
(y ∓ 1) dy =
k
2
−∞ y
Z ∞
nπy 1
nπ
sin2
dy =
2
y
2
4
−∞
Therefore
hp2 i =
4n~2 nπ 2
4n~2
In = 2
=
2
a
a
4
nπ~
a
Z ∞
nπy 1
sin2
dy.
k
2
−∞ y
Z ∞
sin2 u
nπ 2
du =
.
2
4
−∞ u
2
(same as Problem 2.4).
Problem 3.35
1
Φ(p, 0) = √
2π~
Z ∞
−ipx/~
e
−∞
1
Ψ(x, 0) dx = √
2 nπ~λ
Z nλ
ei(2π/λ−p/~)x dx
−nλ
nλ
ei(2π/λ−p/~)x
ei2πn e−ipnλ/~) − e−i2πn eipnλ/~)
1
1
= √
= √
i(2π/λ − p/~)
2 nπ~λ i(2π/λ − p/~) −nλ
2 nπ~λ
r
~λ sin(npλ/~)
=
.
nπ (pλ − 2π~)
|Ψ(x, 0)|2 =
1
2nλ
(−nλ < x < nλ);
|^| 2
-nh
|Φ(p, 0)|2 =
λ~ sin2 (npλ/~)
.
nπ (pλ − 2π~)2
|\| 2
nh
x
2/h/h
p
The width of the |Ψ|2 graph is wx = 2nλ. The |Φ|2 graph is a maximum at 2π~/λ, and goes to zero on either
1
2π~
2π~
side at
1±
, so wp =
. As n → ∞, wx → ∞ and wp → 0; in this limit the particle has a
λ
2n
nλ
well-defined momentum, but a completely indeterminate position. In general,
wx wp = (2nλ)
2π~
= 4π~ > ~/2,
nλ
so the uncertainty principle is satisfied (using the widths as a measure of uncertainty). If we try to check the
uncertainty principle more rigorously, using standard deviation as the measure, we get an uninformative result,
because
Z
λ~ ∞ 2 sin2 (npλ/~)
2
p
hp i =
dp = ∞.
nπ −∞ (pλ − 2π~)2
(At large |p| the integrand is approximately (1/λ2 ) sin2 (npλ/~), so the integral blows up.) Meanwhile hpi is finite
(2π~/λ), so σp = ∞, and the uncertainty principle tells us nothing. The source of the problem is the discontinuity
96
CHAPTER 3. FORMALISM
in Ψ at the end points; here p̂ Ψ = −i~ dΨ/dx picks up a delta function, and hΨ|p̂2 Ψi = hp̂ Ψ|p̂ Ψi → ∞ because
the integral of the square of the delta function blows up. In general, if you want σp to be finite, you cannot
allow discontinuities in Ψ.
Problem 3.36
(a)
2
Z ∞
1
dx = 2|A|2
2 + a2 )2
(x
−∞
1 = |A|
=
∞
1
x
1
2 1
−1 x
dx = 2|A|
+ tan
(x2 + a2 )2
2a2 x2 + a2
a
a
0
0
r
2a
⇒
A=a
.
π
Z ∞
1
1
π
|A|2 tan−1 (∞) = 3 |A|2
a2
a
2a
(b)
hxi = A2
Z ∞
x
−∞
Z ∞
(a2 + x2 )2
dx = 0.
x2
a
√
, x = a y, dx = √ dy.]
2
a
2 y
0
√
√
1/2
2
2
2 Z ∞
y
2a Γ(3/2)Γ(1/2)
2a ( π/2)( π)
2a
dy =
=
= a2 .
=
π 0 (1 + y)2
π
Γ(2)
π
1
p
σx = hx2 i − hxi2 = a.
hx2 i = 2A2
x2
(a2 + x2 )2
dx. [Let y ≡
(c)
Z ∞
A
2π~
px px 1
−ipx/~
dx.
[But
e
=
cos
−
i
sin
, and sine is odd.]
x2 + a2
~
~
−∞
r
Z ∞
a −|p|a/~
cos(px/~)
2A π −|p|a/~ 2A
√
dx
=
e
=
e
.
= √
x2 + a2
~
2π~ 0
2π~ 2a
Φ(p, 0) = √
e−ipx/~
Z ∞
a
|Φ(p, 0)| dp =
~
−∞
2
Z ∞
−2|p|a/~
e
−∞
2a
dp =
~
−2pa/~ ∞
e
= 1. X
−2a/~ 0
(d)
a
hpi =
~
Z ∞
a
hp i = 2
~
2
(e)
pe−2|p|a/~ dp = 0.
−∞
Z ∞
2 −2pa/~
p e
0
2a
dp =
2
~
√ ~
~
~
σx σp = a √
= 2 > . X
2
2
2a
~
2a
3
=
~2
.
2a2
σp =
p
~
hp2 i − hpi2 = √ .
2a
CHAPTER 3. FORMALISM
97
Problem 3.37
d
i
Equation 3.73 ⇒ hxpi = h[H, xp]i; Eq. 3.65 ⇒ [H, xp] = [H, x]p + x[H, p]; Problem 3.15 ⇒ [H, x] =
dt
~
∂V
i~p
−
; Problem 3.18(d) ⇒ [H, p] = i~
. So
m
∂x
d
i
i~ 2
∂V
p2
∂V
∂V
hxpi =
− hp i + i~hx
i = 2h
i − hx
i = 2hT i − hx
i. QED
dt
~
m
∂x
2m
∂x
∂x
In a stationary state all expectation values (at least, for operators that do not depend explicitly on t) are
time-independent (see item 1 on p. 26), so dhxpi/dt = 0, and we are left with Eq. 3.113.
For the harmonic oscillator:
V =
1
dV
dV
mω 2 x2 ⇒
= mω 2 x ⇒ x
= mω 2 x2 = 2V ⇒ 2hT i = 2hV i ⇒ hT i = hV i. QED
2
dx
dx
In Problem 2.11(c) we found that hT i = hV i = 41 ~ω (for n = 0); hT i = hV i = 34 ~ω (for n = 1). X
In Problem 2.12 we found that hT i = 21 n+ 12 ~ω, while hx2 i = (n+ 21 )~/mω, so hV i = 12 mω 2 hx2 i = 21 (n+ 12 )~ω,
and hence hT i = hV i for all stationary states. X
Problem 3.38
1
Ψ(x, t) = √ ψ1 e−iE1 t/~ + ψ2 e−iE2 t/~ ;
2
hΨ(x, t)|Ψ(x, 0)i = 0 ⇒
1 iE1 t/~
e
hψ1 |ψ1 i + eiE1 t/~ hψ1 |ψ2 i + eiE2 t/~ hψ2 |ψ1 i + eiE2 t/~ hψ2 |ψ2 i
2
=
1 iE1 t/~
e
+ eiE2 t/~ = 0, or eiE2 t/~ = −eiE1 t/~ , so ei(E2 −E1 )t/~ = −1 = eiπ .
2
Thus (E2 − E1 )t/~ = π (orthogonality also at 3π, 5π, etc., but this is the first occurrence).
∴ ∆t ≡
t
~
1
=
. But ∆E = σH = (E2 − E1 ) (Problem 3.20).
π
E2 − E1
2
So ∆t ∆E =
~
. X
2
Problem 3.39
r
√
~
~mω
a+ |ni = √n + 1 |n + 1i,
Equation 2.70: x =
(a+ + a− ), p = i
(a+ − a− ); Eq. 2.67 :
a− |ni = n |n − 1i.
2mω
2
r
r
√
~
~ √ 0
0
0
hn|x|n i =
hn|(a+ + a− )|n i =
n + 1 hn|n0 + 1i + n0 hn|n0 − 1i
2mω
2mω
r
r
√
√
~ √ 0
~ √
0
0
0
=
n + 1 δn,n +1 + n δn,n −1 =
n δn0 ,n−1 + n0 δn,n0 −1 .
2mω
2mω
r
√
m~ω √
0
hn|p|n i = i
n δn0 ,n−1 − n0 δn,n0 −1 .
2
r
98
CHAPTER 3. FORMALISM
Noting that n and n0 run from zero to infinity, the matrices are:
√


1 √0 0 0 0
√0
 1 0
2 √0 0 0 
r

√
.
~ 
2 √0
3 √0 0 :
 0
;
X=
2mω 
3 √0
4 √0 
 0 0

 0 0 0
4 0
5 
···
√


0
0
0
0
√0 − 1 √
 1 0 − 2 0
0
0 
r

√
√
.
m~ω 
2 √0 − 3 √
0
0 :
 0
.
P=i
2 
0
3 √0 − 4 √
0 
 0

 0
0
0
4
− 5 
···
Squaring these matrices:

1
0
 0
3
√
~ 
 1·2 0
X2 =
2mω 
 0 √2 · 3

√
−1
0
 0
−3
√
m~ω 
 1·2 0
P2 = −

√
2 
0
2·3

1·2 √0
0
0

2·3 √0
0
0
.
5
0
3 · 4 √ 0 :
;
0
7
0
4·5 
···
√

1·2 √0
0
0

2·3 √0
0
0
.
−5
0
3 · 4 √ 0 :
.
0
−7
0
4·5 
···
So the Hamiltonian, in matrix form, is
1 2 mω 2 2
P +
X
2m
2
√


−1
0
1·2 √0
0
0
 0

−3
0
2·3 √0
0
√
~ω 
.

=−
1·2 0
−5
0
3 · 4 √ 0 :

4 
 0 √2 · 3 0
−7
0
4·5 
···


√


1 0 0 0
1
0
1·2 √0
0
0
0 3 0 0

 0

3
0
2·3 √0
0



√
~ω 
~ω
.


0 0 5 0
 1·2 0
=
+

.
5
0
3
·
4
0
:


√
√


4 
2
0 0 0 7

0
2·3 0
7
0
4·5 
..
···
.
H=
It’s plainly diagonal, and the nonzero elements are Hnn = n + 12 ~ω, as they should be.
CHAPTER 3. FORMALISM
99
Problem 3.40
Using Equation 3.114,
!∗
Z
X
hxi =
cn ψn e−iEn t/~
=
x
cn0 ψn0 e−iEn0 t/~
dx =
XX
n0
n
r
!
X
n
c∗n cn0 ei(En −En0 )t/~ hn|x|n0 i
n0
√
√
~ XX ∗
n0 δn,n0 −1 + n δn0 ,n−1
cn cn0 ei(En −En0 )t/~
2mω n 0
n
r
=
r
=
~
2mω
~
2mω
X h√
n + 1 c∗n cn+1 ei(En −En+1 )t/~ +
√
n c∗n cn−1 ei(En −En−1 )t/~
i
n
"∞
X√
n=0
n + 1 c∗n cn+1 e−iωt +
∞
X
√
#
n c∗n cn−1 eiωt
.
n=0
In the last step I used
1
1
3
1
~ω − n −
~ω = ~ω, En − En+1 = n +
~ω − n +
~ω = −~ω.
En − En−1 = n +
2
2
2
2
√
Noting that n kills the n = 0 term in the second sum, let n → n + 1 and run it from 0 to ∞:
r
∞
~ X√
hxi =
n + 1 c∗n cn+1 e−iωt + c∗n+1 cn eiωt
2mω n=0
!
!
r
r
∞
∞
X
√
~ X√
~
=
n + 1 c∗n cn+1 e−iωt +
n + 1 cn c∗n+1 eiωt
2mω n=0
2mω n=0
1
1
Ceiφ e−iωt +
Ce−iφ eiωt = C cos(ωt − φ). X
=
2
2
√
√
In the case of Problem 2.40, c0 = 3/5, c1 = −2 2/5, c2 = 2 2/5 (and all the rest are zero) so
r
r
r
√
√
√ !
√ ∗ 2~ ∗
2~
(−2 2) 3 √ 2 2 (−2 2)
28
~
−iφ
Ce
=
c1 c0 + 2 c2 c1 =
+ 2
=−
.
mω
mω
5
5
5
5
25 mω
28
C=
25
r
~
,
mω
φ = π.
Problem 3.41
√
√
Evidently Ψ(x, t) = c0 ψ0 (x)e−iE0 t/~ + c1 ψ1 (x)e−iE1 t/~ , with |c0 |2 = |c1 |2 = 1/2, so c0 = eiθ0 / 2, c1 = eiθ1 / 2,
for some real θ0 , θ1 .
hpi = |c0 |2 hψ0 |pψ0 i + |c1 |2 hψ1 |pψ1 i + c∗0 c1 ei(E0 −E1 )t/~ hψ0 |pψ1 i + c∗1 c0 ei(E1 −E0 )t/~ hψ1 |pψ0 i.
But E1 − E0 = ( 32 ~ω) − ( 12 ~ω) = ~ω, and (Problem 2.11) hψ0 |pψ0 i = hψ1 |pψ1 i = 0, while (Eqs. 2.70 and 2.67)
r
hψ0 |pψ1 i = i
r
r
r
i
√
√
~mω
~mω h
~mω
~mω
hψ0 |(a+ − a− )ψ1 i = i
hψ0 | 2ψ2 i − hψ0 | 1ψ0 i = −i
; hψ1 |pψ0 i = i
.
2
2
2
2
100
CHAPTER 3. FORMALISM
!
!
r
r
1 −iθ0 1 iθ1 −iωt
~mω
1 −iθ1 1 iθ0 iωt
~mω
√ e e
√ e e
hpi = √ e
−i
+√ e
i
2
2
2
2
2
2
r
r
i
i ~mω h −i(ωt−θ1 +θ0 )
~mω
=
−e
+ ei(ωt−θ1 +θ0 ) = −
sin(ωt + θ0 − θ1 ).
2
2
2
p
~mω/2; it occurs at t = 0 ⇔ sin(θ0 − θ1 ) = −1, or θ1 = θ0 + π/2. We might as well pick
The maximum is
θ0 = 0, θ1 = π/2; then
i
1 h
1
Ψ(x, t) = √ ψ0 e−iωt/2 + ψ1 eiπ/2 e−3iωt/2 = √ e−iωt/2 ψ0 + iψ1 e−iωt .
2
2
Problem 3.42
r
(a) hxi = hα|xαi =
~
hα|(a+ + a− )αi =
2mω
r
~
(ha− α|αi + hα|a− αi) =
2mω
r
~
(α + α∗ ).
2mω
~
a2 + a+ a− + a− a+ + a2− . But a− a+ = [a− , a+ ] + a+ a− = 1 + a+ a− (Eq. 2.55).
2mω +
~
a2 + 2a+ a− + 1 + a2− .
=
2mω +
x2 =
~
~
hα| a2+ + 2a+ a− + 1 + a2− αi =
ha2− α|αi + 2ha− α|a− αi + hα|αi + hα|a2− αi
2mω
2mω
~ ~ ∗ 2
(α ) + 2(α∗ )α + 1 + α2 =
1 + (α + α∗ )2 .
=
2mω
2mω
hx2 i =
r
hpi = hα|pαi = i
p2 = −
r
r
~mω
~mω
~mω
hα|(a+ − a− )αi = i
(ha− α|αi − hα|a− αi) = −i
(α − α∗ ).
2
2
2
~mω 2
~mω 2
a+ − a+ a− − a− a+ + a2− = −
a+ − 2a+ a− − 1 + a2− .
2
2
~mω
~mω
hα| a2+ − 2a+ a− − 1 + a2− αi = −
ha2− α|αi − 2ha− α|a− αi − hα|αi + hα|a2− αi
2
2
~mω ∗ 2
~mω = −
(α ) − 2(α∗ )α − 1 + α2 =
1 − (α − α∗ )2 .
2
2
hp2 i = −
(b)
~ ~
1 + (α + α∗ )2 − (α + α∗ )2 =
;
2mω
2mω
r
r
~mω
~mω ~
~mω
~
2
2
2
∗ 2
∗ 2
σp = hp i − hpi =
1 − (α − α ) + (α − α ) =
. σx σp =
= . QED
2
2
2mω
2
2
σx2 = hx2 i − hxi2 =
CHAPTER 3. FORMALISM
101
(c) Using Eq. 2.68 for ψn :
1
1
1
αn
cn = hψn |αi = √ h(a+ )n ψ0 |αi = √ hψ0 |(a− )n αi = √ αn hψ0 |αi = √ c0 . X
n!
n!
n!
n!
(d) 1 =
∞
X
|cn |2 = |c0 |2
n=0
(e) |α(t)i =
∞
X
|α|2n
n=0
∞
X
n!
cn e−iEn t/~ |ni =
n=0
= |c0 |2 e|α|
2
⇒
2
c0 = e−|α| /2 .
n
∞
∞
X
X
αe−iωt
2
2
1
αn
√ e−|α| /2 e−i(n+ 2 )ωt |ni = e−iωt/2
√
e−|α| /2 |ni.
n!
n!
n=0
n=0
Apart from the overall phase factor e−iωt/2 (which doesn’t affect its status as an eigenfunction of a− , or
its eigenvalue), |α(t)i is the same as |αi, but with eigenvalue α(t) = e−iωt α. X
p
(f ) From (a), hxi = ~/2mω[α(t) + α∗ (t)]. From (e), α(t) = e−iωt α. So
r
r
r
~
~
mω iφ −iωt
mω −iφ iωt
−iωt
∗ iωt
αe
+α e
=
e e
+C
e e
hxi =
C
2mω
2mω
2~
2~
1 = C e−i(ωt−φ + ei(ωt−φ) = C cos(ωt − φ) .
2
r
r
~
. So the expectation value of x oscillates at the classical frequency (but this is
2mω
true for all states of the harmonic oscillator—see Problem 3.40) and the wave packet maintains a constant
width.
From (b), σx =
(g) Equation 2.59 says a− |ψ0 i = 0, so yes, it is a coherent state, with eigenvalue α = 0.
Problem 3.43
(a) Equation 3.60 becomes |z|2 = [Re(z)]2 + [Im(z)]2 =
2 2
σA
σB ≥
1
(z + z ∗ )
2
2
+
2
1
(z − z ∗ ) ; Eq. 3.61 generalizes to
2i
2 2
1
1
(hf |gi + hg|f i) +
(hf |gi − hg|f i) .
2
2i
But hf |gi − hg|f i = h[Â, B̂]i (p. 106), and, by the same argument,
hf |gi + hg|f i = hÂB̂i − hAihBi + hB̂ Âi − hAihBi = hÂB̂ + B̂ Â − 2hAihBii = hDi.
2 2
So σA
σB ≥
(b) If B̂ = Â,
1
hDi2 + hCi2 . X
4
then Ĉ = 0,
D̂ = 2 Â2 − hAi2 ;
2
hDi = 2 hÂ2 i − hAi2 = 2σA
.
2 2
4
4
σA
σA ≥ (1/4)4σA
= σA
, which is true, but not very informative.
So Eq. 3.115 says
102
CHAPTER 3. FORMALISM
Problem 3.44
First find the eigenvalues and eigenvectors of the Hamiltonian. The characteristic equation says
(a − E)
0
b
0
(c − E)
0
= (a − E)(c − E)(a − E) − b2 (c − E) = (c − E) (a − E)2 − b2 = 0,
b
0
(a − E)
Either E = c, or else (a − E)2 = b2 ⇒ E = a ± b. So the eigenvalues are
E1 = c,
E2 = a + b,
E3 = a − b.
To find the corresponding eigenvectors, write

 
 
a0 b
α
α
 0 c 0   β  = En  β  .
b 0a
γ
γ
(1)

aα + bγ = cα ⇒ (a − c)α + bγ = 0; 
cβ = cβ
(redundant)
; ⇒ (a − c)2 − b2 α = 0.

bα + aγ = cγ ⇒ (a − c)γ + bα = 0.
So (excluding the degenerate case a − c = ±b) α = 0, and hence also γ = 0.
(2)
aα + bγ = (a + b)α ⇒
α − γ = 0;
cβ = (a + b)β ⇒
β = 0;
bα + aγ = (a + b)γ
(redundant).
So α = γ and β = 0.
(3)
aα + bγ = (a − b)α ⇒
α + γ = 0;
cβ = (a − b)β ⇒
β = 0;
bα + aγ = (a − b)γ
(redundant).
So α = −γ and β = 0.
Conclusion: The (normalized) eigenvectors of H are
 
 
0
1
1
|s1 i = 1 , |s2 i = √ 0 ,
2 1
0
 
1
1  
0 .
|s3 i = √
2 −1
(a) Here |S(0)i = |s1 i, so
 
0
|S(t)i = e−iE1 t/~ |s1 i = e−ict/~ 1 .
0
CHAPTER 3. FORMALISM
103
(b)
1
|S(0)i = √ (|s2 i + |s3 i) .
2

 
 
1
1
1 −iE2 t/~
1
1
1
|S(t)i = √ e
|s2 i + e−iE3 t/~ |s3 i = √ e−i(a+b)t/~ √ 0 + e−i(a−b)t/~ √  0 
2
2
2 1
2 −1


 −ibt/~

cos(bt/~)
e
+ eibt/~
1 −iat/~ 
.
 = e−iat/~ 
0
0
= e
2
−ibt/~
ibt/~
−i sin(bt/~)
e
−e
Problem 3.45
hn|x̂|S(t)i = hn|x̂
∞
X
!
0
0
|n ihn | |S(t)i =
n0 =0
∞
X
hn|x̂|n0 i hn0 |S(t)i.
n0 =0
√
p
√
n0 δn,n0 −1 + n δn0 ,n−1 , so
From Equation 3.114, hn|x̂|n0 i = ~/2mω
hn|x̂|S(t)i =
∞
X
n0 =0
r
√
~ √ 0
n δn,n0 −1 + n δn0 ,n−1 cn0 (t) =
2mω
r
√
~ √
n + 1 cn+1 (t) + n cn−1 (t) .
2mω
Problem 3.46
(a) H:
E1 = ~ω, E2 = E3 = 2~ω;
 
 
 
1
0
0
|h1 i = 0 , |h2 i = 1 , |h3 i = 0 .
0
0
1
A:
−a λ
0
λ −a
0
= a2 (2λ − a) − (2λ − a)λ2 = 0 ⇒ a1 = 2λ, a2 = λ, a3 = −λ.
0 0 (2λ − a)


 
 
010
α
α
 λβ = aα
λα = aβ
λ 1 0 0 β  = a β  ⇒

002
γ
γ
2λγ = aγ
(1)

λβ = 2λα ⇒ β = 2α, 
λα = 2λβ ⇒ α = 2β,

2λγ = 2λγ;
α = β = 0;
 
0
|a1 i = 0 .
1
104
CHAPTER 3. FORMALISM
(2)

λβ = λα ⇒ β = α, 
λα = λβ ⇒ α = β,

2λγ = λγ; ⇒ γ = 0.
 
1
1
|a2 i = √ 1 .
2 0
(3)

λβ = −λα ⇒ β = −α, 
λα = −λβ ⇒ α = −β,

2λγ = −λγ; ⇒ γ = 0.
 
1
1  
−1 .
|a3 i = √
2
0
B:
(2µ − b) 0 0
0
−b µ = b2 (2µ − b) − (2µ − b)µ2 = 0 ⇒ b1 = 2µ, b2 = µ, b3 = −µ.
0
µ −b

 
 

α
α
200
 2µα = bα
µγ = bβ
µ  0 0 1  β  = b  β  ⇒

γ
γ
µβ = bγ
010
(1)

2µα = 2µα,

µγ = 2µβ ⇒ γ = 2β,

µβ = 2µγ ⇒ β = 2γ;
β = γ = 0;

2µα = µα ⇒ α = 0, 
µγ = µβ ⇒ γ = β,

µβ = µγ; ⇒ β = γ.
 
0
1
|b2 i = √ 1 .
2 1
 
1
|b1 i = 0 .
0
(2)
(3)

2µα = −µα ⇒ α = 0, 
µγ = −µβ ⇒ γ = −β,

µβ = −µγ; ⇒ β = −γ.
 
0
1  
1 .
|b3 i = √
2 −1
(b)

 
1 0 0
c1
hHi = hS(0)|H|S(0)i = ~ω c∗1 c∗2 c∗3 0 2 0 c2  = ~ω |c1 |2 + 2|c2 |2 + 2|c3 |2 .
0 0 2
c3

 
c1
0 1 0
hAi = hS(0)|A|S(0)i = λ c∗1 c∗2 c∗3 1 0 0 c2  = λ c∗1 c2 + c∗2 c1 + 2|c3 |2 .
0 0 2
c3

 
c1
2 0 0
hBi = hS(0)|B|S(0)i = µ c∗1 c∗2 c∗3 0 0 1 c2  = µ 2|c1 |2 + c∗2 c3 + c∗3 c2 .
0 1 0
c3
CHAPTER 3. FORMALISM
105
(c)
|S(0)i = c1 |h1 i + c2 |h2 i + c3 |h3 i
⇒
|S(t)i = c1 e−iE1 t/~ |h1 i + c2 e−iE2 t/~ |h2 i + c3 e−iE3 t/~ |h3 i = c1 e−iωt |h1 i + c2 e−2iωt |h2 i + c3 e−2iωt |h3 i

 
 
 
 iωt 
1
0
0
c1 e
= e−2iωt c1 eiωt 0 + c2 1 + c3 0 = e−2iωt  c2  .
0
0
1
c3
H: h1 = ~ω, probability |c1 |2 ;
h2 = h3 = 2~ω, probability (|c2 |2 + |c3 |2 ).
 iωt 
c1 e
A: a1 = 2λ, ha1 |S(t)i = e−2iωt 0 0 1  c2  = e−2iωt c3 ⇒ probability |c3 |2 .
c3
 iωt 
c1 e
1
1
a2 = λ, ha2 |S(t)i = e−2iωt √ 1 1 0  c2  = √ e−2iωt c1 eiωt + c2 ⇒
2
2
c3
1
1 ∗ −iωt
probability =
c1 e
|c1 |2 + |c2 |2 + c∗1 c2 e−iωt + c∗2 c1 eiωt .
+ c∗2 c1 eiωt + c2 =
2
2
 iωt 
c
1e
1
1
a3 = −λ, ha3 |S(t)i = e−2iωt √ 1 −1 0  c2  = √ e−2iωt c1 eiωt − c2 ⇒
2
2
c3
1 ∗ −iωt
1
probability =
c1 e
− c∗2 c1 eiωt − c2 =
|c1 |2 + |c2 |2 − c∗1 c2 e−iωt − c∗2 c1 eiωt .
2
2
Note that the sum of the probabilities is 1.
 iωt 
c1 e
B: b1 = 2µ, hb1 |S(t)i = e−2iωt 1 0 0  c2  = e−iωt c1 ⇒ probability |c1 |2 .
c3
 iωt 
c1 e
1
1
b2 = µ, hb2 |S(t)i = e−2iωt √ 0 1 1  c2  = √ e−2iωt (c2 + c3 ) ⇒
2
2
c3
1
1
probability = (c∗2 + c∗3 ) (c2 + c3 ) =
|c2 |2 + |c3 |2 + c∗2 c3 + c∗3 c2 .
2
2
 iωt 
c1 e
1
1
b3 = −µ, hb3 |S(t)i = e−2iωt √ 0 1 −1  c2  = √ e−2iωt (c2 − c3 ) ⇒
2
2
c3
1
1
|c2 |2 + |c3 |2 − c∗2 c3 − c∗3 c2 .
probability = (c∗2 − c∗3 ) (c2 − c3 ) =
2
2
Again, the sum of the probabilities is 1.
Problem 3.47
(a)
p̂
p̂
p̂2
i
p̂2
+W
i√
+W =
+√
(W p̂ − p̂W ) + W 2 =
+ V1 .
Ĥ1 = Â Â = −i √
2m
2m
2m
2m
2m
†
106
CHAPTER 3. FORMALISM
i
[W, p̂]. To calculate the commutator, use a test function f (x):
2m
df
d
df
dW
df
dW
dW
[W, p̂]f (x) = W (−i~
+ i~ (W f ) = i~ −W
+
f +W
= i~
f ⇒ [W, p̂] = i~
.
dx
dx
dx
dx
dx
dx
dx
So V1 = W 2 + √
V1 = W 2 − √
~ dW
.
2m dx
p̂
p̂
p̂2
i
p̂2
Ĥ2 = ÂÂ = i √
+W
−i √
+W =
−√
(W p̂ − p̂W ) + W 2 =
+ V2 .
2m
2m
2m
2m
2m
†
So V2 = W 2 − √
i
~ dW
[W, p̂], and hence V2 = W 2 + √
.
2m
2m dx
(b)
Ĥ1 ψn(1) = En(1) ψn(1) ⇒ † Âψn(1) = En(1) ψn(1) ⇒ Ĥ2 Âψn(1) = † Âψn(1) = ÂEn(1) ψn(1) = En(1) Âψn(1) . X
Ĥ2 ψn(2) = En(2) ψn(2) ⇒ † ψn(2) = En(2) ψn(2) ⇒ Ĥ1 † ψn(2) = † † ψn(2) = † En(2) ψn(2) = En(2) † ψn(2) . X
(c)
(1)
Âψ0 (x) = 0 ⇒
(d)
~ d
√
+W
2m dx
(1)
(1)
ψ0 (x) = 0 ⇒ W (x) = − √
~ d h (1) i
~ (dψ0 /dx)
√
=
−
ln ψ0
.
(1)
2m
2m dx
ψ0
r
√
mα
m α
~ d
~ mα
mα
W (x) = − √
sign(x) =
sign(x).
ln
− 2 |x| = √
2
dx
~
~
~
2 ~
2m
2m
~ dW
mα2
~
V2 = W + √
=
+√
2
2~
2m dx
2m
2
I used the fact that
r
m α d
mα2
(sign(x)) =
+ αδ(x).
2 ~ dx
2~2
(
−1,
sign(x) = −1 + 2θ(x) =
+1,
(x < 0)
(x > 0)
and dθ/dx = δ(x) (Problem 2.23(b)).
Problem 3.48
(a) Integrate by parts:
∗
Z a ∗
Z a
a
df
dg
dg
g ∗ −i~
dx = −i~ g ∗ f + i~
f dx =
−i~
f dx = hp̂g|f i. X
dx
dx
0
0
0 dx
0
Z a
hg|p̂f i =
The boundary term drops out because f (0) = f (a) = 0, regardless of what g(x) does at the end points. No.
It’s not self-adjoint, because the domain of p̂† (the set of allowable g’s) is not the same as that of p̂ (the allowed
f ’s).
CHAPTER 3. FORMALISM
107
(b) For the boundary term to vanish we need
g∗ f
a
= g ∗ (a) f (a) − g ∗ (0) f (0) = g ∗ (a)λf (0) − g ∗ (0) f (0) = [λg ∗ (a) − g ∗ (0)] f (0) = 0.
0
But f (0) is not necessarily zero—else we are back to part (a)—so we need
λg ∗ (a) = g ∗ (0),
or
g(a) =
1
g(0).
λ∗
The two domains will be identical if 1/λ∗ = λ, or |λ|2 = 1, which is to say, λ = eiφ for some real number φ.
(c) For f (x) and g(x) to be in L2 (0, ∞), f (∞) = g(∞) = 0, so to kill the boundary term we need g ∗ (0) f (0) = 0.
Either f (0) = 0, in which case g(0) can be anything, or else g(0) = 0, in which case f (0) can be anything; there
is no way to make the two domains equal. No, there is no self-adjoint momentum operator on the semi-infinite
interval.
Problem 3.49
(a) For the free particle, V (x) = 0, so the time-dependent Schrödinger equation reads
Z ∞
~2 ∂ 2 Ψ
1
∂Ψ
√
=−
.
Ψ(x,
t)
=
i~
eipx/~ Φ(p, t) dp ⇒
∂t
2m ∂x2
2π~ −∞
Z ∞
Z ∞ 2
1
∂2Ψ
1
∂Ψ
∂Φ
p
√
=√
dp,
=
eipx/~
− 2 eipx/~ Φ dp. So
2
∂t
∂t
∂x
~
2π~ −∞
2π~ −∞
2 Z ∞
Z ∞
1
∂Φ
1
p
ipx/~
ipx/~
√
√
Φ dp.
e
i~
dp =
e
∂t
2m
2π~ −∞
2π~ −∞
But two functions with the same Fourier transform are equal (as you can easily prove using Plancherel’s
theorem), so
i~
∂Φ
p2
=
Φ.
∂t
2m
1
ip2
dΦ = −
dt
Φ
2m~
⇒
2
Φ(p, t) = e−ip t/2m~ Φ(p, 0).
(b)
2
Ψ(x, 0) = Ae−ax eilx ,
1
Φ(p, 0) = √
2π~
Φ(p, t) =
2a
π
A=
1/4 Z ∞
2a
π
1/4
(Problem 2.42(a)).
2
e−ipx/~ e−ax eilx dx =
−∞
2
2
1
e−(l−p/~) /4a e−ip t/2m~ ;
2
1/4
(2πa~ )
2
1
e−(l−p/~) /4a
2
1/4
(2πa~ )
|Φ(p, t)|2 = √
(Problem 2.42(b)).
2
1
e−(l−p/~) /2a .
2πa ~
108
CHAPTER 3. FORMALISM
(c)
Z ∞
2
1
pe−(l−p/~) /2a dp
2πa
~
−∞
−∞
[Let y ≡ (p/~) − l, so p = ~(y + l) and dp = ~ dy.]
Z ∞
2
~
= √
(y + l)e−y /2a dy [but the first term is odd]
2πa −∞
r
Z ∞
2~l
2~l
πa
−y 2 /2a
= √
e
dy = √
= ~l [as in Problem 2.42(d)].
2
2πa 0
2πa
Z ∞
hpi =
p|Φ(p, t)|2 dp = √
Z ∞
Z ∞
2
1
~2
2 −(l−p/~)2 /2a
p |Φ(p, t)| dp = √
p e
dp = √
(y 2 + 2yl + l2 )e−y /2a dy
hp i =
2πa ~ −∞
2πa −∞
−∞
Z ∞
Z ∞
2
2
2~2
e−y /2a dy
y 2 e−y /2a dy + l2
= √
2πa 0
0
"
r #
r 3
√
2~2
a
πa
= √
2 π
+ l2
= (a + l2 )~2 [as in Problem 2.42(d)].
2
2
2πa
2
Z ∞
2
2
p2
1 2
~2 2
1
~2 a
1 2
~2 a
; hHi =
hp i =
(l +a) =
hpi2 +
. But hHi0 =
hp i0 =
(Problem 2.21(d)).
2m
2m
2m
2m
2m
2m
2m
1
So hHi =
hpi2 + hHi0 . QED Comment: The energy of the traveling gaussian is the energy of the
2m
same gaussian at rest, plus the kinetic energy (hpi2 /2m) associated with the motion of the wave packet
as a whole.
(d) H =
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
109
Chapter 4
Quantum Mechanics in Three
Dimensions
Problem 4.1
(a)
[x, y] = xy − yx = 0, etc., so [ri , rj ] = 0.
~ ∂
[px , py ]f =
i ∂x
~ ∂f
i ∂y
~ ∂
−
i ∂y
~ ∂f
i ∂x
= −~
2
∂2f
∂2f
−
∂x∂y ∂y∂x
=0
(by the equality of cross-derivatives), so [pi , pj ] = 0.
~
[x, px ]f =
i
∂
~
∂f
∂f
∂f
−
(xf ) =
−x
− f = i~f,
x
x
∂x ∂x
i
∂x
∂x
so [x, px ] = i~ (likewise [y, py ] = i~ and [z, pz ] = i~).
~
∂f
∂
~
∂f
∂f
∂y
[y, px ]f =
y
−
(yf ) =
y
−y
= 0 (since
= 0). So [y, px ] = 0,
i
∂x ∂x
i
∂x
∂y
∂x
and same goes for the other “mixed” commutators. Thus [ri , pj ] = −[pj , ri ] = i~δij .
(b) The derivation of Eq. 3.73 is identical in three dimensions, so
dhxi
i
= h[H, x]i;
dt
~
p2
1 2
1 2
+ V, x =
[p + p2y + p2z , x] =
[p , x]
2m
2m x
2m x
1
1
~
=
(px [px , x] + [px , x]px ) =
[(−i~)px + (−i~)px ] = −i px .
2m
2m
m
[H, x] =
dhxi
i
∴
=
dt
~
~
1
dhri
1
−i hpx i = hpx i. The same goes for y and z, so:
= hpi.
m
m
dt
m
110
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
2
dhpx i
i
p
∂V
= h[H, px ]i; [H, px ] =
+ V, px = [V, px ] = i~
(Eq. 3.65)
dt
~
2m
∂x
i
∂V
∂V
dhpi
= (i~)
= −
. Same for y and z, so:
= h−∇V i.
~
∂x
∂x
dt
(c) From Eq. 3.62: σx σpx ≥
1
1
~
~
h[x, px ]i =
i~ = . Generally, σri σpj ≥ δij .
2i
2i
2
2
Problem 4.2
~2 ∂ 2 ψ ∂ 2 ψ ∂ 2 ψ
+
+
= Eψ (inside the box). Separable solutions: ψ(x, y, z) =
2m ∂x2
∂y 2
∂z 2
X(x)Y (y)Z(z). Put this in, and divide by XY Z:
(a) Equation 4.8 ⇒ −
1 d2 Y
1 d2 Z
2m
1 d2 X
+
+
= − 2 E.
2
2
2
X dx
Y dy
Z dz
~
The three terms on the left are functions of x, y, and z, respectively, so each must be a constant. Call the
separation constants kx2 , ky2 , and kz2 (as we’ll soon seen, they must be positive).
d2 X
= −kx2 X;
dx2
d2 Y
= −ky2 Y ;
dy 2
d2 Z
= −kz2 Z,
dz 2
with
E=
~2 2
(k + ky2 + kz2 ).
2m x
Solution:
X(x) = Ax sin kx x + Bx cos kx x;
Y (y) = Ay sin ky y + By cos ky y;
Z(z) = Az sin kz z + Bz cos kz z.
But X(0) = 0, so Bx = 0; Y (0) = 0, so By = 0; Z(0) = 0, so Bz = 0. And X(a) = 0 ⇒ sin(kx a) = 0 ⇒
kx = nx π/a (nx = 1, 2, 3, . . .). [As before (page 32), nx 6= 0, and negative values are redundant.] Likewise
ky = ny π/a and kz = nz π/a. So
ψ(x, y, z) = Ax Ay Az sin
n π n π n π y
z
x
x sin
y sin
z ,
a
a
a
E=
We might as well normalize X, Y, and Z separately: Ax = Ay = Az =
3/2
n π n π n π 2
x
y
z
ψ(x, y, z) =
sin
x sin
y sin
z ;
a
a
a
a
E=
p
~2 π 2 2
(n + n2y + n2z ).
2m a2 x
2/a. Conclusion:
π 2 ~2 2
(n + n2y + n2z );
2ma2 x
nx , ny , nz = 1, 2, 3, . . .
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
111
(b)
nx ny nz (n2x + n2y + n2z )
1 1 1
3
Energy
1 1 2
1 2 1
2 1 1
6
6
6
1 2 2
2 1 2
2 2 1
9
9
9
1 1 3
1 3 1
3 1 1
11
11
11
2 2 2
12
1
1
2
2
3
3
2
3
1
3
1
2
3
2
3
1
2
1
14
14
14
14
14
14
Degeneracy
E1 = 3
π 2 ~2
;
2ma2
d=1
E2 = 6
π 2 ~2
;
2ma2
d = 3.
E3 = 9
π 2 ~2
;
2ma2
d = 3.
E4 = 11
π 2 ~2
;
2ma2
d = 3.
E5 = 12
π 2 ~2
;
2ma2
d = 1.
E6 = 14
π 2 ~2
;
2ma2
d = 6.
(c) The next combinations are: E7 (322), E8 (411), E9 (331), E10 (421), E11 (332), E12 (422), E13 (431), and
E14 (333 and 511). The degeneracy of E14 is 4. Simple combinatorics accounts for degeneracies of 1
(nx = ny = nz ), 3 (two the same, one different), or 6 (all three different). But in the case of E14 there is
a numerical “accident”: 32 + 32 + 32 = 27, but 52 + 12 + 12 is also 27, so the degeneracy is greater than
combinatorial reasoning alone would suggest.
Problem 4.3
(a) Using Equations 4.8 and 4.13,
−
~2
2m
1 d
r2 dr
dψ
r2
+ V ψ = Eψ,
dr
~2
−
2m
At r → ∞ we get
E=−
1 d
r2 dr
r2
dψ
dr
1
2
ψ − ψ + V ψ = Eψ
a2
ra
~2
,
2ma2
so
V (r) = −
=
⇒
2
1
dψ
d2 ψ 2 dψ
1
2
2d ψ
r
+
2r
=
+
= 2 ψ − ψ.
2
2
2
r
dr
dr
dr
r dr
a
ra
−
~2
~2
+
+ V (r) = E.
2ma2
mar
~2
.
mar
(b) This time
dψ
2r
= − 2 ψ,
dr
a
d2 ψ
2
=− 2 ψ+
dr2
a
2r
a2
2
ψ,
112
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
so
1 d
r2 dr
At r = 0 we get
E=
3~2
,
ma2
r2
dψ
dr
so
4r2
2
6
4r2
2
2r
= − 2ψ + 4 ψ +
− 2 ψ = − 2 ψ + 4 ψ.
a
a
r
a
a
a
~2
4r2
6
−
− 2 + 4 + V (r) = E.
2m
a
a
V (r) =
2~2 2
r .
ma4
Problem 4.4
1
1
Eq. 4.32 ⇒ Y00 = √ P00 (cos θ); Eq. 4.27 ⇒ P00 (x) = P0 (x); Eq. 4.28 ⇒ P0 (x) = 1. Y00 = √ .
4π
4π
r
p
d
5 1 iφ 1
1
Y2 =
e P2 (cos θ); P21 (x) = − 1 − x2 P2 (x);
4π 3 · 2
dx
2
2
1 2
1
1
d
1 d 2
P2 (x) =
x2 − 1 =
2(x − 1)2x =
x − 1 + x(2x) =
3x2 − 1 ;
4 · 2 dx
8 dx
2
2
r
p
p
d 3 2 1
15 iφ
1
1
1
2
2
P2 (x) = − 1 − x
x −
= − 1 − x 3x; P2 (cos θ) = −3 cos θ sin θ. Y2 = −
e sin θ cos θ.
dx 2
2
8π
Z π
Z 2π ZZ
1
1
Normalization:
|Y00 |2 sin θ dθ dφ =
(2)(2π) = 1. X
sin θ dθ
dφ =
4π 0
4π
0
ZZ
Z
Z 2π
Z
15 π 2
15 π
1 2
2
|Y2 | sin θ dθ dφ =
sin θ cos θ sin θ dθ
dφ =
cos2 θ(1 − cos2 θ) sin θ dθ
8π 0
4 0
0
π
15 2 2
15
cos3 θ cos5 θ
5 3
=
+
−
−
= − =1X
=
4
3
5
4
3
5
2 2
0
r Z π
Z 2π
ZZ
1
15
0∗ 1
Orthogonality:
Y0 Y2 sin θ dθ dφ = − √
sin θ cos θ sin θ dθ
eiφ dφ = 0. X
4π 8π 0
|
{z
} | 0 {z }
(sin3 θ)/3|π
0 =0
(eiφ )/i|2π
0 =0
Problem 4.5
dΘ
A
1
A
1
A
d
dΘ
d
2
=
sec (θ/2) =
=
. Therefore
sin θ
=
(A) = 0.
dθ
tan(θ/2) 2
2 sin(θ/2) cos(θ/2)
sin θ
dθ
dθ
dθ
d
dΘ
With ` = m = 0, Eq. 4.25 reads:
sin θ
= 0. So A ln[tan(θ/2)] does satisfy Eq. 4.25. However,
dθ
dθ
π
Θ(0) = A ln(0) = A(−∞); Θ(π) = A ln tan
= A ln(∞) = A(∞). Θ blows up at θ = 0 and at θ = π.
2
Rπ
2
[In truth, this is not as decisive as it appears, since 0 [(ln[tan(θ/2)]) sin θ dθ = π 2 /6, which is perfectly
finite; Θ itself blows up at 0 and π, but sin θ tames it.]
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
113
Problem 4.6
s
(2` + 1) (` + m)! −imφ −m
(` − m)! m
e
P` (cos θ); P`−m (cos θ) = (−1)m
P (cos θ);
4π (` − m)!
(` + m)! `
s
(2` + 1) (` − m)! −imφ m
∗
Y`−m = (−1)m
e
P` (cos θ) = (−1)m (Y`m ) . QED
4π (` + m)!
Y`−m =
Problem 4.7
s
`
(2` + 1) 1 i`φ `
d
P` (x).
e P` (cos θ). P`` (x) = (−1)` (1 − x2 )`/2
4π (2`)!
dx
`
2`
1
d
(−1)`
d
2
`
`
2 `/2
P` (x) = `
(x − 1) , so P` (x) = ` (1 − x )
(x2 − 1)` .
2 `! dx
2 `!
dx
Y`` =
Now (x2 − 1)` = x2` + · · · , where all the other terms involve powers of x less than 2`, and hence give zero when
differentiated 2` times. So
2`
n
d
d
(−1)`
(2`)!
2 `/2
2`
`
x . But
xn = n!, so P`` (x) = (−1)` ` (1 − x2 )`/2 .
P` (x) = ` (1 − x )
2 `!
dx
dx
2 `!
s
∴ Y`` = (−1)`
r
Y32 =
r
Y32 =
(2` + 1) i`φ (2`)!
1
e
(sin θ)` =
`
4π(2`)!
2 `!
`!
7 1 2iφ 2
· e P3 (cos θ);
4π 5!
r
(2` + 1)!
4π
P32 (cos θ) = 15 sin2 θ cos θ;
7 1
1
15e2iφ cos θ sin2 θ =
4π 5!
4
Check that Y`` satisfies Eq. 4.18: Let
r
1
− eiφ sin θ
2
`
.
so
105 2iφ 2
e sin θ cos θ.
2π
1
`!
r
(2` + 1)!
4π
`
1
−
≡ A, so
2
Y`` = A(eiφ sin θ)` .
∂Y `
∂Y``
= Aei`φ `(sin θ)`−1 cos θ; sin θ ` = ` cos θY`` ;
∂θ
∂θ
∂
∂Y``
∂Y``
∂ 2 Y``
sin θ
sin θ
= ` cos θ sin θ
− ` sin2 θY`` = `2 cos2 θ − ` sin2 θ Y`` .
= −`2 Y`` .
∂θ
∂θ
∂θ
∂φ2
So the left side of Eq. 4.18 is `2 (1 − sin2 θ) − ` sin2 θ − `2 Y`` = −`(` + 1) sin2 θ Y`` , which matches the right
side.
r
1 105
2
Check that Y3 satisfies Eq. 4.18: Let B ≡
, so Y32 = Be2iφ sin2 θ cos θ.
4 2π
∂Y32
∂
∂Y32
∂
3
2iφ
2
= Be
2 sin θ cos θ − sin θ ; sin θ
sin θ
= Be2iφ sin θ
2 sin2 θ cos2 θ − sin4 θ
∂θ
∂θ
∂θ
∂θ
= Be2iφ sin θ 4 sin θ cos3 θ − 4 sin3 θ cos θ − 4 sin3 θ cos θ = 4Be2iφ sin2 θ cos θ cos2 θ − 2 sin2 θ
114
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
= 4(cos2 θ − 2 sin2 θ)Y32 .
∂ 2 Y32
= −4Y32 . So the left side of Eq. 4.18 is
∂φ2
4(cos2 θ − 2 sin2 θ − 1)Y32 = 4(−3 sin2 θ)Y32 = −`(` + 1) sin2 θ Y32 ,
where ` = 3, so it fits the right side of Eq. 4.18.
Problem 4.8
Z 1
1
1
P` (x)P`0 (x)dx = ` `0 0
2 `! 2 ` !
−1
Z 1 "
−1
d
dx
`
2
(x − 1)
`
# "
d
dx
`0
#
2
(x − 1)
`0
dx.
If ` 6= `0 , we may as well let ` be the larger of the two (` > `0 ). Integrate by parts, pulling successively each
derivative off the first term onto the second:
#1
" # " 0
Z 1
`−1
`
d
d
`
`0 0
2
`
2
`0
P` (x)P`0 (x)dx =
2 `!2 ` !
(x − 1)
(x − 1)
dx
dx
−1
−1
# " 0
#
Z 1 " `−1
` +1
d
d
2
`
2
`0
−
(x − 1)
(x − 1) dx
dx
dx
−1
`0 +`
Z 1
0
d
(x2 − 1)` dx.
= . . . (boundary terms) . . . + (−1)`
(x2 − 1)`
dx
−1
`0 +`
0
0
But (d/dx)
(x2 − 1)` = 0, because (x2 − 1)` is a polynomial whose highest power is 2`0 , so more than 2`0
derivatives will kill it, and `0 + ` > 2`0 . Now, the boundary terms are of the form:
# " 0
" # +1
`−n
` +n−1
d
d
2
`
2
`0
(x − 1)
(x − 1)
, n = 1, 2, 3, . . . , `.
dx
dx
−1
Look at the first term: (x2 − 1)` = (x2 − 1)(x2 − 1) . . . (x2 − 1); ` factors. So 0, 1, 2, . . . , ` − 1 derivatives
will still leave at least one overall factor of (x2 − 1). [Zero derivatives leaves ` factors; one derivative leaves
` − 1 : d/dx(x2 − 1)` = 2`x(x2 − 1)`−1 ; two derivatives leaves ` − 2 : d2 /dx2 (x2 − 1)` = 2`(x2 − 1)`−1 + 2`(` −
R1
1)2x2 (x2 − 1)`−2 , and so on.] So the boundary terms are all zero, and hence −1 P` (x)P`0 (x)dx = 0.
This leaves only the case ` = `0 . Again the boundary terms vanish, but this time the remaining integral
does not:
2`
Z 1
Z 1
d
(x2 − 1)` dx
(2` `!)2
[P` (x)]2 dx = (−1)`
(x2 − 1)`
dx
−1
−1
|
{z
}
(d/dx)2` (x2` )=(2`)!
`
Z 1
2
`
(x − 1) dx = 2(2`)!
= (−1) (2`)!
−1
Let x ≡ cos θ, so dx = − sin θ dθ,
Z 1
(1 − x2 )` dx =
0
Z 0
π/2
=
Z 1
(1 − x2 ) = sin2 θ,
(sin θ)2` (− sin θ)dθ =
(1 − x2 )` dx.
0
Z π/2
θ : π/2 → 0. Then
(sin θ)2`+1 dθ
0
(2)(4) · · · (2`)
(2` `!)2
(2` `!)2
=
=
.
(1)(3)(5) · · · (2` + 1)
1 · 2 · 3 · · · · (2` + 1)
(2` + 1)!
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Z 1
(2` `!)2
1
2
[P` (x)] dx = ` 2 2(2`)!
=
. So
∴
(2 `!)
(2` + 1)!
2` + 1
−1
2
Z 1
P` (x)P`0 (x)dx =
−1
115
2
δ``0 . QED
2` + 1
Problem 4.9
(a)
n1 (x) = −(−x)
1 d cos x
cos x sin x
=− 2 −
.
x dx
x
x
x
2
1 d
cos x
1 d
1 d cos x
n2 (x) = −(−x)2
= −x2
x dx
x
x dx
x dx
x
d 1 −x sin x − cos x
d sin x cos x
= −x
=
x
+
·
dx x
x2
dx x2
x3
2
x cos x − 2x sin x −x3 sin x − 3x2 cos x
=x
+
x4
x6
cos x
sin x sin x 3 cos x
3
1
3
=
−2 2 − 2 −
=
−
−
cos x − 2 sin x.
x
x
x
x3
x3
x
x
(b) Letting sin x ≈ x and cos x ≈ 1, and keeping only the lowest power of x:
1
1
1
n1 (x) ≈ − 2 − x ≈ − 2 .
x
x
x
As x → 0, this blows up.
3
1
3
3
n2 (x) ≈ − 3 −
− 2 x ≈ − 3 , which again blows up at the origin.
x
x
x
x
Problem 4.10
(a)
sin(kr) cos(kr)
A sin(kr)
u = Arj1 (kr) = A
−
=
− cos(kr) .
k2 r
k
k
(kr)
du
A k 2 r cos(kr) − k sin(kr)
cos(kr) sin(kr)
=
+
k
sin(kr)
=
A
−
+
sin(kr)
.
dr
k
(kr)2
kr
(kr)2
2
d2 u
−k r sin(kr) − k cos(kr) k 3 r2 cos(kr) − 2k 2 r sin(kr)
=A
−
+ k cos(kr)
dr2
(kr)2
(kr)4
sin(kr) cos(kr) cos(kr)
sin(kr)
= Ak −
−
−
+2
+ cos(kr)
2
2
(kr)
(kr)
(kr)
(kr)3
2
2
1
= Ak 1 −
sin(kr) .
cos(kr) +
−
(kr)2
(kr)3
(kr)
With V = 0 and ` = 1, Eq. 4.37 reads:
d2 u
2
2mE
− 2 u = − 2 u = −k 2 u. In this case the left side is
dr2
r
~
116
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
2
2
1
sin(kr)
2
Ak 1 −
cos(kr) +
−
sin(kr) −
− cos(kr)
(kr)2
(kr)3
(kr)
(kr)2
(kr)
sin(kr)
= Ak cos(kr) −
= −k 2 u. So this u does satisfy Eq. 4.37.
kr
cos z
z
= 0, or tan z = z. For high z (large n,
(b) Equation 4.48 ⇒ j1 (z) = 0, where z = ka. Thus sin
z2 − z
if n = 1, 2, 3, . . . counts the allowed energies in increasing order), the intersections occur slightly below
z = (n + 12 )π.
~2 π 2
~2 k 2
~2 z 2
=
∴E=
=
2m
2ma2
2ma2
2
1
n+
. QED
2
z
tan z
//2
3//2
5//2
z
Problem 4.11
For r ≤ a, u(r) = A sin(kr), with k ≡
state) E < 0 ⇒:
p
2m(E + V0 )/~. For r ≥ a, Eq. 4.37 with ` = 0, V = 0, and (for a bound
√
d2 u
2m
= − 2 Eu = κ2 u, with κ ≡ −2mE/~ ⇒ u(r) = Ceκr + De−κr .
dr2
~
But the Ceκr term blows up as r → ∞, so u(r) = De−κr .
1
1
κ
Continuity of u at r = a : A sin(ka) = De−κa
divide: tan(ka) = − , or − cot ka = .
Continuity of u0 at r = a : Ak cos(ka) = −Dκe−κa
k
κ
k
p
√
p
2mV0 a2 /~2 − z 2
κ
2mV0
Let ka ≡ z;
=
. Let z0 ≡
a. − cot z = (z0 /z)2 − 1. This is exactly the
k
z
~
same transcendental equation we encountered in Problem 2.29—see graph there. There is no solution if z0 < π/2,
which is to say, if 2mV0 a2 /~2 < π 2 /4, or V0 a2 < π 2 ~2 /8m. Otherwise, the ground state energy occurs somewhere
between z = π/2 and z = π:
E + V0 =
~2 π 2
~2 k 2 a2
~2 2
~2 π 2
=
z
,
so
<
(E
+
V
)
<
0
0
2ma2
2ma2
8ma2
2ma2
(precise value depends on V0 ).
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
117
Problem 4.12
R30 (n = 3, ` = 0) : Eq. 4.62 ⇒ v(ρ) =
j
j=0 cj ρ .
P
2(1 − 3)
c0 = −2c0 ;
(1)(2)
2(2 − 3)
1
2
2(3 − 3)
c1 = − c1 = c0 ; c3 =
c2 = 0.
(2)(3)
3
3
(3)(4)
r
1
1 r −r/3a
r
2 r 2
Eq. 4.73 ⇒ ρ =
; Eq. 4.75 ⇒ R30 = ρe−ρ v(ρ) =
e
c0 − 2c0
+ c0
3a
r
r 3a
3a 3
3a
c 2 r
2 r 2 −r/3a
0
1−
+
e
.
R30 =
3a
3 a
27 a
Eq. 4.76 ⇒ c1 =
R31 (n = 3, ` = 1) : c1 =
R31 =
c2 =
2(2 − 3)
1
c0 = − c0 ;
(1)(4)
2
c2 =
2(3 − 3)
c1 = 0.
(2)(5)
c 1 r
1 r −r/3a
1 r 2 −r/3a
0
.
e
r
e
c0 − c0
=
1
−
r 3a
2 3a
9a2
6 a
R32 (n = 3, ` = 2) : c1 =
2(3 − 3)
c0 = 0.
(1)(6)
R32 =
c 1 r 3 −r/3a
0
r2 e−r/3a .
e
(c0 ) =
r 3a
27a3
Problem 4.13
(a)
Z ∞
c r −r/2a
r
0
1−
e
. Let z ≡ .
2a
2a
a
0
Z Z
∞
c 2
1 4 −z
c20 a
z 2 −z 2
c20 a ∞ 2
24
a
0
3
3
1=
z − z + z e dz =
a
1−
e z dz =
2−6+
= c20 .
2a
2
4
4
4
4
2
0
0
r
2
1
∴ c0 =
. Eq. 4.74 ⇒ ψ200 = R20 Y00 . Table 4.3 ⇒ Y00 = √ .
a
4π
r
r −r/2a
1 1 r −r/2a
1
2 1 1−
e
⇒ ψ200 = √
1−
e
.
∴ ψ200 = √
2a
2a
4π a 2a
2πa 2a
Eq. 4.31 ⇒
|R|2 r2 dr = 1. Eq. 4.82 ⇒ R20 =
(b)
Z ∞
r
c20 a
3 2
2
1=
a
z e dz =
24 = ac0 , so c0 =
.
16
2
3a
0
!
r
1 1
1 1
3
1 1
−r/2a
−r/2a
±iφ
R21 = √
re
; ψ21±1 = √
re
∓
sin θe
= ∓√
re−r/2a sin θe±iφ ;
2
2
8π
πa 8a2
6a 2a
6a 2a
!
r
1 1
3
1
1
−r/2a
ψ210 = √
re
cos θ = √
re−r/2a cos θ.
4π
6a 2a2
2πa 4a2
c0
R21 = 2 re−r/2a ;
4a
c 2
0
4a2
5
4 −z
118
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Problem 4.14
(a)
L0 = ex e−x = 1.
L1 = ex
d −x e x = ex e−x − e−x x = 1 − x.
dx
2
ex d
ex d
L2 =
2xe−x − e−x x2
e−x x2 =
2 dx
2 dx
ex
=
2e−x − 2xe−x + e−x x2 − 2xe−x = 1 − 2x + 21 x2 .
2
3
ex d 2
ex d
e−x x3 =
−e−x x3 + 3x2 e−x
6 dx
6 dx
ex d −x 3
e x − 3x2 e−x − 3x2 e−x + 6xe−x
=
6 dx
ex
=
−e−x x3 + 3x2 e−x + 6x2 e−x − 12xe−x − 6xe−x + 6e−x
6
L3 =
= 1 − 3x + 32 x2 − 16 x3 .
(b)
v(ρ) = L52 (2ρ);
L52 (x) = (−1)5
ex
L7 (x) =
7!
d
dx
7
ex
7!
d
dx
5
ex
=
7!
d
dx
4
ex
7!
d
dx
3 =
=
7 −x
x e
ex
=
7!
d
dx
5
d
dx
L7 (x).
6
7x6 e−x − x7 e−x
42x5 e−x − 7x6 e−x − 7x6 e−x + x7 e−x
d
dx
2
210x4 e−x − 42x5 e−x − 84x5 e−x + 14x6 e−x + 7x6 e−x − x7 e−x
840x3 e−x − (210 + 630)x4 e−x
+ (126 + 126)x5 e−x − (21 + 7)x6 e−x + x7 e−x
ex
=
7!
2520x2 e−x − (840 + 3360)x3 e−x
+(840 + 1260)x4 e−x − (252 + 168)x5 e−x + (28 + 7)x6 e−x − x7 e−x
ex d
=
5040xe−x − (2520 + 12600)x2 e−x + (4200 + 8400)x3 e−x
7! dx
− (2100 + 2100)x4 e−x + (420 + 210)x5 e−x − (35 + 7)x6 e−x + x7 e−x .
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
119
ex
L7 (x) =
5040e−x − (5040 + 30240)xe−x + (15120 + 37800)x2 e−x
7!
− (12600 + 8400 + 8400)x3 e−x + (2100 + 2100 + 3150)x4 e−x
5 −x
6 −x
7 −x
− (630 + 252)x e + (42 + 7)x e − x e
=
1
5040 − 35280x + 52920x2 − 29400x3 + 7350x4 − 882x5 + 49x6 − x7 .
7!
1
L52 = −
d
dx
5
−882x5 + 49x6 − x7
7!
1 =−
−882(5 · 4 · 3 · 2) + 49(6 · 5 · 4 · 3 · 2)x − 7 · 6 · 5 · 4 · 3x2
7!
60 =
(882 × 2) − (49 × 12)x + 42x2 = 21 − 7x + 21 x2 .
5040
v(ρ) = 21 − 14ρ + 2ρ2 .
(c)
Eq. 4.62 ⇒ v(ρ) =
∞
X
cj ρj . Eq. 4.76 ⇒ c1 =
j=0
c2 =
2(4 − 5)
1
2
c1 = − c1 =
c0 ;
(2)(7)
7
21
c3 =
2
2(3 − 5)
c0 = − c0 .
(1)(6)
3
2(5 − 5)
c2 = 0.
(3)(8)
2
2
c0
v(ρ) = c0 − c0 ρ + c0 ρ2 =
21 − 14ρ + 2ρ2 .
3
21
21
X
Problem 4.15
(a)
ψ=√
hri =
1
πa3
4
a3
e
−r/a
Z ∞
,
1
so hr i =
πa3
n
r3 e−2r/a dr =
0
Z
n −2r/a
r e
4 a 4
3
3!
= a;
a3
2
2
4π
r sin θ dr dθ dφ =
πa3
2
hr2 i =
4
a3
Z ∞
hx2 i =
0
1 2
hr i = a2 .
3
(c)
1 1
ψ211 = R21 Y11 = − √
re−r/2a sin θeiφ
πa 8a2
rn+2 e−2r/a dr.
0
r4 e−2r/a dr =
(b)
hxi = 0;
Z ∞
(Problem 4.13(b)).
4 a 5
4!
= 3a2 .
a3
2
120
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Z 1
1
r2 e−r/a sin2 θ r2 sin2 θ cos2 φ r2 sin θ dr dθ dφ
2
2
πa (8a )
Z ∞
Z π
Z 2π
1
5
6 −r/a
r e
dr
sin θ dθ
cos2 φ dφ
=
64πa5 0
0
0
2·4
1
1
7
6!a
2
·
2π
= 12a2 .
=
64πa5
1·3·5
2
hx2 i =
Problem 4.16
1
4
4
e−r/a ; P = |ψ|2 4πr2 dr = 3 e−2r/a r2 dr = p(r) dr; p(r) = 3 r2 e−2r/a .
a
a
πa3
r
dp
2 −2r/a
4
8r
−2r/a
2
+r − e
= 0 ⇒ r = a.
= 3 2re
= 3 e−2r/a 1 −
dr
a
a
a
a
ψ=√
Problem 4.17
1
[z, [H, z]] = z(Hz − zH) − (Hz − zH)z = 2zHz − z 2 H − Hz 2 , ⇒ zHz =
[z, [H, z]] + z 2 H + Hz 2 ,
2
2
p
1 2
1 2
1
i~
+ V, z =
(px + p2y + p2z ), z =
[pz , z] =
(pz [pz , z] + [pz , z]pz ) = − pz .
[H, z] =
2m
2m
2m
2m
m
2
2
i~
i~
~
1 ~
[z, [H, z]] = − [z, pz ] = − (i~) =
, so zHz =
+ Hz 2 + z 2 H . Using Problem 4.18(b) :
m
m
m
2 m
~2
~2
~2 2
1 ~2
+ hψ|Hz 2 |ψi + hψ|z 2 H|ψi =
+ E1 hz 2 i =
−
hzHzi =
a = 0.
2 m
2m
2m 2ma2
Problem 4.18
~2
1 E1
1
=−
.
(a) Ψ(r, t) = √ ψ2 1 1 e−iE2 t/~ + ψ2 1 −1 e−iE2 t/~ = √ (ψ2 1 1 + ψ2 1 −1 ) e−iE2 t/~ ; E2 =
4
8ma2
2
2
From Problem 4.13(b):
1 1
i
ψ2 1 1 + ψ2 1 −1 = − √
re−r/2a sin θ sin φ.
re−r/2a sin θ eiφ − e−iφ = − √
2
πa 8a
πa 4a2
Ψ(r, t) = − √
i
re−r/2a sin θ sin φ e−iE2 t/~ .
2πa 4a2
(b)
Z 1
e2 1
1
e2
2
2
3
2 −r/a
hV i =
|Ψ| −
d r=
−
r
e
sin
θ
sin
φ
r2 sin θ dr dθ dφ
4π0 r
(2πa)(16a4 )
4π0
r
Z ∞
Z π
Z 2π
4
1
~2
~2
3
2
3 −r/a
4
=
−
r
e
dr
sin
θ
dθ
sin
φ
dφ
=
−
3!a
(π)
32πa5
ma
32πma6
3
0
0
0
Z
= −
2
~2
1
1
= E1 = (−13.6eV) = −6.8eV
2
4ma
2
2
(independent of t).
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
121
Problem 4.19
En (Z) = Z 2 En ;
E1 (Z) = Z 2 E1 ;
R(Z) = Z 2 R.
a(Z) = a/Z;
Lyman lines
range from
range
from ni = 2 to ni = ∞ (with nf = 1);the wavelengths
1
1
3
4
1
1
1
=R 1−
= R ⇒ λ2 =
down to
=R 1−
= R ⇒ λ1 = .
λ2
4
4
3R
λ1
∞
R
1
1
1
For Z = 2 : λ1 =
=
= 2.28 × 10−8 m to λ2 =
= 3.04 × 10−8 m, ultraviolet.
4R
4(1.097 × 107 )
3R
1
4
For Z = 3 : λ1 =
= 1.01 × 10−8 m to λ2 =
= 1.35 × 10−8 m, also ultraviolet.
9R
27R
Problem 4.20
(a) V (r) = −G
Mm
.
r
So
(b) Equation 4.72: a =
=
e2
→ GM m translates hydrogen results to the gravitational analogs.
4π0
4π0
e2
~2
,
m
so
~2
GM m2
ag =
(1.0546 × 10−34 Js)2
(6.6726 × 10−11 m3 /kg · s2 )(1.9892 × 1030 kg)(5.98 × 1024 kg)2
(c) Equation 4.70 ⇒ En = −
h m
2~2
(GM m)2
i 1
n2
= 2.34 × 10−138 m.
.
1
Mm
mv 2
GM m
Mm
1
mv 2 − G
. But G 2 =
⇒ mv 2 =
, so
2
ro
ro
ro
2
2ro
r
h m
i 1
GM m2
ro
ro
GM m
2
2
n
=
=−
(GM
m)
⇒
n
=
r
=
⇒
.
Ec = −
o
2
2
2
2ro
2~
n
~
ag
ag
r
1.496 × 1011
11
ro = earth-sun distance = 1.496 × 10 m ⇒ n =
= 2.53 × 1074 .
2.34 × 10−138
Ec =
(d)
G2 M 2 m3
∆E = −
2~2
So
1
1
− 2 .
(n + 1)2
n
1
1
1
= 2
≈ 2
(n + 1)2
n (1 + 1/n)2
n
1
1
1
2
2
− 2 ≈ 2 1 − − 1 = − 3;
(n + 1)2
n
n
n
n
∆E =
∆E =
2
1−
n
.
G2 M 2 m3
.
~2 n 3
(6.67 × 10−11 )2 (1.99 × 1030 )2 (5.98 × 1024 )3
= 2.09 × 10−41 J.
(1.055 × 10−34 )2 (2.53×74 )3
Ep = ∆E = hν =
hc
.
λ
λ = (3 × 108 )(6.63 × 10−34 )/(2.09 × 10−41 ) = 9.52 × 1015 m.
But 1 ly = 9.46 × 1015 m. Is it a coincidence that λ ≈ 1 ly? No: From part (c), n2 = GM m2 ro /~2 , so
!
r
3/2
ch
~2 n 3
2π~3
GM m2 ro
ro3
λ=
= c2π~ 2 2 3 = c 2 2 3
= c 2π
.
∆E
G M m
G M m
~2
GM
122
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
p
But (from (c)) v = GM/ro = 2πro /T , where T is the period of the orbit (in this case one year), so
p
T = 2π ro3 /GM , and hence λ = cT (one light year). [Incidentally, the same goes for hydrogen: The
wavelength of the photon emitted in a transition from a highly excited state to the next lower one is equal
to the distance light would travel in one orbital period.]
Problem 4.21
hf |L± gi = hf |Lx gi ± ihf |Ly gi = hLx f |gi ± ihLy f |gi = h(Lx ∓ iLy )f |gi = hL∓ f |gi, so (L± )† = L∓ .
Now, using Eq. 4.112, in the form L∓ L± = L2 − L2z ∓ ~Lz :
hf`m |L∓ L± f`m i = hf`m |(L2 − L2z ∓ ~Lz )f`m i = hf`m | ~2 `(` + 1) − ~2 m2 ∓ ~2 m f`m i
= ~2 [`(` + 1) − m(m ± 1)] hf`m |f`m i = ~2 [`(` + 1) − m(m ± 1)] = hL± f`m |L± f`m i
Upper signs:
p
m+1
m+1
m
2
~2 [`(` + 1) − m(m + 1)] = hL+ f`m |L+ f`m i = hAm
|Am
i = |Am
` f`
` f`
` | ⇒ A` = ~ `(` + 1) − m(m + 1).
Lower signs:
p
~2 [`(` + 1) − m(m − 1)] = hL− f`m |L− f`m i = hB`m f`m−1 |B`m f`m−1 i = |B`m |2 ⇒ B`m = ~ `(` + 1) − m(m − 1).
At the top of the ladder (m = `) we get A`` = 0, so there is no higher rung; at the bottom of the ladder (m = −`)
we get B`−` = 0, so there is no lower rung.
Problem 4.22
(a)
[Lz , x] = [xpy − ypx , x] = [xpy , x] − [ypx , x] = 0 − y[px , x] = i~y. X
[Lz , y] = [xpy − ypx , y] = [xpy , y] − [ypx , y] = x[py , y] − 0 = −i~x. X
[Lz , z] = [xpy − ypx , z] = [xpy , z] − [ypx , z] = 0 − 0 = 0. X
[Lz , px ] = [xpy − ypx , px ] = [xpy , px ] − [ypx , px ] = py [x, px ] − 0 = i~py . X
[Lz , py ] = [xpy − ypx , py ] = [xpy , py ] − [ypx , py ] = 0 − px [y, py ] = −i~px . X
[Lz , pz ] = [xpy − ypx , pz ] = [xpy , pz ] − [ypx , pz ] = 0 − 0 = 0. X
(b)
[Lz , Lx ] = [Lz , ypz − zpy ] = [Lz , ypz ] − [Lz , zpy ] = [Lz , y]pz − [Lz , py ]z
= −i~xpz + i~px z = i~(zpx − xpz ) = i~Ly .
(So, by cyclic permutation of the indices, [Lx , Ly ] = i~Lz .)
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
123
(c)
[Lz , r2 ] = [Lz , x2 ] + [Lz , y 2 ] + [Lz , z 2 ] = [Lz , x]x + x[Lz , x] + [Lz , y]y + y[Lz , y] + 0
= i~yx + xi~y + (−i~x)y + y(−i~x) = 0.
[Lz , p2 ] = [Lz , p2x ] + [Lz , p2y ] + [Lz , p2z ] = [Lz , px ]px + px [Lz , px ] + [Lz , py ]py + py [Lz , py ] + 0
= i~py px + px i~py + (−i~px )py + py (−i~px ) = 0.
(d) It follows from (c) that all three components of L commute with r2 and p2 . Lz also commutes with r, as
we can show using a test function f (r):
∂
∂
∂(rf )
∂(rf )
∂f
∂f
∂r
∂r
[Lz , r]f = −i~ x
− y , r f = −i~ x
−y
− rx
+ ry
= −i~ x
−y
f.
∂y
∂x
∂y
∂x
∂y
∂x
∂y
∂x
But
∂ p 2
∂r
x
=
x + y2 + z2 = ,
∂x
∂x
r
so
x
∂r
∂r
−y
∂y
∂x
= 0,
and hence [Lz , r] = 0 (and the same goes for the other two components). So L commutes with H =
p2 /2m + V (r). QED
Problem 4.23
(a)
Equation 3.73 ⇒
dhLx i
i
1 2
= h[H, Lx ]i. [H, Lx ] =
[p , Lx ] + [V, Lx ].
dt
~
2m
The first term is zero (Problem 4.22(c)); the second would be too if V were a function only of r = |r|, but
in general
[H, Lx ] = [V, ypz − zpy ] = y[V, pz ] − z[V, py ]. Now (Problem 3.14(c)):
[V, pz ] = i~
Thus
∂V
∂V
∂V
∂V
and [V, py ] = i~
. So [H, Lx ] = yi~
− zi~
= i~[r × (∇V )]x .
∂z
∂y
∂z
∂y
dhLx i
= −h[r × (∇V )]x i, and the same goes for the other two components:
dt
dhLi
= h[r × (−∇V )]i = hNi. QED
dt
(b)
If V (r) = V (r), then ∇V =
∂V
dhLi
r̂, and r × r̂ = 0, so
= 0. QED
∂r
dt
124
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Problem 4.24
(a)
∂
∂
∂f
∂f
−iφ
L+ L− f = −~ e
+ i cot θ
e
− i cot θ
∂θ
∂φ
∂θ
∂φ
2
2
∂ f
2 ∂f
2 iφ
−iφ ∂ f
− i − csc θ
= −~ e
e
+ cot θ
∂θ2
∂φ
∂θ ∂φ
2
∂f
∂f
∂ f
∂2f
−iφ
−iφ
+i cot θ −ie
− i cot θ
+e
− i cot θ 2
∂θ
∂φ
∂φ ∂θ
∂φ
2
2
2
∂ f
∂ f
∂f
∂2f
2
2 ∂f
2 ∂f
2 ∂ f
= −~
+ i csc θ
− i cot θ
+ cot θ
− i cot θ
+ i cot θ
+ cot θ 2
∂θ2
∂φ
∂θ ∂φ
∂θ
∂φ
∂φ ∂θ
∂φ
2
2
∂
∂
∂
∂
= −~2
+ cot θ
+ cot2 θ 2 + i(csc2 θ − cot2 θ)
f, so
2
∂θ
∂θ
∂φ
∂φ
2
∂
∂
∂
∂2
2
+
cot
θ
+
cot
+i
. QED
L+ L− = −~2
θ
2
2
∂θ
∂θ
∂φ
∂φ
2 iφ
∂
, Eq. 4.112 ⇒ L2 = L+ L− + L2z − ~Lz , so, using (a):
∂φ
2
2
∂
∂
∂2
∂
~ ∂
2
2
2
2 ∂
L = −~
+ cot θ
+ cot θ 2 + i
−~
−~
∂θ2
∂θ
∂φ
∂φ
∂φ2
i ∂φ
2
2
2
∂
∂
∂
∂
∂
∂
∂
1 ∂2
2
2
2
= −~
+ cot θ
+ (cot θ + 1) 2 + i
−i
= −~
+ cot θ
+
∂θ2
∂θ
∂φ
∂φ
∂φ
∂θ2
∂θ sin2 θ ∂φ2
1 ∂
∂
1 ∂2
= −~2
sin θ
+
. QED
sin θ ∂θ
∂θ
sin2 θ ∂φ2
(b) Equation 4.129 ⇒ Lz = −i~
Problem 4.25
(a) L+ Y`` = 0 (top of the ladder).
(b)
Lz Y`` = ~`Y`` ⇒
~ ∂ `
Y = ~`Y`` ,
i ∂φ `
so
∂Y``
= i`Y`` , and hence Y`` = f (θ)ei`φ .
∂φ
[Note: f (θ) is the “constant” here—it’s constant with respect to φ . . . but still can depend on θ.]
∂ df i`φ
∂
+ i cot θ
f (θ)ei`φ = 0, or
e + if cot θ i` ei`φ = 0, so
L+ Y`` = 0 ⇒ ~eiφ
∂θ
∂φ
dθ
Z
Z
df
df
df
cos θ
= ` cot θf ⇒
= ` cot θdθ ⇒
=`
dθ ⇒ ln f = ` ln(sin θ) + constant.
dθ
f
f
sin θ
f
f
`
ln f = ln(sin θ) + K ⇒ ln
=K⇒
= constant ⇒ f (θ) = A sin` θ.
`
sin θ
sin` θ
Y`` (θ, φ) = A(eiφ sin θ)` .
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
125
(c)
1=A
2
Z
2`
2
Z π
sin θ sin θ dθ dφ = 2πA
sin(2`+1) θ dθ = 2πA2 2
0
2
= 4πA2
(2 · 4 · 6 · · · · · 2`)
(2` `!)2
= 4πA2
,
1 · 2 · 3 · 4 · 5 · · · · · (2` + 1)
(2` + 1)!
so
(2 · 4 · 6 · · · · · (2`))
1 · 3 · 5 · · · · · (2` + 1)
r
(−1)` (2` + 1)!
A = `+1
.
2 `!
π
The phase factor (−1)` is arbitrary, of course; I have chosen it to be consistent with Problem 4.7 and
Equation 4.32.
Problem 4.26
#
" r
∂
∂
15
−
L+ Y21 = ~eiφ
+ i cot θ
sin θ cos θeiφ
∂θ
∂φ
8π
r
15 iφ iφ
cos θ
2
2
iφ
=−
~e
e (cos θ − sin θ) + i
sin θ cos θ ie
8π
sin θ
r
r
2
15 2iφ
15
=−
~e
cos2 θ − sin2 θ − cos2 θ =
~ eiφ sin θ
8π
8π
r
√
2
1 15 iφ
= ~ 2 · 3 − 1 · 2 Y22 = 2~Y22 .
∴ Y22 =
e sin θ .
4 2π
Problem 4.27
L2
(a) Classically, H = 12 Iω 2 and L = Iω; so H =
. But we know the eigenvalues of L2 : ~2 `(` + 1). Or, since
2I
we usually label energies with n:
En =
~2 n(n + 1)
(n = 0, 1, 2, . . .).
2I
(b) ψnm (θ, φ) = Ynm (θ, φ), the ordinary spherical harmonics. The degeneracy of the nth energy level is the
number of m-values for given n: 2n + 1.
(c) As in Equation 4.92, the frequency of an emitted photon, in a transition from ni to nf , is given by
Eγ = hν = 2π~ν = Ei − Ef =
~2
[ni (ni + 1) − nf (nj + 1)] .
2I
The term in square brackets is the difference of two even numbers, so it is necessary even. It cannot be
zero, but it does hit every positive even integer, since if ni = nf + 1, then [ni (ni + 1) − nj (nj + 1)] =
~j
2(nf + 1) = 2, 4, 6, . . . . Calling this number 2j, νj =
, j = 1, 2, 3, . . . .
2πI
126
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
(d) ∆ν = 68−27
= 3.73 cm−1 = 3.73 × (3 × 1010 )Hz = 11 × 1010 /s.
11
m1 m2
mC = 12 u = 2.00 × 10−26 kg, mO = 16 u = (4/3) mC ;
= 1.14 × 10−26 kg.
m1 + m2
1/2
~
1.05 × 10−34
m1 m2 2
m = 1.15 × 10−10 m.
a =
⇒ a=
I=
m1 + m2
2π(∆ν)
2π(11 × 1010 )(1.14 × 10−26 )
(The accepted value is 1.128 × 10−10 ).
Problem 4.28
(1.6 × 10−19 )2
= 2.81 × 10−15 m.
4π(8.85 × 10−12 )(9.11 × 10−31 )(3.0 × 108 )2
2 2 v 2
1
mr
L = ~ = Iω =
= mrv so
2
5
r
5
rc =
v=
(5)(1.055 × 10−34 )
5~
=
= 5.15 × 1010 m/s.
4mr
(4)(9.11 × 10−31 )(2.81 × 10−15 )
Since the speed of light is 3 × 108 m/s, a point on the equator would be going more than 100 times the speed
of light. Nope : This doesn’t look like a very realistic model for spin.
Problem 4.29
(a)
~2
0 1
0 −i
0 −i
0 1
[Sx , Sy ] = Sx Sy − Sy Sx =
−
1 0
i 0
i 0
1 0
4
2
2
~ 2i 0
~ 1 0
~
i 0
−i 0
−
=
= i~
= i~Sz . X
=
0 −i
0 i
4
4 0 −2i
2 0 −1
(b)
σx σx =
σx σy =
σy σx =
10
= 1 = σy σy = σz σz ,
01
i 0
= iσz ;
0 −i
−i 0
= −iσz ;
0 i
σy σz =
so σj σj = 1 for j = x, y, or z.
0 i
i 0
σz σy =
= iσx ;
σz σx =
0 −i
= −iσx ;
−i 0
Equation 4.153 packages all this in a single formula.
X
0 1
= iσy ;
−1 0
σx σz =
0 −1
= −iσy .
1 0
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Problem 4.30
(a)
χ† χ = |A|2 (9 + 16) = 25|A|2 = 1 ⇒ A = 1/5.
(b)
0 1
4
1 ~
~
~
3i
−3i 4
−3i 4
hSx i = χ Sx χ =
(−12i + 12i) = 0.
=
=
1 0
3i
4
25 2
50
50
†
0 −i
−4i
12
1 ~
~
~
3i
−3i 4
−3i 4
hSy i = χ Sy χ =
(−12 − 12) = − ~.
=
=
i
0
−3
4
25 2
50
50
25
1 0
3i
1 ~
7
~
~
3i
−3i 4
−3i 4
hSz i = χ† Sz χ =
(9 − 16) = − ~.
=
=
0 −1
−4
4
25 2
50
50
50
†
(c)
hSx2 i = hSy2 i = hSz2 i =
~2
~2
~
(always, for spin 1/2), so σS2 x = hSx2 i − hSx i2 =
− 0, σSx = .
4
4
2
σS2 y = hSy2 i − hSy i2 =
~2
−
4
12
25
2
σS2 z = hSz2 i − hSz i2 =
~2
−
4
7
50
2
~2 =
7
~2
49 2
(625 − 576) =
~ , σSy =
~.
2500
2500
50
~2 =
576 2
12
~2
(625 − 49) =
~ , σSz =
~.
2500
2500
25
(d)
σSx σSy =
~ 7
~ 7 ? ~
· ~ ≥ |hSz i| = · ~ (right at the uncertainty limit). X
2 50
2
2 50
σSy σSz =
7
12 ? ~
~ · ~ ≥ |hSx i| = 0
50
25
2
σSz σSx =
12
~ ? ~
~ 12
~ · ≥ |hSy i| = · ~ (right at the uncertainty limit). X
25
2
2
2 25
(trivial). X
Problem 4.31
hSx i =
~ ∗ ∗ 0 1
~ ∗ ∗ b
~
a
a b
a b
=
= (a∗ b + b∗ a) = ~ Re(ab∗ ).
10
b
a
2
2
2
~ ∗ ∗ 0 −i
~ ∗ ∗ −ib
a
a b
a b
hSy i =
=
i 0
b
ia
2
2
=
~
~
(−ia∗ b + iab∗ ) = i(ab∗ − a∗ b) = −~ Im(ab∗ ).
2
2
127
128
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
hSz i =
S2x =
~ ∗ ∗ 1 0
~
~
~ ∗ ∗ a
a
a b
a b
= (a∗ a − b∗ b) = (|a|2 − |b|2 ).
=
0 −1
−b
b
2
2
2
2
~2
4
01
10
~2
~2 1 0
01
;
=
=
10
4 01
4
~2
S2z =
~2
1 0
1 0
;
=
0 −1
4 0 −1
4
hSx2 i + hSy2 i + hSz2 i =
S2y =
~2
4
~2
0 −i
0 −i
;
=
i 0
i 0
4
so hSx2 i = hSy2 i = hSz2 i =
~2
.
4
3 2 ?
1 1
3
~ = s(s + 1)~2 = ( + 1)~2 = ~2 = hS 2 i. X
4
2 2
4
Problem 4.32
(a)
Sy =
~
2
~
2
0 −i
;
i 0
~2
~
−λ −i~/2
= λ2 −
⇒ λ = ± (of course).
i~/2 −λ
4
2
~ α
0 −i
α
⇒ −iβ = ±α;
=±
i 0
β
2 β
1
;
2 i
1
(y)
χ+ = √
1
(y)
χ− = √
2
1
|α|2 + |β|2 = 1 ⇒ |α|2 + |α|2 = 1 ⇒ α = √ .
2
1
.
−i
(b)
c+ =
c− =
(y)
χ+
(y)
χ−
†
a
1
1
√
1 −i
χ=
= √ (a − ib);
b
2
2
†
a
1
1
χ= √ 1 i
= √ (a + ib);
b
2
2
~
1
+ , with probability |a − ib|2 .
2
2
~
1
− , with probability |a + ib|2 .
2
2
1 ∗
[(a + ib∗ )(a − ib) + (a∗ − ib∗ )(a + ib)]
2
1 2
=
|a| − ia∗ b + iab∗ + |b|2 + |a|2 + ia∗ b − iab∗ + |b|2 = |a|2 + |b|2 = 1. X
2
P+ + P− =
(c)
~2
, with probability 1.
4
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
129
Problem 4.33
Sr = S · r̂ = Sx sin θ cos φ + Sy sin θ sin φ + Sz cos θ
~
0
sin θ cos φ
0
−i sin θ sin φ
cos θ
0
=
+
+
sin θ cos φ
0
i sin θ sin φ
0
0 − cos θ
2
~
~
cos θ
sin θ(cos φ − i sin φ)
cos θ e−iφ sin θ
.
=
=
− cos θ
2 sin θ(cos φ + i sin φ)
2 eiφ sin θ − cos θ
~2
~2
( ~2 cos θ − λ) ~2 e−iφ sin θ
= − cos2 θ + λ2 −
sin2 θ = 0 ⇒
~
~ iφ
e
sin
θ
(−
cos
θ
−
λ)
4
4
2
2
~
~2
~2
(sin2 θ + cos2 θ) =
⇒ λ = ± (of course).
4
4
2
~
~ α
cos θ e−iφ sin θ
α
=±
⇒ α cos θ + βe−iφ sin θ = ±α;
iφ
β
2 e sin θ − cos θ
2 β
λ2 =
β = eiφ
(±1 − cos θ)
α.
sin θ
sin(θ/2)
α. Normalizing:
cos(θ/2)
1
θ
θ
sin2 (θ/2) 2
cos(θ/2)
(r)
iφ
=
|α| = |α|2
⇒
α
=
cos
,
β
=
e
sin
,
.
1 = |α|2 +|β|2 = |α2 |+ 2
χ
+
eiφ sin(θ/2)
cos (θ/2)
cos2 (θ/2)
2
2
Upper sign: Use 1 − cos θ = 2 sin2 θ2 , sin θ = 2 sin θ2 cos θ2 .
Then β = eiφ
θ
cos2 (θ/2) 2
cos(θ/2)
1
Lower sign: Use 1 + cos θ = 2 cos2 , β = −eiφ
α; 1 = |α|2 +
|α| = |α|2 2
.
2
2
sin(θ/2)
sin (θ/2)
sin (θ/2)
−iφ
e
sin(θ/2)
(r)
.
Pick α = e−iφ sin(θ/2); then β = − cos(θ/2), and χ− =
− cos(θ/2)
Problem 4.34
 
 
 
0
0
1
There are three states: χ+ = 0 , χ0 = 1 , χ− = 0 .
0
0
1
Sz χ+ = ~χ+ , Sz χ0 = 0, Sz χ− = −~χ− ,


1 0 0
⇒ Sz = ~ 0 0 0  .
0 0 −1
From Eq. 4.136:




√
√
0 1 0
0 0 0
√
√
S+ χ+ = 0,√
S+ χ0 = ~√2χ+ , S+ χ− = ~ 2χ0
⇒ S+ = 2~ 0 0 1 , S− = 2~ 1 0 0 .
S− χ+ = ~ 2χ0 , S− χ0 = ~ 2χ− , S− χ− = 0
0 0 0
0 1 0


010
1
~ 
1 0 1 ,
Sx = (S+ + S− ) = √
2
2 010


0 −1 0
1
i~ 
1 0 −1 .
Sy = (S+ − S− ) = √
2i
2 0 1 0
130
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Problem 4.35
(a) Using Eqs. 4.151 and 4.163:
cos α eiγB0 t/2
1 h
α iγB0 t/2
1
α −iγB0 t/2 i
(x)
(x)†
2
√
=
cos
.
c+ = χ+ χ = √ 1 1
e
+
sin
e
sin α2 e−iγB0 t/2
2
2
2
2
ih
i
1h
α
α
α
α
(x)
(x)
P+ (t) = |c+ |2 =
cos e−iγB0 t/2 + sin eiγB0 t/2 cos eiγB0 t/2 + sin e−iγB0 t/2
2
2
2
2
2
i
1h 2α
α
α iγB0 t
2 α
−iγB0 t
=
cos
+ sin
+ sin cos
e
+e
2
2
2
2
2
i
1h
α
1
α
=
1 + 2 sin cos cos(γB0 t) =
[1 + sin α cos(γB0 t)] .
2
2
2
2
1 1
(y)
(b) From Problem 4.32(a): χ+ = √
.
2 i
cos α eiγB0 t/2
1 h
α iγB0 t/2
α −iγB0 t/2 i
1
(y)
(y)†
2
√
=
cos
e
−
i
sin
e
c+ = χ+ χ = √ 1 −i
;
sin α2 e−iγB0 t/2
2
2
2
2
ih
i
1h
α
α
α
α
(y)
(y)
cos e−iγB0 t/2 + i sin eiγB0 t/2 cos eiγB0 t/2 − i sin e−iγB0 t/2
P+ (t) = |c+ |2 =
2
2
2
2
2
i
α
α iγB0 t
1h 2α
2 α
−iγB0 t
+ sin
+ i sin cos
e
−e
=
cos
2
2
2
2
2
i
1h
α
α
1
=
1 − 2 sin cos sin(γB0 t) =
[1 − sin α sin(γB0 t)] .
2
2
2
2
(c)
(z)
χ+ =
1
;
0
(z)
c+ = 1 0
cos α2 eiγB0 t/2
sin α2 e−iγB0 t/2
= cos
α iγB0 t/2
e
;
2
(z)
(z)
P+ (t) = |c+ |2 = cos2
Problem 4.36
(a)
γB0 ~
1 0
cos ωt
H = −γB · S = −γB0 cos ωt Sz = −
.
0 −1
2
(b)
1
α(t)
χ(t) =
, with α(0) = β(0) = √ .
β(t)
2
∂χ
γB0 ~
γB0 ~
α̇
1 0
α
α
i~
= i~
= Hχ = −
cos ωt
=−
cos ωt
.
0 −1
β
−β
β̇
∂t
2
2
α
.
2
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
α̇ = i
γB0
2
dα
cos ωt α ⇒
=i
α
α(t) = Aei(γB0 /2ω) sin ωt ;
β̇ = −i
γB0
2
γB0
2
cos ωt dt ⇒ ln α =
1
α(0) = A = √ ,
2
131
iγB0 sin ωt
+ constant.
2
ω
1
so α(t) = √ ei(γB0 /2ω) sin ωt .
2
1
cos ωt β ⇒ β(t) = √ e−i(γB0 /2ω) sin ωt .
2
1
χ(t) = √
2
ei(γB0 /2ω) sin ωt
.
e−i(γB0 /2ω) sin ωt
(c)
(x)
i(γB /2ω) sin ωt i
0
1
1 h i(γB0 /2ω) sin ωt
e
− e−i(γB0 /2ω) sin ωt
(1 − 1) −i(γB0 /2ω) sin ωt =
e
e
2
2
γB0
γB0
(x)
(x)
= i sin
sin ωt . P− (t) = |c− |2 = sin2
sin ωt .
2ω
2ω
(x)†
c− = χ− χ =
(d) The argument of sin2 must reach π/2 (so P = 1) ⇒
γB0
π
πω
= , or B0 =
.
2ω
2
γ
Problem 4.37
(a)
1
(2) 1
(1)
S− |1 0i = (S− + S− ) √ (↑↓ + ↓↑) = √ [(S− ↑) ↓ +(S− ↓) ↑ + ↑ (S− ↓)+ ↓ (S− ↑)] .
2
2
√
√
1
But S− ↑= ~ ↓, S− ↓= 0 (line above Eq. 4.146), so S− |10i = √ [~ ↓↓ +0 + 0 + ~ ↓↓] = 2~ ↓↓= 2~|1−1i.X
2
(b)
1
(1)
(2) 1
S± |0 0i = (S± + S± ) √ (↑↓ − ↓↑) = √ [(S± ↑) ↓ −(S± ↓) ↑ + ↑ (S± ↓)− ↓ (S± ↑)] .
2
2
1
1
S+ |0 0i = √ (0 − ~ ↑↑ +~ ↑↑ −0) = 0; S− |0 0i = √ (~ ↓↓ −0 + 0 − ~ ↓↓) = 0. X
2
2
(c)
h
i
S 2 |1 1i = (S (1) )2 + (S (2) )2 + 2S(1) · S(2) ↑↑
= (S 2 ↑) ↑ + ↑ (S 2 ↑) + 2 [(Sx ↑)(Sx ↑) + (Sy ↑)(Sy ↑) + (Sz ↑)(Sz ↑)]
3
3
~ ~
i~ i~
~ ~
= ~2 ↑↑ + ~2 ↑↑ +2
↓ ↓+ ↓
↓+ ↑ ↑
4
4
2 2
2
2
2 2
2 3
~
= ~2 ↑↑ +2
↑↑ = 2~2 ↑↑= 2~2 |1 1i = (1)(1 + 1)~2 |1 1i, as it should be.
2
4
132
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
i
h
S 2 |1 − 1i = (S (1) )2 + (S (2) )2 + 2S(1) · S(2) ↓↓
3~2
3~2
↓↓ +
↓↓ +2 [(Sx ↓)(Sx ↓) + (Sy ↓)(Sy ↓) + (Sz ↓)(Sz ↓)]
4
4
3
~
i~
i~
~
~
~
= ~2 ↓↓ +2
↑
↑ + − ↑
− ↑ + − ↓
− ↓
2
2
2
2
2
2
2
2
3
~
= ~2 ↓↓ +2
↓↓= 2~2 ↓↓= 2~2 |1 − 1i. X
2
4
=
Problem 4.38
(a) 1/2 and 1/2 gives 1 or zero; 1/2 and 1 gives 3/2 or 1/2; 1/2 and 0 gives 1/2 only. So baryons can have
spin 3/2 or spin 1/2 (and the latter can be acheived in two distinct ways). [Incidentally, the lightest
baryons do carry spin 1/2 (proton, neutron, etc.) or 3/2 (∆, Ω− , etc.); heavier baryons can have higher
total spin, but this is because the quarks have orbital angular momentum as well.]
(b) 1/2 and 1/2 gives spin 1 or spin 0. [Again, these are the observed spins for the lightest mesons: π’s and
K’s have spin 0, ρ’s and ω’s have spin 1.]
Problem 4.39
Reading down the vertical columns of the (1/2 × 1/2) table:
|11i = | 21 12 12 12 i = | ↑↑i X
1
1
1
1
√ | 12 12 −1
√ (| ↑↓i + | ↓↑i)
|10i = √ | 12 12 12 −1
2 i+
2 2i =
2
2
2
−1
i
=
|
↓↓i
X
|1 − 1i = | 12 12 −1
2 2
1 1 1 1 −1
1
1
1
√ (| ↑↓i − | ↓↑i)
|00i = √ | 2 2 2 2 i − √ | 12 12 −1
2 2i =
2
2
2
X
X
Problem 4.40
(a) From the 2 × 1 Clebsch-Gordan table we get
r
r
r
1
8
6
|2 2i|1 − 1i +
|2 1i|1 0i +
|2 0i|1 1i,
|3 1i =
15
15
15
so you might get 2~ (probability 1/15), ~ (probability 8/15), or 0 (probability 6/15).
(b) From the 1× 12 table: |1 0i| 12 − 12 i =
q
2 3
1
3 | 2 − 2 i+
15/4~2 and 3/4~2 , respectively. Thus you get
q
1 1
1
3 | 2 − 2 i.
So the total is 3/2 or 1/2, with l(l+1)~2 =
15 2
3
~ (probability 2/3), or ~2 (probability 1/3).
4
4
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
133
Problem 4.41
(1)
2
(1)
2
(1)
(1)
Using Eq. 4.177: [S 2 , Sz ] = [S (1) , Sz ] + [S (2) , Sz ] + 2[S(1) · S(2) , Sz ]. But [S 2 , Sz ] = 0 (Eq. 4.102), and
anything with superscript (2) commutes with anything with superscript (1). So
n
o
[S 2 , Sz(1) ] = 2 Sx(2) [Sx(1) , Sz(1) ] + Sy(2) [Sy(1) , Sz(1) ] + Sz(2) [Sz(1) , Sz(1) ]
o
n
= 2 −i~Sy(1) Sx(2) + i~Sx(1) Sy(2) = 2i~(S(1) × S(2) )z.
[S 2 , Sz(1) ] = 2i~(Sx(1) Sy(2) − Sy(1) Sx(2) ), and [S 2 , S(1) ] = 2i~(S(1) × S(2) ). Note that [S 2 , S(2) ] = 2i~(S(2) × S(1) ) =
−2i~(S(1) × S(2) ), so [S 2 , (S(1) + S(2) )] = 0.]
Problem 4.42
(a)
i
dhri
= h[H, r]i.
dt
~
1
1 2
H=
(p − qA) · (p − qA) + qϕ =
p − q(p · A + A · p) + q 2 A2 + qϕ.
2m
2m
1 2
q
[H, x] =
[p , x] −
[(p · A + A · p), x].
2m
2m
[p2 , x] = [(p2x + p2y + p2z ), x] = [p2x , x] = px [px , x] + [px , x]px = px (−i~) + (−i~)px = −2i~px .
Start with Eq. 3.73:
[p · A, x] = [(px Ax + py Ay + pz Az ), x] = [px Ax , x] = px [Ax , x] + [px , x]Ax = −i~Ax .
[A · p, x] = [(Ax px + Ay py + Az pz ), x] = [Ax px , x] = Ax [px , x] + [Ax , x]px = −i~Ax .
1
q
i~
i~
[H, x] =
(−2i~px ) −
(−2i~Ax ) = − (px − qAx ); [H, r] = − (p − qA).
2m
2m
m
m
dhri
1
= h(p − qA)i. QED
dt
m
(b)
1
dhvi
i
∂v
∂v
q ∂A
(p − qA);
= h[H, v]i + h i;
=−
.
m
dt
~
∂t
∂t
m ∂t
m
1
1
H = mv 2 + qϕ ⇒ [H, v] = [v 2 , v] + q[ϕ, v]; [ϕ, v] = [ϕ, p].
2
2
m
∂ϕ
i~
[ϕ, px ] = i~
(Eq. 3.66), so [ϕ, p] = i~∇ϕ, and [ϕ, v] = ∇ϕ.
∂x
m
[v 2 , vx ] = [(vx2 + vy2 + vz2 ), vx ] = [vy2 , vx ] + [vz2 , vx ] = vy [vy , vx ] + [vy , vx ]vy + vz [vz , vx ] + [vz , vx ]vz .
We define the operator v ≡
1
q
[(py − qAy ), (px − qAx )] = − 2 ([Ay , px ] + [py , Ax ])
m2 m
∂Ay
∂Ax
i~q
q
i~q
= − 2 i~
− i~
= − 2 (∇ × A)z = − 2 Bz .
m
∂x
∂y
m
m
1
q
[vz , vx ] = 2 [(pz − qAz ), (px − qAx )] = − 2 ([Az , px ] + [pz , Ax ])
m
m
q
∂Az
∂Ax
i~q
i~q
= − 2 i~
− i~
= 2 (∇ × A)y = 2 By .
m
∂x
∂z
m
m
[vy , vx ] =
134
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
i~q
i~q
(−vy Bz − Bz vy + vz By + By vz ) = 2 [−(v × B)x + (B × v)x ] .
m2
m
i~q
[v 2 , v] = 2 [(B × v) − (v × B)] . Putting all this together:
m
dhvi
qi~
i
m i~q
q ∂A
(B
×
v
−
v
×
B)
+
=
∇ϕ
− h
i.
2
dt
~
2 m
m
m ∂t
q
∂A
q
dhvi
= h(v × B) − (B × v)i + q −∇ϕ −
= h(v × B − B × v)i + qhEi. Or, since
[F] m
dt
2
dt
2
∴ [v 2 , vx ] =
1
1
q
[(p − qA) × B − B × (p − qA)] =
[p × B − B × p] −
[A × B − B × A] .
m
m
m
q
q2
dhvi
= qhEi +
hp × B − B × pi − hA × Bi. QED
m
dt
2m
m
v×B−B×v =
(c) Go back to Eq. F, and use hEi = E, hv × Bi = hvi × B; hB × vi = B × hvi = −hvi × B. Then
m
dhvi
= qhvi × B + qE. QED
dt
Problem 4.43
(a)
E = −∇ϕ = −2Kz k̂.
B=∇×A=
î
ĵ
k̂
∂/∂x ∂/∂y ∂/∂z = B0 k̂.
−B0 y/2 B0 x/2 0
(b) For time-independent potentials Eq. 4.191 separates in the usual way:
1
(−i~∇ − qA) · (−i~∇ − qA) ψ + qϕψ = Eψ,
2m
−
~2 2
iq~
q2 2
∇ ψ+
[∇ · (Aψ) + A · (∇ψ)]+
A ψ+qϕψ = Eψ.
2m
2m
2m
−
~2 2
iq~
∇ ψ+
[2A · (∇ψ) + (∇ · A)ψ] +
2m
2m
But
or
∇·(Aψ) = (∇·A)ψ+A·(∇ψ),
so
q2 2
A + qϕ ψ = Eψ.
2m
This is the time-independent Schrödinger equation for electrodynamics. In the present case
∂ψ
∂ψ
B2
B0
x
−y
, A2 = 0 x2 + y 2 , ϕ = Kz 2 .
∇ · A = 0, A · (∇ψ) =
2
∂y
∂x
4
∂
∂
~2 2
qB0
q 2 B02 2
But Lz = −i~ x
−y
, so −
∇ ψ−
Lz ψ +
x + y 2 + qKz 2 ψ = Eψ.
∂y
∂x
2m
2m
8m
Since Lz commutes with H, we may as well pick simultaneous eigenfunctions of both:
Lz ψ = m̄~ψ,
where m̄ = 0, ±1, ±2, . . . (with the overbar to distinguish the magnetic quantum number from the mass).
Then
~2 2 (qB0 )2 2
qB0 ~
2
2
∇ +
m̄ ψ.
−
x + y + qKz ψ = E +
2m
8m
2m
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
135
p
Now let ω1 ≡ qB0 /m, ω2 ≡ 2Kq/m, and use cylindrical coordinates (r, φ, z) (you can also do it in
Cartesian coordinates):
1
~2 1 ∂
1
∂ψ
1 ∂2ψ ∂2ψ
1
2
2
2
2 2
−
+
+ mω1 x + y + mω2 z ψ = E + m̄~ω1 ψ.
r
+ 2
2m r ∂r
∂r
r ∂φ2
∂z 2
8
2
2
1
1
∂
∂2ψ
= − 2 L2z ψ = − 2 m̄2 ~2 ψ = −m̄2 ψ. Use separation of variables: ψ(r, φ, z) =
, so
∂φ
∂φ2
~
~
R(r)Φ(φ)Z(z) :
d2 Z
1
~2
1 d
dR
m̄2
1
1
2 2
2 2
−
ΦZ
r
− 2 RΦZ + RΦ 2 +
mω1 r + mω2 z RΦZ = E + m̄~ω1 RΦZ.
2m
r dr
dr
r
dz
8
2
2
But Lz = −i~
Divide by RΦZ and collect terms:
1
1
~2
1 d
dR
m̄2
~2 1 d2 Z
1
2 2
2 2
+ mω2 z
−
r
− 2 + mω1 r + −
= E + m̄~ω1 .
2m rR dr
dr
r
8
2m Z dz 2
2
2
The first term depends only on r, the second only on z, so they’re both constants; call them Er and Ez :
dR
m̄2
1
~2 d2 Z 1
1
~2 1 d
r
− 2 R + mω12 r2 R = Er R; −
+ mω22 z 2 Z = Ez Z; E = Er +Ez − m̄~ω1 .
−
2m r dr
dr
r
8
2m dz 2 2
2
The z equation is a one-dimensional harmonic oscillator, and we can read off immediately that Ez =
(n2 + 1/2)~ω2 , with
√ n2 = 0, 1, 2, . . . . The r equation is actually a two-dimensional harmonic oscillator; to
get Er , let u(r) ≡ r R, and follow the method of Sections 4.1.3 and 4.2.1:
√
u
dR
u0
u
dR √ 0
u
d
dR
u
R= √ ,
= √ − 3/2 , r
= ru − √ ,
r
= r u00 + 3/2 ,
dr
dr
dr
dr
r
r 2r
2 r
4r
00
1 d
u
~2
1 u 1
m̄2 u
u
u
dR
u00
u
1
√ +
√ − 2 √
r
= √ + 5/2 ; −
+ mω12 r2 √ = Er √
r dr
dr
2m
r
8
r 4r
r 4 r2 r
r
r
r
1
u
~2 d2 u
1
−
+
− m̄2
+ mω12 r2 u = Er u.
2m dr2
4
r2
8
This is identical to the equation we shall encounter in Problem 4.47 (the three-dimentional harmonic
oscillator), only with ω → ω1 /2, E → Er , and `(` + 1) → m̄2 − 1/4, which is to say, `2 + ` + 1/4 = m̄2 , or
(` + 1/2)2 = m̄2 , or ` = |m̄| − 1/2. [Our present equation depends only on m̄2 , and hence is the same for
either sign, but the solution to Problem 4.47 assumes ` + 1/2 ≥ 0 (else u is not normalizable), so we need
|m| here.] Quoting that solution:
E = (jmax + ` + 3/2)~ω → Er = (jmax + |m̄| + 1)~ω1 /2,
where
jmax = 0, 2, 4, . . . .
E = (jmax + |m̄| + 1)~ω1 /2 + (n2 + 1/2)~ω2 − m̄~ω1 /2 = (n1 + 12 )~ω1 + (n2 + 21 )~ω2 ,
where n1 = 0, 1, 2, . . . (if m̄ ≥ 0, then n1 = jmax /2; if m̄ < 0, then n1 = jmax /2 − m̄).
[A more elegant solution runs as follows: Let
1
qB0
P ≡ √ (px + py ) −
(x − y) ,
2
2
1
1
1
Q≡ √
(px − py ) + (x + y) .
2
2 qB0
136
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Show (a) P and Q satisfy the canonical commutation relations (Eq. 2.52): [Q, P ] = i~, and (b) the
Hamiltonian (square brackets in Eq. 4.191) can be written as
H=
1 2 1
P + mω12 Q2
2m
2
+
1 2 1
2 2
p + mω2 z .
2m z 2
The problem is therefore mathematically identical to two independent harmonic oscillators, with the same
mass, and frequencies ω1 and ω2 ; the total energy is just the sum: (n1 + 21 )~ω1 + (n2 + 12 )~ω2 .]
Problem 4.44
Assume that Ψ satisfies the Schrödinger equation (4.191):
∂Ψ
1
2
i~
=
(−i~∇ − qA) + qϕ Ψ.
∂t
2m
Let Ψ0 = eiqΛ/~ Ψ. Then
1
∂Ψ0
iq ∂Λ iqΛ/~
∂Λ
2
iqΛ/~ ∂Ψ
iqΛ/~
i~
= i~ e
+
e
Ψ =e
(−i~∇ − qA) + qϕ − q
Ψ.
∂t
∂t
~ ∂t
2m
∂t
[Note that ∇ does not act on eiqΛ/~ , so it’s important to keep that factor on the left.] But ϕ = ϕ0 + ∂Λ/∂t and
A = A0 − ∇Λ (Equation 4.196), so
∂Ψ0
1
2
iqΛ/~
0
0
i~
=e
(−i~∇ − qA + q(∇Λ)) + qϕ Ψ.
∂t
2m
Now we need to move eiqΛ/~ through to the right. First note that
−i~∇ eiqΛ/~ f = −i~eiqΛ/~ ∇f + q(∇Λ)eiqΛ/~ f = eiqΛ/~ [−i~∇ + q(∇Λ)] f,
for any function f (r, t), so
eiqΛ/~ [−i~∇ − qA0 + q(∇Λ)] f = (−i~∇ − qA0 ) eiqΛ/~ f .
Then
2
eiqΛ/~ [−i~∇ − qA0 + q(∇Λ)] Ψ = eiqΛ/~ [−i~∇ − qA0 + q(∇Λ)] e−iqΛ/~ eiqΛ/~ [−i~∇ − qA0 + q(∇Λ)] Ψ
{z
}
|
f
0
h
= (−i~∇ − qA ) e
iqΛ/~ −iqΛ/~
e
0
(−i~∇ − qA ) e
= (−i~∇ − qA0 )2 Ψ0 .
Therefore,
∂Ψ0
1
0 2
0
i~
=
(−i~∇ − qA ) + qϕ Ψ0 . QED
∂t
2m
iqΛ/~
Ψ
i
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
137
Problem 4.45
(a) Giving H a test function f to act upon:
1
(−i~∇ − qA) · (−i~∇f − qAf ) + qϕf
2m 1
2
2
=
− ~ ∇ · (∇f ) + iq~ ∇ · (Af ) +iq~A · (∇f ) + q A · Af + qϕf.
| {z }
2m
(∇·A)f +A·(∇f )
Hf =
But ∇ · A = 0 and ϕ = 0 (see comments after Eq. 4.198), so
1 2 2
1 2 2
Hf =
−~ ∇ f + 2iq~A · ∇f + q 2 A2 f , or H =
−~ ∇ + q 2 A2 + 2iq~A · ∇ . QED
2m
2m
(b) Apply
(−i~∇ − qA) ·
to both sides of Eq. 4.210:
2
(−i~∇ − qA) Ψ = (−i~∇ − qA) · −i~eig ∇Ψ0 = −~2 ∇ · (eig ∇Ψ0 ) + iq~eig A · ∇Ψ0 .
q
But ∇ · (eig ∇Ψ0 ) = ieig (∇g) · (∇Ψ0 ) + eig ∇ · (∇Ψ0 ) and ∇g = A, so the right side is
~
q
−i~2 eig A · ∇Ψ0 − ~2 eig ∇2 Ψ0 + iq~eig A · ∇Ψ0 = −~2 eig ∇2 Ψ0 . QED
~
Problem 4.46
(a)
−
~2
2m
∂2ψ ∂2ψ ∂2ψ
+
+
∂x2
∂y 2
∂z 2
1
+ mω 2 x2 + y 2 + z 2 ψ = Eψ.
2
Let ψ(x, y, z) = X(x)Y (y)Z(z); plug it in, divide by XY Z, and collect terms:
1
~2 1 d 2 Y
1
~2 1 d 2 Z
1
~2 1 d 2 X
2 2
2 2
2 2
−
+
mω
x
+
−
+
mω
y
+
−
+
mω
z
= E.
2m X dx2
2
2m Y dy 2
2
2m Z dz 2
2
The first term is a function only of x, the second only of y, and the third only of z. So each is a constant
(call the constants Ex , Ey , Ez , with Ex + Ey + Ez = E). Thus:
−
~2 d 2 X
1
~2 d2 Y
1
~2 d 2 Z
1
2 2
2 2
+
mω
x
X
=
E
X;
−
+
mω
y
Y
=
E
Y
;
−
+ mω 2 z 2 Z = Ez Z.
x
y
2m dx2
2
2m dy 2
2
2m dz 2
2
Each of these is simply the one-dimensional harmonic oscillator (Eq. 2.45). We know the allowed energies
(Eq. 2.62):
Ex = (nx + 21 )~ω; Ey = (ny + 12 )~ω; Ez = (nz + 12 )~ω; where nx , ny , nz = 0, 1, 2, 3, . . . .
So E = (nx + ny + ny + 23 )~ω = (n + 32 )~ω,
with n ≡ nx + ny + nz .
(b) The question is: “How many ways can we add three non-negative integers to get sum n?”
If nx = n, then ny = nz = 0; one way.
If nx = n − 1, then ny = 0, nz = 1, or else ny = 1, nz = 0; two ways.
If nx = n − 2, then ny = 0, nz = 2, or ny = 1, nz = 1, or ny = 2, nz = 0; three ways.
And so on. Evidently d(n) = 1 + 2 + 3 + · · · + (n + 1) =
(n + 1)(n + 2)
.
2
138
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Problem 4.47
~2 d 2 u
1
~2 `(` + 1)
2 2
Eq. 4.37: −
+ mω r +
u = Eu.
2m dr2
2
2m r2
r
mω
1
~2 mω d2 u
~2 mω `(` + 1)
2 ~
2
+ mω
u = Eu,
Following Eq. 2.72, let ξ ≡
r. Then −
ξ +
~
2m ~ dξ 2
2
mω
2m ~
ξ2
d2 u
`(` + 1)
2E
2
or
= ξ +
− K u, where K ≡
(as in Eq. 2.74).
dξ 2
ξ2
~ω
At large ξ,
2
d2 u
≈ ξ 2 u, and u ∼ ( )e−ξ /2 (see Eq. 2.78).
dξ 2
At small ξ,
d2 u
`(` + 1)
≈
u, and u ∼ ( )ξ `+1 (see Eq. 4.59).
2
dξ
ξ2
2
So let u(ξ) ≡ ξ `+1 e−ξ /2 v(ξ).
[This defines the new function v(ξ).]
2
2
2
du
= (` + 1)ξ ` e−ξ /2 v − ξ `+2 e−ξ /2 v + ξ `+1 e−ξ /2 v 0 .
dξ
2
2
2
2
d2 u
= `(` + 1)ξ `−1 e−ξ /2 v − (` + 1)ξ `+1 e−ξ /2 v + (` + 1)ξ ` e−ξ /2 v 0 − (` + 2)ξ `+1 e−ξ /2 v
2
dξ
2
2
2
2
2
+ ξ `+3 e−ξ /2 v − ξ `+2 e−ξ /2 v 0 + (` + 1)ξ ` e−ξ /2 v 0 − ξ `+2 e−ξ /2 v 0 + ξ `+1 e−ξ /2 v 00
(
2
2
2
(2(
−ξ
/2
((
(`−1
e
`(`(+(1)ξ
e−ξ /2 v − (2` + 3)ξ `+1 e−ξ /2 v + ξ `+3
v + 2(` + 1)ξ ` e−ξ /2 v 0
=(
(
2
2
2
2
(2(
−ξ
/2
((
(`−1
e
`(`(+(1)ξ
e−ξ /2 v − Kξ `+1 e−ξ /2 v.
− 2ξ `+2 e−ξ /2 v 0 + ξ `+1 e−ξ /2 v 00 = ξ `+3
v +(
2
Cancelling the indicated terms, and dividing off ξ `+1 e−ξ /2 , we have:
`+1
v 00 + 2v 0
− ξ + (K − 2` − 3) v = 0.
ξ
Let v(ξ) ≡
∞
X
j=0
∞
X
j=2
aj ξ j , so v 0 =
∞
X
jaj ξ j−1 ;
v 00 =
j=0
j(j − 1)aj ξ j−2 + 2(` + 1)
∞
X
∞
X
j(j − 1)aj ξ j−2 . Then
j=2
jaj ξ j−2 − 2
j=1
∞
X
jaj ξ j + (K − 2` − 3)
∞
X
aj ξ j = 0.
j=0
j=1
In the first two sums, let j → j + 2 (rename the dummy index):
∞
∞
∞
∞
X
X
X
X
(j + 2)(j + 1)aj+2 ξ j + 2(` + 1)
(j + 2)aj+2 ξ j − 2
jaj ξ j + (K − 2` − 3)
aj ξ j = 0.
j=0
j=0
j=0
j=0
Note: the second sum should start at j = −1; to eliminate this term (there is no compensating one in ξ −1 ) we
must take a1 = 0. Combining the terms:
∞
X
j=0
[(j + 2)(j + 2` + 3)aj+2 + (K − 2j − 2` − 3)aj ] = 0, so aj+2 =
(2j + 2` + 3 − K)
aj .
(j + 2)(j + 2` + 3)
Since a1 = 0, this gives us a single sequence: a0 , a2 , a4 , . . .. But the series must terminate (else we get the
wrong behavior as ξ → ∞), so there occurs some even number jmax such that ajmax 6= 0 but ajmax +2 = 0. Thus
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
139
1
~ωK, so E = jmax + ` + 23 ~ω. Or, letting jmax + ` ≡ n,
2
En = (n + 23 )~ω, and n can be any nonnegative integer; this agrees with Equation 4.216.
K = 2jmax + 2` + 3.
But E =
Since v(ξ) is a polynomial of degree jmax in ξ, or jmax /2 in ξ 2 , the number of nodes (roots of v(ξ) is jmax /2 =
N − 1, so n = jmax + ` ⇒ n = 2N + ` − 2. Suppose n is even; then (since jmax is even) ` = 0, 2, 4, . . . , n. For
each ` there are (2` + 1) values for m. So
n
X
d(n) =
(2` + 1). Let j = `/2;
`=0,2,4,...
( n2 )( n2 + 1)
=4
2
n/2
n/2
n/2
X
X
X
(4j + 1) = 4
j+
1
then d(n) =
j=0
+ ( n2 + 1) = ( n2 + 1)(n + 1) =
j=0
j=0
(n + 1)(n + 2)
, as before (Problem 4.46(b)).
2
n=7
n=6
n=5
n=4
n=3
n=2
n=1
n=0
N=4
N=3
N=2
N=1
l= 0
l=1
l= 2
l= 3
l= 4
l= 5
Problem 4.48
(a)
d
i
hr · pi = h[H, r · p]i.
dt
~
[H, r · p] =
3
X
i=1
[p2 , ri ] =
[H, ri pi ] =
3
X
([H, ri ]pi + ri [H, pi ]) =
i=1
3 X
1
i=1
2m
2
[p , ri ]pi + ri [V, pi ] .
3
3
3
X
X
X
[pj pj , ri ] =
(pj [pj , ri ] + [pj , ri ]pj ) =
[pj (−i~δij ) + (−i~δij )pj ] = −2i~pi .
j=1
j=1
j=1
3 X
∂V
∂V
1
(Problem 3.13(c)). [H, r · p] =
(−2i~)pi pi + ri i~
∂ri
2m
∂ri
i=1
2
p
d
p2
= i~ − + r · ∇V .
hr · pi = h − r · ∇V i = 2hT i − hr · ∇V i.
m
dt
m
d
For stationary states hr · pi = 0, so 2hT i = hr · ∇V i. QED
dt
[V, pi ] = i~
140
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
(b)
V (r) = −
e2 1
e2 1
e2 1
r̂
⇒
r
·
∇V
=
⇒ ∇V =
= −V. So 2hT i = −hV i.
4π0 r
4π0 r2
4π0 r
But hT i + hV i = En , so hT i − 2hT i = En , or hT i = −En ; hV i = 2En . QED
(c)
V =
1
mω 2 r2 ⇒ ∇V = mω 2 r r̂ ⇒ r · ∇V = mω 2 r2 = 2V. So 2hT i = 2hV i, or hT i = hV i.
2
But hT i + hV i = En , so hT i = hV i =
1
En . QED
2
Problem 4.49
i~ i~ ∇Ψ · ∇Ψ∗ + Ψ(∇2 Ψ∗ ) − ∇Ψ∗ · ∇Ψ − Ψ∗ (∇2 Ψ) =
Ψ(∇2 Ψ∗ ) − Ψ∗ (∇2 Ψ) .
2m
2m
~2 2
∂Ψ
=−
∇ Ψ + V Ψ, so
But the Schrödinger equation says i~
∂t
2m
2m
∂Ψ
2m
∂Ψ∗
2
2 ∗
∗
∇ Ψ = 2 V Ψ − i~
, ∇ Ψ = 2 V Ψ + i~
. Therefore
~
∂t
~
∂t
(a) ∇ · J =
∂Ψ∗
∂Ψ
i~ 2m
∗
∗
Ψ V Ψ + i~
− Ψ V Ψ − i~
∇·J =
2m ~2
∂t
∂t
∗
i
∂Ψ
∂Ψ
∂
∂
= i~ Ψ
+ Ψ∗
= − (Ψ∗ Ψ) = − |Ψ|2 . X
~
∂t
∂t
∂t
∂t
(b) From Problem 4.13(b),
∇Ψ =
1 1
Ψ2 1 1 = − √
re−r/2a sin θeiφ e−iE2 t/~ .
πa 8a2
∂Ψ
1 ∂Ψ
1 ∂Ψ
r̂ +
θ̂ +
φ̂,
∂r
r ∂θ
r sin θ ∂φ
In spherical coordinates,
so
1 1 r −r/2a
1
∇Ψ2 1 1 = − √
1−
e
sin θeiφ e−iE2 t/~ r̂ + re−r/2a cos θeiφ e−iE2 t/~ θ̂
2
2a
r
πa 8a
1
r
i
1
+
re−r/2a sin θ ieiφ e−iE2 t/~ φ̂ = 1 −
r̂ + cot θ θ̂ +
φ̂ Ψ2 1 1 .
r sin θ
2a
sin θ
r
Therefore
i~ r i
r i
1
J =
1−
r̂ + cot θ θ̂ −
φ̂ − 1 −
r̂ − cot θ θ̂ −
φ̂ |Ψ2 1 1 |2
2m
2a
sin θ
2a
sin θ
r
=
i~ (−2i)
~
~ 1 1 r2 e−r/a sin2 θ
|Ψ2 1 1 |2 φ̂ =
φ̂ =
re−r/a sin θ φ̂.
2m r sin θ
m πa 64a4
r sin θ
64πma5
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
141
~
r2 e−r/a sin θ r̂ × φ̂ , while r̂ × φ̂ = −θ̂ and ẑ · θ̂ = − sin θ, so
5
64πma
~
r × Jz =
r2 e−r/a sin2 θ, and hence
64πma5
Z ~
2
2 −r/a
r
e
sin
θ
r2 sin θ dr dθ dφ
Lz = m
64πma5
Z 2π
Z π
Z ∞
4
~
~
3
4 −r/a
5
dφ
=
sin
θ
dθ
=
r
e
dr
4!a
(2π) = ~,
64πa5 0
64πa5
3
0
0
(c) Now r × J =
as it should be, since (Eq. 4.133) Lz = ~m, and m = 1 for this state.
Problem 4.50
(a)
ψ=√
1
πa3
e−r/a ⇒ φ(p) =
1
1
√
3/2
(2π~)
πa3
Z
e−ip·r/~ e−r/a r2 sin θ dr dθ dφ.
With axes as suggested, p · r = pr cos θ. Doing the (trivial) φ integral:
Z π
Z ∞
1
2π
2 −r/a
−ipr cos θ/~
√
φ(p) =
r e
e
sin θ dθ dr.
(2πa~)3/2 π 0
0
Z π
pr 2~
~ ipr/~
~ −ipr cos θ/~ π
e
=
e
− e−ipr/~ =
sin
.
e−ipr cos θ/~ sin θ dθ =
ipr
ipr
pr
~
0
0
Z ∞
pr 1
1
2~
−r/a
φ(p) = √
dr.
re
sin
~
π 2 (a~)3/2 p 0
Z ∞
Z ∞
Z ∞
pr 1
−r/a
−r/a ipr/~
−r/a −ipr/~
re
sin
dr =
re
e
dr −
re
e
dr
~
2i 0
0
0
"
#
1
1
1
(2ip/a~)2
1
=
−
i2
2 = 2i h
2i (1/a − ip/~)2
2
2
(1/a + ip/~)
(1/a) + (p/~)
=
r
φ(p) =
(2p/~)a3
2.
[1 + (ap/~)2 ]
2 1 1 2pa3
1
1
=
2
3/2
2
~ a πp ~ [1 + (ap/~) ]
π
2a
~
3/2
1
[1 + (ap/~)2 ]
2.
(b)
Z
|φ|2 d3 p = 4π
Z ∞
p2 |φ|2 dp = 4π
0
Z ∞
1
π2
2a
~
3 Z ∞
0
p2
4 dp.
[1 + (ap/~)2 ]
x2
π −5/2
dx =
m
, so
2 )4
(m
+
x
32
0
8
−5
3 Z
3
Z ∞
p2
~
π ~
π ~
32 a 3 π ~
2 3
dp
=
=
;
|φ|
d
p
=
= 1. X
4
a
32 a
32 a
π ~ 32 a
[1 + (ap/~)2 ]
0
From math tables:
142
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
(c)
Z
2
hp i =
Z ∞
0
1
p |φ| d p = 2
π
2
2
3
2a
~
3
Z ∞
p4
0
[1 + (ap/~)2 ]
4π
π
x4
4
dx
=
m−3/2 . So hp2 i =
2
4
[m + x ]
32
π
4 dp.
2a
~
m
2~2
From math tables:
3 8
−3
~2
~
π ~
= 2.
a
32 a
a
(d)
hT i =
1 2
~2 m2
1 ~2
=
hp i =
2m
2m a2
2m ~4
e2
4π0
2
=
e2
4π0
2
= −E1 ,
which is consistent with Eq. 4.218.
Problem 4.51
(
−V0
V (r) =
0
(r ≤ a),
(r > a).
The Laplacian in polar coordinates (r, θ) is
∇2 =
1 ∂
r ∂r
r
∂
∂r
+
1 ∂2
.
r2 ∂θ2
The Schrödinger equation says
~2 1 ∂
∂ψ
1 ∂2ψ
r
+ 2 2 + V ψ = Eψ.
−
2m r ∂r
∂r
r ∂θ
Solve by separation of variables, ψ(r, θ) = R(r) Θ(θ):
~2 Θ d
dR
R d2 Θ
−
r
+ 2 2 + V RΘ = ERΘ ⇒
2m r dr
dr
r dθ
−
~2 r d
dR
1 d2 Θ
r
+
+ r2 V (r) = r2 E.
2m R dr
dr
Θ dθ2
1 d2 Θ
= −`2 (a constant) ⇒ Θ = ei`θ , ` = 0, ±1, ±2, . . . . But for the ground state we want ` = 0 (circular
Θ dθ2
symmetry). This leaves the radial equation:
(
κ2 R
(r > a),
~2 1 d
dR
d2 R 1 dR
2m
−
r
+ V R = ER;
+
=
−
(E
−
V
)R
=
2m r dr
dr
dr2
r dr
~2
−l2 R (r ≤ a),
p
√
where κ ≡ −2mE/~ and l ≡ 2m(E + V0 )/~. The solutions are Bessel functions: R(r) = AJ0 (lr), for r ≤ a
(the other solution blows up at the origin) and R(r) = BK0 (κr) for r > a (the other solution blows up as
r → ∞).
The boundary conditions are AJ0 (la) = BK0 (κa) and AlJ00 (la) = BκK00 (κa); dividing,
J 0 (la)
K 0 (κa)
l 0
=κ 0
.
J0 (la)
K0 (κa)
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
143
√
This determines the allowed energies. Borrowing the notation of Equation 2.158, z ≡ la, z0 ≡ a 2mV0 /~, κa =
p
z02 − z 2 , and using the fact that J00 (x) = −J1 (x) and K00 (x) = −K1 (x):
J1 (z)
z
=
J0 (z)
q
p
K1 ( z02 − z 2 )
p
z02 − z 2
.
K0 ( z02 − z 2 )
[F]
Plotting both sides (as in Figure 2.17), we look for intersections:
��������
Plotx BesselJ[1, x] BesselJ[0, x],
144 - x2 BesselK1,
144 - x2 BesselK0,
144 - x2 ,
{x, 0, 12}, PlotRange -> {0, 15}
14
12
10
8
��������
6
4
2
0
2
4
6
8
10
12
The graph is for z0 = 12; as z0 (the “strength” of the potential well) decreases, the left side of Equation F
(the rising vertical curves, on the graph) does not change, but the right side (the falling curve) shrinks down as
the zero crossing moves left along the horizontal axis. There always remains one intersection, however, so like
the one-dimensional case (but unlike the 3-dimensional case) there is always (at least) one bound state for the
circular well.
Problem 4.52
(a) From Tables 4.3 and 4.7,
ψ321 = R32 Y21 =
" r
#
1 r 2 −r/3a
15
1
1
iφ
e
−
sin θ cos θe
= −√
r2 e−r/3a sin θ cos θeiφ .
3/2
a
8π
π 81a7/2
81 30 a
4
√
(b)
Z
Z 1
1
r4 e−2r/3a sin2 θ cos2 θ r2 sin θ dr dθ dφ
2
7
π (81) a
Z ∞
Z π
1
6 −2r/3a
=
2π
r
e
dr
(1 − cos2 θ) cos2 θ sin θ dθ
π(81)2 a7
0
" 0 # π
7
2
3a
cos3 θ cos5 θ
−
+
=
6!
(81)2 a7
2
3
5
0
7 7
2
3 a 2 2
3·5 4
−
=
·
= 1. X
= 8 76 · 5 · 4 · 3 · 2 7
3 a
2
3 5
4
15
|ψ|2 d3 r =
144
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
(c)
2
Z ∞
4
1 1
rs+6 e−2r/3a dr
7
81
30
a
0
0
s+7
s
s
8
3a
3a
1
(s + 6)! 3a
=
(s
+
6)!
=
.
(s
+
6)!
=
15(81)2 a7
2
2
720
6!
2
hrs i =
Z ∞
rs |R32 |2 r2 dr =
Finite for s > −7 .
Problem 4.53
(a) From Tables 4.3 and 4.7,
1 r 3 −r/4a
1
√
e
ψ433 = R43 Y33 =
768 35 a3/2 a
r
−
35
sin3 θe3iφ
64π
!
= −
1
√
r3 e−r/4a sin3 θe3iφ .
6144 πa9/2
(b)
Z
hri =
r|ψ|2 d3 r =
1
(6144)2 πa9
Z
r r6 e−r/2a sin6 θ r2 sin θ dr dθ dφ
Z ∞
Z π
Z 2π
1
7
9 −r/2a
r
e
dr
sin
θ
dθ
dφ
(6144)2 πa9 0
0
0
1
2·4·6
10
=
9!(2a)
2
(2π) = 18a.
(6144)2 πa9
3·5·7
=
(c) Using Eq. 4.133: L2x + L2y = L2 − L2z = 3(4)~2 − (3~)2 = 3~2 , with probability 1.
Problem 4.54
(a)
b
Z b
4π
4
2r
a 2 −2r/a a3 −2r/a
−2r/a 2
P = |ψ| d r =
e
r dr = 3 − r e
+ e
− −1
πa3 0
a
2
4
a
0
b
2
2
b
2r 2r
2b
=− 1+
+ 2 e−2r/a = 1 − 1 +
+ 2 2 e−2b/a .
a
a
a
a
0
Z
2
3
(b)
1
1
2
3
P = 1 − 1 + + 2 e− ≈ 1 − 1 + + 2
1−+
−
2
2
2
3!
2
3
3
2
3
1 1 1
2
3
≈1−1+−
+
−+ −
−
+
=
− +
2
6
2
2
2
6 2 2
3
3
1 2b
4 b
=
=
.
6 a
3 a
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
145
(c)
|ψ(0)|2 =
4
4
1
1
⇒ P ≈ πb3 3 =
πa3
3
πa
3
3
b
. X
a
(d)
P =
4
3
10−15
0.5 × 10−10
3
=
3
4
4
32
2 × 10−5 = · 8 × 10−15 =
× 10−15 = 1.07 × 10−14 .
3
3
3
Problem 4.55
(a) Equation 4.75 ⇒ Rn(n−1) =
So
1 n −ρ
ρ e v(ρ), where
r
2 2
2
Z ∞
|R| r dr = (Nn )
1=
Eq. 4.76 ⇒ c1 =
Rn(n−1) = Nn rn−1 e−r/na , where
v(ρ) = c0 , and hence
Z ∞
r
;
na
ρ≡
2n −2r/na
r e
0
0
Nn ≡
2(n − n)
c0 = 0.
(1)(2n)
c0
.
(na)n
n s
2n+1
na
2
2
dr = (Nn ) (2n)!
; Nn =
.
2
na
na(2n)!
2
(b)
hrl i =
Z ∞
|R|2 rl+2 dr = Nn2
0
hri =
hr2 i =
r2n+l e−2r/na dr.
0
2
na
2n+1
2n+1
Z ∞
2
na
2n+2
na
1
1
(2n + 1)!
= n+
na.
(2n)!
2
2
2n+3
2
1
na
na
1
(2n + 2)!
= (2n + 2)(2n + 1)
= n+
(n + 1)(na)2 .
(2n)!
2
2
2
(c)
σr2 = hr2 i − hri2 =
=
n+
2
1
1
(n + 1)(na)2 − n +
(na)2
2
2
1
1
1
n+
(na)2 =
hri2 ;
2
2
2(n + 1/2)
R10
a
Maxima occur at:
σr = √
R 32
r
6a
hri
.
2n + 1
R 26 25
r
650a
r
dRn,n−1
1 n−1 −r/na
= 0 ⇒ (n − 1)rn−2 e−r/na −
r
e
= 0 ⇒ r = na(n − 1).
dr
na
146
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Problem 4.56
Here are a couple of examples: {32, 28} and {224,56}; {221, 119} and {119, 91}. For further discussion see
D. Wyss and W. Wyss, Foundations of Physics 23, 465 (1993).
Problem 4.57
(a) Using Eqs. 3.64 and 4.122: [A, B] = [x2 , Lz ] = x[x, Lz ] + [x, Lz ]x = x(−i~y) + (−i~y)x = −2i~xy.
2
1
2 2
Equation 3.62 ⇒ σA
σB ≥
(−2i~)hxyi = ~2 hxyi2 ⇒ σA σB ≥ ~|hxyi|.
2i
(b) Equation 4.118 ⇒ hBi = hLz i = m~;
hB 2 i = hL2z i = m2 ~2 ;
so σB = m2 ~2 − m2 ~2 = 0.
(c) Since the left side of the uncertainty principle is zero, the right side must also be: hxyi = 0, for eigenstates
of Lz .
Problem 4.58
(a) 1 = |A|2 (1 + 4 + 4) = 9|A|2 ;
(b)
A = 1/3.
5
~
4
~
5~ 4
, with probability ; − , with probability . hSz i =
+
2
9
2
9
92 9
−
~
2
=
~
.
18
(c) From Eq. 4.151,
1 − 2i
1 1
9+4
3 − 2i
13
1
(x)
√
11
χ=
=
.
= √ (1 − 2i + 2) = √ ; |c+ |2 =
2
3 2
9
·
2
18
3 2
3 2
†
1 − 2i
1 1
1
1 + 2i
1+4
5
(x)
(x)
(x)
c− = χ −
χ = √ 1 −1
= √ (1 − 2i − 2) = − √ ; |c− |2 =
=
.
2
3 2
9
·
2
18
3 2
3 2
(x)
c+ =
(x)
χ+
†
~
13
~
5
, with probability
; − , with probability
.
2
18
2
18
13 ~
5
hSx i =
+
18 2 18
~
−
2
=
2~
.
9
(d) From Problem 4.32(a),
†
1 − 2i
1 1
1
1 − 4i
1 + 16
17
(y)
(y)
(y)
c+ = χ+
χ = √ 1 −i
= √ (1 − 2i − 2i) = √ ; |c+ |2 =
=
.
2
3 2
9·2
18
3 2
3 2
†
1 − 2i
1 1
1
1
1
1
(y)
(y)
(y)
c− = χ−
χ= √ 1 i
= √ (1 − 2i + 2i) = √ ; |c− |2 =
=
.
2
3 2
9·2
18
3 2
3 2
~
17
~
1
, with probability
; − , with probability
.
2
18
2
18
17 ~
1
hSy i =
+
18 2 18
~
4~
−
=
.
2
9
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
147
Problem 4.59
(1)
(1)
(2)
We may as well choose axes so that â lies along the z axis and b̂ is in the xz plane. Then Sa = Sz , and Sb
(2)
(2)
(1) (2)
cos θ Sz + sin θ Sx . h0 0|Sa Sb |0 0i is to be calculated.
=
i
1 h
(2)
Sa(1) Sb |0 0i = √ Sz(1) (cos θ Sz(2) + sin θ Sx(2) ) (| ↑↓i − | ↓↑i)
2
1
= √ [(Sz | ↑i)(cos θ Sz | ↓i + sin θ Sx | ↓i) − (Sz | ↓i)(cos θ Sz | ↑i + sin θ Sx | ↑i)]
2
1
~
~
~
~
~
~
=√
| ↑i cos θ − | ↓i + sin θ
| ↑i − − | ↓i cos θ
| ↑i + sin θ
| ↓i
(using Eq. 4.145)
2
2
2
2
2
2
2
1
1
~2
1
~2
cos θ √ (−| ↑↓i + | ↓↑i) + sin θ √ (| ↑↑i + | ↓↓i) =
− cos θ|0 0i + sin θ √ (|1 1i + |1 − 1i) .
=
4
4
2
2
2
2
~
~2
1
(2)
(2)
so hSa(1) Sb i = h0 0|Sa(1) Sb |0 0i =
h0 0| − cos θ|0 0i + sin θ √ (|1 1i + |1 − 1i) = − cos θh0 0|0 0i
4
4
2
(2)
(by orthogonality), and hence hSa(1) Sb i = −
~2
cos θ. QED
4
Problem 4.60
(a) First note from Eq. 4.136 and line after Eq. 4.146 that
1
[S+ |s mi + S− |s mi]
2
i
p
~ hp
s(s + 1) − m(m + 1)|s m + 1i + s(s + 1) − m(m − 1)|s m − 1i
=
2
Sx |s mi =
1
[S+ |s mi − S− |s mi]
2i
i
p
~ hp
s(s + 1) − m(m + 1)|s m + 1i − s(s + 1) − m(m − 1)|s m − 1i
=
2i
Sy |s mi =
Now, using Eqs. 4.177 and 4.147:
(1) (2)
(1) (2)
(1) (2)
1 1
1
1
1
1
2
(1) 2
(2) 2
S |s mi = (S ) + (S ) + 2(Sx Sx + Sy Sy + Sz Sz ) A| 2 2 i|s2 m − 2 i + B| 2 − 2 i|s2 m + 2 i
= A S 2 | 21 12 i |s2 m − 12 i + | 12 12 i S 2 |s2 m − 21 i
+2 Sx | 12 12 i Sx |s2 m − 12 i + Sy | 12 12 i Sy |s2 m − 12 i + Sz | 12 12 i Sz |s2 m − 21 i
+B S 2 | 12 − 12 i |s2 m + 12 i + | 12 − 12 i S 2 |s2 m + 21 i
1
1
1
1
1
1
1
1
1
+2 Sx | 2 − 2 i Sx |s2 m + 2 i + Sy | 2 − 2 i Sy |s2 m + 2 i + Sz | 2 − 2 i Sz |s2 m + 2 i
= A 34 ~2 | 12 12 i|s2 m − 12 i + ~2 s2 (s2 + 1)| 12 12 i|s2 m − 12 i
148
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
q
~ 1
1 ~
+2 2 | 2 − 2 i 2
s2 (s2 + 1) − (m − 12 )(m + 12 )|s2 m + 21 i
q
1
3
3
+ s2 (s2 + 1) − (m − 2 )(m − 2 )|s2 m − 2 i
q
i~
1
1 ~
+ 2 | 2 − 2 i 2i
s2 (s2 + 1) − (m − 12 )(m + 21 )|s2 m + 12 i
q
1
3
3
~ 1 1
1
1
− s2 (s2 + 1) − (m − 2 )(m − 2 )|s2 m − 2 i + 2 | 2 2 i~(m − 2 )|s2 m − 2 i
+B 34 ~2 | 12 − 12 i|s2 m + 12 i + ~2 s2 (s2 + 1)| 21 − 21 i|s2 m + 21 i
q
q
~ 1 1 ~
3
3
1
1
1
1
+2 2 | 2 2 i 2
s2 (s2 + 1) − (m + 2 )(m + 2 )|s2 m + 2 i + s2 (s2 + 1) − (m + 2 )(m − 2 )|s2 m − 2 i
q
~
+ −i~
s2 (s2 + 1) − (m + 12 )(m + 23 )|s2 m + 32 i
| 12 12 i 2i
2
q
1
1
1
1
−~
1
1
1
− s2 (s2 + 1) − (m + 2 )(m − 2 )|s2 m − 2 i + 2 | 2 − 2 i~(m + 2 )|s2 m + 2 i
q
= ~2 A 34 + s2 (s2 + 1) + m − 12 + B s2 (s2 + 1) − m2 + 41 | 12 12 i|s2 m − 21 i
q
+~2 B 34 + s2 (s2 + 1) − m − 12 + A s2 (s2 + 1) − m2 + 41 | 12 − 12 i|s2 m + 21 i
= ~2 s(s + 1)|s mi = ~2 s(s + 1) A| 12 12 i|s2 m − 21 i + B| 21 − 12 i|s2 m + 12 i .

 q
 A s2 (s2 + 1) + 1 + m +B s2 (s2 + 1) − m2 + 1 = s(s + 1)A, 
4
4
q
or
 B s2 (s2 + 1) + 1 − m +A s2 (s2 + 1) − m2 + 1 = s(s + 1)B, 
4
4
 
q
 A s2 (s2 + 1) − s(s + 1) + 1 + m +B s2 (s2 + 1) − m2 + 1 = 0, 
A(a + m) +Bb = 0
4
4
q
or
,
B(a − m) +Ab = 0
 B s2 (s2 + 1) − s(s + 1) + 1 − m +A s2 (s2 + 1) − m2 + 1 = 0, 
4
4
q
where a ≡ s2 (s2 + 1) − s(s + 1) + 41 , b ≡ s2 (s2 + 1) − m2 + 41 . Multiply by (a − m) and b, then subtract:
A(a2 − m2 ) + Bb(a − m) = 0; Bb(a − m) + Ab2 = 0 ⇒ A(a2 − m2 − b2 ) = 0 ⇒ a2 − b2 = m2 , or:
2
s2 (s2 + 1) − s(s + 1) + 14 − s2 (s2 + 1) + m2 − 14 = m2 ,
2
2
s2 (s2 + 1) − s(s + 1) + 14 = s22 + s2 + 41 = s2 + 12 , so
s2 (s2 + 1) − s(s + 1) + 14 = ± s2 + 12 ; s(s + 1) = s2 (s2 + 1) ∓ s2 + 21 + 14 .
Add 14 to both sides:
1 2
s2 + s + 14 = s + 2
1
= s2 (s2 + 1) ∓ s2 + 2 + 12 =
 2
 s2 + s2 − s2 − 21 + 12 = s22


 2

s2 + s2 + s2 + 21 + 12 = (s2 + 1)2


s2 − 21




⇒ s = ±s2 − 12 =
 s + 12 = ±s2

−s2− 12
So
.
1
s2 + 2 



 s + 12 = ±(s2 + 1) ⇒ s = ±(s2 + 1) − 12 =
3 
−s2 − 2
.
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
But s ≥ 0, so the possibilities are s = s2 ± 1/2.
149
Then:
1
1
1
s2 ± + 1 +
s2 ±
2
2
4
1
1 1 1
1
1
1
2
2
= s2 + s2 − s2 ∓ s2 − s2 ∓ s2 − ∓ + = ∓s2 ∓ = ∓ s2 +
.
2
2
4 2 4
2
2
s
s
s
2
1
1
1
1
2
2
2
−m =
b=
s2 + s2 +
−m =
s2 +
s2 + + m
s2 + − m .
4
2
2
2
a = s22 + s2 −
q
s2 + 21 + m s2 + 12 − m
∴ A ∓ s2 + 12 + m = ∓A s2 + 12 ∓ m = −Bb = −B
q
q
⇒ A s2 + 12 ∓ m = ±B s2 + 12 ± m. But |A|2 + |B|2 = 1, so
2
2
|A| + |A|
s2 + 12 ∓ m
s2 + 12 ± m
s
⇒ A=
|A|2
1
1
(2s2 + 1)
=
s2 + ± m + s2 + ∓ m =
|A|2 = 1.
2
2
(s2 + 21 ± m)
(s2 + 12 ± m)
s2 ± m + 12
.
2s2 + 1
q
B = ±A q
s2 + 21 ∓ m
s
= ±
s2 + 21 ± m
s2 ∓ m + 21
.
2s2 + 1
(b) Here are four examples:
(i) From the 1/2 × 1/2 table (s2 = 1/2), pick s = 1 (upper signs), m = 0. Then
q1
q1
1
1
2 +0+ 2
2 −0+ 2
√1 ;
=
= √12 .
A=
B
=
1+1
1+1
2
(ii) From the 1 × 1/2 table (s2 = 1), pick s = 3/2 (upper signs), m = 1/2. Then
q 1 1
q 1 1
q
1+ 2 + 2
1− 2 + 2
2
A=
=
;
B
=
= √13 .
2+1
3
2+1
(iii) From the 3/2 × 1/2 table (s2 = 3/2), pick s = 1 (lower signs), m = −1. Then
q3
q3
√
1
1
3
2 +1+ 2
2 −1+ 2
A=
=
;
B
=
−
= − 12 .
3+1
2
3+1
(iv) From the 2 × 1/2 table (s2 = 2), pick s = 3/2 (lower signs), m = 1/2. Then
q 1 1
q 1 1
q
q
2− 2 + 2
2+ 2 + 2
2
A=
= 5; B = −
= − 35 .
4+1
4+1
These all check with the values on Table 4.8, except that the signs (which are conventional) are reversed
in (iii) and (iv). Normalization does not determine the sign of A (nor, therefore, of B).
Problem 4.61
 
 
 
 
1
0
0
0








0
1
0
3 1
3 −1
3 −3
 
 
0

| 32 23 i = 
0 ; | 2 2 i = 0 ; | 2 2 i = 1 ; | 2 2 i = 0 .
0
0
0
1

√
3 1
 S+ | 32 23 i = 0, S+ | 32 12 i = 3~| 23 32 i, S+ | 32 −1
2 i = 2~| 2 2 i,

S− | 32 32 i =
√
3~| 32 12 i,
S− | 32 12 i = 2~| 23 −1
2 i,
S− | 32 −1
2 i=
√
Equation 4.136 ⇒
S+ | 32 −3
2 i=
3~| 23 −3
2 i,
√
3~| 32 −1
2 i;
S− | 23 −3
2 i = 0.
150
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
√
 √





3 0 0
0 30 0
√0 0 0 0
√0
0 0 2 0 
 3 0 0 0

1
~


 3 0 2 √0  .
√ 
So: S+ = ~ 
0 0 0 3 ; S− = ~  0 2 0 0 ; Sx = 2 (S+ + S− ) = 2  0 2 0

3
√
√
0 0 0 0
0 0 3 0
0 0
3 0
√
√
−λ
3 0 0
√
−λ 2 √0
3 2 √0
√
3 −λ 2 √0
3
= −λ 2 −λ
−
3
0
−λ
3
√
√
0 2 −λ
3
√
0
3 −λ
0
3 −λ
3 −λ
0 0
√ i
√ h√
= −λ −λ3 + 3λ + 4λ − 3 3λ2 − 3 3 = λ4 − 7λ2 − 3λ2 + 9 = 0,
or λ4 − 10λ2 + 9 = 0;
(λ2 − 9)(λ2 − 1) = 0;
λ = ±3, ±1. So the eigenvalues of Sx are
1
3
2 ~, 2 ~,
− 12 ~, − 23 ~.
Problem 4.62
From Eq. 4.135, Sz |s mi = ~m|s mi. Since s is fixed, here, let’s just identify the states by the value of m (which
runs from −s to +s). The matrix elements of Sz are
Snm = hn|Sz |mi = ~mhn|mi = ~mδn m .
It’s a diagonal matrix, with elements m~, ranging from m = s in the upper left corner to m = −s in the lower
right corner:


s 0
0 ··· 0
0 s − 1 0 · · · 0 




Sz = ~ 0 0 s − 2 · · · 0  .
 .. ..
.. . . .. 
. .
. . 
.
0 0
0 · · · −s
From Eq. 4.136,
p
p
S± |s mi = ~ s(s + 1) − m(m ± 1) |s (m ± 1)i = ~ (s ∓ m)(s ± m + 1) |s (m ± 1)i.
p
(S+ )nm = hn|S+ |mi = ~ (s − m)(s + m + 1) hn|m + 1i = ~bm+1 δn (m+1) = ~bn δn (m+1) .
All nonzero elements have row index (n) one greater than the column index (m), so they are on the diagonal
just above the main diagonal (note that the indices go down, here: s, s − 1, s − 2. . . , −s):


0 bs 0
0 ···
0
0 0 bs−1 0 · · ·
0 


0 0 0 bs−2 · · ·
0 


S+ = ~  . . .
.. . .
..  .
 .. .. ..
.
.
. 


0 0 0
0 · · · b−s+1 
0 0 0
0 ···
0
Similarly
p
(S− )nm = hn|S− |mi = ~ (s + m)(s − m + 1) hn|m − 1i = ~bm δn (m−1) .
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
151
This time the nonzero elements are on the diagonal just below the main diagonal:


0 0 0 ···
0 0
bs 0 0 · · ·
0 0


 0 bs−1 0 · · ·
0
0
S− = ~ 
.
 .. .. .. . .
..
.. 
. . . .
.
.
0 0 0 · · · b−s+1 0
1
To construct Sx = 12 (S+ + S− ) and Sy = 2i
(S+ − S− ), simply add and subtract the matrices S+ and S− :




0 bs
0
0 ···
0
0
0
bs
0
0 ···
0
0
bs 0 bs−1 0 · · ·
−bs
0
0 
0
bs−1
0 ···
0
0 





 0 bs−1 0 bs−2 · · ·

0
0 
0 −bs−1
0 bs−2 · · ·
0
0 



~  0
~
0
0 
0 −bs−2 0 · · ·
0
0 
Sx =  0 0 bs−2 0 · · ·
 ; Sy =
.

2  .. ..
2i  ..
..
.. . .
..
.. 
..
..
.. . .
..
.. 


. .
 .
.
.
.
.
.
.
.
.
.
.
.




0 0
 0
0
0 ···
0 b−s+1 
0
0
0 ···
0
b−s+1 
0 0
0
0 · · · b−s+1 0
0
0
0
0 · · · −b−s+1 0
Problem 4.63
p
L+ Y`m = ~ `(` + 1) − m(m + 1)Y`m+1 (Eqs. 4.120 and 121). Equation 4.130 ⇒
p
∂
∂
+ i cot θ
K`m eimφ P`m (cos θ) = ~ `(` + 1) − m(m + 1)K`m+1 ei(m+1)φ P`m+1 (cos θ).
~eiφ
∂θ
∂φ
p
d
− m cot θ P`m (cos θ) = `(` + 1) − m(m + 1)K`m+1 P`m+1 (cos θ).
K`m
dθ
p
d
cos θ
x
d
dx d
d
Let x ≡ cos θ; cot θ =
=√
;
=
= − sin θ
= − 1 − x2 . So, using 4.226:
2
sin θ
dθ
dθ
dx
dx
dx
1−x
p
x
1
dP m
d
− m√
P`m (x) = −K`m √
(1 − x2 ) ` + mxP`m = K`m P`m+1
K`m − 1 − x2
dx
dx
1 − x2
1 − x2
=
p
`(` + 1) − m(m + 1)K`m+1 P`m+1 (x).
⇒
1
K`m .
K`m+1 = p
`(` + 1) − m(m + 1)
Now `(` + 1) − m(m + 1) = (` − m)(` + m + 1), so (assuming m is non-negative)
K`m+1 = √
1
1
√
K`m ⇒ K`1 = √ √
K`0 ;
`−m `+1+m
` `+1
K`2 = √
1
1
√
p
K`1 = p
K`0 ;
`−1 `+2
`(` − 1) (` + 1)(` + 2)
1
1
√
K`2 = p
K`0 , etc.
`−2 `+3
(` + 3)(` + 2)(` + 1)`(` − 1)(` − 2)
s
(` − m)!
The quantity inside the square root is [(` + m)!/(` − m)!], so K`m =
C(`) (where C(`) ≡ K`0 ).
(` + m)!
K`3 = √
If m is negative, either run the same argument using L− , or else rewrite the recursion formula as
p
p
K`m = `(` + 1) − m(m + 1) K`m+1 , so K`−1 = (` + 1)` K`0 ,
152
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
K`−2 =
p
(` + 2)(` − 1)K`−1 =
p
(` + 1)(` + 2)`(` − 1)K`0 ,
p
p
K`−3 = (` + 3)(` − 2)K`−2 = (` + 3)(` + 2)(` + 1)`(` − 1)(` − 2)K`0 , . . .
p
which leads to K`−m = (` + m)!/(` − m)! K`0 , so in fact the same formula holds regardless of the sign of m.
Now, Problem 4.25 says:
(−1)`
Y`` = `
r
2 `!
P`` (x) = (−1)` (1 − x2 )`/2
(2` + 1)! iφ
(e sin θ)` = K`` ei`φ P`` (cos θ).
4π
d
dx
`
1
`
2 `!
d
dx
`
2
`
(x − 1)
But (Equation 4.27)
(1 − x2 )`/2
= (−1)`
2` `!
d
dx
|
2`
(x2` − . . .)
{z
}
(2`)!
(2`)!
(2`)!
= (−1)` ` (1 − x2 )`/2 , so P`` (cos θ) = (−1)` ` (sin θ)` . Therefore
2 `!
2 `!
(−1)`
2` `!
s
But K`` =
r
(2`)!
1
(2` + 1)! iφ
(e sin θ)` = K`` ei`φ (−1)` ` (sin θ)` ⇒ K`` =
4π
2 `!
(2`)!
1
C(`),
(2`)!
r
so C(`) =
2` + 1
,
4π
s
and hence
K`m =
r
(2` + 1)!
=
4π
s
(2` + 1) (` − m)!
, in agreement with
4π (` + m)!
Equation 4.32.
Problem 4.64
(a) For both terms, ` = 1, so
(b) 0, P =
~2 (1)(2) = 2~2 , P = 1.
1
2
, or ~, P = .
3
3
(c)
3 2
~ , P = 1.
4
(d)
~
1
~
2
, P = , or − , P = .
2
3
2
3
(e) From the 1 × 12 Clebsch-Gordan table (or Problem 4.60):
q
q
hq
i q h
i
2 1 −1
2 3 1
2 √1 3 1
2 1 1
√1 | 1 1 i|1 0i +
√1
√1 | 1 1 i +
|
i|1
1i
=
|
i
−
|
i
+
|
i
3 2 2
3 2 2
3
3 2 2
3 2 2
3
3 2 2
3 2 2
√ 15
8
3
1
= 2 32 | 32 12 i + 13 | 12 12 i. So j = 23 or 12 .
~2 , P = , or ~2 , P = .
4
9
4
9
(f )
1
~, P = 1.
2
(2` + 1)
.
4π(2`)!
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
153
(g)
|ψ|2 = |R21 |2
√ 1 02 †
2 0∗ 1 †
2
|Y1 | (χ+ χ+ ) +
Y1 Y1 (χ+ χ− ) +Y11∗ Y10 (χ†− χ+ ) + |Y11 |2 (χ†− χ− )
3
3
| {z }
| {z } 3
| {z }
| {z }
1
0
0
1 1 1 r2 −r/a 3
3
1
· 3 · 2e
cos2 θ + 2
sin2 θ
= |R21 |2 |Y10 |2 + 2|Y11 |2 = ·
3
3 24 a a
4π
8π
=
1
[Tables 4.3, 4.7]
1
3
1
r2 e−r/a ·
r2 e−r/a .
(cos2 θ + sin2 θ) =
5
3 · 24 · a
4π
96πa5
(h)
1
r2 |R21 |2
3
Z
|Y10 |2 sin θ dθ dφ =
r2
1
1 4 −r/a
1 4 −r/a
r e
=
r e
.
|R21 |2 = ·
3
3 24a3
72a5
Problem 4.65
Quartet.
| 32 32 i = | ↑↑↑i. For one particle (Equation 4.146), S− | ↑i = ~| ↓i, S− | ↓i = 0. For all three,
S = S1 + S2 + S3 ⇒ S− = S1− + S2− + S3− . So (using Equations 4.120 and 4.121):
3/2
S− | 32 32 i = S− | ↑↑↑i = ~ (| ↓↑↑i + | ↑↓↑i + | ↑↑↓i) = B3/2 | 23 12 i =
√
3~| 23 12 i
1
⇒ | 32 12 i = √ (| ↓↑↑i + | ↑↓↑i + | ↑↑↓i) .
3
~
2~
S− | 32 12 i = √ (| ↓↓↑i + | ↓↑↓i + | ↓↓↑i + | ↑↓↓i + | ↓↑↓i + | ↑↓↓i) = √ (| ↓↓↑i + | ↓↑↓i + | ↑↓↓i)
3
3
1
1/2
3 −1
3 −1
√ (| ↓↓↑i + | ↓↑↓i + | ↑↓↓i) .
= B3/2 | 32 −1
2 i = 2~| 2 2 i ⇒ | 2 2 i =
3
√
√
~
−1/2 3 −3
√ (| ↓↓↓i + | ↓↓↓i + | ↓↓↓i) = 3~| ↓↓↓i = B3/2
S− | 32 −1
| 2 2 i = 3~| 23 −3
2 i=
2 i
3
⇒ | 32 −3
2 i = | ↓↓↓i.
Doublet 1. | 12 12 i1 = √12 (| ↑↓↑i − | ↓↑↑i) .
~
~
S− | 12 12 i1 = √ (| ↓↓↑i + | ↑↓↓i − | ↓↓↑i − | ↓↑↓i) = √ (| ↑↓↓i − | ↓↑↓i)
2
2
1
1/2
1 −1
1 −1
√ (| ↑↓↓i − | ↓↑↓i) .
= B1/2 | 21 −1
2 i1 = ~| 2 2 i1 ⇒ | 2 2 i1 =
2
Doublet 2. | 21 12 i2 = a| ↑↑↓i + b| ↑↓↑i + c| ↓↑↑i. Make it orthogonal to | 21 12 i1 and | 23 21 i:
1
11 11
1 h 2 2 | 2 2 i2 = √
1
(h↑↓↑ | − h↓↑↑ |)(a| ↑↑↓i + b| ↑↓↑i + c| ↓↑↑i) = √ (b − c) = 0 ⇒ c = b.
2
2
154
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
1
31 11
h 2 2 | 2 2 i2 = √
3
1
(h↓↑↑ | + h↑↓↑ | + h↑↑↓ |)(a| ↑↑↓i + b| ↑↓↑i + c| ↓↑↑i) = √ (a + b + c) = 0 ⇒ a = −2b.
3
1
1
2
Normalization: |a|2 + |b|2 + |c|2 = 1 ⇒ 4|b|2 + |b|2 + |b|2 = 6|b|2 = 1 ⇒ b = √ , c = √ , a = − √ .
6
6
6
1
| 12 21 i2 = √ (−2| ↑↑↓i + | ↑↓↑i + | ↓↑↑i).
6
~
S− | 12 12 i2 = √ (−2| ↓↑↓i − 2| ↑↓↓i + | ↓↓↑i + | ↑↓↓i + | ↓↓↑i + | ↓↑↓i)
6
~
1/2
1 −1
= √ (−| ↓↑↓i − | ↑↓↓i + 2| ↓↓↑i) = B1/2 | 21 −1
2 i2 = ~| 2 2 i2 .
6
1
√ (−| ↓↑↓i − | ↑↓↓i + 2| ↓↓↑i) .
| 12 −1
2 i2 =
6
Problem 4.66
a
From Problem 4.31 we know that in the generic state χ =
(with |a|2 + |b|2 = 1),
b
hSz i =
~
|a|2 − |b|2 ,
2
hSx i = ~Re(ab∗ ),
hSy i = −~Im(ab∗ );
hSx2 i = hSy2 i =
~2
.
4
Writing a = |a|eiφa , b = |b|eiφb , we have ab∗ = |a||b|ei(φa −φb ) = |a||b|eiθ , where θ ≡ φa − φb is the phase
difference between a and b. Then
hSx i = ~Re(|a||b|eiθ ) = ~|a||b| cos θ,
σS2 x = hSx2 i − hSx i2 =
We want σS2 x σS2 y =
~2
− ~2 |a|2 |b|2 cos2 θ;
4
hSy i = −~Im(|a||b|eiθ ) = −~|a||b| sin θ.
σS2 y = hSy2 i − hSy i2 =
~2
− ~2 |a|2 |b|2 sin2 θ.
4
~2
hSz i2 , or
4
~2
~2 ~2
2
~2
1 − 4|a|2 |b|2 cos2 θ
1 − 4|a|2 |b|2 sin2 θ =
|a|2 − |b|2 .
4
4
4 4
1 − 4|a|2 |b|2 cos2 θ + sin2 θ + 16|a|4 |b|4 sin2 θ cos2 θ = |a|4 − 2|a|2 |b|2 + |b|4 .
1 + 16|a|4 |b|4 sin2 θ cos2 θ = |a|4 + 2|a|2 |b|2 + |b|4 = |a|2 + |b|2
2
=1
⇒ |a|2 |b|2 sin θ cos θ = 0.
So either θ = 0 or π, in which case a and b are relatively real, or else θ = ±π/2, in which case a and b are
relatively imaginary (these two options subsume trivially the solutions a = 0 and b = 0).
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
155
Problem 4.67
(a) S = S1 + S2 + S3 ⇒ S 2 = S12 + S22 + S32 + 2S1 · S2 + 2S1 · S3 + 2S2 · S3 , so H = 21 J(S 2 − S12 − S22 − S32 ).
But Si2 = ~2 s(s + 1) = 43 ~2 (because they all have spin 1/2), so H = 21 J S 2 − 94 ~2 .
(b) S 2 = s(s + 1)~2 , where s is the total spin. Combine two spin 1/2 states and you get spin 1 or spin 0; add
spin 1/2 to spin 1 and you get 1 + 12 = 32 or 1 − 12 = 21 ; add 12 to 0 and you get 21 . So the total spin is 3/2 (four
states) or 1/2 (two states)—and the latter occurs in two different ways. There are 23 = 8 states in all (each of
the three could be spin up or spin down), 4 of which carry spin 1/2 (the ground state). The degeneracy of the
ground state is 4, and its energy is 12 J 12 ( 32 )~2 − 94 ~2 = − 34 ~2 J. (c) S 2 = S12 + S22 + S32 + S42 + 2S1 · S2 +
2S1 · S3 + 2S1 · S4 + 2S2 · S3 + 2S2 · S4 + 2S3 · S4 .
S 2 − (S1 + S3 )2 − (S2 + S4 )2 = S12 + S22 + S32 + S42 + 2S1 · S2 + 2S
2S
1 · S3 + 2S1 · S4 + 2S2 · S3 + 2 · S4 +
2
2
2
2
2S3 · S4 − S1 − S3 − 2S
2S
1 · S3 − S2 − S4 −
2 · S4 = 2 (S1 · S2 + S1 · S4 + S2 · S3 + S3 · S4 ) = 2H/J,
so H = 21 J S 2 − (S1 + S3 )2 − (S2 + S4 )2 . QED
Write H = 12 ~2 J [s(s + 1) − s13 (s13 + 1) − s24 (s24 + 1)], where s is the total spin, s13 is the spin of the 1,3
combination, and s24 that of the 2,4 combination. For the ground state we want s as small as possible, and
s13 , s24 as large as possible: s = 0, s13 = s24 = 1 (the triplet combinations). Then E = 12 ~2 J(0 − 2 − 2) =
−2~2 J.
[Are we sure we can construct such a state? From 1 and 3 we can make singlet or triplet combinations;
likewise 2 and 4: 21 ⊗ 12 = 1, 0. If both are in the triplet state we can get a total spin of 2, 1, or 0: 1 ⊗ 1 = 2, 1, 0.
So yes: here is the total spin 0 with each pair having total spin 1. By the way, the other possibilities are
1 ⊗ 0 = 1, 0 ⊗ 1 = 1, and 0 ⊗ 0 = 0. Thus the total spin could be
s=2
(2s + 1) = 5 states,
s = 1 (three ways)
3 × 3 = 9 states,
s = 0 (two ways)
2 × 1 = 2 states,
a total of 5 + 9 + 2 = 16 states, which is just right: 24 = 16. But the ground state is unique—there’s exactly
one combination that puts 1,3 and 2,4 in the triplet configuration and gets a total spin of 0.]
Problem 4.68
(a) The effective potential is (Equation 4.53):
Now
Vj = V (rj ), rj = j∆r, ∆r =
V (r) = −
e2 1
~2 `(` + 1)
+
.
4π0 r
2me r2
b
~2
, λ=
,
N +1
2me (∆r)2
so
2me
~2 `(` + 1)(N + 1)2
Vj
b2
e2 (N + 1)
= 2
+
−
λ
~ (N + 1)2
4π0
jb
2me
j 2 b2
2me e2
b
1 `(` + 1)
2b
1 `(` + 1)
2β
`(` + 1)
=−
+
=−
+
=−
+
. X
4π0 ~2 (N + 1) j
j2
a(N + 1) j
j2
j
j2
vj ≡
b
b
b ⇒ 1 (1/β) N + 1 ∼ N. X Note that u(0) = 0 (Equation
N +1
(N + 1)β
4.59), and u(b) = 0, so we have the same boundary conditions as Problem 2.61.
(b) ∆r a b ⇒
156
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
`=0
h = Table If i == j, 2 - 1
25 j
, 0 , {i, 1000}, {j, 1000}
k = Table[If[i ⩵ j + 1, - 1, 0], {i, 1000}, {j, 1000}]
m = Table[If[i ⩵ j - 1, - 1, 0], {i, 1000}, {j, 1000}]
p = Table[h[[i, j]] + k[[i, j]] + m[[i, j]], {i, 1000}, {j, 1000}]
EIG = Eigenvalues[N[p]]
EVE = Eigenvectors[N[p]]
ListLinePlot[EVE[[994]], PlotRange -> {0, 0.11}]
0.10
0.08
0.06
0.04
0.02
0
200
400
600
800
1000
ListLinePlot[EVE[[997]], PlotRange -> {- 0.05, 0.07}]
0.06
0.04
0.02
200
400
600
800
1000
-0.02
2
ProbD65L=0.nb
-0.04
ListLinePlot[EVE[[999]], PlotRange -> {- 0.05, 0.07}]
0.06
0.04
0.02
200
400
600
800
1000
-0.02
-0.04
EIG[[994]]
- 0.00039996
EIG[[997]]
- 0.0000999873
EIG[[999]]
- 0.0000399639
2
2
2
(N +1)
To get the actual energy (see Equation 2.199), multiply by λ = ~ 2m
= 2m~e a2 β12 = −E1 /β 2 ; in other words,
2
eb
we must multiply by 1/β 2 = 2500, and the result will be the energy in units of 13.6 eV (the exact answer, of
course, would be −1/n2 = −1, −0.25, −0.1111):
−0.00039996×(2500) = 0.9999E1 , −0.000099987×(2500) = 0.24997E1 , −0.000039964×(2500) = 0.09991E1 .
The first two are pretty good; the third is a bit off (but notice from the graph that the wave function has not
dropped to zero well before r = b).
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
157
`=1
h = Table If i == j, 2 - 1
25 j
+ 2
j^2
, 0 , {i, 1000}, {j, 1000}
k = Table[If[i ⩵ j + 1, - 1, 0], {i, 1000}, {j, 1000}]
m = Table[If[i ⩵ j - 1, - 1, 0], {i, 1000}, {j, 1000}]
p = Table[h[[i, j]] + k[[i, j]] + m[[i, j]], {i, 1000}, {j, 1000}]
EIG = Eigenvalues[N[p]]
EVE = Eigenvectors[N[p]]
ListLinePlot[EVE[[997]], PlotRange -> {0, 0.065}]
0.06
0.05
0.04
0.03
0.02
0.01
0
200
400
600
800
1000
ListLinePlot[EVE[[999]], PlotRange -> {- 0.04, 0.06}]
0.06
0.04
0.02
200
2
400
600
800
1000
-0.02
ProbD65L=1.nb
-0.04
ListLinePlot[EVE[[1000]], PlotRange -> {- 0.06, 0.05}]
0.04
0.02
200
400
600
800
1000
-0.02
-0.04
-0.06
EIG[[997]] * 2500
- 0.249991
EIG[[999]] * 2500
- 0.103278
EIG[[1000]] * 2500
0.0158618
For ` = 1 the lowest state is n = 2, so the exact energies (in units of E1 ) are 0.25, 0.1111, and 0.0625
0.24999E1 ,
0.10328E1 ,
−0.015862E1 .
The first is good; the second is a bit off (but notice from the graph that the wave function has not dropped to
zero well before r = b), and the third is terrible (wrong sign).
158
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
`=2
h = Table If i == j, 2 - 1
25 j
+ 6
j^2
, 0 , {i, 1000}, {j, 1000}
k = Table[If[i ⩵ j + 1, - 1, 0], {i, 1000}, {j, 1000}]
m = Table[If[i ⩵ j - 1, - 1, 0], {i, 1000}, {j, 1000}]
p = Table[h[[i, j]] + k[[i, j]] + m[[i, j]], {i, 1000}, {j, 1000}]
EIG = Eigenvalues[N[p]]
EVE = Eigenvectors[N[p]]
ListLinePlot[EVE[[999]], PlotRange -> {0, 0.065}]
0.06
0.05
0.04
0.03
0.02
0.01
0
200
400
600
800
1000
ListLinePlot[EVE[[1000]], PlotRange -> {- 0.05, 0.05}]
0.04
0.02
200
400
600
800
1000
-0.02
2
ProbD65L=2.nb
-0.04
ListLinePlot[EVE[[998]], PlotRange -> {- 0.06, 0.05}]
0.04
0.02
200
400
600
800
1000
-0.02
-0.04
-0.06
EIG[[999]] * 2500
- 0.107958
EIG[[1000]] * 2500
- 0.0112765
EIG[[998]] * 2500
0.135962
For ` = 2 the lowest state is n = 3, so the exact energies (in units of E1 ) are 0.1111, 0.0625, and 0.04.
0.10796E1 ,
0.011277E1 ,
−0.13596E1 .
The first is OK; the second is a way off, and the third is terrible (wrong sign). But in all three cases the wave
function has not dropped to zero well before r = b. To improve the results we would a substantially larger b.
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Problem 4.69
`=0
Plot Evaluate u[x] /. NDSolve
u ''[x] - 1 - 2 * 0.9999
x + 0.000001
MaxSteps → 10 000
u[x] ⩵ 0, u[0] ⩵ 0, u '[0] ⩵ 1 , u[x], x, 10 ^ - 8 , 10 ,
, {x, 0.01, 10}, PlotRange → {- 0.02, 0.5}
0.5
0.4
0.3
0.2
0.1
2
4
6
8
Plot Evaluate u[x] /. NDSolve
x + 0.000001
MaxSteps → 10 000
10
u ''[x] - 1 - 2 * 1.0001
u[x] ⩵ 0, u[0] ⩵ 0, u '[0] ⩵ 1 , u[x], x, 10 ^ - 8 , 10 ,
, {x, 0.01, 10}, PlotRange → {- 0.1, 0.5}
0.5
0.4
0.3
0.2
0.1
2
2
4
6
8
10
-0.1
ProbD66L=0.nb
So the lowest n is between 0.9999 and 1.0001.
Plot Evaluate u[x] /. NDSolve
x + 0.000001
MaxSteps → 10 000
u ''[x] - 1 - 2 * 2.0001
u[x] ⩵ 0, u[0] ⩵ 0, u '[0] ⩵ 1 , u[x], x, 10 ^ - 8 , 15 ,
, {x, 0.01, 15}, PlotRange → {- 0.4, 0.2}
0.2
0.1
2
4
6
8
10
12
14
-0.1
-0.2
-0.3
-0.4
Plot Evaluate u[x] /. NDSolve
x + 0.000001
MaxSteps → 10 000
u ''[x] - 1 - 2 * 1.9999
u[x] ⩵ 0, u[0] ⩵ 0, u '[0] ⩵ 1 , u[x], x, 10 ^ - 8 , 15 ,
, {x, 0.01, 15}, PlotRange → {- 0.4, 0.2}
0.2
0.1
2
4
6
8
10
12
-0.1
-0.2
-0.3
-0.4
So the next n is between 1.9999 and 2.0001.
14
159
160
CHAPTER 4. QUANTUM MECHANICS
IN THREE DIMENSIONS
ProbD66L=0.nb
3
Plot Evaluate u[x] /. NDSolve
x + 0.000001
MaxSteps → 10 000
u ''[x] - 1 - 2 * 2.9999
u[x] ⩵ 0, u[0] ⩵ 0, u '[0] ⩵ 1 , u[x], x, 10 ^ - 8 , 17 ,
, {x, 0.01, 17}, PlotRange → {- 0.2, 0.3}
0.3
0.2
0.1
5
10
15
-0.1
-0.2
Plot Evaluate u[x] /. NDSolve
x + 0.000001
MaxSteps → 10 000
u ''[x] - 1 - 2 * 3.00001
u[x] ⩵ 0, u[0] ⩵ 0, u '[0] ⩵ 1 , u[x], x, 10 ^ - 8 , 20 ,
, {x, 0.01, 20}, PlotRange → {- 0.2, 0.3}
0.3
0.2
0.1
5
10
15
20
-0.1
-0.2
So the third n is between 2.9999 and 3.00001.
`=1
2
ProbD66L=1.nb
This time there is no solution for n = 1; the lowest state (no nodes) is around n = 2:
Plot Evaluate u[x] /. NDSolve
2
u ''[x] - 1 -
x ^ 2 + 0.000000001
2 * 1.999
x + 0.000001
+
u[x] ⩵ 0, u[1] ⩵ 1, u '[0] ⩵ 0 ,
u[x], x, 10 ^ - 8 , 10 , MaxSteps → 10 000
, {x, 0.1, 10}
1.4
1.2
1.0
0.8
0.6
0.4
0.2
2
4
6
Plot Evaluate u[x] /. NDSolve
2
8
10
u ''[x] - 1 -
x ^ 2 + 0.000000001
2 * 2.0001
, {x, 0.1, 15}
1.5
1.0
0.5
2
4
6
8
10
x + 0.000001
u[x] ⩵ 0, u[1] ⩵ 1, u '[0] ⩵ 0 ,
u[x], x, 10 ^ - 8 , 15 , MaxSteps → 10 000
12
14
-0.5
-1.0
-1.5
So the lowest n is between 1.999 and 2.0001.
+
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS ProbD66L=1.nb
Plot Evaluate u[x] /. NDSolve
2
u ''[x] - 1 -
x ^ 2 + 0.000000001
2 * 2.9999
x + 0.000001
+
u[x] ⩵ 0, u[1] ⩵ 1, u '[0] ⩵ 0 ,
u[x], x, 10 ^ - 8 , 18 , MaxSteps → 10 000
, {x, 0.1, 18}
1.0
0.5
5
10
15
-0.5
-1.0
-1.5
Plot Evaluate u[x] /. NDSolve
2
u ''[x] - 1 -
x ^ 2 + 0.000000001
2 * 3.0001
x + 0.000001
+
u[x] ⩵ 0, u[1] ⩵ 1, u '[0] ⩵ 0 ,
u[x], x, 10 ^ - 8 , 20 , MaxSteps → 10 000
, {x, 0.1, 20}
3
2
1
5
10
15
20
-1
4
ProbD66L=1.nb
So the next n is between 2.9999 and 3.0001.
Plot Evaluate u[x] /. NDSolve
2
u ''[x] - 1 -
x ^ 2 + 0.000000001
2 * 3.9999
x + 0.000001
+
u[x] ⩵ 0, u[1] ⩵ 1, u '[0] ⩵ 0 ,
u[x], x, 10 ^ - 8 , 20 , MaxSteps → 10 000
, {x, 0.1, 20}
2
1
5
10
15
20
-1
Plot Evaluate u[x] /. NDSolve
2
u ''[x] - 1 -
x ^ 2 + 0.000000001
2 * 4.0001
u[x], x, 10 ^ - 8 , 20 , MaxSteps → 10 000
, {x, 0.1, 20}
2
1
5
10
x + 0.000001
u[x] ⩵ 0, u[1] ⩵ 1, u '[0] ⩵ 0 ,
15
-1
So the third n is between 3.9999 and 4.0001.
20
+
3
161
162
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
ProbD66L=2.nb
3
`=2
This time there
is no solution for n = 1 or n = 2; xthe
lowest state (no nodes) is around n = 3:
Plot Evaluate u[x] /. NDSolve u ''[x] - 1 - 2 * 2.999
+ 0.000001 +
6
x ^ 2 + 0.000000001
u[x] ⩵ 0, u[1] ⩵ 1, u '[0] ⩵ 0 ,
u[x], x, 10 ^ - 8 , 15 , MaxSteps → 10 000
, {x, 0.1, 15}
�������������� � ��� ��������� ������� ���� ��� �������� ���������� ��� ����������� ���-������������ ��� �������� ���������� ���
��� �� ��������� �� �������� ����� � ��������� ��� �������� �������� ��� ����� ��� �������� ���������� ������� 15
10
5
2
4
6
8
Plot Evaluate u[x] /. NDSolve
6
10
12
14
u ''[x] - 1 -
x ^ 2 + 0.000000001
2 * 3.0001
x + 0.000001
+
u[x] ⩵ 0, u[1] ⩵ 1, u '[0] ⩵ 0 ,
u[x], x, 10 ^ - 8 , 15 , MaxSteps → 10 000
, {x, 0.1, 15}
�������������� � ��� ��������� ������� ���� ��� �������� ���������� ��� ����������� ���-������������ ��� �������� ���������� ���
��� �� ��������� �� �������� ����� � ��������� ��� �������� �������� ��� ����� ��� �������� ���������� ������� 12
10
8
6
4
2
2
4
4
6
8
10
12
14
ProbD66L=2.nb
-2
So the lowest n is between 2.999 and 3.0001.
Plot Evaluate u[x] /. NDSolve
6
u ''[x] - 1 -
x ^ 2 + 0.000000001
2 * 3.999
x + 0.000001
+
u[x] ⩵ 0, u[1] ⩵ 1, u '[0] ⩵ 0 ,
u[x], x, 10 ^ - 8 , 15 , MaxSteps → 10 000
, {x, 0.1, 15}
�������������� � ��� ��������� ������� ���� ��� �������� ���������� ��� ����������� ���-������������ ��� �������� ���������� ���
��� �� ��������� �� �������� ����� � ��������� ��� �������� �������� ��� ����� ��� �������� ���������� ������� 2
1
2
4
6
8
10
12
14
-1
-2
-3
-4
Plot Evaluate u[x] /. NDSolve
x + 0.000001
6
u ''[x] - 1 -
2 * 4.0001
+
x ^ 2 + 0.000000001
u[x] ⩵ 0, u[1] ⩵ 1, u '[0] ⩵ 0 ,
u[x], x, 10 ^ - 8 , 20 , MaxSteps → 10 000
, {x, 0.1, 20}
�������������� � ��� ��������� ������� ���� ��� �������� ���������� ��� ����������� ���-������������ ��� �������� ���������� ���
��� �� ��������� �� �������� ����� � ��������� ��� �������� �������� ��� ����� ��� �������� ���������� ������� 4
2
5
10
15
-2
-4
So the next n is between 3.9999 and 4.0001.
20
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Plot Evaluate u[x] /. NDSolve
6
u ''[x] - 1 -
x ^ 2 + 0.000000001
2 * 4.999
x + 0.000001
ProbD66L=2.nb
5
163
+
u[x] ⩵ 0, u[1] ⩵ 1, u '[0] ⩵ 0 ,
u[x], x, 10 ^ - 8 , 25 , MaxSteps → 10 000
, {x, 0.1, 25}
�������������� � ��� ��������� ������� ���� ��� �������� ���������� ��� ����������� ���-������������ ��� �������� ���������� ���
��� �� ��������� �� �������� ����� � ��������� ��� �������� �������� ��� ����� ��� �������� ���������� ������� 20
15
10
5
5
10
15
Plot Evaluate u[x] /. NDSolve
6
20
u ''[x] - 1 -
x ^ 2 + 0.000000001
25
2 * 5.0001
x + 0.000001
+
u[x] ⩵ 0, u[1] ⩵ 1, u '[0] ⩵ 0 ,
u[x], x, 10 ^ - 8 , 25 , MaxSteps → 10 000
, {x, 0.1, 25}
�������������� � ��� ��������� ������� ���� ��� �������� ���������� ��� ����������� ���-������������ ��� �������� ���������� ���
��� �� ��������� �� �������� ����� � ��������� ��� �������� �������� ��� ����� ��� �������� ���������� ������� 5
5
10
15
20
25
-5
-10
So the third n is between 4.9999 and 5.0001.
Problem 4.70
(a) All of them. (If Sz was measured at t1 , everything was spin up; if Sx was measured, half of them were spin
down, but the second measurement did nothing, and these were thrown away.)
(b)
P++ = cos2 (θab /2) cos2 (θbc /2),
P+− = cos2 (θab /2) sin2 (θbc /2),
P−+ = sin2 (θab /2) cos2 (θ̄bc /2),
P−− = sin2 (θab /2) sin2 (θ̄bc /2),
where θ̄ = π − θ, so cos(θ̄bc /2) = cos((π/2) − θbc /2) = sin(θbc /2); sin(θ̄bc /2) = sin((π/2) − θbc /2) = cos(θbc /2).
Therefore
P−+ = sin2 (θab /2) sin2 (θbc /2),
f=
P−− = sin2 (θab /2) cos2 (θbc /2).
P++
cos2 (θab /2) cos2 (θbc /2) + sin2 (θab /2) sin2 (θbc /2)
, or f −1 =
= 1 + tan2 (θab /2) tan2 (θbc /2).
P++ + P−+
cos2 (θab /2) cos2 (θbc /2)
164
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Problem 4.71
(a)
1
r sin θ cos φ −r/2a
r −r/2a
1
ψ2px = √
e
e
=√
3
3
a
a
32 π a
24 a
= R21 (r)
r
3
ei φ + e−i φ
sin θ
4π
2
−Y11 (θ, φ) + Y1−1 (θ, φ)
1
√
= √ (−ψ211 + ψ21−1 ).
2
2
A similar calculation gives
i
ψ2py = √ (ψ211 + ψ21−1 ) .
2
r cos θ −r/2a
1 r −r/2a
e
e
ψ2pz = √
=√
3
a
32πa
24a3 a
1
r
3
cos θ = R21 (r)Y10 (θ, φ) = ψ210 .
4π
(b) L̂z ψ2pz = L̂z ψ210 = 0, since m = 0. By cyclic permutation (z → x → y) the same goes for the other two;
all three wave functions are eigenstates of the respective components of angular momentum, and the eigenvalue
in each case is 0.
(c) Setting a = 1/2, define the three functions:
�������
psiX[x_, y_, z_] := x ⅇ-
x^2 +y^2 +z^2
π
�������
psiY[x_, y_, z_] := y ⅇ-
x^2 +y^2 +z^2
π
�������
psiZ[x_, y_, z_] := z ⅇ-
x^2 +y^2 +z^2
π
�������
ContourPlot3D
�������
psiX[x, y, z] ^ 2 ⩵ .03 ^ 2, {x, - 5, 5}, {y, - 5, 5}, {z, - 5, 5}
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
2
4P71.nb
�������
ContourPlot3D
psiY[x, y, z] ^ 2 ⩵ .03 ^ 2, {x, - 5, 5}, {y, - 5, 5}, {z, - 5, 5}
�������
�������
ContourPlot3D
165
4P71.nb
psiZ[x, y, z] ^ 2 ⩵ .03 ^ 2, {x, - 5, 5}, {y, - 5, 5}, {z, - 5, 5}
�������
Problem 4.72
(a)
1
B = ∇ × A = − ∇ ×(r × B0 )
2
1
1
= − [(B0 · ∇) r −(r · ∇) B0 + r (∇ · B0 ) − B0 (∇ · r)] = − [B0 + 0 + 0 − 3 B0 ] = B0 . X
2
2
(b) The Hamiltonian (Equations 4.188 and 4.158) is
H=
1
p2
q
q2 2
2
(p − q A) + q ϕ − γ S · B0 =
−
(p · A + A · p) +
A + q ϕ − γ S · B0 .
2m
2m 2m
2m
Now
i
1h 2 2
1
1 1
2
(r × B0 ) ·(r × B0 ) = [r · (B0 × (r × B0 ))] = r · rB02 − B0 (r · B0 ) =
r B0 −(r · B0 ) .
4
4
4
4
To evaluate the middle term we use a test function f (r):
A2 =
1
i~ [ ∇ ·(r × B0 ) f + 2 (r × B0 ) · ∇f ]
2
1
= i~ [ B0 · (∇
×
r) f − r · (∇
×
B
0 ) f − 2 B0 ·(r × ∇ f )] = B0 · L f .
2
p ·(A f ) + A · p f = −i~ [(∇ · A) f + 2 A · ∇f ] =
Combining these expressions gives the desired result.
Problem 4.73
First we compute the velocity
d
i
1
hzi = h[H, z]i =
hpz i .
dt
~
m
Then
d2
1 d
1 i
1 i
γ i
hzi =
hpz i =
h[H, pz ]i =
h[−γ B · S, pz ]i = −
h[B, pz ]i · S
dt2
m dt
m ~
m ~
m ~
Dh
iE
γ i
γ i
γ
=−
−α x ı̂ +(B0 + α z) k̂, pz · S = −
α h[z, pz ]i k̂ · S = α Sz .
m ~
m~
m
3
166
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
Problem 4.74
(a) Write S in terms of raising and lowering operators:
p2
S+ + S−
S+ − S−
p2
1
H=
− γ Bx
+ By
+ Bz Sz =
+ γ α x (S+ + S− ) − γ(B0 + α z) Sz .
2m
2
2i
2m 2
Then
p2
p2
1
1
Ψ+ (r, t) χ+ +
Ψ− (r, t) χ− + γ α x Ψ+ (r, t) ~ χ− + γ α x Ψ− (r, t) ~ χ+
2m
2m
2
2
~
~
− γ(B0 + α z) Ψ+ (r, t) χ+ − γ(B0 + α z) Ψ− (r, t) − χ−
2
2
2
2
p
~
~
p
~
~
=
Ψ+ + γ α x Ψ− − γ(B0 + α z) Ψ+ χ+ +
Ψ− + γ α x Ψ+ + γ(B0 + α z) Ψ− χ− ,
2m
2
2
2m
2
2
HΨ=
while
i~
∂Ψ+
∂Ψ−
∂
Ψ = i~
χ+ + i ~
χ− .
∂t
∂t
∂t
Since χ+ and χ− are orthogonal, the coefficients on either side of Schrödinger’s equation must match and we
have
p2
~
~
∂Ψ+
Ψ+ + γ α x Ψ− − γ (B0 + α z) Ψ+ = i ~
2m
2
2
∂t
~
~
∂Ψ−
p2
Ψ− + γ α x Ψ+ + γ (B0 + α z) Ψ− = i ~
.
2m
2
2
∂t
(b)
∂Ψ±
i
∂
= ± γ B0 Ψ± + e±i γ B0 t/2
Ψ̃± .
∂t
2
∂t
Plugging into the results from (a), and cancelling the exponential factor:
∂ Ψ̃+
p2
~
~
Ψ̃+ + γ α x e−i γ B0 t Ψ̃− − γ α z Ψ̃+ = i ~
2m
2
2
∂t
2
p
~
~
∂ Ψ̃−
Ψ̃− + γ α x ei γ B0 t Ψ̃+ + γ α z Ψ̃− = i ~
.
2m
2
2
∂t
(c) The potential V± = ∓γα ~z/2 is a linear potential that decreases in the positive-z direction for spin up
states, and in the negative-z direction for spin down. Therefore if the particle starts in a state where Ψ+ = Ψ− ,
once it enters the magnet Ψ+ will propagate in the positive-z direction and Ψ− will propagate in the negative-z
direction.
Problem 4.75
We simply need to plug into the Schrödinger equation. Taking the time derivative
i~
~2
∂Ψ
= ~ f 0 (t) Ψ(t) =
∂t
2 m b2
2
q Φ(t)
n−
Ψ
2π~
CHAPTER 4. QUANTUM MECHANICS IN THREE DIMENSIONS
167
and then evaluating the right side of Schrödinger’s equation
"
#
2
1
~2 ∂ 2
q Φ(t)
~ q Φ(t) ∂
HΨ=
− 2
+
+i
Ψ
2m
b ∂φ2
2πb
π b2 ∂φ
#
"
2
2
q Φ(t)
~2
q Φ(t)
~2 2
~ q Φ(t)
1
n
+
n
Ψ
=
n
−
−
Ψ,
=
2 m b2
2πb
π b2
2 m b2
2π~
we see that these are equal.
Problem 4.76
(a) A changing flux introduces an EMF given by Faraday’s law:
I
dΦ
dΦ
⇒ E 2πb = −
E = E · dl = −
dt
dt
⇒
E=−
1 dΦ
φ̂.
2πb dt
The work done when the charge moves a distance dl = v dt around the ring is
dW = qE dl = −
qv
q dΦ
v dt = −
dΦ.
2 π b dt
2πb
dW
qω
=− . X
dΦ
2π
(b) The mechanical angular momentum is (Equation 4.192 and foonote 62)
Φ
Φ
Lmechanical = (r × p) − q (r × A) = L − q r
(r̂ × φ̂) = L − q
k̂ .
2πr
2π
But v/b is the angular velocity ω, so
Therefore the z component of the mechanical angular momentum is
[Lmechanical ]z = Lz −
qΦ
d
qΦ
= −i~
−
.
2π
dφ
2π
Acting on ψn = Aeinφ (Equation 4.203),
L[mechanical]z ψn =
−i~
d
qΦ
−
dφ
2π
ψn =
we see that ψn is an eigenstate with eigenvalue
qΦ
~ n−
.
2π~
(c) Taking the derivative of En (Equation 4.206), we get
dEn
~2
q Lmechanical q
qΦ =
2
n
−
−
=−
,
dΦ
2 m b2
2π~
2π~
I
2π
where I = mb2 is the moment of inertia, and since L = I ω we have
dEn
ω
= −q
. X
dΦ
2π
qΦ
~n −
ψn ,
2π
168
CHAPTER 5. IDENTICAL PARTICLES
Chapter 5
Identical Particles
Problem 5.1
(a)
(m1 + m2 )R = m1 r1 + m2 r2 = m1 r1 + m2 (r1 − r) = (m1 + m2 )r1 − m2 r
r1 = R +
⇒
m2
µ
r=R+
r. X
m1 + m2
m1
(m1 + m2 )R = m1 (r2 + r) + m2 r2 = (m1 + m2 )r2 + m1 r ⇒ r2 = R −
m1
µ
r=R−
r. X
m1 + m2
m2
Let R = (X, Y, Z), r = (x, y, z).
∂
∂x ∂
∂X ∂
+
=
∂x1
∂x1 ∂X
∂x1 ∂x
m1
∂
µ
∂
=
+ (1)
=
(∇R )x + (∇r )x ,
m1 + m2 ∂X
∂x
m2
∂X ∂
∂x ∂
∂
=
+
(∇2 )x =
∂x
∂x2 ∂X
∂x2 ∂x
2
m2
∂
∂
µ
=
− (1)
=
(∇R )x − (∇r )x ,
m1 + m2 ∂X
∂x
m1
(∇1 )x =
so ∇1 =
µ
∇R + ∇ r . X
m2
so ∇2 =
µ
∇R − ∇ r . X
m1
(b)
µ
∇21 ψ = ∇1 · (∇1 ψ) = ∇1 ·
∇ R ψ + ∇r ψ
m2
µ
µ
µ
=
∇R ·
∇ R ψ + ∇r ψ + ∇r ·
∇R ψ + ∇ r ψ
m2
m2
m2
2
µ
µ
=
∇2R ψ + 2
(∇r · ∇R )ψ + ∇2r ψ.
m2
m2
Likewise,
∇22 ψ =
µ
m1
2
∇2R ψ − 2
µ
(∇r · ∇R ) + ∇2r ψ.
m1
CHAPTER 5. IDENTICAL PARTICLES
169
~2 2
~2 2
∴ Hψ = −
∇1 ψ −
∇ ψ + V (r1 , r2 )ψ
2m1
2m2 2
~2
µ2
2µ
1 2
µ2
2µ
1 2
2
2
=−
∇
+
∇
·
∇
+
∇
+
∇
−
∇
·
∇
+
∇
ψ
r
R
r
R
2 m1 m22 R m1 m2
m1 r m2 m21 R m2 m1
m2 r
~2
µ2
1
1
1
1
2
+ V (r)ψ = −
+
∇R +
+
∇2r ψ + V (r)ψ = Eψ.
2 m1 m2 m2
m1
m1
m2
But
−
1
1
+
m1 m2
m1 + m2
1
µ2
=
= , so
m1 m2
µ
m1 m2
1
1
+
m2 m1
~2 2
~2
∇2R ψ −
∇ ψ + V (r)ψ = Eψ.
2(m1 + m2 )
2µ r
=
µ
m1 m2
1
=
=
.
m1 m2
m1 m2 (m1 + m2 )
m1 + m2
X
(c) Put in ψ = ψr (r)ψR (R), and divide by ψr ψR :
~2
1 2
~2 1 2
−
∇R ψR + −
∇r ψr + V (r) = E.
2(m1 + m2 ) ψR
2µ ψr
The first term depends only on R, the second only on r, so each must be a constant; call them ER and
Er , respectively. Then:
−
~2
∇2 ψR = ER ψR ;
2(m1 + m2 )
−
~2 2
∇ ψr + V (r)ψr = Er ψr ,
2µ
with
ER + Er = E.
Problem 5.2
∆E1
∆m
m−µ
m(m + M ) M
m
=
=
=
−
=
.
E1
µ
µ
mM
M
M
The fractional error is the ratio of the electron mass to the proton mass:
(a) From Eq. 4.77, E1 is proportional to mass, so
9.109 × 10−31 kg
= 5.44 × 10−4 . The percent error is 0.054% (pretty small).
1.673 × 10−27 kg
(b) From Eq. 4.94, R is proportional to m, so
So (in magnitude) ∆λ/λ = ∆µ/µ.
nuclear mass.
∆µ =
=
But
∆(1/λ)
∆R
∆µ
(1/λ2 )∆λ
∆λ
=
=
=−
=−
.
(1/λ)
R
µ
(1/λ)
λ
µ = mM/(m + M ), where m = electron mass, and M =
mmp
mmp
m(2mp )
−
=
(2m + 2mp − m − 2mp )
m + 2mp
m + mp
(m + mp )(m + 2mp )
m2 mp
mµ
=
.
(m + mp )(m + 2mp )
m + 2mp
∆λ
∆µ
m
m
m
=
=
≈
, so ∆λ =
λh , where λh is the hydrogen wavelength.
λ
µ
m + 2mp
2mp
2mp
170
CHAPTER 5. IDENTICAL PARTICLES
1
=R
λ
∴ ∆λ =
(c) µ =
1 1
−
4 9
=
5
36
36
R⇒λ=
=
m = 6.563 × 10−7 m.
36
5R
5(1.097 × 107 )
9.109 × 10−31
(6.563 × 10−7 )m = 1.79 × 10−10 m.
2(1.673 × 10−27 )
mm
m
= , so the energy is half what it would be for hydrogen: (13.6/2)eV = 6.8 eV.
m+m
2
mp mµ
mp mµ mp + me
mµ (mp + me )
; R ∝ µ, so R is changed by a factor
·
=
, as compared with
mp + mµ
mp + mµ mp me
me (mp + mµ )
hydrogen. For hydrogen, 1/λ = R(1 − 1/4) = 34 R ⇒ λ = 4/(3R) = 4/[3(1.097 × 107 )] m = 1.215 × 10−7 m,
and λ ∝ 1/R, so for muonic hydrogen the Lyman-alpha line is at
(d) µ =
λ=
me (mp + mµ )
1 (1.673 × 10−27 + 206.77 × 9.109 × 10−31 )
(1.215 × 10−7 m) =
(1.215 × 10−7 m)
mµ (mp + me )
206.77
(1.673 × 10−27 + 9.109 × 10−31 )
= 6.54 × 10−10 m.
Problem 5.3
The energy of the emitted photon, in a transition from vibrational state ni to state nf , is
Ep = (ni + 12 )~ω − (nf + 12 )~ω = n~ω, (where n ≡ ni − nf ). The frequency of the photon is
s
k
Ep
nω
n
ν=
=
=
. The splitting of this line is given by
h
2π
2π µ
s
1 n
n√
1
k ∆µ
1 ∆µ
∆ν =
k − 3/2 ∆µ =
= ν
.
2π
2 2π µ µ
2 µ
2µ
Now
µ=
mh mc
1
−1
= 1
2
1 ⇒ ∆µ = mh + mc
+
1
1
mc
mh
mc + mh
∆ν =
−
1
∆mc
m2c
=
µ2
∆mc .
m2c
1 µ∆mc
1 (∆mc /mc )
.
ν
= ν
2
m2c
2
1 + mc
mh
Using the average value (36) for mc , we have ∆mc /mc = 2/36, and mc /mh = 36/1, so
∆ν =
Problem 5.4
(a)
Z
1=
|ψ± |2 d3 r1 d3 r2
1 (1/18)
1
ν=
ν = 7.51 × 10−4 ν.
2 (1 + 36)
(36)(37)
CHAPTER 5. IDENTICAL PARTICLES
= |A|2
2
Z
∗
[ψa (r1 )ψb (r2 ) ± ψb (r1 )ψa (r2 )] [ψa (r1 )ψb (r2 ) ± ψb (r1 )ψa (r2 )] d3 r1 d3 r2
Z
= |A|
Z
±
171
|ψa (r1 )|2 d3 r1
∗
3
Z
|ψb (r2 )|2 d3 r2 ±
Z
ψb (r1 ) ψa (r1 )d r1
∗
3
Z
ψa (r1 )∗ ψb (r1 )d3 r1
Z
ψa (r2 ) ψb (r2 )d r2 +
2 3
Z
ψb (r2 )∗ ψa (r2 )d3 r2
Z
|ψb (r1 )| d r1
2 3
|ψa (r2 )| d r2
√
= |A|2 (1 · 1 ± 0 · 0 ± 0 · 0 + 1 · 1) = 2|A|2 =⇒ A = 1/ 2.
(b)
1 = |A|2
= 4|A|2
Z
Z
∗
[2ψa (r1 )ψa (r2 )] [2ψa (r1 )ψa (r2 )] d3 r1 d3 r2
|ψa (r1 )|2 d3 r1
Z
|ψa (r2 )|2 d3 r2 = 4|A|2 .
A = 1/2.
Problem 5.5
(a)
−
~2 ∂ 2 ψ
~2 ∂ 2 ψ
−
= Eψ
2m ∂x21
2m ∂x22
(for 0 ≤ x1 , x2 ≤ a, otherwise ψ = 0).
√ πx 2
2πx2
πx1 2πx1
2
ψ=
sin
sin
− sin
sin
a
a
a
a
a
#
√ " 2
πx πx 2πx2
d2 ψ
2
π 2
2π
2πx1
1
2
−
sin
=
sin
+
sin
sin
dx21
a
a
a
a
a
a
a
#
√ " 2
πx πx d2 ψ
2
2π
2πx2
π 2
2πx1
1
2
=
−
sin
sin
+
sin
sin
dx22
a
a
a
a
a
a
a
"
#
2
π 2 2π 2
d ψ d2 ψ
π2
=
−
+
ψ
=
−5
+
ψ,
dx21
dx22
a
a
a2
~2
−
2m
2
5π 2 ~2
d ψ d2 ψ
+
=
ψ = Eψ,
dx21
dx22
2ma2
with E =
5π 2 ~2
= 5K. X
2ma2
172
CHAPTER 5. IDENTICAL PARTICLES
(b) Distinguishable:
ψ22 = (2/a) sin (2πx1 /a) sin (2πx2 /a) , with E22 = 8K (nondegenerate).
ψ13 = (2/a) sin (πx1 /a) sin (3πx2 /a)
, with E13 = E31 = 10K (doubly degenerate).
ψ31 = (2/a) sin (3πx1 /a) sin (πx2 /a)
Identical Bosons:
ψ22 = (2/a) sin (2πx1 /a) sin (2πx2 /a), E22 = 8K (nondegenerate).
√
ψ13 = ( 2/a) [sin (πx1 /a) sin (3πx2 /a) + sin (3πx1 /a) sin (πx2 /a)], E13 = 10K
Identical Fermions:
√
ψ13 = ( 2/a) sin πxa 1 sin
3πx2
a
− sin
3πx1
a
sin
πx2
a
√
ψ23 = ( 2/a) sin
3πx2
a
− sin
3πx1
a
sin
2πx2
a
2πx1
a
sin
, E13 = 10K
(nondegenerate).
(nondegenerate).
, E23 = 13K
(nondegenerate).
Problem 5.6
(a) Use Eq. 5.23 and Problem 2.4, with hxin = a/2 and hx2 in = a2
1
1
3 − 2(nπ)2
.
1
1
1
1
− 2
+ 2
2
6 2π
n
m
h
i
Ra
R
(m−n)π
1 a
nπ
(b) hximn = a2 0 x sin mπ
x − cos (m+n)π
x dx
a x sin a x dx = a 0 x cos
a
a
2
a
ax
= a1
cos (m−n)π
x + (m−n)π
sin (m−n)π
x
(m−n)π
a
a
h(x1 − x2 )2 i = a2
−
a
(m+n)π
2
1
1
3 − 2(nπ)2
+ a2
cos
(m+n)π
x
a
−
1
1
3 − 2(mπ)2
ax
(m+n)π
sin
− 2 · a2 · a2 = a2
(m+n)π
x
a
.
a
0
= a1
a
(m−n)π
2
(cos[(m − n)π] − 1) −
a
(m+n)π
2
(cos[(m + n)π] − 1) .
But cos[(m ± n)π] = (−1)m+n , so
a hximn = 2 (−1)m+n − 1
π
1
1
−
(m − n)2
(m + n)2
So Eq. 5.25 ⇒ h(x1 − x2 )2 i = a2
1
1
− 2
6 2π
(
=
1
1
+ 2
2
n
m
a(−8mn)
π 2 (m2 −n2 )2 ,
0,
−
if m and n have opposite parity,
if m and n have same parity.
128a2 m2 n2
.
π 4 (m2 − n2 )4
(The last term is present only when m, n have opposite parity.)
1
1
1
128a2 m2 n2
2 1
2
(c) Here Eq. 5.25 ⇒ h(x1 − x2 ) i = a
− 2
+
+
.
6 2π
n2
m2
π 4 (m2 − n2 )4
(Again, the last term is present only when m, n have opposite parity.)
CHAPTER 5. IDENTICAL PARTICLES
173
Problem 5.7
(a) From Equations 2.60 and 2.63:
ψ0 (x) =
mω 1/4
π~
e
2
− mω
2~ x
r
(i) ψd (x1 , x2 ) = ψ0 (x1 ) ψ1 (x2 ) =
(ii) and (iii)
2
�������
ψ1 (x) =
π~
mω
2
x e− 2~ x .
~
2
2
mω
2 mω
x1 e− 2~ (x1 +x2 ) .
π ~
1
ψ± (x1 , x2 ) = √ [ψ0 (x1 ) ψ1 (x2 ) ± ψ1 (x1 ) ψ0 (x2 )] =
2
2
�������
,
mω 1/4 r 2mω
r
2
2
mω
1 mω
(x1 ± x2 ) e− 2~ (x1 +x2 ) .
π ~
2
Plot3D 2 x2 ⅇ-x ⅇ-y , {x, - 2, 2}, {y, - 2, 2}
2
Plot3D 2 x2 ⅇ-x ⅇ-y , {x, - 2, 2}, {y, - 2, 2}
�������
�������
2
�������
Plot3D
�������
Plot3D
2
2
2
2
2
2
x2 + y2 + 2 x y ⅇ-x ⅇ-y , {x, - 2,
2}, {y, -x2,
2 2}2
+ y - 2 x y ⅇ-x ⅇ-y , {x, - 2, 2}, {y, - 2, 2}
������� Plot3D
x2 + y2 + 2 x y ⅇ-x ⅇ-y , {x, - 2, 2}, {y, - 2, 2}
�������
�������
�������
(b) From Problem 2.11:
hxi0 = 0,
hxi1 = 0,
hx2 i0 =
~
,
2mω
hx2 i1 =
3~
.
2mω
From Equation (5.24):
r
Z
hxi01 =
x ψ0 (x)ψ1 (x) dx =
mω
π~
r
2mω
~
Z ∞
2
2 − mω
~ x
x e
−∞
r
dx =
2 mω √
4 π
π
~
1
2
r
~
mω
!3
r
=
~
.
2mω
174
CHAPTER 5. IDENTICAL PARTICLES
(i) h(x1 − x2 )2 id = hx2 i0 + hx2 i1 − 2hxi0 hxi1 =
(ii) and (iii)
h(x1 − x2 )2 i± =
2~
.
mω
2~
~
~
3~
∓2
=
(bosons),
(fermions).
mω
2mω
mω
mω
(c)
r
x1 = R + ,
2
r
ψd (R, r) =
r
x2 = R − ,
2
r − mω (R2 + r2 )
2 mω 4 .
R+
e ~
π ~
2
r
ψ− (R, r) =
x21 + x22 = 2R2 +
r
ψ+ (R, r) =
r2
.
2
2
mω
r2
1 2mω
R e− ~ (R + 4 ) .
π ~
1 mω − mω (R2 + r2 )
4 .
re ~
π ~
Z
2mω 2
4 mω 2 − mω r2 ∞
r2
2
2~
Pd =
e
e− ~ R dR
R + Rr +
π
~
4
−∞
Z ∞
Z
4 mω 2 − mω r2
r2 ∞ − 2mω R2
R2
2 − 2mω
2~
~
~
=
e
2
dR + 0 +
e
dR
R e
π
~
4 0
 0

!3
!
r
r
1
~
1
~
8 mω 2 − mω r2  √
r2 √

e 2~
2 π
π
=
+
π
~
2 2mω
4
2 2mω
mω 2
1 mω 3/2 2
~
= √
r +
e− 2~ r .
~
mω
2π
2
Z ∞
mω 2
2mω 2
2 2mω
P+ =
e− 2~ r
R2 e− ~ R dR
π
~
−∞

!3 
r
r
1
~
2mω − mω r2
8 mω 2 − mω r2  √

2~
e
2 2 π
=
e 2~ .
=
π
~
2 2mω
π~
Z
2 mω 2 2 − mω r2 ∞ − 2mω R2
P− =
r e 2~
e ~
dR
π
~
−∞
"
!#
r
r 2 mω 2 2 − mω r2 √
1
~
2 mω 3/2 2 − mω r2
=
r e 2~ 2
π
=
r e 2~ .
π
~
2 2mω
π
~
As we see from the graphs below, the bosons tend to be closer together, and the fermions farther apart.
CHAPTER 5. IDENTICAL PARTICLES
�������
Plot
1
x2 + 1 ⅇ-x
2π
2
2
175
, {x, 0, 5}
0.5
0.4
0.3
�������
0.2
0.1
1
�������
Plot
2
ⅇ-x
2
2
2
3
4
5
3
4
5
3
4
5
, {x, 0, 5}
π
0.8
0.6
������� 0.4
0.2
1
�������
Plot
2
x2 ⅇ-x
2
2
2
, {x, 0, 5}
π
0.6
0.5
0.4
������� 0.3
0.2
0.1
1
2
(d)
Z Z
hn(x)i =
|ψ(x1 , x2 )|2 [δ(x − x1 ) + δ(x − x2 )] dx1 dx2 =
Z
|ψ(x, x2 )|2 dx2 +
Z
Z
2 mω 2
2 − mω
(x2 +x22 )
2 − mω
(x21 +x2 )
~
~
hn(x)id =
x e
dx2 + x1 e
dx1
π
~
Z
Z
∞
2 mω 2 − mω x2 2 ∞ − mω x22
2 − mω
x21
~
~
~
=
e
x
e
dx2 +
x1 e
dx1
π
~
−∞
−∞

!
!3 
r
r
2
√
√
mω 2
2 mω
1
~
1
~

=
e− ~ x 2 x2 π
+2 π
π
~
2 mω
2 mω
mω 2
2 mω 3/2 2
~
= √
x +
e− ~ x .
~
2mω
π
Z
|ψ(x1 , x)|2 dx1 .
176
CHAPTER 5. IDENTICAL PARTICLES
Z
Z
1 mω 2
(x2 +x22 )
2 − mω
(x21 +x2 )
2 − mω
~
~
hn(x)i± =
dx2 + (x1 ± x) e
dx1
(x ± x2 ) e
π
~
Z
mω 2
2 mω 2 − mω x2 ∞ 2
e ~
(x ± 2x x2 + x22 )e− ~ x2 dx2
=
π
~
−∞
Z ∞
Z ∞
2 mω 2 − mω x2
− mω
2
x22
2 − mω
x22
~
~
~
e
e
x
dx2 ± 0 +
x2 e
dx2
=
π
~
−∞
−∞
mω 2
~
2 mω 3/2 2
x +
= √
e− ~ x .
~
2mω
π
Remarkably, the density is exactly the same for all three cases! Identical bosons tend to be closer together, and
fermions farther apart, but this is not reflected in the expectation value of the number density.
��������
2
Plot
x2 +
π
1
2
2
ⅇ-x , {x, - 4, 4}
0.7
0.6
0.5
0.4
��������
0.3
0.2
0.1
-4
2
-∞
π
x2 +
��������
��������
2
-2
∞
1
2
4
2
ⅇ-x ⅆ x
2
(Just as a check, I integrated the density over all x: the result is 2, of course, because there are two particles
here.)
Problem 5.8
(a) ψ(x1 , x2 , x3 ) = ψa (x1 )ψb (x2 )ψc (x3 ).
(b) ψ(x1 , x2 , x3 ) =
√1 [ψa (x1 )ψb (x2 )ψc (x3 ) + ψa (x1 )ψc (x2 )ψb (x3 ) + ψb (x1 )ψa (x2 )ψc (x3 )
6
.
(c) ψ(x1 , x2 , x3 ) =
√1 [ψa (x1 )ψb (x2 )ψc (x3 ) − ψa (x1 )ψc (x2 )ψb (x3 ) − ψb (x1 )ψa (x2 )ψc (x3 )
6
.
+ψb (x1 )ψc (x2 )ψa (x3 ) + ψc (x1 )ψb (x2 )ψa (x3 ) + ψc (x1 )ψa (x2 )ψb (x3 )]
+ψb (x1 )ψc (x2 )ψa (x3 ) − ψc (x1 )ψb (x2 )ψa (x3 ) + ψc (x1 )ψa (x2 )ψb (x3 )]
CHAPTER 5. IDENTICAL PARTICLES
177
Problem 5.9
(a) Distinguishable:
ground state: ψ11 |s mi, where |s mi could be |00i, |11i, |10i, or |1 − 1i. E = 2K, d = 4.
1st excited: ψ12 |s mi or ψ21 |s mi, where |s mi could be any of the same four spin states. E = 5K, d = 8.
2nd excited: ψ22 |s mi, where |s mi could be |00i, |11i, |10i, or |1 − 1i. E = 8K, d = 4.
3rd excited: ψ13 |s mi or ψ31 |s mi, where |s mi could be any of the same four spin states. E = 10K, d = 8.
Identical:
ground state: ψ11 |00i. E = 2K, d = 1.
1st excited: √12 (ψ12 +ψ21 )|00i or √12 (ψ12 −ψ21 )|s mi, where |s mi could be |11i, |10i, or |1−1i. E = 5K, d = 4.
2nd excited: ψ22 |00i. E = 8K, d = 1.
3rd excited: √12 (ψ13 +ψ31 )|00i or √12 (ψ13 −ψ31 )|s mi, where |s mi could be |11i, |10i, or |1−1i. E = 10K, d = 4.
(b) Two particles of spin 1 can combine to make spin 0, spin 1, or spin 2. From the 1 × 1 C-G table:
n
singlet: |00i = √13 |111 − 1i − √13 |1100i + √13 |11 − 11i, (symmetric).

√1
√1
(antisymmetric),

|11i = 2 |1110i − 2 |1101i,
1
1
√
√
(antisymmetric),
triplet: |10i = 2 |111 − 1i − 2 |11 − 11i,


|1 − 1i = √12 |110 − 1i − √12 |11 − 10i, (antisymmetric).


|22i = |1111i,
(symmetric),



1
1


|1101i,
(symmetric),
|21i = √2 |1110i + √2 q

1
2
1
quintuplet: |20i = √ |111 − 1i + 3 |1100i + √ |11 − 11i, (symmetric),
6
6




√1 |110 − 1i + √1 |11 − 10i,
|2
−
1i
=
(symmetric),

2
2


|2 − 2i = |11 − 1 − 1i,
(symmetric).
Distinguishable:
ground state: ψ11 |s mi, where |s mi could be any of the nine spin states. E = 2K, d = 9.
1st excited: ψ12 |s mi or ψ21 |s mi, where |s mi could be any of the nine spin states. E = 5K, d = 18.
2nd excited: ψ22 |s mi, where |s mi could be any of the nine spin states. E = 8K, d = 9.
3rd excited: ψ13 |s mi or ψ31 |s mi, where |s mi could be any of the nine spin states. E = 10K, d = 18.
Identical:
ground state: ψ11 |smi, where |smi is any of the singlet or quintuplet spin states. E = 2K, d = 6.
1st excited: √12 (ψ12 + ψ21 )|smi, where |smi is any of the singlet or quintuplet spin states,
or √12 (ψ12 − ψ21 )|s mi, where |s mi is any of the triplet spin states. E = 5K, d = 9.
2nd excited: ψ22 |smi, where |smi is any of the singlet or quintuplet spin states. E = 8K, d = 6.
3rd excited: √12 (ψ13 + ψ31 )|smi, where |smi is any of the singlet or quintuplet spin states,
or √12 (ψ13 − ψ31 )|s mi, where |s mi is any of the triplet spin states. E = 10K, d = 9.
178
CHAPTER 5. IDENTICAL PARTICLES
Problem 5.10
(a) χ(2, 1, 3) = a| ↑↑↑i + b| ↑↑↓i + c| ↓↑↑i + d| ↓↑↓i + e| ↑↓↑i + f | ↑↓↓i + g| ↓↓↑i + h| ↓↓↓i.
χ(2, 1, 3) = −χ(1, 2, 3) ⇒ a = b = g = h = 0, e = −c, f = −d.
(Because the individual states are orthogonal, the associated coefficients must be equal.)
χ(1, 3, 2) = a| ↑↑↑i + b| ↑↓↑i + c| ↑↑↓i + d| ↑↓↓i + e| ↓↑↑i + f | ↓↓↑i + g| ↓↑↓i + h| ↓↓↓i.
χ(1, 3, 2) = −χ(1, 2, 3) ⇒ a = d = e = h = 0, c = −b, g = −f.
So all 8 coefficients are zero, and χ(1, 2, 3) = 0.
(b) You can’t put all 3 in ψ1 because this would be a symmetric combination, and we just showed that there is
no antisymmetric spin state to go with it.
ψ1 (x1 )| ↑i1
ψ1 (x2 )| ↑i2
ψ1 (x3 )| ↑i3
ψ1 (x1 )| ↓i1
ψ1 (x2 )| ↓i2
ψ1 (x3 )| ↓i3
ψ2 (x1 )| ↑i1
ψ2 (x2 )| ↑i2
ψ2 (x3 )| ↑i3
= ψ1 (x1 )ψ1 (x2 )ψ2 (x3 )| ↑↓↑i + ψ1 (x1 )ψ2 (x2 )ψ1 (x3 )| ↓↑↑i + ψ2 (x1 )ψ1 (x2 )ψ1 (x3 )| ↑↑↓i
− ψ2 (x1 )ψ1 (x2 )ψ1 (x3 )| ↑↓↑i − ψ1 (x1 )ψ2 (x2 )ψ1 (x3 )| ↑↑↓i − ψ1 (x1 )ψ1 (x2 )ψ2 (x3 )| ↓↑↑i.
π 2 ~2 2 2 2
3π 2 ~2
(1
+1
+2
)
=
. Degeneracy is d = 2, (we could have started with ψ2 (x1 )| ↓i1
2ma2
ma2
instead of ψ2 (x1 )| ↑i1 ).
√
(c) For normalization, divide by 6 (since the terms are orthonormal):
The energy is E =
1 Φ(1, 2, 3) = √ ψ1 (x1 )ψ1 (x2 )ψ2 (x3 )(| ↑↓↑i − | ↓↑↑i) + ψ1 (x1 )ψ1 (x3 )ψ2 (x2 )(| ↓↑↑i − | ↑↑↓i)
6
+ ψ1 (x2 )ψ1 (x3 )ψ2 (x1 )(| ↑↑↓i − | ↑↓↑i)
1 h
= √ ψ1 (x1 )ψ1 (x2 )(| ↑i1 | ↓i2 − | ↓i1 | ↑i2 ) ψ2 (x3 )| ↑i3 + ψ1 (x1 )ψ1 (x3 )(| ↓i1 | ↑i3 − | ↑i1 | ↓i3 ) ψ2 (x2 )| ↑i2
6
i
+ ψ1 (x2 )ψ1 (x3 )(| ↑i2 | ↓i3 − | ↓i2 | ↑i3 ) ψ2 (x1 )| ↑i1
i
1 h
= √ Φ(1, 2) φ(3) − Φ(1, 3) φ(2) + Φ(2, 3) φ(1) . X
3
i
1 h
Φ(2, 1, 3) = √ Φ(2, 1) φ(3) − Φ(2, 3) φ(1) + Φ(1, 3) φ(2)
3
i
1 h
= √ − Φ(1, 2) φ(3) − Φ(2, 3) φ(1) + Φ(1, 3) φ(2) = −Φ(1, 2, 3),
3
X
and the same goes for the other two permutations.
Problem 5.11
(a) The exponential factor is obviously symmetric; the product term is
q
[(z1 − z2 )(z1 − z3 ) · · · (z1 − zN )(z2 − z3 ) · · · (z2 − zN ) · · · (zN −1 − zN )] .
Under 1 ↔ 2 the first factor switches sign, while the others simply change places: (z1 − z3 ) ↔ (z2 − z3 ), etc.,
so the term in square brackets picks up a minus sign. Since q is an odd positive integer, the whole thing (and
hence also ψ itself) is antisymmetric. X
CHAPTER 5. IDENTICAL PARTICLES
179
2
1
2
2
(b) ψ(z1 , z2 , z3 ) = A(z1 − z2 )(z1 − z3 )(z2 − z3 )e− 2 (|z1 | +|z2 | +|z3 | ) . Now
z12 z1 1
z22 z2 1 ,
z32 z3 1
(z1 − z2 )(z1 − z3 )(z2 − z3 ) = z12 z2 + z22 z3 + z32 z1 − z12 z3 − z22 z1 − z32 z2 =
so
2
1
1
z12 e− 2 |z1 |
2
1
ψ(z1 , z2 , z3 ) = A z22 e− 2 |z2 |
2
1
z32 e− 2 |z3 |
2
1
z1 e− 2 |z1 |
2
1
z2 e− 2 |z2 |
2
1
z3 e− 2 |z3 |
2
e− 2 |z1 |
2
1
e− 2 |z2 | . X
2
1
e− 2 |z3 |
The occupied single-particle states are
2
1
1
ψa (z) = Aa e− 2 |z| ,
2
ψb (z) = Ab ze− 2 |z| ,
1
2
ψc (z) = Ac z 2 e− 2 |z| ,
for appropriate normalization constants Aa , Ab , Ac .
2
1
2
(c) ψ(z1 , z2 ) = A(z1 − z2 )3 e− 2 (|z1 | +|z2 | ) .
(z1 − z2 )3 = z13 − 3z12 z2 + 3z1 z22 − z23 =
so
(
ψ(z1 , z2 ) = A
1
2
z13 e− 2 |z1 |
2
1
z23 e− 2 |z2 |
1
2
z z2
z13 1
+ 3 1 12 ,
3
z2 z2
z2 1
1
2
z e− 2 |z1 |
e− 2 |z1 |
+ 3 1 − 1 (z2 |2
− 12 |z2 |2
z2 e 2
e
1
2
z12 e− 2 |z1 |
2
1
z22 e− 2 |z2 |
)
. X
Problem 5.12
(a)
ψ± = A ψ(r1 , r2 , r3 , . . . , rZ ) ± ψ(r2 , r1 , r3 , . . . , rZ ) + ψ(r2 , r3 , r1 , . . . , rZ ) + etc. ,
where “etc.” runs over all permutations of the arguments r1 , r2 , . . . , rZ , with a + sign for all even permutations
(even number of transpositions ri ↔ rj , starting from r1 , r2 , . . . , rZ ), and ± for all odd permutations (+ for
symmetric,
√ − for antisymmetric). At the end of the process, normalize the result to determine A. (Typically
A = 1/ Z!, but this may not be right if the starting function is already symmetric under some interchanges.)
If ψ is symmetric in the first two arguments (or any other pair), the antisymmetric combination is zero. For
example, if Z = 3,
ψ− = ψ(r1 , r2 , r3 ) − ψ(r2 , r1 , r3 ) + ψ(r2 , r3 , r1 ) − ψ(r3 , r2 , r1 ) + ψ(r3 , r1 , r2 ) − ψ(r1 , r3 , r2 ) = 0
(they cancel in pairs).
(b) Since there are only two possible one-particle states, | ↑i and | ↓i, a configuration of three or more
electrons will have to put at least two of them in the same state, which will then be symmetric under their
interchange, and from that no completely antisymmetric combination can be constructed.
Problem 5.13
(a) The energy of each electron is E = Z 2 E1 /n2 = 4E1 /4 = E1 = −13.6eV, so the total initial energy is
2 × (−13.6) eV= −27.2 eV. One electron drops to the ground state Z 2 E1 /1 = 4E1 , so the other is left
with 2E1 − 4E1 = −2E1 = 27.2 eV.
180
CHAPTER 5. IDENTICAL PARTICLES
(b) He+ has one electron; it’s a hydrogenic ion (Problem 4.19) with Z = 2, so the spectrum is
1/λ = 4R 1/n2f − 1/n2i , where R is the hydrogen Rydberg constant, and ni , nf are the initial and final
quantum numbers (1, 2, 3, . . . ).
Problem 5.14
(a) The ground state (Eq. 5.41) is spatially symmetric, so it goes with the symmetric (triplet) spin configuration. Thus the ground state is orthohelium, and it is triply degerate. The excited states (Eq. 5.43) come
in ortho (triplet) and para (singlet) form; since the former go with the symmetric spatial wave function,
the orthohelium states are higher in energy than the corresponding (nondegenerate) para states.
(b) The ground state (Eq. 5.41) and all excited states (Eq. 5.43) come in both ortho and para form. All are
quadruply degenerate (or at any rate we have no way a priori of knowing whether ortho or para are higher
in energy, since we don’t know which goes with the symmetric spatial configuration).
Problem 5.15
(a)
1
|r1 − r2 |
=
8
πa3
2 Z "Z
p
|
Z ∞
= 2π
e
−4(r1 +r2 )/a
"Z
0
1
r1 r2
r12 + r22 − 2r1 r2 cos θ2
{z
π
0
F
1
[(r1 + r2 ) − |r1 − r2 |] =
=
r1 r2
1
r1
Z r1
0
1
r1
Z r1
}
p
dθ2 r22 dr2
r12 + r22 − 2r1 r2 cos θ2
{z
}
0
d r2 d3 r1
r22 e−4r2 /a dr2 =
1
r1 r2
q
r12 + r22 + 2r1 r2 −
q
r12 + r22 − 2r1 r2
2/r1 (r2 < r1 )
2/r2 (r2 > r1 )
r22 e−4r2 /a dr2 +
0
#
sin θ2
q
π
r12 + r22 − 2r1 r2 cos θ2 =
= 4πe−4r1 /a
#
3
|
F=
e−4(r1 +r2 )/a
Z ∞
r2 e−4r2 /a dr2 .
r1
r1
1
a 2 −4r2 /a a a 2 −4r2 /a
4r2
− r2 e
+
e
−
−1
r1
4
2 4
a
0
CHAPTER 5. IDENTICAL PARTICLES
181
a
ar1 −4r1 /a a2 −4r1 /a a2
2 −4r1 /a
=−
r e
+
e
+ e
−
.
4r1 1
2
8
8
Z ∞
r2 e−4r2 /a dr2 =
a 2
4
r1
= 4π
=
=
32
= 4
a
∞
4r2
ar1 −4r1 /a a2 −4r1 /a
−
=
−1
e
+ e
.
a
4
16
r1
a2
a3
ar1
a2 −8r1 /a
a3 −4r1 /a
ar1
e
−
−
+
+
e
+ −
32r1
4
8
32r1
4
16
πa2
8
1
|r1 − r2 |
32
a4
e−4r2 /a
a −4r1 /a
a
e
− 2+
e−8r1 /a .
r1
r1
=
8
· 4π
πa4
Z ∞
0
a −4r1 /a
a
e
− 2+
e−8r1 /a r12 dr1
r1
r1
Z ∞
Z ∞
Z ∞
a
r1 e−4r1 /a dr1 − 2
r12 e−8r1 /a dr1 − a
r1 e−8r1 /a dr1
0
a·
0
a 2
4
−2·2
a 3
8
−a·
0
a 2 8
32
=
a
1
1
1
−
−
16 128 64
=
5
.
4a
(b)
Vee ≈
e2
4π0
1
|r1 − r2 |
=
5m
5 e2 1
=
4 4π0 a
4 ~2
e2
4π0
2
=
5
5
(−E1 ) = (13.6 eV) = 34 eV.
2
2
E0 + Vee = (−109 + 34)eV = −75 eV, which is pretty close to the experimental value (−79 eV).
Problem 5.16
ψ0 (r1 , r2 , r3 ) = ψ100 (r1 )ψ100 (r2 )ψ200 (r3 ), where ψn`m is the hydrogenic wave function with “Bohr radius”
ā = a/Z = a/3. The energy (expressed in terms of Bohr levels) is E = Z 2 (E1 + E1 + E2 ) = 9(2E1 + E41 ) =
81
E1 = −275 eV. In the notation of Problem 5.10:
4
ψ1 (r) = ψ100 (r) = √
1
πā3
e−r/ā ;
1
r −r/2ā 1
√ ,
ψ2 (r) = ψ200 (r) = R20 Y00 = √ ā−3/2 1 −
e
2ā
2
4π
182
CHAPTER 5. IDENTICAL PARTICLES
27 −3(ri +rj )/a
e
|00iij
πa3
(where |00iij denotes the singlet configuration of spins i and j), and
Φ(i, j) ≡ ψ1 (ri )ψ1 (rj )|00iij =
φ(j) ≡ ψ2 (rj )| 12 12 ij =
1
√
2 2π
3/2 3
3rj
1−
e−3rj /2a | 12 12 ij .
a
2a
The complete ground state, then, is (quoting the result of Problem 5.10(c)):
i
1 h
Φ(1, 2, 3) = √ Φ(1, 2) φ(3) − Φ(1, 3) φ(2) + Φ(2, 3) φ(1) .
3
It is doubly degenerate, since for φ(j) we could have used | 21 − 12 i.
Problem 5.17
(a) Hydrogen: (1s); helium: (1s)2 ; lithium: (1s)2 (2s); beryllium: (1s)2 (2s)2 ;
boron: (1s)2 (2s)2 (2p); carbon: (1s)2 (2s)2 (2p)2 ; nitrogen: (1s)2 (2s)2 (2p)3 ;
oxygen: (1s)2 (2s)2 (2p)4 ; fluorine: (1s)2 (2s)2 (2p)5 ; neon: (1s)2 (2s)2 (2p)6 .
These values agree with those in Table 5.1—no surprises so far.
(b) Hydrogen: 2 S1/2 ;
helium: 1 S0 ;
lithium: 2 S1/2 ;
beryllium 1 S0 . (These four are unambiguous,
because the orbital angular momentum is zero in all cases.) For boron, the spin (1/2) and orbital (1)
angular momenta could add to give 3/2 or 1/2, so the possibilities are 2 P3/2 or 2 P1/2 . For carbon, the
two p electrons could combine for orbital angular momentum 2, 1, or 0, and the spins could add to 1 or 0:
1
S0 , 3 S1 , 1 P1 , 3 P2 , 3 P1 , 3 P0 , 1 D2 , 3 D3 , 3 D2 , 3 D1 . For nitrogen, the 3 p electrons can add to orbital angular
momentum 3, 2, 1, or 0, and the spins to 3/2 or 1/2:
2
4
S1/2 , 4 S3/2 , 2 P1/2 , 2 P3/2 , 4 P1/2 , 4 P3/2 , 4 P5/2 , 2 D3/2 , 2 D5/2 ,
D1/2 , 4 D3/2 , 4 D5/2 , 4 D7/2 , 2 F5/2 , 2 F7/2 , 4 F3/2 , 4 F5/2 , 4 F7/2 , 4 F9/2 .
Problem 5.18
(a) Orthohelium should have lower energy than parahelium, for corresponding states (which is true).
(b) Hund’s first rule says S = 1 for the ground state of carbon. But this (the triplet) is symmetric, so the
orbital state will have to be antisymmetric. Hund’s second rule favors L = 2, but this is symmetric, as
you can see most easily by going to the “top of the ladder”: |2 2i = |1 1i1 |1 1i2 . So the ground state of
carbon will be S = 1, L = 1. This leaves three possibilities: 3 P2 , 3 P1 , and 3 P0 .
(c) For boron there is only one electron in the 2p subshell (which can accommodate a total of 6), so Hund’s
third rule says the ground state will have J = |L − S|. We found in Problem 5.17(b) that L = 1 and
S = 1/2, so J = 1/2, and the configuration is 2 P1/2 .
(d) For carbon we know that S = 1 and L = 1, and there are only two electrons in the outer subshell, so
Hund’s third rule says J = 0, and the ground state configuration must be 3 P0 .
For nitrogen Hund’s first rule says S = 3/2, which is symmetric (the top of the ladder is | 32 32 i =
Hund’s second rule favors L = 3, but this is also symmetric. In fact, the only
| 21 12 i1 | 12 12 i2 | 12 12 i3 ).
CHAPTER 5. IDENTICAL PARTICLES
183
antisymmetric orbital configuration here is L = 0. [You can check this directly by working out the
Clebsch-Gordan coefficients, but it’s easier to reason as follows: Suppose the three outer electrons are in
the “top of the ladder” spin state, so each one has spin up (| 12 12 i); then (since the spin states are all the
same) the orbital states have to be different: |1 1i, |1 0i, and |1−1i. In particular, the total z-component of
orbital angular momentum has to be zero. But the only configuration that restricts Lz to zero is L = 0.]
The outer subshell is exactly half filled (three electrons with n = 2, l = 1), so Hund’s third rule says
J = |L − S| = |0 − 23 | = 3/2. Conclusion: The ground state of nitrogen is
4
S3/2 . (Table 5.1 confirms
this.)
Problem 5.19
S = 2; L = 6; J = 8. (1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (3d)10 (4s)2 (4p)6 (4d)10 (5s)2 (5p)6 (4f )10 (6s)2 .
|
{z
}|
{z
}
definite (36 electrons)
likely (30 electrons)
Problem 5.20
Divide Eq. 5.56 by Eq. 5.54, using Eq. 5.53:
Etot /N q
3
~2 (3π 2 N q)5/3 1
2m
= .
=
EF
5
10π 2 mV 2/3 N q ~2 (3π 2 N q/V )2/3
Problem 5.21
~2
Nq
N
atoms
moles
gm
NA
(3ρπ 2 )2/3 . ρ =
=
=
×
×
=
· d, where NA is Avogadro’s
2m
V
V
mole
gm
volume
M
23
number (6.02 × 10 ), M = atomic mass = 63.5 gm/mol, d = density = 8.96 gm/cm3 .
(a) EF =
ρ=
(6.02 × 1023 )(8.96 gm/cm3 )
= 8.49 × 1022 /cm3 = 8.49 × 1028 /m3 .
(63.5 gm)
EF =
(1.055 × 10−34 J · s)(6.58 × 10−16 eV · s)
(3π 2 8.49 × 1028 /m3 )2/3 = 7.04 eV.
(2)(9.109 × 10−31 kg)
(b)
7.04 eV =
1
v2
14.08
v
(0.511 × 106 eV/c2 )v 2 ⇒ 2 =
= 2.76 × 10−5 ⇒ = 5.25 × 10−3 ,
6
2
c
.511 × 10
c
so it’s nonrelativistic.
v = (5.25 × 10−3 ) × (3 × 108 ) = 1.57 ×106 m/s.
(c)
T =
7.04 eV
= 8.17 × 104 K.
8.62 × 10−5 eV/K
P =
(3π 2 )2/3 ~2 5/3
(3π 2 )2/3 (1.055 × 10−34 )2
2
ρ
=
(8.49 × 1028 )5/3 N/m = 3.84 × 1010 N/m2 .
5m
5(9.109 × 10−31 )
(d)
184
CHAPTER 5. IDENTICAL PARTICLES
Problem 5.22
EF
~2
ρm
mass
=
(3ρπ 2 )2/3 (Equation 5.54). The number density ρ = atoms
mass × volume = m , where ρm is the
kB
2mkB
mass density and m is the atomic mass—3 times the mass of the proton: m = 3mp (well, OK, two protons and
a neutron, and two electrons, and some binding energy . . . but 3 protons is close enough). So
TF =
TF =
~2
6mp kB
π 2 ρm
mp
2/3
=
(1.055 × 10−34 )2
6(1.673 × 10−27 )(1.381 × 10−23 )
π 2 (82)
1.673 × 10−27
2/3
5
5
AV −5/3 = P.
3
3
= 4.95 K.
Problem 5.23
P =
(3π 2 )2/3 ~2
5m
Nq
V
5/3
= AV −5/3 ⇒ B = −V
dP
= −V A
dV
−5
3
V −5/3−1 =
For copper, B = 53 (3.84 × 1010 N/m2 ) = 6.4 × 1010 N/m2 .
Problem 5.24
(a) Equations 5.66 and 5.70 ⇒ ψ = A sin kx + B cos kx;
A sin ka = eiqa − cos ka B.
So
iqa
A sin ka
A
cos kx = iqa
e sin kx − sin kx cos ka + cos kx sin ka
(eiqa − cos ka)
(e − cos ka)
Aeiqa
.
= C sin kx + e−iqa sin[k(a − x)] , where C ≡ iqa
e − cos ka
ψ = A sin kx +
(b) If z = ka = jπ, then sin ka = 0, Eq. 5.71 ⇒ cos qa = cos ka = (−1)j ⇒ sin qa = 0, so eiqa = cos qa +
i sin qa = (−1)j , and the constant C involves division by zero. In this case we must go back to Eq. 5.70,
which is a tautology (0=0) yielding no constraint on A or B, Eq. 5.68 holds automatically, and Eq. 5.69
gives
2mα
kA − (−1)j k A(−1)j − 0 = 2 B ⇒ B = 0. So ψ = A sin kx.
~
Here ψ is zero at each delta spike, so the wave function never “feels” the potential at all.
Problem 5.25
We’re looking for a solution to Eq. 5.73 with β = 10 and z . π: f (z) = cos z + 10
Mathematica gives z = 2.62768.
So
E=
sin z
= 1.
z
~2 k 2
~2 z 2
z2 α
(2.62768)2
=
=
=
eV = 0.345 eV.
2
2m
2ma
2β a
20
CHAPTER 5. IDENTICAL PARTICLES
185
Problem 5.26
Positive-energy solutions. These are the same as before, except that α (and hence also β) is now a negative
number.
Negative-energy solutions. On 0 < x < a we have
d2 ψ
= κ2 ψ,
dx2
√
where
κ≡
−2mE
~
⇒
ψ(x) = A sinh κx + B cosh κx.
According to Bloch’s theorem the solution on −a < x < 0 is
ψ(x) = e−iqa [A sinh κ(x + a) + B cosh κ(x + a)] .
Continuity at x = 0 ⇒
B = e−iqa [A sinh κa + B cosh κa] ,
or A sinh κa = B eiqa − cosh κa .
[F]
The discontinuity in ψ 0 (Eq. 2.128) ⇒
κA − e
−iqa
2mα
2mα
−iqa
−iqa
κ [A cosh κa + B sinh κa] = 2 B, or A 1 − e
cosh κa = B
+e
sinh κa .
~
~2 κ
[]
Plugging F into and cancelling B:
eiqa − cosh κa
2mα
1 − e−iqa cosh κa = 2 sinh κa + e−iqa sinh2 κa.
~ κ
2mα
sinh κa.
~2 κ
mα
cos qa = cosh κa + 2 sinh κa.
~ κ
eiqa − 2 cosh κa + e−iqa cosh2 κa − e−iqa sinh2 κa =
eiqa + e−iqa = 2 cosh κa +
2ma
sinh κa,
~2 κ
This is the analog to Eq. 5.71. As before, we let β ≡ mαa/~2 (but remember it’s now a negative number), and
this time we define z ≡ −κa, extending Eq. 5.72 to negative z, where it represents negative-energy solutions.
In this region we define
sinh z
f (z) = cosh z + β
.
[FF]
z
In the figure below I have plotted f (z) for β = −1.5, using Eq. 5.73 for positive z and FF for negative
z. As before, allowed energies are restricted to the range −1 ≤ f (z) ≤ 1, and occur at intersections of f (z)
with the N horizontal lines cos qa = cos(2πn/N a), with n = 0, 1, 2 . . . N − 1. Evidently the first band (partly
negative, and partly positive) contains N states, as do all the higher bands.
1
0
-1
0
/
2/
3/
4/
186
CHAPTER 5. IDENTICAL PARTICLES
Problem 5.27
2πn
n
Equation 5.63 says q =
⇒ qa = 2π ; on page 223 we found that n = 0, 1, 2, . . . , N − 1. Each value of
Na
N
n corresponds to a distinct state. To find the allowed energies we draw N horizontal lines on Figure 5.5, at
heights cos qa = cos(2πn/N ), and look for intersections with f (z). The point is that almost all of these lines
come in pairs—two different n’s yielding the same value of cos qa:
N = 1 ⇒ n = 0 ⇒ cos qa = 1. Nondegenerate.
N = 2 ⇒ n = 0, 1 ⇒ cos qa = 1, −1. Nondegenerate.
N = 3 ⇒ n = 0, 1, 2 ⇒ cos qa = 1, − 12 , − 12 . The first is nondegenerate, the other two are degenerate.
N = 4 ⇒ n = 0, 1, 2, 3 ⇒ cos qa = 1, 0, −1, 0. Two are nondegenerate, the others are degenerate.
Evidently they are doubly degenerate (two different n’s give same cos qa) except when cos qa = ±1, i.e., at
the top or bottom of a band. The Bloch factors eiqa lie at equal angles in the complex plane, starting with 1
(see figure below, drawn for the case N = 8); by symmetry, there is always one with negative imaginary part
symmetrically opposite each one with positive imaginary part; these two have the same real part (cos qa). Only
points which fall on the real axis have no twins.
sin(qa)
n=2
n=3
n=1
n=0
n=4
n=5
n=6
cos(qa)
n=7
Problem 5.28
�������
ContourPlot Cos[x] ⩵ Cos
2y
+ Sin
2y
2 y , {x, - π, π}, {y, 0, 50}
50
40
30
�������
20
10
0
-3
-2
-1
0
1
2
3
CHAPTER 5. IDENTICAL PARTICLES
187
Problem 5.29
(a) Each particle has 3 possible states: 3 × 3 × 3 = 27.
(b) All in same state: aaa, bbb, ccc ⇒ 3.
2 in one state: aab, aac, bba, bbc, cca, ccb ⇒ 6 (each symmetrized).
3 different states: abc (symmetrized) ⇒ 1.
Total: 10.
(c) Only abc (antisymmetrized) =⇒ 1.
Problem 5.30
!
n2y
~2 k 2
πnx πny
π 2 ~2 n2x
Equation 5.50 ⇒ Enx ny =
+ 2 =
, with k =
,
. Each state is represented by an
2m
lx2
ly
2m
lx
ly
intersection on a grid in “k-space”—this time a plane—and each state occupies an area π 2 /lx ly = π 2 /A (where
A ≡ lx ly is the area of the well). Two electrons per state means
Nq
1 2
πk =
4 F
2
π2
A
, or kF =
2π
Nq
A
1/2
= (2πσ)1/2 ,
where σ ≡ N q/A is the number of free electrons per unit area.
∴ EF =
~2
π~2 σ
~2 kF2
=
2πσ =
.
2m
2m
m
ky
kx
188
CHAPTER 5. IDENTICAL PARTICLES
Problem 5.31
(a) The energy of a particle in the ground state of an infinite cubical well of side a is (Equation5.50)
π 2 ~2 1
1
1
3π 2 ~2
3π 2 ~2
E111 =
+
+
=
. So the energy of N such particles is Ea = N
.
2
2
2
2
2m a
a
a
2ma
2ma2
(b) Now we put the N electrons in a single cubical box of volume V ; according to Equation 5.56 the total energy
is
~2 (3π 2 N )5/3 −2/3
V
. Here V is the total volume of N cubes, each of volume a3 , so V = N a3 , and hence
Eb =
10 π 2 m
2/3
π 2 ~2 3 3
~2 (3π 2 N )5/3
=
N
Eb =
.
2ma2 5 π
10π 2 m(N a3 )2/3
"
2/3 #
∆E
Ea − Eb
3π 2 ~2
1 3
(c)
=
=
1−
.
N
N
2ma2
5 π
"
2/3 #
π 2 ~2
∆E
9
1 3
(d) In Problem 2.58(d) we found that
= 1.6 eV, so
=
1−
× 1.6 eV = 5.8 eV.
3ma2
N
2
5 π
Problem 5.32
(a) Repeat the calculation of Section 5.3.1 separately for N+ and N− (setting N d → N± ), and then add the
energies.
before 5.52, with no 1/2 on the right side:
Startwith the
equation
1 4 3
N±
π3
πkF± = N±
⇒ kF± = (6ρ± π 2 )1/3 , where ρ± =
. So Equation 5.54 becomes
8 3
V
V
[ 1 πk 2 dk]
~2
V 2
EF± =
(6ρ± π 2 )2/3 . The number of states in the shell is 2 3
=
k dk, Equation 5.55 becomes
2m
(π /V )
2π 2
~2 k 2 V 2
dE =
k dk, and
2m 2π 2
~2 V
Etot± =
4mπ 2
Z kF
±
k 4 dk =
0
i
~2 V 5
~2 (6π 2 N± )5/3 −2/3
~2 (6π 2 )5/3 −2/3 h 5/3
5/3
k
=
V
;
E
=
V
N
+
N
.
tot
F
+
−
20π 2 m ±
20π 2 m
20π 2 m
If N+ = N− = (N d)/2, this reduces to Etot =
~2 (6π 2 )5/3 −2/3
V
2
20π 2 m
Nd
2
5/3
=
~2
(3π 2 N d)5/3 V −2/3 , in
10π 2 m
agreement with Equation 5.56.
1
N
∆N
(b) Let N ≡ (N+ + N− ), ∆N ≡ (N+ − N− ), so N± = (N ± ∆N ) = (1 ± ), where ≡
1. Expanding
2
2
N
in powers of :
5/3
5/3
N+ + N−
=
i N 5/3 (1 + )5/3 + (1 − )5/3 ≈
1 + 53 + 21
2
5/3 N
5
+ 1 − 53 + 21 53 23 2 = 2
1 + 2 .
2
9
N
2
5/3 h
5
3
2
3
2
Therefore
~2 (6π 2 )5/3 −5/3
Etot
=
V
2
V
20π 2 m
N
2
5/3 "
5
1+
9
∆N
N
2 #
~2
=
10π 2 m
3π 2 N
V
5/3 "
2 #
5 ∆N
1+
,
9
N
CHAPTER 5. IDENTICAL PARTICLES
189
N
MV
∆N
MV
M
or, using ρ =
and ∆N = −
⇒
=−
=−
,
V
µB
N
µB ρV
µB ρ
(Note that M is the z-component of M, not its magnitude.)
"
2 #
Etot
~2 (3π 2 ρ)5/3
5
M
=
1+
.
V
10π 2 m
9 ρµB
X
Problem 5.33
"
2 #
5
Etot
~2 (3π 2 ρ)5/3
M
1+
− BM . The
(a) The extra energy is µB B(N+ − N− ) = −BM V , so
=
V
10π 2 m
9 ρµB
minimum energy is given by
1 ∂Etot
~2 (3π 2 ρ)5/3
=
V ∂M
10π 2 m
(b) χ = µ0
3ρ
π
1/3 1/3 10π 2 m
9
3ρ
5
2M
mµ2B
2
−
B
=
0
⇒
M
=
B
B
.
(ρµ
)
=
B
9 (ρµB )2
π
π~2
~2 (3π 2 ρ)5/3 10
mµ2B
π~2
. The numbers are
µ0 = 4π × 10−7 N/A2 ,
ρ = 18.1 × 1028 m−3 ,
m = 9.11 × 10−31 kg,
~ = 1.055 × 10−34 J s,
µB = 5.788 × 10−5 eV/T = 9.272 × 10−24 J/T.
Putting them into the formula, χ = 15.7 × 10−6 . This is not wildly far from the experimental value (16, as
compared to 22).
Problem 5.34
2
2
(a) In the notation of Problem 5.32: N 2 = (N+ + N− )2 = N+
+ 2N+ N− + N−
,
2
2
,
− 2N+ N− + N−
(∆N )2 = (N+ − N− )2 = N+
"
2 #
U 2
M
and hence
∆E =
ρ −
V 2.
4
µB
so
4N+ N− = N 2 − (∆N )2 = ρ2 V 2 −
MV
µB
2
(b) Quoting the result of Problem 5.32(b):
"
"
2 #
2 #
Etot
~2 (3π 2 ρ)5/3
5
M
UV
M
2
=
1+
+
ρ −
= A(1 + 59 x2 ) + B(1 − x2 ),
V
10π 2 m
9 ρµB
4
µB
where
A≡
Thus
~2 (3π 2 ρ)5/3
,
10π 2 m
x≡
M
,
ρ µB
B ≡ 14 U V ρ2 .
Etot
= (A + B) + ( 59 A − B)x2 ).
V
,
190
CHAPTER 5. IDENTICAL PARTICLES
If the coefficient of x2 is positive, then the minimum occurs x = 0, and M = 0. To get spontaneous magnetization
we need B > 95 A, which is to say,
2π~2
4 5 ~2 (3π 2 ρ)5/3
=
U>
V ρ2 9 10π 2 m
mV
π
3ρ
1/3
.
Problem 5.35
(a)
V =
4 3
πR ,
3
so E =
~2 (3π 2 N q)5/3
10π 2 m
4 3
πR
3
−2/3
=
2~2
15πmR2
9
πN q
4
5/3
.
(b) Imagine building up a sphere by layers. When it has reached mass m, and radius r, the work necessary
to bring in the next increment dm is: dW = −(Gm/r) dm. In terms of the mass density ρ, m = 34 πr3 ρ,
and dm = 4πr2 drρ, where dr is the resulting increase in radius. Thus:
dr
16π 2 2 4
4
=−
ρ Gr dr,
dW = −G πr3 ρ 4πr2 ρ
3
r
3
and the total energy of a sphere of radius R is therefore
Z R
16π 2 2
16π 2 ρ2 R5
NM
Egrav = −
ρ G
r4 dr = −
G. But ρ =
, so
3
3
15
4/3πR
0
Egrav = −
16π 2 R5 9N 2 M 2
3 N 2M 2
G
=
−
G
.
15
16π 2 R6
5
R
(c)
A
B
− ,
R2
R
Etot =
where A ≡
2~2
15πm
9
πN q
4
5/3
and
B≡
3
GN 2 M 2 .
5
5/3
B
2A
9
5
dEtot
2A
4~2
= − 3 + 2 = 0 ⇒ 2A = BR, so R =
=
πN q
.
dR
R
R
B
15πm 4
3GN 2 M 2
" # 2/3
5/3
9π
9π
N 5/3
~2
~2
q 5/3
4
5/3
R=
q
=
.
2
2
2
9π
4
N
GmM
4
GmM N 1/3
R=
9π
4
2/3
(1.055 × 10−34 J · s)2 (1/2)5/3
N −1/3
(6.673 × 10−11 Nm2 /kg2 )(9.109 × 10−31 kg)(1.674 × 10−27 kg)2
= (7.58 × 1025 m)N −1/3 .
(d) Mass of sun:
1.989 × 1030 kg,
so N =
1.989 × 1030
= 1.188 × 1057 ; N −1/3 = 9.44 × 10−20 .
1.674 × 10−27
R = (7.58 × 1025 )(9.44 × 10−20 ) m = 7.16 × 106 m (slightly larger than the earth).
CHAPTER 5. IDENTICAL PARTICLES
191
(e)
From Eq. 5.54:
EF =
~2
2m
3π 2
Nq
4/3πR3
2/3
=
~2
2mR2
9π
Nq
4
2/3
.
Numerically:
2/3
(1.055 × 10−34 J · s)2
9π
57 1
EF =
(1.188 × 10 )
= 3.102 × 10−14 J,
2(9.109 × 10−31 kg)(7.16 × 106 m)2 4
2
or, in electron volts:
EF =
3.102 × 10−14
eV = 1.94 × 105 eV.
1.602 × 10−19
Erest = mc2 = 5.11 × 105 eV, so the Fermi energy (which is the energy of the most energetic electrons) is
comparable to the rest energy, so they are getting fairly relativistic.
Problem 5.36
(a)
V 2
~cV
k dk ⇒ Etot = 2
2
π
π
dE = (~ck)
Z kF
k 3 dk =
0
~cV 4
k ;
4π 2 F
kF =
3π 2 N q
V
4/3
1/3
.
~c
(3π 2 N q)4/3 V −1/3 .
4π 2
So Etot =
(b)
~c
4
(3π 2 N q)4/3
V = πR3 ⇒ Edeg =
3
4π 2 R
4π
3
−1/3
~c
=
3πR
9
πN q
4
.
Adding in the gravitational energy, from Problem 5.35(b),
A B
Etot =
− ,
R R
~c
where A ≡
3π
9
πN q
4
4/3
and B ≡
3
GN 2 M 2 .
5
dEtot
(A − B)
=−
= 0 ⇒ A = B,
dR
R2
but there is no special value of R for which Etot is minimal. Critical value: A = B(Etot = 0) ⇒
4/3
~c 9
3
πN q
= GN 2 M 2 , or
3π 4
5
15 √
Nc =
5π
16
= 2.04 × 1057 .
~c
G
3/2
q2
15 √
=
5π
3
M
16
1.055 × 10−34 J · s × 2.998 × 108 m/s
6.673 × 10−11 N · m2 /kg2
3/2
(1/2)2
(1.674 × 10−27 kg)3
(About twice the value for the sun—Problem 5.35(d).)
(c) Same as Problem 5.35(c), with m → M and q → 1, so multiply old answer by (2)5/3 m/M :
R = 25/3
(9.109 × 10−31 )
(7.58 × 1025 m)N −1/3 = (1.31 × 1023 m)N −1/3 . Using N = 1.188 × 1057 ,
(1.674 × 10−27 )
192
CHAPTER 5. IDENTICAL PARTICLES
R = (1.31 × 1023 m)(9.44 × 10−20 ) = 12.4 km. To get EF , use Problem 5.35(e) with q = 1, the new R,
and the neutron mass in place of m:
2/3
EF = 2
7.16 × 106
1.24 × 104
2 9.11 × 10−31
1.67 × 10−27
(1.94 × 105 eV) = 5.60 × 107 eV = 56.0 MeV.
The rest energy of a neutron is 940 MeV, so a neutron star is reasonably nonrelativistic.
Problem 5.37
√
(a) From Equation 5.71, cos(qa) = cos(ka) ⇒ q = ±k = ±
2mE
;
~
E=
1
1 m
m~
1
√
Gfree (E) =
=
=
2
Na
π ~2 |q|
π~
π~ 2mE
(b) −a sin(qa) =
r
~2 q 2
dE
~2 q
⇒
=
.
2m
dq
m
m
. X
2E
√
dk
mα
a
dk
2m dE
1
−a sin(ka) + 2 − 2 sin(ka) + cos(ka)
. Equation (5.65):
= √
.
~
k
k
dq
dq
2~ E dq
m dE
mα 1
sin(ka) − cos(ka)
.
sin(qa) = sin(ka) + 2
~ k ka
~2 k dq
dE
sin(qa)
~2 k
=
.
mα
1
dq
sin(ka) + ~2 k ka sin(ka) − cos(ka) m
r
h
i2
mα
1 − cos(ka) + 2 sin(ka)
~ k
mα 2
2mα
sin2 (ka)
= 1 − cos2 (ka) − 2 cos(ka) sin(ka) − 2
~ k
~ k
r
mα 2
2mα
= sin(ka) 1 − 2 cot(ka) − 2
.
~ k
~ k
p
sin(qa) = 1 − cos2 (qa) =
r
1
1
1 m | sin(ka) + ~mα
2k
ka sin(ka) − cos(ka) |
q
G(E) =
Na
π ~2 k
mα 2
sin(ka) 1 − 2mα
~2 k cot(ka) − ~2 k
1
|1 + ~mα
m
2k
ka − cot(ka) |
q
=
.
π~2 k
2mα
mα 2
1 − ~2 k cot(ka) − ~2 k
(c) With m = ~ = a = 1, the expressions in (a) and (b) reduce to


√
1
1
1
1  1 + 2E − √2E cot( 2E) 
√ ;
√
q
α=0:
α=1:
.
√
1
π 2E
π 2E
1 − 2E
− √22E cot( 2E)
In the graph, the lower curve is for α = 0, the upper one (with the breaks) is for α = 1.
CHAPTER 5. IDENTICAL PARTICLES
1
Plot
π
2x
1
Plot
π
2x
193
, {x, 0, 50}, PlotRange → {0, .5}
1+
1
2x
-
1
2x
Cot
2x
1-
1
2x
-
2
2x
Cot
2x
,
{x, 0, 50}, PlotRange → {0, .5}
Show[%3, %4]
0.5
0.4
0.3
0.2
0.1
0
10
20
30
40
50
Problem 5.38
(a) Consider
N
N
1 X i 2 π j r/N
1 X i 2 π r/N j
1 e2 i π r − 1
e
=
e
=
N j=1
N j=1
N 1 − e−2 i π r/N
For any integer r the numerator vanishes and the sum must be zero unless the denominator also vanishes. The
denominator vanishes for r = m N where m is any integer; when r = m N it is easy to see that the expression
evaluates to 1. Now, to prove the first identity, k − k 0 = m N only when m = 0 and hence k = k 0 . In the second
case, k + k 0 = m N only when m = 1 and k 0 = N − k.
(b) We first consider


s 0 
r
ωk
ωk
∂
∂
1 X i 2 π(j 0 k0 −j k)/N 1 
e
−
xj ,
+
, xj 0
[ak− , ak0 + ] =

N
2 
ωk 0
∂xj 0
ωk ∂xj
0
j, j
where I’ve dropped the two terms involving commutators of two coordinates or two derivatives. The remaining
commutators are −δjj 0 and δjj 0 respectively and we have


r
r
X
0
1
ωk
ωk 0
i 2 π j (k −k)/N  1

[ak− , ak0 + ] =
e
+
.
N j
2
ωk 0
ωk
194
CHAPTER 5. IDENTICAL PARTICLES
We can now do the sum on j using the result from part (a), and that gives the first commutator. Next we take


s 0 
r
ωk
1 X −i 2 π(j 0 k0 +j k)/N 1  ωk
∂
∂
[ak− , ak0 − ] =
xj ,
+
e
, xj 0

N
2  ωk 0
∂xj 0
ωk ∂xj
0
j, j
Evaluating the two commutators we have
 

s 
r
X
0
ωk0 
1
ωk
1
e−i 2 π j (k +k)/N  −
[ak− , ak0 − ] = 
+
N j
2
ωk 0
ωk
Doing the sum on j now gives the result that k 0 = N − k (or k 0 = k if k = N ) but since ωN −k = ωk this
vanishes, proving the second result. The final commutator works out the same way.
(c)
(
"r
#
r
N
mωk
~
∂
1 X −i2πjk/N
xj +
(ak− + aN −k+ ) = √
e
2~
2mωk ∂xj
N j=1
s
"r
#)
mωN −k
~
∂
i2πj(N −k)/N
+e
xj −
.
2~
2mωN −k ∂xj
But ωN −k = ωk , and ei2πj(N −k)/N = ei2πj e−i2πjk/N = e−i2πjk/N , so
N
2 X
(ak− + aN −k+ ) = √
N j=1
N −1 r
1 X
√
N k=1
=
=
1
N
k=1
0
ei2π(j−j )k/N xj 0 =
j 0 =1 k=1
j 0 =1
mωk −i2πjk/N
e
xj .
2~
N −1 r
~
1 X
(ak− + aN −k+ )ei2πjk/N = √
2mωk
N
−1
N N
X
X
N
X
r
N
1 X i2π(j−j 0 )k/N
e
N
k=1
1
N
!
xj 0 −
N
X
N
X
j 0 =1
k=1
N
~
2 X
√
2mωk N 0
r
0
0
mωk −i2πj 0 k/N
e
xj 0 ei2πjk/N
2~
!
j =1
ei2π(j−j )k/N − ei2π(j−j )
xj 0
N
N
X
1 X
xj 0 =
δjj 0 xj 0 − R = xj − R.
N 0
0
j =1
j =1
That proves the first relation; to prove the second, we repeat these steps, starting with
(
"r
#
r
N
mωk
~
∂
1 X −i2πjk/N
e
xj +
(ak− − aN −k+ ) = √
2~
2mωk ∂xj
N j=1
s
#)
"r
mω
~
∂
N
−k
− ei2πj(N −k)/N
xj −
2~
2mωN −k ∂xj
N r
2 X
~
∂
=√
e−i2πjk/N
.
∂xj
N j=1 2mωk
CHAPTER 5. IDENTICAL PARTICLES
N −1 r
1 X
√
N k=1
195
N −1 r
mωk
1 X
(ak− − aN −k+ )ei2πjk/N = √
2~
N
N
mωk 2 X
√
2~
N 0
r
0
~
∂ i2πjk/N
e−i2πj k/N
e
2mωk
∂xj 0
j =1
k=1
!
N
X
0
0
∂
ei2π(j−j )k/N − ei2π(j−j )
∂xj 0
N N −1
N
1 X X i2π(j−j 0 )k/N ∂
1 X
=
=
e
N 0
∂xj 0
N 0
j =1 k=1
j =1 k=1
!
N
N
N
N
N
X
X
X
0
1 X ∂
1 X ∂
∂
∂
∂
1 ∂
1
−
=
−
=
=
ei2π(j−j )k/N
δjj 0
−
.
0
0
0
0
N
∂x
N
∂x
∂x
N
∂x
∂x
N
∂R
j
j
j
j
j
0
0
0
0
j =1
j =1
k=1
In the last step I used
j =1
j =1
N
X
∂
∂xj
∂xj ∂
=
, and
= 1 (for instance, if N = 2, R = 12 (x1 + x2 ), and
∂R
∂R
∂x
∂R
j
j=1
r = 21 (x1 − x2 ), so x1 = R + r and x2 = R − r ⇒ ∂xi /∂R = 1).
(d)
N −1 r
1 X
(xj+1 − xj ) = √
N k=1
~
(ak− + aN −k+ )ei2πjk/N ei2πk/N − 1 ,
2mωk
N −1
(xj+1 − xj )2 =
N
X
1
j=1
2
mω 2 (j+1 −xj )2 =
=
1 X
~
i2πj(k+k0 )/N
0 + aN −k 0 )e
(ak− + aN −k+ )(ak−
√
+
0
N
2m ωk ωk
k,k0
0
× ei2πk/N − 1 ei2πk /N − 1 ,
N
−1
X
~ω 2
0 + aN −k 0 )δk 0 ,N −k
(ak− + aN −k+ )(ak−
√
+
4 ωk ωk0
k,k0
0
× ei2πk/N − 1 ei2πk /N − 1
N
−1
X
k=1
=
N
−1
X
k=1
~ω 2
ω2
(ak− + aN −k+ )(aN −k− + ak+ ) k2
4ωk
ω
~ωk
ak− ak+ + aN −k+ ak+ + ak− aN −k− + aN −k+ aN −k− .
4
In the penultimate line I used
ei2πk/N − 1 ei2π(N −k)/N − 1 = eiπk/N eiπk/N − e−iπk/N e−iπk/N e−iπk/N − eiπk/N
ω 2
k
2
.
= (2i sin(πk/N )) (−2i sin(πk/N )) = [2 sin(πk/N )] =
ω
Meanwhile,
N −1 ~2 ∂ 2
1 X
~2 m √
i2πj(k+k0 )/N
0 − aN −k 0 )e
−
=
−
ωk ωk0 (ak− − aN −k+ )(ak−
+
2m ∂x2j
N
2m
2~
k,k0
"
N −1 r
~2 1
∂ X mωk
−
ak− − aN −k+ ei2πjk/N
3/2
2m N
∂R
2~
k=1
#
r
N
−1
X
i2πjk/N ∂
mωk
~2 1 ∂ 2
+
ak− − aN −k+ e
−
.
2~
∂R
2m N 2 ∂R2
k=1
196
CHAPTER 5. IDENTICAL PARTICLES
The middle terms vanish when summed over j
N −1 N
X
~2 X ∂ 2
=
−
2m j=1 ∂x2j
k=1
=
N
−1
X
k=1
~ωk
−
4
P
ak− − aN −k+
, leaving
N
i2πjk/N
= N δk,0 = 0
j=1 e
~2 ∂ 2
aN −k− − ak+ −
2N m ∂R2
~ωk
~2 ∂ 2
.
ak− ak+ − aN −k+ ak+ − ak− aN −k− + aN −k+ aN −k− −
4
2N m ∂R2
Adding the two results:
Ĥ = −
N −1
N −1
k=1
k=1
X ~ωk
X ~ωk
~2 ∂ 2
~2 ∂ 2
+
ak− ak+ + aN −k+ aN −k− = −
+
ak− ak+ + ak+ ak−
2
2
2N m ∂R
2
2N m ∂R
2
(the change in the final two subscripts just means adding these terms in reverse order, given that ωN −k = ωk ).
Thus
N
−1
X
~2
∂2
1
Ĥ = −
+
~ωk ak+ ak− +
. X
2(N m) ∂R2
2
k=1
Problem 5.39
(a)
ψ(x + lx , y, z) = p
1
1
ei[kx (x+lx )+ky y+kz z] = ψ(x, y, z) = p
ei(kx x+ky y+kz z) ⇒ eikx lx = 1,
lx ly lz
lx ly lz
so kx lx = 0, ±2π, ±4π, . . . = 2nx π; likewise ky ly = 2ny π, kz lz = 2nz π.
∆kx ∆ky ∆kz =
2π 2π 2π
8π 3
.
=
lx ly lz
V
(b) Because nx , ny , and nz run negative as well as positive, we want the whole sphere, not just one octant:
4 3
8π 3 N d
πkF =
⇒ kF =
3
V 2
3π 2 N d
V
1/3
= (3ρπ 2 )1/3 ,
~2
~2 2
kF =
(3ρπ 2 )2/3 (same as Equation 5.54). A
2m
2m
shell of thickness dk contains a volume 4πk 2 dk of k-space, so the number of electron states in the shell is
where ρ ≡ N d/V (same as Equation 5.52). EF =
4πk 2 dk
V 2
~2 (3π 2 N d)5/3 −2/3
=
k
dk
(same
as
in
the
text),
so
E
=
V
(same as Equation 5.56). Each
tot
8π 3 /V
π2
10π 2 m
state occupies an 8-times larger volume of k-space (because of the 2’s in part (a)), but the k-space itself is 8
times larger (because negative n’s are allowed, so with periodic boundary conditions all 8 octants are accessible).
2
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
197
Chapter 6
Symmetries and Conservation Laws
Problem 6.1
(a) A mirror reflection across the x-y plane produces
M̂ ψ(x, y, z) = ψ 0 (x, y, z) = ψ(x, y, −z) .
A 180◦ rotation in the x-y plane has the effect
R̂ ψ(x, y, z) = ψ 0 (x, y, z) = ψ(−x, −y, z) .
Composing these two operations gives
M̂ R̂ ψ(x, y, z) = ψ(−x, −y, −z) = Π̂ ψ(x, y, z) .
(b) In polar coordinates
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ .
Replacing θ with π − θ and φ with φ + π gives
x → r sin(π − θ) cos(φ + π) = −x
| {z } | {z }
sin θ
− cos φ
y → r sin(π − θ) sin(φ + π) = −y
| {z } | {z }
sin θ
− sin φ
z → r cos(π − θ) = −z
| {z }
− cos θ
so this transformation is equivalent to changing the sign of each Cartesian coordinate. One could also
solve this problem by drawing a careful picture in spherical coordinates and showing that the transformed
angles identify the point −r.
(c) The Hydrogenic orbitals may be written as
ψn`m (r, θ, φ) = Rn` (r) Y`m (θ, φ)
198
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
so that
Π̂ ψn`m (r, θ, φ) = Rn ` (r) Y`m (π − θ, φ + π) .
The spherical harmonics are defined in Equation 4.32,
Y`m (θ, φ) = K`m ei m φ P`m (cos θ)
where K`m is the normalization constant, and therefore
Y`m (π − θ, φ + π) = K`m (−1)
m
ei m φ P`m (− cos θ) .
Furthermore, from Equation 4.27 we see that
m
d
m
m
2 m/2
P` (−x) = (−1) 1 − x
−
P` (−x) ,
dx
`
and from Equation 4.28 that P` (−x) = (−1) P` (x), so that
P`m (−x) = (−1)
`+m
P`m (x)
(which you could also infer from Table 4.2). Combining these results,
`
Π̂ ψn`m (r, θ, φ) = (−1) ψn`m (r, θ, φ) .
Problem 6.2
We define the exponential of an operator by its power series so that
Û = 1 + i Ω̂ −
1
i
Ω̂ Ω̂ − Ω̂ Ω̂ Ω̂ + . . .
2
3!
Taking the adjoint, we complex conjugate all numbers, reverse the order of all operators and replace each one
with its adjoint. Therefore
1 † †
i
Ω̂ Ω̂ + Ω̂† Ω̂† Ω̂† + . . .
2
3!
i
1
= 1 − i Ω̂ − Ω̂ Ω̂ + Ω̂ Ω̂ Ω̂ + . . .
2
3!
= e−i Ω̂
Û † = 1 − i Ω̂† −
where we’ve used the fact that Ω̂ is hermitian in the second line. Now, we know (from Problem 3.29) that
e eB̂ = eÂ+B̂ if  and B̂ commute. Since i Ω̂ clearly commutes with −i Ω̂ we may then write
Û Û † = ei Ω̂ e−i Ω̂ = ei Ω̂−i Ω̂ = e0 = 1 .
(One could also multiply the two series directly if you haven’t seen Problem 3.29).
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
199
Problem 6.3
We consider how the operator p̂0 acts on an arbitrary function f (x).
~ d
T̂ (a) f (x)
i dx
~ d
= T̂ † (a)
f (x − a)
i dx
~
= T̂ (−a) f 0 (x − a)
i
~ 0
= f (x)
i
~ d
=
f (x)
i dx
p̂0 = T̂ † (a)
(An alternative proof would quote Equation 6.3 and prove that p̂ must commute with T̂ from there.)
Problem 6.4
We start by writing
T̂ † Ω̂ T̂ =
X
amn T̂ † x̂m p̂n T̂ .
mn
The idea is that we’re then free to insert copies of T̂ T̂ † = 1 wherever necessary. For example, consider the term
T̂ † x̂3 p̂2 T̂ = |T̂ †{z
x̂ T̂} T̂ † x̂ T̂ T̂ † x̂ T̂ T̂ † p̂ T̂ T̂ † p̂ T̂ = x̂03 p̂03 .
x̂0
It is then clear that
n X
m X
m
n
T̂ † p̂ T̂
=
amn (x̂0 ) (p̂0 ) .
T̂ † Ω̂ T̂ =
amn T̂ † x̂ T̂
mn
mn
so that
Ω̂0 (x̂, p̂) = T̂ † Ω̂(x̂, p̂) T̂ = Ω̂(x̂0 , p̂0 ) .
Problem 6.5
We insert ψ(x) = ei q x u(x) into Equation 6.11 to get
ei q(x−a) u(x − a) = e−i q a ei q x u(x)
u(x − a) = u(x) .
This shows that u(x) is in fact a periodic function.
200
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
Problem 6.6
(a) First note that
d2 i q x
d2
ψ
=
e
u(x)
dx2
dx2
d i q ei q x u(x) + ei q x u0 (x)
=
dx
= −q 2 ei q x u(x) + 2 i q ei q x u0 (x) + ei q x u00 (x) .
Plugging this into the time-independent Schrödinger equation we get
~2 d2 ψ
+ V (x) ψ = E ψ
2 m dx2
~2 d2 u ~ q ~ du
~2 q 2
−
+
+
V
(x)
u
=
E
−
u
2 m dx2
m i dx
2m
−
after cancelling the common factor of ei q x . This is the desired result.
(b) First we will non-dimensionalize our equation. Let x = a s where s will range from 0 to 1. Then we get a
proper dimensionless Equation
d2 u
1 du 2 m a
2ma
2
+
− 2 + 2 (q a)
V (x) = E
−(q a) u .
ds
i ds
~2
~2
The numerical solution to this equation for the requested values of n and q follows.
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
201
Set the number of samples:
In[! ]:=
Num = 100; dx =
1.
Num + 1
;
Define the potential function:
In[! ]:=
Potential[x_] := If[x > .25 && x < .75, - 20 , 0];
The first derivative operator (with appropriate periodic boundary conditions)
In[! ]:=
D1 =
1
2 dx
Table[If[i ⩵ j + 1, 1, If[i ⩵ j - 1, - 1, 0]], {i, 1, Num}, {j, 1, Num}];
1
;
2 dx
1
D1[[Num, 1]] = ;
2 dx
D1[[1, Num]] =
The second derivative operator
In[! ]:=
D2 =
1
Table[
dx2
If[i ⩵ j, - 2, If[i ⩵ j + 1, 1, If[i ⩵ j - 1, 1, 0]]], {i, 1, Num}, {j, 1, Num}];
1
D2[[1, Num]] = D2[[Num, 1]] =
;
dx2
The potential operator
In[! ]:=
V = Table[If[i ⩵ j, Potential[i dx], 0], {i, 1, Num}, {j, 1, Num}];
Solve the effective Schrodinger equation to get the nth solution for a given choice of q
In[! ]:=
Energy[q_, n_] := q ^ 2 - Sort[Eigenvalues[D2 + 2. I q D1 - V] // Chop][[Num + 1 - n]]
In[! ]:=
TableForm[Table[Energy[q, n], {n, 1, 2}, {q, {- Pi, - Pi / 2, 0, Pi / 2, Pi}}],
TableHeadings → {{1, 2}, {- Pi, - Pi / 2, 0, Pi / 2, Pi}}]
Out[! ]//TableForm=
1
2
π
2
-π
-
- 6.77261
5.84137
- 10.0067
14.1147
0
π
2
π
- 11.9549
29.6386
- 10.0067
14.1147
- 6.77261
5.84137
(c) Running this program for 100 different values of q generates this plot.
202
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
In[! ]:=
Show
ListPlot[Table[{q, Energy[q, 1]}, {q, - Pi, Pi, Pi / 100}], Joined → True],
ListPlot[Table[{q, Energy[q, 2]}, {q, - Pi, Pi, Pi / 100}], Joined → True],
PlotRange → All, AxesLabel → "qa", "E
2 m a2
ℏ2
"

2 m a2
E
ℏ2
30
20
Out[! ]=
10
0
-3
-2
1
-1
2
3
qa
-10
Problem 6.7
(a) We can expand T̂ in a Taylor series just as we did in the single-particle case:
T̂ (a) ψ(x1 , x2 ) = ψ(x1 − a, x2 − a)
X 1
∂ m+n
m+n
=
.
m ∂xn ψ(x1 , x2 ) (−a)
m!
n!
∂x
1
2
m,n
We can write this in terms of the two momenta as
X 1 i a m+n
n
T̂ (a) ψ(x1 , x2 ) =
−
p̂m
1 p̂2 ψ(x1 , x2 )
m!
n!
~
mn
and then split up the sums as
X 1 i a m
X 1 i a n
m
T̂ (a) ψ(x1 , x2 ) =
−
p̂1
−
p̂n2 ψ(x1 , x2 )
m!
~
n!
~
m
n
= e−i a p̂1 /~ e−i a p̂2 /~ ψ(x1 , x2 ) .
Finally, since p̂1 and p̂2 commute we can write
T̂ (a) = e−i a p̂1 /~ e−i a p̂2 /~ = e−i a(p̂1 +p̂2 )/~ = e−i a P̂ /~ .
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
203
(b) We first need to show that the Hamiltonian is translationally invariant. Acting on a test function f (x1 , x2 ),
Ĥ 0 f (x1 , x2 ) = T̂ † (a) Ĥ T̂ (a) f (x1 , x2 )
~2 ∂ 2
~2 ∂ 2
−
= T̂ (−a) −
2
2 m ∂x1
2 m ∂x22
+V (|x1 − x2 |)] f (x1 − a, x2 − a)
~2 ∂ 2
~2 ∂ 2
= −
−
02
2 m ∂x1
2 m ∂x02
2
+V (|x1 + a −(x2 + a)|)] f (x1 + a − a, x2 + a − a)
where x01 = x1 + a and x02 = x2 + a. Since
∂x1 ∂
∂
∂
=
=
0
0
∂x1
∂x1 ∂x1
∂x1
(and similarly for x02 ) we have
~2 ∂ 2
~2 ∂ 2
Ĥ 0 f (x) = −
−
+
V
(|x
−
x
|)
f (x1 , x2 )
1
2
2 m ∂x21
2 m ∂x22
so that Ĥ 0 = Ĥ. Of course, this is all rather pedantic. Just as in the single-particle case the momentum
(and therefore kinetic energy) is invariant under a translation and since the potential in this case depends
only on the distance between particles, it is invariant as well.
Since the Hamiltonian has a continuous translational symmetry, it is invariant under an infinitessimal
transformation
T̂ † (δ) Ĥ T̂ (δ) = Ĥ
δ
δ
Ĥ + i P̂ Ĥ − i Ĥ P̂ = Ĥ
~
~
h
i
Ĥ, P̂ = 0
and therefore, according to the generalized Ehrenfest theorem, the total momentum is conserved.
Problem 6.8
(a) We consider
D
Z
E
f Π̂g =
h
i
f ∗ (x) Π̂ g(x) dx
Z ∞
=
f ∗ (x) g(−x) dx
−∞
Z −∞
=
∞
f ∗ (−x) g(x) (− dx)
204
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
where we have changed the integration variable from x to −x in then last line. Then using the minus sign
to reverse the limits of integration we have
E Z ∞
D
∗
[f (−x)] g(x) dx
f Π̂g =
−∞
Z ∞h
i∗
=
Π̂f (x) g(x) dx
−∞
E
D
= Π̂f g ,
showing that the parity operator is hermitian.
(b) Because the operator is hermitian and its own inverse, it must also be unitary. A hermitian operator has
eigenvalues λ that are real and a unitary operator has eigenvalues λ such that |λ| = 1. The only way
to satisfy both these conditions is if the eigenvalues are ±1. Clearly both cases occur: all even functions
are eigenfunctions of the parity operator with eigenvalue 1 and all odd functions are eigenfunctions of the
parity operator with eigenvalue −1.
Problem 6.9
(a) For a “true” scalar operator,
Π̂† fˆ Π̂ = fˆ
Π̂
Π̂}† fˆ Π̂ = Π̂ fˆ
| {z
1
h
fˆ Π̂ = Π̂ fˆ
i
Π̂, fˆ = 0 ,
and for a pseudoscalar operator:
Π̂† fˆ Π̂ = −fˆ
Π̂
Π̂}† fˆ Π̂ = −Π̂ fˆ
| {z
1
fˆ Π̂ = −Π̂ fˆ
n
o
Π̂, fˆ = 0 .
(b) For a “true” vector operator
Π̂† V̂ Π̂ = −V̂
Π̂
Π̂}† V̂ Π̂ = −Π̂ V̂
| {z
1
V̂ Π̂ = −Π̂ V̂
n
o
Π̂, V̂ = 0 ,
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
and for a pseudvector operator:
Π̂† V̂ Π̂ = V̂
Π̂
Π̂}† V̂ Π̂ = Π̂ V̂
| {z
1
h
V̂ Π̂ = Π̂ V̂
i
Π̂, V̂ = 0 .
Problem 6.10
Applying x̂0 to a test function we have
x̂0 f (x) = Π̂† x̂ Π̂ f (x) = Π̂ x f (−x) = (−x) f (x) = −x̂ f (x)
so that x̂0 = x̂. Applying p̂0 to a test function we have
d
p̂0 f (x) = Π̂† p̂ Π̂ f (x) = Π̂ −i ~
f (−x)
dx
d
d
f (x) = −i ~ f (x) = −p̂ f (x)
= i~
d(−x)
dx
so that p̂0 = −p̂.
The same calculation works in three dimensions.
x̂0 f (x, y, z) = Π̂† x̂ Π̂ f (x, y, z) = Π̂ x f (−x, −y, −z)
= (−x) f (x, y, z) = −x̂ f (x, y, z)
ŷ 0 f (x, y, z) = Π̂† ŷ Π̂ f (x, y, z) = Π̂ y f (−x, −y, −z)
= (−y) f (x, y, z) = −ŷ f (x, y, z)
0
ẑ f (x, y, z) = Π̂† ẑ Π̂ f (x, y, z) = Π̂ y f (−x, −y, −z)
= (−z) f (x, y, z) = −ẑ f (x, y, z)
so that r̂0 = −r̂. And
∂
−i ~
f (−x, −y, −z)
∂x
∂
∂
= i~
f (x, y, z) = −i ~
f (x, y, z) = −p̂x f (x, y, z)
∂(−x)
∂x
∂
p̂0y f (x, y, z) = Π̂† p̂y Π̂ f (x, y, z) = Π̂ −i ~
f (−x, −y, −z)
∂y
∂
∂
= i~
f (x, y, z) = −i ~
f (x, y, z) = −p̂y f (x, y, z)
∂(−y)
∂y
∂
0
†
p̂z f (x, y, z) = Π̂ p̂z Π̂ f (x, y, z) = Π̂ −i ~
f (−x, −y, −z)
∂z
∂
∂
= i~
f (x, y, z) = −i ~
f (x, y, z) = −p̂z f (x, y, z)
∂(−z)
∂z
p̂0x f (x, y, z) = Π̂† p̂x Π̂ f (x, y, z) = Π̂
so that p̂0 = −p̂.
205
206
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
Problem 6.11
From Equation 6.24 we have
Π̂† L̂ Π̂ = L̂ .
Sandwiching this statement in between two parity eigenstates we have
E
D
E D
n0 `0 m0 Π̂† L̂ Π̂ n`m = n0 `0 m0 L̂ n`m
E
E D
D
`0
`
n0 `0 m0 (−1) L̂ (−1) n`m = n0 `0 m0 L̂ n`m
so that
E
h
iD
`+`0
1 −(−1)
n0 `0 m0 L̂ n`m = 0 .
Therefore, the matrix element vanishes unless ` + `0 is even (equivalently, it vanishes unless the two states have
the same parity).
Problem 6.12
We need to find a Hermitian 2 × 2 matrix Π that commutes with each component of S. So let
a b
Π= ∗
b c
with a and b real. Then
~ 01
a b ~ 01
a b
[Π, Sx ] = ∗
−
b∗ c
b c 2 10
2 10
~ b − b∗ a − c
=
.
2 −a + c −b + b∗
Since this must be zero we see that a = c and b = b∗ (so that b is real) if we are to have Π and Sx commute.
Next
~ 0 −i
a b
a b ~ 0 −i
−
[Π, Sy ] =
b a 2 i 0
b a
2 i 0
~ 2b 0
=i
.
0 −2 b
2
so that, in addition, b must be zero if Π is to commute with Sx . That means that Π = a 1, which clearly
commutes with the remaining spin matrix Sz .
Problem 6.13
(a) The ground state is non-degenerate and, by Equation 6.26,
hpe i = h100 | pe | 100i = 0 .
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
207
√
(b) The n = 2 state is degenerate. Consider a state with a mixture of ` values: (|200i + |210i) / 2. For this
state,
h200| + h210|
|200i + |210i
√
√
hpe i =
pe
2
2
1
= (h200 | pe | 200i + 2 Re [h210 | pe | 200i] + h210 | pe | 210i) ,
2
and by Equation 6.26:
hpe i = Re [h210 | pe | 200i] .
The parity selection rule tells us nothing about the remaining matrix element. We can go ahead and
actually compute it:
hpe i = Re [h210 | pe | 200i]
Z
∗
=Re −e
ψ210
(r) r sin θ cos φ ψ200 (r) d3 r ı̂
Z
∗
− e ψ210
(r) r sin θ sin φ ψ200 (r) d3 r ̂
Z
∗
−e
ψ210
(r) r cos θ ψ200 (r) d3 r k̂
= 3 e a k̂ .
Problem 6.14
(a) We first compute the eigenvalues of M.
0 = |M − λ 1| =
ϕ 2
1 − λ −ϕ/N
= 1 − λ2 +
ϕ/N 1 − λ
N
so that λ = 1 ± i ϕ/N . The corresponding eigenvectors are
M v± = (1 ± i ϕ/N ) v±
1 −ϕ/N
a
a
= (1 ± i ϕ/N )
ϕ/N
1
b
b
a − ϕ/N b
a ± i ϕ/N a
=
x
x
so that b = ∓i a. The matrix S−1 is then
1
1 1
S= √
2 −i i
and the matrix M0 is
1
1
1 i
1 −ϕ/N
1 1
0
√
M =√
ϕ/N
1
2 1 −i
2 −i i
1 + i ϕ/N
0
=
.
0
1 − i ϕ/N
208
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
(b) Now
N
1±i
ϕ N X
=
N
k=0
N
k
(±i ϕ)
N
ϕ k X
N!
=
±i
N
(N − k)!N k
k!
k
k=0
The coefficient
k−1
Y
N!
i
=
1
−
(N − k)!N k
N
i=0
becomes 1 in the limit that N → ∞ and
∞
k
ϕ N X (±i ϕ)
lim 1 ± i
=
= e±i ϕ .
N →∞
N
k!
k=0
Therefore
0 N
lim (M )
N →∞
iϕ
e
0
=
.
0 e−i ϕ
(c) Finally we transform back to the original basis to get
N
lim MN = S−1 lim (M0 ) S
N →∞
iϕ
1
1
1 1
1 i
e
0
√
=√
0 e−i ϕ
2 −i i
2 1 −i
cos ϕ − sin ϕ
=
.
sin ϕ cos ϕ
N →∞
Problem 6.15
Consider an infinitesimal rotation applied to the scalar operator fˆ. To first order in δ we have
iδ
iδ
ˆ
ˆ
n · L̂ f 1 −
n · L̂
R̂n (δ) f R̂n (δ) = 1 +
~
~
iδ
iδ ˆ
= fˆ +
n · L̂ fˆ −
f n · L̂
~
~
h
i
iδ
= fˆ +
n · L̂, f
~
†
so that if fˆ commutes with L then it is invariant under a rotation by an infinitesimal angle δ and by extension,
for any finite angle ϕ as well.
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
Problem 6.16
For an infinitesimal rotation about the y axis
iδ
iδ
V̂0 = 1 +
Ly V 1 −
Ly
~
~


 
[Ly , Vx ]
Vx
i
δ
 [Ly , Vy ] 
=  Vy  +
~
[Ly , Vz ]
Vz
 


Vx
−i ~ Vz
i
δ
 0 
=  Vy  +
~
Vz
i ~ Vx


Vx + δ Vz

Vy
=
Vz − δVx
so that the matrix for an infinitesimal rotation about the y axis is


1 0 δ
D(δ) =  0 1 0  .
−δ 0 1
Compare to Equations 6.30 and 6.31. This is clearly the infinitesimal version of


cos ϕ 0 sin ϕ
1 0 .
D(ϕ) =  0
− sin ϕ 0 cos ϕ
Problem 6.17
Consider the infinitesimal rotation of the state ψn`m
iδ
Rn (δ) ψn`m = 1 −
n · L ψn `m
~
iδ
= ψn`m −
(nx Lx + ny Ly + nz Lz ) ψn`m
~ i δ nx − i ny
nx + i ny
= ψn`m −
L+ +
L− + nz Lz ψn`m .
~
2
2
Then from Equations 4.121 we have
Rn (δ) ψn`m =(1 − i δ m) ψn `m
nx − i ny p
− iδ
` (` + 1) − m (m + 1) ψn`m+1
2
nx + i ny p
− iδ
` (` + 1) − m (m − 1) ψn`m−1
2
X
≡
Dm0 m ψn`m0
m0
209
210
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
so that
Dm0 m = (1 − i δ m) δm0 ,m
nx − i ny p
− iδ
` (` + 1) − m (m + 1) δm0 ,m+1
2
p
nx + i ny
` (` + 1) − m (m − 1) δm0 ,m−1 .
− iδ
2
Problem 6.18
(a) Consider operating with the commutator on a test function f (x):
h
i
Π̂, T̂ f (x) = Π̂ T̂ f (x) − T̂ Π̂ f (x)
= Π̂ f (x − a) − T̂ f (−x)
= f (−x − a) − f (−(x − a))
= f (−x − a) − f (−x + a) .
So these two operators don’t commute.
(b) Acting on one of the states fp (x) we have
Π̂ fp (x) = Π̂ √
1
1
ei p x/~ = √
e−i p x/~ = f−p (x) .
2π~
2π~
(c) The translation operator applied to the first state gives
p x
1
p (x − a)
1
T̂ √
cos
cos
=√
~
~
π~
π~
p a 1
p x
√
= cos
cos
~
~
π~
p a 1
p x
√
+ sin
sin
~
~
π~
and
p x
1
1
p (x − a)
T̂ √
sin
=√
sin
~
~
π~
π~
p a 1
p x
√
= cos
sin
~
~
π~
p a 1
p x
√
− sin
cos
.
~
~
π~
Therefore, the translation operator mixes these two states together.
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
211
Probelm 6.19
(a) First,
[Lz , V± ] = [Lz , Vx ] ± i [Lz , Vy ] = i ~ Vy ± i (−i ~ Vx )
= ~ (± Vx + i Vy ) = ±~ (Vx ± i Vy ) = ±~ V± .
If you’re comfortable working with index notation, the next part could be quicker, but I’ll do it the long
way. Here we go:
2
L , V± = L2x + L2y + L2z , Vx ± iVy
= L2x , Vx ± i L2x , Vy + L2y , Vx ± i L2y , Vy + L2z , V±
= 0 ± i Lx [Lx , Vy ] ± i [Lx , Vy ] Lx + Ly [Ly , Vx ] + [Ly , Vx ] Ly
+ 0 + Lz [Lz , V± ] + [Lz , V± ] Lz
= ± i Lx i ~ Vz ± i (i ~ Vz ) Lx + Ly (−i ~ Vz ) +(−i ~ Vz ) Ly
+ Lz (±~ V± ) ± ~ V± Lz
where we’ve used the result proved above. Finally we need to get all L’s to the right of all V ’s:
2
L , V± = ∓ ~ Vz Lx ∓ ~ [Lx , Vz ] ∓ ~ Vz Lx − i ~ Vz Ly − i ~ [Ly , Vz ]
− i ~ Vz Ly ± ~ V± Lz ± ~ [Lz , V± ] ± ~ V± Lz
= ~ Vz (∓2 Lx − 2 i Ly ) ∓ ~(−i ~ Vy ) − i ~ i ~ Vx
± 2 ~ V± Lz ± ~ (±~ V± )
= ∓ 2 ~ Vz (Lx ± i Ly ) + ~2 (Vx ± i Vy ) ± 2 ~ V± Lz + ~2 V±
= 2 ~2 V± ± 2 ~ V± Lz ∓ 2 ~ Vz L± .
This is the desired result.
(b) Acting on a state ψn`` with each of the commutator relations form part a we get
[Lz , V+ ] ψn`` = ~ V+ ψn``
Lz V+ ψn`` − V+ Lz ψn`` = ~ V+ ψn``
| {z }
` ~ ψn``
Lz (V+ ψn`` ) = (` + 1) ~ (V+ ψn`` )
and
2
L , V+ ψn`` = 2 ~2 V+ + 2 ~ V+ Lz − 2 ~ Vz L+ ψn``
L2 V+ ψn`` − V+
L2 ψn``
| {z }
`(`+1) ~2 ψn``
2
= 2 ~2 V+ ψn`` + 2 ~ V+ Lz ψn`` −2 ~ Vz L+ ψn``
| {z }
| {z }
` ~ ψn``
0
2
L (V+ ψn`` ) = [` (` + 1) + 2 + 2 `] ~ (V+ ψn`` )
= (` + 1) (` + 2) ~2 (V+ ψn`` ) .
From these results, the state V+ ψn`` has an m value of ` + 1 and an ` value of ` + 1.
212
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
Problem 6.20
Sandwiching the commutator between two angular momentum eigenstates gives
hn0 `0 m0 | [L− , f ] | n`mi = 0
hn0 `0 m0 | L− f | n`mi − hn0 `0 m0 | f L− | n`mi = 0
0
0 0 0
m
0 0 0
Am
`0 hn ` m + 1 | f | n`mi − B` hn ` m | f | n`m − 1i = 0
Now we know that these matrix elements vanish unless m0 +1 = m and `0 = `. When those are equal Am−1
= B`m
`
and
hn0 `m | f | n`mi = hn0 `m − 1 | f | n`m − 1i .
Since m is arbitrary, this is equivalent to Equation 6.46.
Problem 6.21
The expectation value of r is
1
(h211| + h21 − 1|) r(|211i + |21 − 1i)
2
1
= h211 | r | 211i + Re [h211 | r | 21 − 1i] + h21 − 1 | r | 21 − 1i
2
hψ | r | ψi =
and applying Equation 6.47 we get
hψ | r | ψi = h21 k r k 21i .
Now we need to actually evaluate the reduced matrix element. We can choose any value of m to do so:
Z
∗
∗
h21 k r k 21i = h210 | r | 210i = ψ210
(r) r ψ210
(r) d3 r = 5 a .
Therefore
hψ | r | ψi = h21 k r k 21i = 5 a .
Problem 6.22
(a)
[Lz , Vz ] = 0
[Lz , V± ] = [Lz , Vx ] ± i [Lz , Vy ] = i ~ Vy ± i (−i ~ Vx )
= ±~ (Vx ± i Vy ) = ± ~ V±
[L± , V± ] = [Lx , Vx ] ± i [Lx , Vy ] ± i [Ly , Vx ] − [Ly , Vy ]
= 0 ± i (i ~ Vz ) ± i (−i ~ Vz ) − 0 = 0
[L± , Vz ] = [Lx , Vz ] ± i [Ly , Vz ] = −i ~ Vy ± i (i ~ Vx )
= ∓~ (Vx ± i Vy ) = ∓ ~ V±
[L± , V∓ ] = [Lx , Vx ] ∓ i [Lx , Vy ] ± i [Ly , Vx ] + [Ly , Vy ]
= 0 ∓ i (i ~ Vz ) ± i (−i ~ Vz ) = ±2 ~ Vz .
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
213
(b) Sandwiching 6.50 in between two angular-momentum eigenstates we get
hn0 `0 m0 | [Lz , Vz ] | n`mi = 0
hn0 `0 m0 | Lz Vz | n`mi − hn0 `0 m0 | Vz Lz | n`mi = 0
m0 hn0 `0 m0 | Vz | n`mi − m hn0 `0 m0 | Vz | n`mi = 0
so that hn0 `0 m0 |Vz |n`mi = 0 unless m = m0 .
Problem 6.23
(a) This is nothing more than Fourier’s trick. Taking an inner product with an arbitrary state from the sum
and using the orthonormality,
X j j J
2
hj1 j2 m01 m02 | J M i =
Cm11 m
hj1 j2 m01 m02 | j1 j2 m1 m2 i
2M
{z
}
|
m ,m
1
2
δm ,m0 δm ,m0
1
1
2
2
j1 j2 ,J
= Cm
0 m0 M .
1
2
(b) Applying J± to Equation 6.64 we get
i
X j j J h (1)
(2)
2
J± |J M i =
Cm11 m
J
|j
j
m
m
i
+
J
|j
j
m
m
i
.
1
2
1
2
1
2
1
2
±
±
2M
m1 ,m2
(a) For the plus case, Equation 4.121 gives
X
m1
j1 j2 J
2
AM
Cm
Aj1 |j1 j2 m1 + 1 m2 i + Am
J |J M + 1i =
j2 |j1 j2 m1 m2 + 1i .
1 m2 M
m1 m2
Next we take the inner product with hj1 j2 m01 m02 | to get
X
m1
j1 j2 J
j1 j2 J
2
AM
Cm
Aj1 δm1 +1,m01 δm2 ,m02 + Am
J Cm0 m0 M +1 =
j2 δm1 ,m01 δm2 +1,m02
1 m2 M
1
2
m1 m2
m0 −1
= Aj 1 1
m0 −1
j1 j2 J
2
Cm
0 −1 m M + Aj
2
2
1
j1 j2 J
Cm
0 m0 −1 M .
1
2
Now Am−1
= Bjm and dropping the primes we have
j
j1 j2 J
j1 j2 J
j1 j2 J
m1
m2
AM
J Cm1 m2 M +1 = Bj1 Cm1 −1 m2 M + Bj2 Cm1 m2 −1 M
which is the first desired equality.
(b) For the minus case, Equation 4.121 gives
X
m1
j1 j2 J
BJM |J M − 1i =
Cm
Bj1 |j1 j2 m1 − 1 m2 i + Bjm2 2 |j1 j2 m1 m2 − 1i .
1 m2 M
m1 m2
Next we take the inner product with hj1 j2 m01 m02 | to get
X
m1
j1 j2 J
j1 j2 J
BJM Cm
Cm
Bj1 δm1 −1,m01 δm2 ,m02 + Bjm2 2 δm1 ,m01 δm2 −1,m02
0 m0 M −1 =
1 m2 M
1
2
m1 m2
m0 +1
= Bj 1 1
m0 +1
j1 j2 J
2
Cm
0 +1 m M + Bj
2
2
1
j1 j2 J
Cm
0 m0 +1 M .
1
2
214
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
Now Bjm+1 = Am
j and dropping the primes we have
j1 j2 J
j1 j2 J
j1 j2 J
m2
1
BJM Cm
= Am
j1 Cm1 +1 m2 M + Aj2 Cm1 m2 +1 M
1 m2 M −1
which is the second desired equality.
Problem 6.24
(a)
i. From Equation 6.52 (upper signs),
hn0 `0 m0 | L+ V+ − V+ L+ | n`mi = 0
0
0 0 0
B`m0 hn0 `0 m0 − 1 | V+ | n`mi = Am
` hn ` m | V+ | n`m + 1i .
ii. From Equation 6.52 (lower signs),
hn0 `0 m0 | L− V− − V− L− | n`mi = 0
0
0 0 0
m
0 0 0
Am
`0 hn ` m + 1 | V− | n`mi = B` hn ` m | V+ | n`m − 1i .
iii. From Equation 6.53 (upper signs),
hn0 `0 m0 | L+ Vz − Vz L+ | n`mi = −~ hn0 `0 m0 | V+ | n`mi
0
0 0 0
0 0 0
B`m0 hn0 `0 m0 − 1 | Vz | n`mi = Am
` hn ` m | Vz | n`m + 1i − ~ hn ` m | V+ | n`mi .
iv. From Equation 6.53 (lower signs),
hn0 `0 m0 | L− Vz − Vz L− | n`mi = ~ hn0 `0 m0 | V− | n`mi
0
0 0 0
m
0 0 0
0 0 0
Am
`0 hn ` m + 1 | Vz | n`mi = B` hn ` m | Vz | n`m − 1i + ~ hn ` m | V− | n`mi .
v. From Equation 6.54 (upper signs),
hn0 `0 m0 | L+ V− − V− L+ | n`mi = 2 ~ hn0 `0 m0 | Vz | n`mi
0
0 0 0
0 0 0
B`m0 hn0 `0 m0 − 1 | V− | n`mi = Am
` hn ` m | V− | n`m + 1i + 2 ~ hn ` m | Vz | n`mi .
vi. From Equation 6.54 (lower signs),
hn0 `0 m0 | L− V+ − V+ L− | n`mi = −2 ~ hn0 `0 m0 | Vz | n`mi
0
0 0 0
m
0 0 0
0 0 0
Am
`0 hn ` m + 1 | V+ | n`mi = B` hn ` m | V+ | n`m − 1i − 2 ~ hn ` m | Vz | n`mi .
(b)
i. Plugging in Equation 6.59 to the result from a.i,
0
0
0
?
` 1`
m `
1`
B`m0 Cm1m
0 −1 = A` Cm+11m0
and applying Equation 6.66,
0
0
?
0
`
1`
1 ` 1`
m `
1`
Am
` Cm+11m0 + A1 Cm2m0 = A` Cm+11m0 .
Since A11 = 0, the equation is satisfied.
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
215
ii. Plugging in Equation 6.60 to the result from a.ii,
0
0
0
?
` 1 `
m `
1 `
Am
`0 Cm−1m0 +1 = B` Cm−1−1m0
and applying Equation 6.66,
0
0
0
?
−
m `
1 `
`
1 `
` 1 `
B`m Cm−1−1m
0 + B1 1 Cm−2m0 = B` Cm−1−1m0 .
Since B1−1 = 0, the equation is satisfied.
iii. Plugging in Equation 6.59 and 6.61 to the result from a.iii,
0
0
0
?
` 1`
m `
1`
B`m0 Cm0m
0 −1 = A` Cm+10m0 +
√
0
` 1`
2 ~ Cm1m
0
and applying Equation 6.66,
0
0
√
0
?
`
1`
0 ` 1`
m `
1`
Am
` Cm+10m0 + A1 Cm1m0 = A` Cm+10m0 +
Since A01 =
√
0
` 1`
2 ~ Cm1m
0 .
2 ~, the equation is satisfied.
iv. Plugging in Equation 6.60 and 6.61 to the result from a.iv,
0
0
0
?
` 1`
m `
1`
Am
`0 Cm0m0 +1 = B` Cm−10m0 +
√
0
` 1 `
2 ~ Cm−1m
0
and applying Equation 6.66,
0
0
0
?
`
1`
0 ` 1 `
m `
1`
B`m Cm−10m
0 + B1 Cm−1m0 = B` Cm−10m0 +
Since B10 =
√
√
0
` 1 `
2 ~ Cm−1m
0 .
2 ~, the equation is satisfied.
v. Plugging in Equation 6.60 and 6.61 to the result from a.v,
0
0
0
?
` 1 `
m `
1 `
B`m0 Cm−1m
0 −1 = A` Cm+1−1m0 +
√
0
` 1`
2 ~ Cm0m
0 ,
and applying Equation 6.66,
0
0
0
?
−1 ` 1`
`
1 `
m `
1 `
Am
` Cm+1−1m0 + A1 Cm0m0 = A` Cm+1−1m0 +
Since A−1
1 =
√
√
0
` 1`
2 ~ Cm0m
0 .
2 ~, the equation is satisfied.
vi. Finally, plugging in Equations 6.59 and 6.61 to the result from a.vi,
0
0
0
?
` 1`
m `
1`
Am
`0 Cm1m0 +1 = B` Cm−11m0 +
√
0
` 1`
2 ~ Cm0m
0 ,
and applying Equation 6.66,
0
0
?
0
`
1`
1 ` 1`
m `
1`
B`m Cm−11m
0 + B1 Cm0m0 = B` Cm−11m0 +
Since B11 =
√
2 ~, the equation is satisfied.
√
0
` 1`
2 ~ Cm0m
0 .
216
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
Problem 6.25
The expectation value of the dipole moment pe is
1
(h211| + h200|) pe (|211i + |200i)
2
1
= (h211 | pe | 211i + 2 Re [h211 | pe | 200i] + h210 | pe | 200i) .
2
hψ | pe | ψi =
Now because of Laporte’s rule, the first and last matrix elements vanish and
hψ | pe | ψi = Re [h211 | pe | 200i] .
We can now write out the operator
h
i
hψ | pe | ψi = −e Re h211 | x | 200i ı̂ + h211 | y | 200i ̂ + h211 | z | 200i k̂ .
Now, according to Equation 6.57 the last term vanishes and we may write r± = x ± i y so that
r+ + r−
r+ − r−
hψ | pe | ψi = −e Re 211
200 ı̂ + 211
200 ̂
.
2
2i
Now according to Equation 6.58, the matrix elements of r− vanish and
ı̂ − i ̂
hψ | pe | ψi = −e Re h211 | r+ | 200i
2
We can then use Equation 6.59 to write
√ 011
ı̂ − i ̂
hψ | pe | ψi = −e Re − 2 C011
h21 k r k 21i
2
ı̂ − i ̂
e
= e Re h21 k r k 20i √
= √ h21 k r k 21i ı̂
2
2
since, as we’ll prove in just a second, the reduced matrix element is real in this case. To evaluate the reduced
matrix element, we choose the simplest instance:
011
h210 | z | 100i = C000
h21 k r k 21i = h21 k r k 21i
or
h21 k r k 20i =
√
2 h210 | z | 200i
√ Z ∗
= 2 ψ210 (r) r cos θ ψ200 (r) d3 r = −3 a
and
3ea
hψ | pe | ψi = − √ .
2
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
217
Problem 6.26
We can follow the same steps as in Example 6.7. We have
pH (t) ψn (x) = U † p U ψn (x)
r
~mω
†
=U i
(a+ − a− ) e−i En t/~ ψn (x)
2
r
√
~ m ω −i En t/~ √
†
n + 1 ψn+1 (x) − n ψn (x)
=U i
e
2
r
√
~ m ω −i En t/~ √
=i
n + 1 ei En+1 t/~ ψn+1 (x) − n ei En−1 t/~ ψn (x)
e
2
r
√
~mω iωt √
=i
n + 1 ψn+1 (x) − e−i ω t n ψn−1 (x)
e
2
r
~mω iωt
=i
e
a+ − e−i ω t a− ψn (x)
2
so that
r
~mω iωt
e
a+ − e−i ω t a−
2
r
~mω
1
1
=i
ei ω t √
(−i p + m ω x) − e−i ω t √
(i p + m ω x)
2
2~mω
2~mω
pH (t) = i
= p cos(ω t) − m ω x sin(ω t)
and
pH (t) = pH (0) cos(ω t) − m ω xH (0) sin(ω t) .
This is identical to the solution of the classical equations of motion with the classical variables x and p replaced
by the quantum operators xH and pH .
Problem 6.27
For the free particle
p2
[p, H] = p,
=0
2m
p2
1
i~
[x, H] = x,
=
([x, p] p + p [x, p]) =
p
2m
2m
m
and
i~
i~
i~
x, H 2 = [x, H] H + H [x, H] =
pH + H
p=2
pH
m
m
m
i~
i~
i~
x, H 3 = [x, H] H 2 + H x, H 2 =
p H2 + H 2
pH = 3
p H2 .
m
m
m
218
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
Now that we see the pattern we can do a proof by induction. Assume that [x, H n ] = n i ~ p H n−1 /m (which
we’ve already verified for n = 1, 2, 3) and show that if it holds for n, then it holds for n + 1. Well
x, H n+1 = H [x, H n ] + [x, H] H n
i~
i~
p H n−1 +
p Hn
=Hn
m
m
i~
= (n + 1)
p Hn
m
so that the proof by induction holds. Now from this
n
∞
X
i
1
[x, H n ]
− t
n!
~
n=0
n
∞
i~ X
i
1
=1+
p
− t
H n−1
m n=1 (n − 1)!
~
n−1
∞
pt X
1
i
H n−1
− t
=1+
m n=1 (n − 1)!
~
|
{z
}
[x, U ] =
U
and since p commutes with H we also have [p, U ] = 0. We are then finally in a position to compute
xH (t) = U † x U = U † (U x + [x, U ]) = x + U †
pt
pt
U =x+
m
m
pH (t) = U † p U = U † (U p + [p, U ]) = p .
Writing everything in terms of Heisenberg picture operators we have
xH (t) = xH (0) +
1
pH (0) t
m
pH (t) = pH (0) .
Problem 6.28
Taylor expanding Ψ(x, t) we have
∂Ψ
+ ...
∂t t0
1
Ĥ(t0 ) Ψ(x, t0 ) + . . .
= Ψ(x, t0 ) + δ
i~
iδ
= 1−
Ĥ(t0 ) + . . . Ψ(x, t0 )
~
Ψ(x, t0 + δ) = Ψ(x, t0 ) + δ
and comparing to Equation 6.75 we may read off
Û (t0 + δ, t0 ) ≈ 1 −
iδ
Ĥ(t0 ) .
~
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
Problem 6.29
We start with
i~
i
d
d h †
Q̂H (t) = i ~
Û (t) Q̂ Û (t)
dt
dt
dU †
dÛ
= i~
Q̂ Û (t) + Û † (t) Q̂ i ~
dt
dt
Since Ĥ is time independent,
d
i
i
dU
= e−i Ĥ t/~ = − Ĥ e−i Ĥ t/~ = − Ĥ Û (t) ,
dt
dt
~
~
and taking the adjoint of this expression
dU †
i
= Û † (t) Ĥ .
dt
~
Ĥ clearly commutes with Û or Û † and therefore
i~
d
Q̂H (t) = −Û † (t) Ĥ Q̂ Û (t) + Û † (t) Q̂ Ĥ Û (t)
dt
= Û † (t) Q̂ Û (t) Ĥ − Ĥ Û † (t) Q̂ Û (t)
= Q̂H (t) Ĥ − Ĥ Q̂H (t)
which is the desired result.
For the operator x̂,
i
i
1 h
d
1 h †
x̂H =
Û x̂ Û , Ĥ
x̂H , Ĥ =
dt
i~ 
i~
=

i
h
h
i
h
i

1 
Û † x̂ Û , Ĥ +Û † x̂, Ĥ Û + U † , Ĥ x̂ Û 

i~ 
| {z }
| {z }
0
0
1 †
p̂2
Û x̂,
+ V Û
=
i~
2m


1 † 1
=
Û
x̂, p̂2 + [x̂, V ] Û
| {z }
i~
2m
0


1 † 1 
=
Û
p̂ [x̂, p̂] + [x̂, p̂] p̂ Û
| {z }
i~
2m
i~
1 †
1
=
Û p̂ Û =
p̂H (t) .
m
m
219
220
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
And for the operator p̂ we reproduce the first two lines of the above calculation to arrive at
d
1 †
p̂2
p̂H =
Û p̂,
+ V Û
dt
i~
2m



p̂2
1 †
+ [p̂, V ]
Û 
p̂,
=
 Û

i~
2m
| {z }
0
1 †
dV
dV
=
Û −i ~
Û = −
(t) .
i~
dx
dx H
These have an identical structure to the classical equations of motion for the dynamic variables x and p:
m
dp̂H
dV
= −
.
dt
dx H
dx̂H
= p̂H (t)
dt
Problem 6.30
(a) We can express Ψ(x, 0) as a linear combination of stationary states
X
Ψ(x, 0) =
cn ψn (x)
n
where
Z
cn =
ψn∗ (x) Ψ(x, 0) dx .
Then we have
Ψ(x, t) =
X
cn ψn (x) e−i En t/~
n
=
X Z
ψn∗ (x0 ) Ψ(x0 , 0) dx0
ψn (x) e−i En t/~
n
=
Z X
ψn∗ (x0 ) e−i En t/~ ψn (x) Ψ(x0 , 0) dx0
n
Z
=
K(x, x0 , t) Ψ(x0 , 0) dx0
which is the desired result.
(b) For the simpler harmonic oscillator, the stationary states are
2
ψn = An Hn (ξ) e−ξ /2
p
where ξ = m ω/~ x. Then
K(x, x0 , t) =
∞
X
n=0
2
2
An Hn (η) e−η /2 e−i(n+1/2) ω t An Hn (ξ) e−ξ /2
[F]
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
221
p
√
1/4
where η = m ω/~ x. We define z = e−i ω t and recalling that An = (m ω/π ~)
1/ 2n n! (Equation
2.86) we have
r
∞
m ω √ −(η2 +ξ2 ) X z n
η 2 +ξ 2 )/2
0
(
K(x, x , t) = e
Hn (ξ) Hn (η)
ze
π~
2n n!
n=0
r
r
2
η + ξ2
mω
z
η2 + ξ2 − 2 η ξ z
=
exp
−
π~
1 − z2
2
1 − z2
"
#
r
r
1 1 + z2 η2 + ξ2 − 4 η ξ z
mω
z
=
exp −
.
π~
1 − z2
2
1 − z2
Lets clean this up a little bit. We note that z z ∗ = 1 so
#
"
r
r
mω
1
1 (z ∗ + z) η 2 + ξ 2 − 4 η ξ
0
K(x, x , t) =
.
exp −
π~
z∗ − z
2
z∗ − z
We then further note that z ∗ − z = 2 i sin(ω t) and z + z ∗ = 2 cos(ω t) and we have
r
K(x, x0 , t) =
1 mω
π 2~
s
"
#
1
m ω cos(ω t) x2 + x02 − 2 xx0
exp −
.
i sin(ω t)
2~
i sin(ω t)
(c) For the given initial state we have
Z ∞
Ψ(x, t) =
K(x, x0 , t) Ψ(x0 , 0) dx0
−∞
s
# "
r
Z ∞
1/4
2
0
1 mω
1
2a
m ω cos(ω t) x2 + x02 − 2 xx0
exp −
e−a(x −x0 ) dx0
=
π 2~
i sin(ω t) −∞
2~
i sin(ω t)
π
s
r
1/4
x2
2a
mω
1 mω
1
− a x20
exp −
=
π 2~
i sin(ω t) π
2 ~ i tan(ω t)
Z ∞
mω
1
mω
x
02
×
exp −
+a x +
+ 2a x0 x0 dx0
2 ~ i tan(ω t)
~ i sin(ω t)
−∞
Even though the constants are complicated, this is nothing more than the Gaussian integral given in the
problem statement and we have
s
r
1/4
1 mω
1
2a
mω
x2
2
exp −
− a x0
Ψ(x, t) =
π 2~
i sin(ω t) π
2 ~ i tan(ω t)

2 
s
mω
x
+
2a
x
0
π
 ~ i sin(ω t)

×
exp  .
mω
1
mω
1
+
a
4 2 ~ i tan(ω t) + a
2 ~ i tan(ω t)
This simplifies to
2
ω
x2 +x20 ) cos(ω t)−2 x x0 + i2m
~ a sin(ω t) x
1/4 exp −a (
ia~
cos(ω t)+ 2m
2a
ω sin(ω t)
q
Ψ(x, t) =
.
π
cos(ω t) + 2mi aω~ sin(ω t)
222
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
Now, Problem 2.49 is a special case of this equation. Plugging in t = 0 we see
Ψ(x, 0) =
m ω 1/4
π~
h mω
i
2
exp −
(x − x0 )
2~
so that this corresponds to choosing a = m ω/2 ~. If we plug a = m ω/2 ~ into our general solution we get
2
2
2
m ω (x +x0 ) cos(ω t)−2 x x0 +i sin(ω t) x
exp
−
m ω 1/4
2~
ei ω t
√
Ψ(x, t) =
π~
ei ω t
m ω 1/4
h mω =
exp −
x2 + x20 1 + e−2 i ω t /2 − 2 x x0 e−i ω t
π~
2~
2 iωt
−2 i ω t
= + 1−e
x /2 +
2
m ω 1/4
i~t
mω
x2
=
exp −
x2 + 0 1 + e−2 i ω t +
− 2 e−i ω t x x0
,
π~
2~
2
m
exactly as in Problem 2.49.
(d) For the free particle
K(x, x0 , t) =
Z ∞
−∞
Z ∞
∗
fp (x0 ) e−i Ep t/~ fp (x) dp
0
e−i p x /~ −i p2 t/(2 m ~) ei p x/~
√
√
e
dp
=
2π~
2π~
−∞
Z ∞
p2 t
p
1
0
exp −i
+ i (x − x ) dp
=
2 π ~ −∞
2m~
~
"
#
r
2
1
(x − x0 ) /~2
π
=
exp −
2 π ~ i t/2 m ~
2 i t/(m ~)
where we have used the general expression given for the Gaussian integral. So, for the free particle
0
r
K(x, x , t) =
m
im
0 2
exp
(x − x ) .
2πi~t
2~t
(e) For this initial state,
Z ∞
Ψ(x, t) =
K(x, x0 , t) Ψ(x0 , 0) dx0
−∞
1/4
02
m
im
2a
2
exp
(x − x0 )
e−a x dx0
2πi~t
2~t
π
−∞
r
1/4
Z ∞
2a
im 2
im
im
m
02
0
exp
x
exp − a −
x −
x x dx0
=
2πi~t π
2~t
2~t
~t
−∞
"
#
r
1/4
r
2
m
2a
im 2
π
(m x/~ t)
=
exp
x
exp −
2πi~t π
2~t
a − i m/(2 ~ t)
4 (a − i m/(2 ~ t))
Z ∞r
=
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
223
where we have again used the general expression for the Gaussian integral. This simplifies to
2a
Ψ(x, t) =
π
1/4 s
1
a
2
exp −
x ,
1 + i 2 a ~ t/m
1 + i 2 a ~ t/m
exactly as found in Problem 2.21.
Problem 6.31
First we need to compute Φ(p).
Z ∞
1
e−i p x/~ ψ(x) dx
Φ(p) = √
2 π ~ −∞
r
Z ∞
λ
e−λ |x|−i p x/~ dx
=
2 π ~ −∞
r
Z 0
Z ∞
λ
(λ−i p/~) x
−(λ+i p/~) x
=
e
dx +
e
dx
2π~
−∞
0
r
λ
1
1
+
=
2 π ~ λ − i p/~ λ + i p/~
r
2λ
λ
=
π ~ (p/~)2 + λ2
Then we may use Equation 6.81 to compute T̂ (a) ψ(x).
Z ∞
T̂ (a) ψ(x) =
−i p a/~ e
e
−∞
i p x/~
√
2π~
r
2λ
λ
dp .
π ~ (p/~)2 + λ2
Let p = q λ ~ so that
√ Z ∞
Z
λ ∞ ei q λ(x−a)
2 λ3
cos [q λ (x − a)]
T̂ (a) ψ(x) =
dq =
dq
2
π −∞ q + 1
π
q2 + 1
0
√ Z
2 λ ∞ cos(q λ |x − a|)
=
dq .
π
q2 + 1
0
√
The remaining integral can be looked up in any standard table (or computed by contour integration if you’re
familiar with the technique) and
√
T̂ (a) ψ(x) =
exactly as required.
λ e−λ |x−a| ,
224
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
Problem 6.32
(a) We have
(a · σ) (b · σ) =
X
aj σj
j
X
bk σ k
k
!
=
XX
j
=
aj bk
δjk + i
X
X
aj bj + i
XXX
n
j
=a·b+i
jkn σn
n
k
X
j
jkn aj bk σn
k
[a × b]n σn
n
= a · b + i (a × b) · σ
where the critical identity in the second line comes from Problem 4.29.
(b) First note that from part (a), for a unit vector n,
2
(n · σ) = n · n + i (n × n) · σ = 1
and therefore
2
(n · σ) =
n · σ n odd
1
n even
We can then write
i
h ϕ
i
h ϕ
exp −i n · S = exp i n · σ
~
2
k
∞
X
1
iϕ
k
=
−
(n · σ)
n!
2
k=0
X 1 i ϕ k
X 1 i ϕ k
=
−
+
−
n·σ
k!
2
k!
2
k even
=
k odd
X (−1)k/2 ϕ k
X (−1)(k−1)/2 ϕ k
2
k odd
ϕ
ϕ
= cos
− i sin
n·σ.
2
2
k even
k!
−i
k!
2
n·σ
I’ve kept everything as sums but it would also be fine to just write out the first half-dozen or so terms in
the series.
(c) In polar coordinates
n = sin θ cos φ ı̂ + sin θ sin φ ̂ + cos θ k̂
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
225
and
ϕ
ϕ
− i sin
[nx σx + ny σy + nz σz ]
Rn = cos
2
2
ϕ 1 0 ϕ 0 1
= cos
− i sin
sin θ cos φ
01
1 0
2
2
0 −i
1 0
= + sin θ sin φ
+ cos θ
i 0
0 −1
ϕ cos θ sin θ e−i φ ϕ 1 0 .
− i sin
= cos
01
sin θ ei φ − cos θ
2
2
(d) The Hermitian conjugate of Rn is
T
ϕ ϕ 10
cos θ sin θ ei φ
R†n = cos
+ i sin
01
sin θ e−i φ − cos θ
2
2
ϕ 1 0 ϕ cos θ sin θ e−i φ = cos
+ i sin
01
sin θ ei φ − cos θ
2
2
which is clearly a rotation by −ϕ/2 about the same axis, so it must be the inverse of Rn . However, lets
check it explicitly as well:
ϕ ϕ 10
1 0
cos
01
0 1
2
2
ϕ 1 0 ϕ cos θ sin θ e−i φ = + i cos
sin
01
sin θ ei φ − cos θ
2
2
ϕ
ϕ 1 0 cos θ sin θ e−i φ
= − i sin
cos
0 1
sin θ ei φ − cos θ
2
2
ϕ
−i φ
cos θ sin θ e
= + sin
sin θ ei φ − cos θ
2
ϕ cos θ sin θ e−i φ
= × sin
sin θ ei φ − cos θ
2
10
2 ϕ
= cos
01
2
ϕ cos2 θ + sin2 θ
0
2
= + sin
0
sin2 θ + cos2 θ
2
10
=
.
01
Rn R†n = cos
(e) For a rotation about the z axis, θ = 0 and
cos(ϕ/2) − i sin(ϕ/2)
0
0
cos(ϕ/2) + i sin(ϕ/2)
−i ϕ/2
e
0
=
.
0
ei ϕ/2
R=
226
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
Then
S0x = R† Sx R
i ϕ/2
−i ϕ/2
~ 01
e
0
e
0
=
0 e−i ϕ/2 2 1 0
0
ei ϕ/2
~
0 ei ϕ
=
.
−i ϕ
e
0
2
This can then be written as
01
0 −i
cos ϕ
− sin ϕ
10
i 0
2
~
S0x =
= Sx cos ϕ − Sy sin ϕ
which is precisely what we expect for the rotation of a vector (Equation 6.31).
(f) For a π rotation about the x axis, θ = π/2, φ = 0, and ϕ = π so that
R=
0 −i
−i 0
.
and
R
1
0
= −i
0
1
which transforms an up spin into a down spin, albeit with a different phase factor.
(g) For a rotation of 2 π about any axis, ϕ = 2 π and sin(ϕ/2) = 0 so that
R(2 π) = −
10
01
.
This is surprising because it is not simply the identity matrix!
Problem 6.33
(a) In polar coordinates x = r cos φ and y = r sin φ the states are
2
aπ
L
bπ
L
ψab (r, φ) =
sin
r cos φ −
sin
r sin φ −
.
L
L
2
L
2
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
227
Rotating the state counter-clockwise produces
R̂ ψab (r, φ) = ψab (r, φ − π/2)
2
aπ
π L
bπ
π L
−
−
=
sin
r cos φ −
sin
r sin φ −
L
L
2
2
L
2
2
2
aπ
L
bπ
L
=
sin
r sin φ −
sin
−r cos φ −
L
L
2
L
2
2
aπ
L
bπ
L
=
sin
y−
sin −
x− +L
L
L
2
L
2
2
bπ
L
aπ
L
= − sin
x−
+ b π sin
y−
L
L
2
L
2
b
π
L
a
π
L
2
b+1
sin
x−
sin
y−
= (−1)
L
L
2
L
2
so that
R̂ ψab = (−1)
b+1
ψba .
(b) Using the result from part (a),
i
1 h
R̂ ψ± = √ R̂ ψab ± R̂ ψba
2
i
1 h
b+1
a+1
= √ (−1)
ψba ±(−1)
ψab
2
h
i
a+1 1
√ ψab ±(−1)a+b ψba .
= ±(−1)
2
Now if a and b are both even or both odd then (−1)
a
∓(−1) .
a+b
= 1 and ψ± is an eigenstate of R̂ with eigenvalue
(c) Plot is shown below. All of the listed symmetries are clearly present.
228
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
In[! ]:=
ψ[a_, b_, x_, y_] :=
In[! ]:=
ψM[x_, y_] :=
2
L
Sin
aπ
L
x -
L
2
 Sin
bπ
L
y -
L
2

ψ[5, 7, x, y] - ψ[7, 5, x, y]
In[! ]:=
L = 1;
In[! ]:=
ContourPlotψM[x, y], x, -
2
1
2
,
1
2
, y, -
1
2
,
1
2

Out[! ]=
Problem 6.34
(a) We first write
M · L = Mx Lx + My Ly + Mz Lz
M+ + M− L+ + L−
M+ − M− L+ − L−
=
+
+ Mz Lz
2
2
2i
2i
M+ L− + M− L+
=
+ Mz Lz .
2
Now we apply this to ψn`` to get
1
1
M+ L− ψn`` + M− L+ ψn`` + Mz Lz ψn`` = 0
2
2
1
(L− M+ + [M+ , L− ]) ψn`` + 0 + ` ~ Mz ψn`` = 0
2
1
1
L− cn` ψn`+1`+1 + [M+ , L− ] ψn`` + ` ~ Mz ψn`` = 0 .
2
2
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
229
The commutator (remember M is a vector operator) is given in Equation 6.54.
1 `
B
cn` ψn`+1` + ~ Mz ψn`` + ` ~ Mz ψn`` = 0
2 `+1
r
`+1
~ cn` ψn``−1 = −(` + 1) ~ Mz ψn``
2
so that
M̂z ψn`` = − p
1
2 (` + 1)
cn` ψn`+1` .
(b) Note that, just as for M · L in part (a),
M·M=
M+ M− + M− M +
+ Mz2 .
2
Then from vi,
#
2
2
2
e
Ĥ L2 + ~2 ψn ` `
+
M 2 ψn ` ` =
4 π 0
m
(
)
2
1
1
e2
2
2
2
M+ M− ψn`` + M− M+ ψn`` + Mz ψn ` ` =
En [` (` + 1) + 1] ~
ψn``
+
2
2
4 π 0
m
2 2 e
1
`2 + ` + 1
[M+ , M− ] ψn`` + M− M+ ψn`` + Mz2 ψn ` ` =
ψn`` .
1−
2
4 π 0
n2
"
We can work out the commutator from v,
[M+ , M− ] = [Mx , Mx ] + i [My , Mx ] − i [Mx , My ] − [My , My ]
2
= −2 i [Mx , My ] = −2 ~ Lz
Ĥ .
m
Therefore we have
2 2 2
e
`2 + ` + 1
2
−~ Lz
Ĥ ψn`` + M− M+ ψn`` + Mz ψn ` ` =
1−
ψn``
m
4 π 0
n2
2 2 e
2
`2 + ` + 1
ψn``
−~ ` ~ En ψn`` + M− M+ ψn`` + Mz2 ψn ` ` =
1−
m
4 π 0
n2
so that
(
M̂− M̂+ ψn`` =
e2
4 π 0
2 "
`+1
1−
n
(c) Finally we note that
Z
2
∗
|cn` | = (M+ ψn`` ) (M+ ψn`` ) d3 r
Z
∗
= ψn``
M− M+ ψn`` d3 r .
2 #
)
− M̂z2
ψn`` .
230
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
From part (b) this is
2 "
2 # Z
`+1
e2
∗
1−
− ψn``
Mz2 ψn`` d3 r
|cn` | =
4 π 0
n
2 2 "
2 # Z
e
`+1
∗
=
1−
− (Mz ψn`` ) Mz ψn`` d3 r
4 π 0
n
2
and from part (a) this is
e2
|cn` | =
4 π 0
2
2 "
2 #
Z
`+1
1
2
∗
1−
|cn` |
ψn`+1`
ψn`+1` d3 r .
−
n
2 (` + 1)
{z
}
|
1
Finally, solving for cn` gives
2` + 3
|cn` | =
2` + 2
2
e2
4 π 0
2 "
`+1
1−
n
2 #
.
As we see, this nonzero except when ` = n − 1.
Problem 6.35
(a) For an infinitesimal velocity δ
x̂0 ≈ Γ̂† x̂ Γ̂
i
i
≈ 1 + δ (t p̂ − m x̂) x̂ 1 − δ (t p̂ − m x̂)
~
~
i
i
≈ x̂ + δ t [p̂, x̂] − δ m [x̂, x̂]
~
~
≈ x̂ − δ t
and
i
i
p̂0 ≈ 1 + δ (t p̂ − m x̂) p̂ 1 − δ (t p̂ − m x̂)
~
~
i
i
≈ x̂ + δ t [p̂, p̂] − δ m [x̂, p̂]
~
~
≈ p̂ + m δ .
The first equation just says that the origins of the two coordinate systems are displaced by an amount δt
at time t and the second equation is simply Galilean velocity addition: in the primed frame, the velocity
of the particle is greater by δ.
(b) Let the operators be
 =
−i v t
p̂
~
B̂ =
imv
x̂
~
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
231
so that
h
i m v2 t
Ĉ = Â, B̂ =
[p̂, x̂] = −i m v 2 t .
~
Then, according the the Baker-Campbell-Hausdorff formula,
Γ̂ = eÂ+B̂ = e eB̂ e−Ĉ/2
i
i m v2
i
m v x̂ exp
t .
= exp − v t p̂ exp
~
~
~ 2
The first exponential is simply the translation operator and
Γ̂ = T̂ (v t) exp
i
m v2
m v x̂ +
t
.
~
2
Now if we simply switch the roles of  and B̂, that will change the sign of Ĉ and again applying the
Baker-Campbell-Hausdorff gives
i
i m v2
i
m v x̂ exp − v t p̂ exp −
t
Γ̂ = exp
~
~
~ 2
or
Γ̂ = exp
i
m v2
T̂ (v t) .
m v x̂ −
t
~
2
(c) Consider Ψ̂0 = Γ Ψ. We have
i~
dΓ
∂Ψ
∂ Γ̂ Ψ = i ~
Ψ + i~Γ
∂t
dt
∂t
and by the assumption that Ψ satisfies the time-dependent Schrödinger equation with hamiltonian H we
have
∂ dΓ
i~
Γ̂ Ψ = i ~
Ψ + Γ Ĥ Ψ
∂t
dt
dΓ †
†
= i~
Γ + Γ Ĥ Γ̂ Γ̂ Ψ .
dt
Therefore Ψ0 satisfies the Schrödinger equation with Hamiltonian
dΓ̂
Ĥγ = Γ̂ Ĥ Γ̂† + i ~
Γ̂
dt
i
m v2
= exp
m v x̂ −
t
T̂ (v t) Ĥ T̂ † (v t)
~
2
i
m v2
dΓ̂ †
× exp −
m v x̂ −
t + i~
Γ̂
~
2
dt
2
i
p̂
i
dΓ̂ †
= exp
m v x̂
+ V (x − v t) exp − m v x̂ + i ~
Γ̂ .
~
2m
~
dt
232
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
First we compute
d
i
m v2
dΓ̂
=
T̂ (v t) exp
m v x̂ +
t
dt
dt
~
2
d
i
m v2
+ T̂ (v t)
exp
m v x̂ +
t
dt
~
2
i
i
m v2
= − v p̂ T̂ (v t) exp
m v x̂ +
t
~
~
2
i m v2
i
m v2
+ T̂ (v t)
exp
m v x̂ +
t
~
2
~
2
2
mv
i
v p̂ −
Γ̂
=−
~
2
and therefore
∂ Γ̂ †
m v2
Γ̂ = v p̂ −
.
∂t
2
Finally we need to compute
2
p̂
ei m v x̂/~
+ V (x − v t) e−i m v x̂/~
2m
1 i m v x̂/~ 2 −i m v x̂/~
e
p e
+ V (x − v t)
=
2m
1 i m v x̂/~ −i m v x̂/~ i m v x̂/~ −i m v x̂/~
=
e
pe
e
pe
+ V (x − v t)
2m
where
n
∞
h
i X
1 imv
ei m v x̂/~ , p̂ =
[x̂n , p̂]
n!
~
n=0
n
∞
X
1 imv
=
i ~ n x̂n−1
n!
~
n=1
n−1
∞
X
imv
1
x̂
= −m v
(n − 1)!
~
n=1
i~
= −m v ei m v x/~ .
That means that
ei m v x̂/~ p̂ e−i m v x̂/~ = p̂ − m v
and putting all of this together we get
m v2
1
2
(p̂ − m v) + V (x − v t) + v p̂ −
2m
2
1 2
=
p̂ + V (x − v t) .
2m
Ĥγ =
(d) Finally, we want to compare to Problem 2.50. In that case, the solution in the frame with the stationary
potential is
Ψ(x, t) = ψ0 (x) e−i E t/~
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
233
where E is the bound-state energy and the solution in the frame with the moving potential is
Ψ0 (x, t) = Γ̂ Ψ
√
m α −m α |x|2 /~2 −i E t/~
i
m v2
m v x̂ −
t
T̂ (v t)
e
e
~
2
~
√
m α −m α |x−v t|2 /~2 −i E t/~
i
m v2
= exp
m v x̂ −
e
t
e
~
2
~
√
m α −m α |x−v t|2 /~2 −i [(E+m v2 /2) t−m v x]/~
=
e
e
~
= exp
as shown in Problem 2.50.
Problem 6.36
(a) On a test function f (x),
x̂0 f = Θ̂−1 x̂ Θ̂ f (x) = Θ̂−1 x̂ f ∗ (x) = Θ̂−1 x f ∗ (x) = x f (x) ,
since x is a real number. And
d
f ∗ (x)
p̂0 f = Θ̂−1 p̂ Θ̂ f (x) = Θ̂−1 p̂ f ∗ (x) = Θ̂−1 −i ~
dx
d
= i~
f (x) = −p̂ f (x) .
dx
(b) We have
Û (δ) Θ̂ Û (δ) = Θ̂
δ
δ
1 − i Ĥ Θ̂ 1 − i Ĥ ≈ Θ̂
~
~
δ
δ
Θ̂ − i Ĥ Θ̂ + Θ̂ −i Ĥ ≈ Θ̂
~
~
δ
δ
−i Ĥ Θ̂ + i Θ̂ Ĥ ≈ 0
~
~
where you have to remember to conjugate
h thei i in the second term when you interchange it with Θ̂.
Therefore the statement is equivalent to Ĥ, Θ̂ = 0.
(c) If Ĥ ψn = En ψn and Ĥ commutes with Θ̂, then
Ĥ ψn∗ = Ĥ Θ̂ ψn = Θ̂ Ĥ ψn = Θ̂ En ψn = En ψn∗
F
since En is of course real. If the energy is non-degenerate, then ψn∗ must be a constant (of magnitude 1
since the functions have the same norm)
times ψn : ψn∗ = c ψn . One can then simply multiply ψn by an
√
appropriate factor to make it real; c ψn will do the job.
(d)
i. For a momentum eigenfunction
Θ̂ fp (x) = Θ̂ √
1
1
ei p x/~ = √
e−i p x/~ = f−p (x) .
2π~
2π~
234
CHAPTER 6. SYMMETRIES AND CONSERVATION LAWS
ii. For a hydrogen wave function (see Equation 4.32 for the Y`m )
Θ̂ ψn`m (r) = Θ̂ Rn` (r) Y`m (θ, φ) = Rn` (r) (−1)
= (−1)
m
m
Y`−m (θ, φ)
ψn`−m (r) .
Both states are clearly degenerate with the corresponding untransformed state and describe particle moving in the opposite direction (flipped p or Lz respectively).
Problem 6.37
(a) Acting on a generic spinor
∗
0 ∗
a
−b
a
−(b0 )
Θ̂2
= Θ̂
≡
Θ̂
=
∗
(a0 )
b
a∗
b0
∗
−(a∗ )
a
=
=−
∗ ∗
(−b )
b
where I defined a0 and b0 to make it easier to follow the manipulations.
(b) Lets assume that they are the same state. Then
Θ̂2 |ψn i = Θ̂ |ψn0 i = Θ̂ c |ψn i = c∗ Θ̂ |ψn i = c∗ |ψn0 i
2
= c∗ c |ψn i = |c| |ψn i
2
but we know from part (a) that Θ2 = −1 and this is a contradiction since, of course, |c| > 0.
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
235
Chapter 7
Time-Independent Perturbation
Theory
Problem 7.1
(a)
r
ψn0 (x) =
En1 =
nπ 2
2
sin
x , so En1 = hψn0 |H 0 |ψn0 i = α
a
a
a
nπ a 2α
nπ 2α
sin2
sin2
=
=
a
a 2
a
2
Z a
0
sin2
nπ a
x δ x−
dx.
a
2
0,
if n is even,
2α/a, if n is odd.
For even n the wave function is zero at the location of the perturbation (x = a/2), so it never “feels” H 0 .
(b) Here n = 1, so we need
2α
0
hψm
|H 0 |ψ10 i =
a
Z
sin
mπ π mπ π 2α
mπ a
2α
x δ x−
sin
x dx =
sin
sin
=
sin
.
a
2
a
a
2
2
a
2
This is zero for even m, so the first three nonzero terms will be m = 3, m = 5, and m = 7. Meanwhile,
π 2 ~2
0
E10 − Em
=
(1 − m2 ), so
2ma2
1
−1
(2α/a) sin(mπ/2) 0
2α 2ma2 −1 0
0
0
ψ +
ψ +
ψ + ...
ψm =
0
E10 − Em
a π 2 ~2 1 − 9 3 1 − 25 5 1 − 49 7
m=3,5,7,...
r 4maα 2 1
3π
1
5π
1
7π
= 2 2
sin
x −
sin
x +
sin
x + ...
π ~
a 8
a
24
a
48
a
r mα
a
3π
1
5π
1
7π
= 2 2
sin
x − sin
x + sin
x + ... .
π ~
2
a
3
a
6
a
ψ11 =
X
236
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
Problem 7.2
p
√
1 3
(a) En = (n + 12 )~ω 0 , where ω 0 ≡ k(1 + )/m = ω 1 + = ω(1 + 12 − 18 2 + 16
· · · ),
√
En = (n + 12 )~ω 1 + = (n + 12 )~ω(1 + 21 − 18 2 + · · · ).
so
(b) H 0 = 21 k 0 x2 − 12 kx2 = 12 kx2 (1 + − 1) = ( 12 kx2 ) = V , where V is the unperturbed potential energy. So
En1 = hψn0 |H 0 |ψn0 i = hn|V |ni, with hn|V |ni the expectation value of the (unperturbed) potential energy
in the nth unperturbed state. This is most easily obtained from the virial theorem (Problem 3.37), but it
can also be derived algebraically. In this case the virial theorem says hT i = hV i. But hT i + hV i = En0 . So
hV i = 21 En0 = 21 (n + 12 )~ω;
En1 = 2 (n + 12 )~ω, which is precisely the 1 term in the power series from
part (a).
Problem 7.3
(a) In terms of the one-particle states (Eq. 2.31) and energies (Eq. 2.30):
πx πx 2
π 2 ~2
1
2
Ground state: ψ10 (x1 , x2 ) = ψ1 (x1 )ψ1 (x2 ) =
sin
.
sin
; E10 = 2E1 =
a
a
a
ma2
First excited state: ψ20 (x1 , x2 ) = √12 [ψ1 (x1 )ψ2 (x2 ) + ψ2 (x1 )ψ1 (x2 )]
√ πx 2πx2
2
πx1 2πx1
2
=
sin
sin
+ sin
sin
;
a
a
a
a
a
E20 = E1 + E2 =
5 π 2 ~2
.
2 ma2
(b)
E11 = hψ10 |H 0 |ψ10 i = (−aV0 )
=−
4V0
a
Z a
0
sin4
πx a
2 Z a Z a
πx πx 2
2
1
sin2
δ(x1 − x2 ) dx1 dx2
sin2
a
a
a
0
0
dx = −
4V0 a
a π
Z π
0
sin4 y dy = −
4V0 3π
3
·
= − V0 .
π
8
2
E21 = hψ20 |H 0 |ψ20 i
ZZ a πx 2
2
2πx2
2πx1
πx1 2
= (−aV0 )
sin
+
sin
sin
sin
δ(x1 − x2 ) dx1 dx2
a2
a
a
a
a
0
Z πx 2
2V0 a
πx
2πx
2πx
=−
sin
sin
+ sin
sin
dx
a 0
a
a
a
a
Z
Z
8V0 a 2 πx 2 2πx
8V0 a π 2
=−
sin
sin
dx = −
·
sin y sin2 (2y) dy
a 0
a
a
a π 0
Z π
Z
8V0
32V0 π
=−
·4
sin2 y sin2 y cos2 y dy = −
(sin4 y − sin6 y) dy
π
π
0
0
32V0 3π 5π
=−
−
= −2V0 .
π
8
16
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
237
Problem 7.4
(a) To get the eigenvalues of
0
Ea + λVaa
λVab
A λVab
H = H0 + H =
=
∗
λVba
Eb0 + λVbb
λVab
B
0
we solve the characteristic equation:
(A − E) λVab
= (A − E)(B − E) − λ2 |Vab |2 = E 2 − E(A + B) + AB − C = 0
∗
λVab
(B − E)
(for short, A ≡ Ea0 + λVaa , B ≡ Eb0 + λVbb , C ≡ λ2 |Vab |2 ). Thus
E± =
i
i
p
p
1h
1h
A + B ± (A + B)2 − 4(AB − C) =
A + B ± (A − B)2 + 4C .
2
2
s
"
#
1
4C
(A + B) ± (B − A) 1 +
. To order λ2 ,
(b) E± =
2
(B − A)2
4C
4λ2 |Vab |2
≈
;
2
(B − A)
(Eb0 − Ea0 )2
E± ≈
E+ ≈
s
1+
1 4λ2 |Vab |2
4C
≈1+
2
(B − A)
2 (Eb0 − Ea0 )2
2|Vab |2
(Ea0 + Eb0 ) + λ(Vaa + Vbb ) ± (Eb0 − Ea0 ) + λ(Vbb − Vaa ) + λ2 0
.
(Eb − Ea0 )
2|Vab |2
1
Ea0 + Eb0 + λ(Vaa + Vbb ) + Eb0 − Ea0 + λ(Vbb − Vaa ) + λ2 0
2
(Eb − Ea0 )
= Eb0 + λVbb + λ2
|Vab |2
|Vab |2
→ E+ ≈ Eb0 + Vbb +
.
0
0
(Eb − Ea
(Eb0 − Ea0 )
Similarly, E− ≈ Ea0 + Vaa −
|Vab |2
. First-order perturbation theory (Equation 7.9) says
(Eb0 − Ea0 )
Vaa Vab
1
10
= Vaa ,
Vba Vbb
0
Vaa Vab
0
Eb1 = hψb0 |H 0 |ψb0 i = 0 1
= Vbb ,
Vba Vbb
1
Ea1 = hψa0 |H 0 |ψa0 i =
confirming the middle terms in the boxed equations. Second-order perturbation theory (Equation 7.15)
says
Vaa Vab
|Vab |2
|hψb0 |H 0 |ψa0 i|2
1
0
0 0
0
1
;
hψ
|H
|ψ
i
=
= Vba ; Ea2 = − 0
,
Ea2 =
b
a
0
0
Vba Vbb
0
(Ea − Eb )
(Eb − Ea0 )
Vaa Vab
|hψa0 |H 0 |ψb0 i|2
|Vab |2
0
0
0 0
1
0
;
hψ
|H
|ψ
i
=
,
Eb2 =
= Vab ; Eb2 =
a
b
0
0
Vba Vbb
1
(Eb − Ea )
(Eb0 − Ea0 )
confirming the third terms in the boxed equations.
238
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
(c) Returning to the boxed equation in (a),
s
"
#
2
1
4|V
|
ab
E± =
(Ea0 + Eb0 ) ± (Eb0 − Ea0 ) 1 +
.
2
(Eb0 − Ea0 )2
√
But the power series expansion for 1 + converges only for || < 1, which means that the perturbation
series converges (in this case) only for
4|Vab |2
< 1,
(Eb0 − Ea0 )2
or
Vab
1
< . X
Eb0 − Ez0
2
Problem 7.5
(a)
2
0
hψm
|H|ψn0 i = α
Z a
sin
a
0
nπ mπ nπ mπ a
2α
x δ x−
sin
x dx =
sin
sin
,
a
2
a
a
2
2
which is zero unless both m and n are odd—in which case it is ±2α/a. So Eq. 7.15 says
X
En2 =
m6=n, odd
En2 =




 2m
2α
a
2α
π~
2
1
.
0
0 )
(En − Em
0,
X
2
m6=n, odd
But Eq. 2.30 says
En0 =
if n is even;
1
, if n is odd.
(n2 − m2 )
1
1
1
1
=
−
.
(n2 − m2 )
2n m + n m − n
X 1 X 1
1
=
−
2 3,5,7,... m + 1 m − 1
To sum the series, note that
for n = 1:
1
=
2
1
=
6
π 2 ~2 2
n , so
2ma2
1 1 1
1 1 1 1
+ + + ··· − − − − ···
4 6 8
2 4 6 8
X 1 X
1
1
for n = 3:
=
−
6 1,5,7,... m + 3 m − 3
1
=
2
Thus,
1
1
−
=− ;
2
4
1 1
1
1 1 1 1 1
1
+ +
+ ··· + − − − − −
···
4 8 10
2 2 4 6 8 10
1
=
6
1
−
6
=−
1
.
36
In general, there is perfect cancellation except for the “missing” term 1/2n in the first sum, so the total
1
1
1
0,
if n is even;
2
is
−
=−
. Therefore: En =
2
−2m (α/π~n) , if n is odd.
2n
2n
(2n)2
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
239
(b)
H0 =
1
kx2 ;
2
hm|x2 |ni =
0
hψm
|H 0 |ψn0 i =
1
khm|x2 |ni. Using Eqs. 2.67 and 2.70:
2
~
hm|(a2+ + a+ a− + a− a+ + a2− )|ni
2mω
i
p
~ hp
(n + 1)(n + 2)hm|n + 2i + nhm|ni + (n + 1)hm|ni + n(n − 1)hm|n − 2i .
2mω
h
i
p
p
1
~
0
So, for m 6= n, hψm
|H 0 |ψn0 i =
k
(n + 1)(n + 2) δm,n+2 + n(n − 1) δm,n−2 .
2
2mω
hp
i2
p
2 X
(n
+
1)(n
+
2)
δ
+
n(n
−
1)
δ
m,n+2
m,n−2
~ω
En2 =
4
(n + 12 )~ω − (m + 21 )~ω
m6=n
=
2 ~ω X [(n + 1)(n + 2) δm,n+2 + n(n − 1) δm,n−2 ]
16
(n − m)
m6=n
2 ~ω (n + 1)(n + 2)
n(n − 1)
2 ~ω
1
1
+
=
− (n + 1)(n + 2) + n(n − 1)
=
16
n − (n + 2)
n − (n − 2)
16
2
2
2 ~ω
2 ~ω
1
2
2
21
=
−n − 3n − 2 + n − n =
(−4n − 2) = − ~ω n +
32
32
8
2
=
(which agrees with the 2 term in the exact solution—Problem 7.2(a)).
Problem 7.6
(a)
En1 = hψn0 |H 0 |ψn0 i = −qEhn|x|ni = 0 (Problem 2.12).
From Eq. 7.15 and Problem 3.39:
En2 = (qE)2
X |hm|x|ni|2
m6=n
(n − m)~ω
√
=
√
(qE)2 ~ X [ n + 1 δm,n+1 + n, δm,n−1 ]2
(qE)2 X [(n + 1) δm,n+1 + n δm,n−1 ]
=
~ω 2mω
(n − m)
2mω 2
(n − m)
m6=n
m6=n
(qE)2
(n + 1)
n
(qE)2
(qE)2
=
+
=
[−(n
+
1)
+
n]
=
−
.
2mω 2 n − (n + 1) n − (n − 1)
2mω 2
2mω 2
1
~2 d2 ψ
2 2
(b) −
+
mω x − qEx ψ = Eψ. With the suggested change of variables,
2m dx2
2
=
2
1
1
qE
qE
2 2
2
0
0
mω x − qEx = mω x +
− qE x +
2
2
mω 2
mω 2
1
qE
1
(qE)2
(qE)2
1
1 (qE)2
2
2
mω 2 x0 + mω 2 x0
+ mω 2 2 4 − qEx0 −
= mω 2 x0 −
.
2
2
2
mω
2
m ω
mω
2
2 mω 2
240
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
So the Schrödinger equation says
~2 d2 ψ
1
1 (qE)2
2 02
ψ,
−
+
mω
x
ψ
=
E
+
2m dx0 2
2
2 mω 2
which is the Schrödinger equation for a simple harmonic oscillator, in the variable x0 . The constant on
the right must therefore be (n + 21 )~ω, and we conclude that
1
1 (qE)2
En = (n + )~ω −
.
2
2 mω 2
The subtracted term is exactly what we got in part (a) using perturbation theory. Evidently all the higher
corrections (like the first-order correction) are zero, in this case.
Problem 7.7
(a)
Z a/2 r
mπx r 2
πx 2
sin
V0
sin
dx
a
a
a
a
0
Z
2V0 a/2 1
(m − 1)πx
(m + 1)πx
=
cos
− cos
dx
a 0
2
a
a
a/2
V0
(m − 1)πx
(m + 1)πx
a
a
=
sin
sin
−
a (m − 1)π
a
(m + 1)π
a
0
1
(m − 1)π
1
(m + 1)π
V0
sin
−
sin
.
=
π (m − 1)
2
(m + 1)
2
0
hψm
|H 0 |ψ10 i =
If m is odd, the result is zero. So the first three nonzero terms are m = 2, m = 4, and m = 6.
V0
π
3π
V0
1
4V0
1
0
0 0
hψ2 |H |ψ1 i =
sin
− sin
=
1+
=
,
π
2
3
2
π
3
3π
V0 1
3π
1
5π
V0
1 1
8V0
hψ40 |H 0 |ψ10 i =
sin
− sin
=
− −
=−
,
π 3
2
5
2
π
3 5
15π
V0 1
5π
7π
1
V0 1 1
12V0
hψ60 |H 0 |ψ10 i =
sin
+
.
− sin
=
=
π 5
2
7
2
π 5 7
35π
Plugging this into Equation 7.13:
r 4V0 2ma2 2
1
2πx
2
4πx
3
6πx
1
ψ1 =
sin
−
sin
+
sin
+ ···
π π 2 ~2 a 3(1 − 4)
a
15(1 − 16)
a
35(1 − 36)
a
r 8V0 ma2 2
1
2πx
2
4πx
3
6πx
=
−
sin
+
sin
−
sin
+
·
·
·
.
π 3 ~2
a
(3)2
a
(15)2
a
(35)2
a
V0
8
4~2
~2
(N + 1)2
,
λ
=
=
V0 . In the left half (j ≤ 50), vj =
=
; in the right half
ma2
2m(∆x)2
8
λ
(101)2
vj = 0. Here is the Mathematica code:
(b) V0 =
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
241
�������
h = Table If i == j, 2 + If i < 51, 8
10 201 , 0 , 0 , {i, 100}, {j, 100}
�������
k = Table[If[i ⩵ j + 1, - 1, 0], {i, 100}, {j, 100}]
�������
m = Table[If[i ⩵ j - 1, - 1, 0], {i, 100}, {j, 100}]
�������
p = Table[h[[i, j]] + k[[i, j]] + m[[i, j]], {i, 100}, {j, 100}]
�������
EVE = Eigenvectors[N[p]]
�������
ListLinePlot[EVE[[100]], PlotRange → {0, 0.15}]
0.14
0.12
h = Table
0.10
If i == j, 2 + If i < 51, 8
10 201 , 0 , 0 , {i, 100}, {j, 100}
k = Table[If[i ⩵ j + 1, - 1, 0], {i, 100}, {j, 100}]
0.08
�������
m = Table[If[i ⩵ j - 1, - 1, 0], {i, 100}, {j, 100}]
0.06
p = Table[h[[i, j]] + k[[i, j]] + m[[i, j]], {i, 100}, {j, 100}]
0.04
EVE = Eigenvectors[N[p]]
0.02
ListLinePlot[EVE[[100]], PlotRange → {0, 0.15}]
0.14 0
20
40
�������
0.12
EIG
= Eigenvalues[N[p]]
�������
EIG[[100]]
�������
0.00132025
60
80
100
0.10
0.08
0.06
2 Sin λ
πx
100 , {x, 0, 100}
The lowest eigenvaluePlot
is 0.00132025
= (10201/8)(0.00132025)
V0 = 1.6835 V0 . First-order perturbation
0.04
2
2 2
π ~
V0
π
1
0.02
theory (Example 7.1)1.4
says it is approximately
+
=
+
V0 = 1.7337 V0 . The agreement is
2ma2
2
8
2
1.2
not too bad, considering
large
perturbation.
0 that this
20 is a relatively
40
60
80
100
1.0
Eigenvalues[N[p]]
(c) In the figures below, EIG
the =centered
graph is the unperturbed ground state, the dashed graph is the per0.8
turbation theory waveEIG[[100]]
function, and the solid line is the numerical solution. (The latter is automatically
normalized by Mathematica.)
The agreement between the latter two is remarkably good.
0.6
0.00132025
0.4
Plot
2
100 Sin π x
100 , {x, 0, 100}
0.2
0.14
0.12
20
40
60
80
100
0.10
0.08
0.06
0.04
0.02
20
40
60
80
100
2
242
7P7.nb
In[27]:=
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
Plot
2 / 100
Sin[π x / 100] +
32
π3
-
1
32
Sin[2 π x / 100] +
2
152
Sin[4 π x / 100] -
{x, 0, 100}, PlotStyle → Dashed
0.15
0.10
Out[27]=
0.05
20
In[29]:=
40
60
80
100
40
60
80
100
Show[%14, %27, %28]
0.14
0.12
0.10
0.08
Out[29]=
0.06
0.04
0.02
0
In[20]:=
20
T = EVE[[100]]
100
In[21]:=
 (T[[i]]) ^ 2
i=1
Out[21]=
1.
Problem 7.8
(a)
0
0
hψ+
|ψ−
i = h(α+ ψa0 + β+ ψb0 )|(α− ψa0 + β− ψb0 )i
∗
∗
∗
∗
= α+
α− hψa0 |ψa0 i + α+
β− hψa0 |ψb0 i + β+
α− hψb0 |ψa0 i + β+
β− hψb0 |ψb0 i
∗
∗
1
= α+
α− + β+
β− . But Eq. 7.27 ⇒ β± = α± (E±
− Waa )/Wab ,
so
3
352
Sin[6 π x / 100]
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
243
1
1
α∗ α− (E+
− Waa )(E−
− Waa )
1
1
= + 2 |Wab |2 + (E+
1+
− Waa )(E−
− Waa ) .
∗
Wab Wab
|Wab |
The term in square brackets is:
√
√
1 1
1
1
2
1
is
[ ] = E+
E− − Waa (E+
+ E−
) + |Wab |2 + Waa
. But Eq. 7.33 ⇒ E±
= 12 [(Waa + Wbb ) ± ], where
1
1
shorthand for the square root term. So E+
+ E−
= Waa + Wbb , and
0
0
∗
hψ+
|ψ−
i = α+
α−
1 1
E+
E− =
Thus
√ 1
1
(Waa + Wbb )2 − ( )2 =
(Waa + Wbb )2 − (Waa − Wbb )2 − 4|Wab |2 = Waa Wbb − |Wab |2 .
4
4
2
[ ] = Waa Wbb − |Wab |2 − Waa (Waa + Wbb ) + |Wab |2 + Waa
= 0,
0
0
so hψ+
|ψ−
i = 0. QED
(b)
0
0
∗
∗
∗
∗
hψ+
|H 0 |ψ−
i = α+
α− hψa0 |H 0 |ψa0 i + α+
β− hψa0 |H 0 |ψb0 i + β+
α− hψb0 |H 0 |ψa0 i + β+
β− hψb0 |H 0 |ψb0 i
∗
∗
∗
∗
= α+
α− Waa + α+
β− Wab + β+
α− Wba + β+
β− Wbb
1
1
1
(E 1 − Waa ) (E−
− Waa )
(E − Waa )
(E − Waa )
∗
+ Wbb + ∗
+ Wba + ∗
= α+
α− Waa + Wab −
Wab
Wab
Wab
Wab
1
1
(E − Waa )(E− − Waa )
∗
1
1
.
= α+
α− Waa + E−
− Waa + E+
− Waa + Wbb +
|Wab |2
But we know from (a) that
1
1
− Waa )
− Waa )(E−
(E+
= −1,
2
|Wab |
0
0
∗
1
1
hψ+
|H 0 |ψ−
i = α+
α− [E−
+ E+
− Waa − Wbb ] = 0.
so
QED
(c)
0
0
∗
∗
∗
∗
hψ±
|H 0 |ψ±
i = α±
α± hψa0 |H 0 |ψa0 i + α±
β± hψa0 |H 0 |ψb0 i + β±
α± hψb0 |H 0 |ψa0 i + β±
β± hψb0 |H 0 |ψb0 i
1
1
(E±
− Waa )
(E±
− Wbb )
2
2
= |α± | Waa + Wab
+ |β± | Wba
+ Wbb
Wab
Wba
(this time I used Eq. 7.29 to express α in terms of β, in the third term).
1
0
0
1
1
1
∴ hψ±
|H 0 |ψ±
i = |α± |2 (E±
) + |β± |2 (E±
) = |α± |2 + |β± |2 E±
= E±
. QED
Problem 7.9
(a) See Problem 2.46.
(b) With a → n, b → −n, we have:
Z
Z
V0 L/2 −x2 /a2
V0 ∞ −x2 /a2
V0 √
Waa = Wbb = −
e
dx ≈ −
e
dx = − a π.
L −L/2
L −∞
L
Z
Z
2
V0 L/2 −x2 /a2 −4πnix/L
V0 ∞ −(x2 /a2 +4πnix/L)
V0 √
Wab = −
e
e
dx ≈ −
e
dx = − a πe−(2πna/L) .
L −L/2
L −∞
L
(We did this integral in Problem 2.21.) In this case Waa = Wbb , and Wab is real, so Eq. 7.33 ⇒
1
E±
= Waa ± |Wab |,
√ V0 a 2
1
or E±
= − π
1 ∓ e−(2πna/L) .
L
244
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
#
" √
2
(E 1 − Waa )
± π(V0 a/L)e−(2πna/L)
√
= ∓α. Evidently, the “good” linear
(c) Equation 7.27 ⇒ β = α
=α
Wab
− π(V0 a/L)e−(2πna/L)2
combinations are:
r
i
1 1 h i2πnx/L
2
2πnx
−i2πnx/L
= i
−e
ψ+ = αψn − αψ−n = √ √ e
sin
and
L
L
2 L
r
ψ− = αψn + αψ−n =
2
cos
L
2πnx
.
L
Z L/2
Using Eq. 7.9, we have :
2πnx
dx,
L
−L/2
Z L/2
2
2πnx
1
0
−x2 /a2
2
E− = hψ− |H |ψ− i = (−V0 )
e
dx.
cos
L
L
−L/2
1
E+
= hψ+ |H 0 |ψ+ i =
2
(−V0 )
L
2
2
e−x /a sin2
sin2 θ = (1 − cos 2θ)/2, and cos2 θ = (1 + cos 2θ)/2, so
Z ∞
Z
Z ∞
V0
V0 ∞ −x2 /a2
4πnx
4πnx
1
−x2 /a2
−x2 /a2
dx = −
dx
E± ≈ −
e
e
1 ∓ cos
dx ∓
e
cos
L −∞
L
L
L
−∞
−∞
But
=−
i
i
√
√ V0 a h
2
2
V0 h√
π a ∓ a πe−(2πna/L) = − π
1 ∓ e−(2πna/L) ,
L
L
same as (b).
(d) Af (x) = f (−x) (the parity operator). The eigenstates are even functions (with eigenvalue +1) and odd
functions (with eigenvalue −1). The linear combinations we found in (c) are precisely the odd and even
linear combinations of ψn and ψ−n .
Problem 7.10
Expanding the exact energies (Equation 7.21) to first order in :
√
√
1
1
1
1
Emn = m +
~ω+ + n +
~ω− = m +
~ω 1 + + n +
~ω 1 − 2
2
2
2
1
1
≈ m+
~ω 1 +
+ n+
~ω 1 −
= (m + n + 1)~ω + (m − n)~ω.
2
2
2
2
2
So for m = 1, n = 0 the first-order correction is +(~ω/2), and for n = 1, m = 0 it is −(~ω/2), confirming
Equation 7.34.
Problem 7.11
Ground state is nondegenerate; Eqs. 7.9 ⇒
3
ZZZ a
π π π 2
a
a
3a
E1 =
a3 V0
sin2
x sin2
y sin2
z δ(x − )δ(y − )δ(z − ) dx dy dz
a
a
a
a
4
2
4
0
π
π
3π
1
1
= 8V0 sin2
sin2
sin2
= 8V0
(1)
= 2V0 .
4
2
4
2
2
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
245
First excited states:
ZZZ
π π 2π
a
a
3a
Waa = 8V0
sin2
x sin2
y sin2
z δ(x − )δ(y − )δ(z − ) dx dy dz
a
a
a
4
2
4
1
= 8V0
(1)(1) = 4V0 .
2
π π 2π
a
a
3a
2
Wbb = 8V0
sin
x sin
y sin2
z δ(x − )δ(y − )δ(z − ) dx dy dz
a
a
a
4
2
4
1
1
= 8V0
(0)
= 0.
2
2
ZZZ
2
π π 2π
a
a
3a
sin2
x sin2
y sin2
z δ(x − )δ(y − )δ(z − ) dx dy dz
a
a
a
4
2
4
1
= 8V0 (1)(1)
= 4V0 .
2
π
π
3π
3π
sin
sin(π) sin
sin
= 0.
Wab = 8V0 sin2
4
2
2
4
π
π
3π
3π
1
1
2 π
√
√
sin
sin
sin
Wac = 8V0 sin
sin
= 8V0
(1)(1)(−1)
= −4V0 .
4
2
2
2
4
2
2
π
π
π
3π
= 0.
Wbc = 8V0 sin
sin
sin(π) sin
sin2
4
2
2
4


1 0 −1
(1 − λ) 0
−1
0
−λ
0
W = 4V0  0 0 0  = 4V0 D; det(D − λ) =
= −λ(1 − λ)2 + λ = 0 ⇒
−1 0 1
−1
0 (1 − λ)
ZZZ
Wcc = 8V0
λ = 0,
or
(1 − λ)2 = 1 ⇒ 1 − λ = ±1 ⇒ λ = 0
or λ = 2.
So the first-order corrections to the energies are 0, 0, 8V0 .
Problem 7.12
 
1
(a) χ1 =  0 , eigenvalue V0 ;
0
 
0
χ2 =  1 , eigenvalue V0 ;
0
(b) Characteristic equation: det(H − λ) =
 
0
χ3 =  0 , eigenvalue 2V0 .
1
[V0 (1 − ) − λ]
0
0
0
[V0 − λ]
V0
= 0;
0
V0 [2V0 − λ]
[V0 (1 − ) − λ][(V0 − λ)(2V0 − λ) − (V0 )2 ] = 0 ⇒ λ1 = V0 (1 − ).
(V0 − λ)(2V0 − λ) − (V0 )2 = 0 ⇒ λ2 − 3V0 λ + (2V02 − 2 V02 ) = 0 ⇒
p
i V p
3V0 ± 9V02 − 4(2V02 − 2 V02 )
V0 h
0
=
3 ± 1 + 42 ≈
3 ± (1 + 22 ) .
λ=
2
2
2
p
p
V0
V0
λ2 =
3 − 1 + 42 ≈ V0 (1 − 2 ); λ3 =
3 + 1 + 42 ≈ V0 (2 + 2 ).
2
2
246
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
(c)


−1 0 0
H0 = V0  0 0 1  ;
0 10

E31 = hχ3 |H 0 |χ3 i = V0
 
−1 0 0
0
0 0 1  0 0 10
0 1 0
1
 
0
= V0 0 0 1  1  = 0 (no first-order correction).
0

 
 
−1 0 0
0
0
X |hχm |H 0 |χ3 i|2
0





1
0
0
0
0
1
0
1
0
0
1  = 0,
;
hχ
|H
E32 =
|χ
i
=
V
=
V
1
3
0
0
0
E30 − Em
m=1,2
0 1 0
1
0

 
0
−1 0 0
hχ2 |H 0 |χ3 i = V0 0 1 0  0 0 1  0  = V0 .
0 1 0
1
So E32 = (V0 )2 /V0 = 2 V0 .
Through second-order, then,
E3 = E30 + E31 + E32 = 2V0 + 0 + 2 V0 = V0 (2 + 2 )
(same as we got for λ3 in (b)).
E30 − E20 = 2V0 − V0 = V0 .
(d)


 
1
−1 0 0
−1
1 0 0  0 0 1   0  = V0 1 0 0  0  = −V0 .
0
0
0 1 0

Waa = hχ1 |H 0 |χ1 i = V0

Wbb = hχ2 |H 0 |χ2 i = V0
Wab = hχ1 |H 0 |χ2 i = V0
 
 
−1 0 0
0
0
0 1 0  0 0 1   1  = V0 0 1 0  0  = 0.
0 1 0
0
1
 
 

0
0
−1 0 0
1 0 0  0 0 1   1  = V0 1 0 0  0  = 0.
1
0
0 1 0
Plug the expressions for Waa , Wbb , and Wab into Eq. 7.33:
q
1
1
1
2
2
E± =
−V0 + 0 ± V0 + 0 = (−V0 ± V0 ) = {0, −V0 }.
2
2
To first-order, then, E1 = V0 − V0 ,
we got in (b).
E2 = V0 , and these are consistent (to first order in ) with what
Problem 7.13
Given a set of orthonormal states {ψj0 } that are degenerate eigenfunctions of the unperturbed Hamiltonian:
Hψj0 = E 0 ψj0 ,
hψj0 |ψl0 i = δjl ,
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
247
construct the general linear combination,
ψ0 =
n
X
αj ψj0 .
j=1
It too is an eigenfunction of the unperturbed Hamiltonian, with the same eigenvalue:
H 0 ψ0 =
n
X
n
X
αj H 0 ψj0 = E 0
j=1
αj ψj0 = E 0 ψ 0 .
j=1
We want to solve the Schrödinger equation Hψ = Eψ for the perturbed Hamiltonian H = H 0 + λH 0 .
Expand the eigenvalues and eigenfunctions as power series in λ:
E = E 0 + λE 1 + λ2 E 2 + . . . ,
ψ = ψ 0 + λψ 1 + λ2 ψ 2 + . . . .
Plug these into the Schrödinger equation and collect like powers:
(H 0 + λH 0 )(ψ 0 + λψ 1 + λ2 ψ 2 + . . .) = (E 0 + λE 1 + λ2 E 2 + . . .)(ψ 0 + λψ 1 + λ2 ψ 2 + . . .)
⇒
H 0 ψ 0 + λ(H 0 ψ 1 + H 0 ψ 0 ) + . . . = E 0 ψ 0 + λ(E 0 ψ 1 + E 1 ψ 0 ) + . . .
The zeroth-order terms cancel; to first order
H 0 ψ1 + H 0 ψ0 = E 0 ψ1 + E 1 ψ0 .
Take the inner product with ψj0 :
hψj0 |H 0 ψ 1 i + hψj0 |H 0 ψ 0 i = E 0 hψj0 |ψ 1 i + E 1 hψj0 |ψ 0 i.
But
hψj0 |H 0 ψ 1 i = hH 0 ψj0 |ψ 1 i = E 0 hψj0 |ψ 1 i, so the first terms cancel, leaving
hψj0 |H 0 ψ 0 i = E 1 hψj0 |ψ 0 i.
Now use
ψ0 =
n
X
αl ψl0 ,
and exploit the orthonormality of {ψl0 }:
l=1
n
X
αl hψj0 |H 0 |ψl0 i = E 1
l=1
n
X
αl hψj0 |ψl0 i = E 1 αj ,
l=1
or, defining
Wjl ≡ hψj0 |H 0 |ψl0 i,
n
X
Wjl αl = E 1 αl .
l=1
This (the generalization of Eq. 7.27 for the case of n-fold degeneracy) is the eigenvalue equation for the matrix
W (whose jlth element, in the {ψj0 } basis, is Wjl ); E 1 is the eigenvalue, and the eigenvector (in the {ψj0 } basis)
is χj = αj . Conclusion: The first-order corrections to the energy are the eigenvalues of W. QED
Problem 7.14
"
m
(a) From Eq. 4.70: En = −
2~2
e2
4π0
2 #
1
1
= − mc2
2
n
2
1 e2
~c 4π0
2
1
α2 mc2
= −
.
2
n
2n2
(b) I have found a wonderful solution—unfortunately, there isn’t enough room on this page for the proof.
248
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
Problem 7.15
Equation 4.218 ⇒ hV i = 2En ,
e2
−
4π0
for hydrogen.
"
2 2 #
e
1
m
1
= −2
2
r
2~
4π0
n2
"
2 2 #
e
e2 1
m
1
.
V =−
; En = −
4π0 r
2~2 4π0
n2
⇒
1
1
1
me2
=
=
2
2
r
4π0 ~
n
an2
So
(Eq. 4.72). QED
Problem 7.16
In Problem 4.52 we found (for n = 3, ` = 2, m = 1) that
(s + 6)!
hr i =
6!
s
3a
2
s
.
6!
(1) = 1 (of course). X
6!
−1
5! 3a
1 2
1
1
1
1
=
= ·
=
. X
s = −1 :
Eq. 7.56 says 2 =
r
6! 2
6 3a
9a
3 a
9a
−2
1
1
4
4! 3a
2
1
2
s = −2 :
=
·
=
=
Eq.
7.57
says
=
. X
r2
6! 2
6 · 5 9a2
135a2
(5/2) · 27 · a2
135a2
−3
1
3! 3a
1
8
1
1
1
=
=
·
=
Eq. 7.66 says
=
.X
s = −3 :
r3
6! 2
6 · 5 · 4 27a3
405a3
2(5/2)3 · 27 · a3
405a3
s = 0 : h1i =
For s = −7 (or smaller) the integral does not converge: h1/r7 i = ∞ in this state; this is reflected in the fact
that (−1)! = ∞.
Problem 7.17
1 2
1
1
Equation 7.55 ⇒ Er1 = −
E − 2EhV i + hV 2 i . Here E = (n + )~ω, V = mω 2 x2
2
2mc
2
2
"
#
2
1
1
1
1
1
Er1 = −
n+
~2 ω 2 − 2 n +
~ω mω 2 hx2 i + m2 ω 4 hx4 i .
2
2mc
2
2
2
4
⇒
1 ~
But Problem 2.12 ⇒ hx2 i = (n + )
, so
2 mω
"
#
2
2
1
1
1
1 2 4 4
mω 4
1
2 2
2 2
Er = −
n+
~ ω − n+
~ ω + m ω hx i = − 2 hx4 i.
2
2mc
2
2
4
8c
From Eq. 2.70:
x4 =
~2
a2 + a+ a− + a− a+ + a2− a2+ + a+ a− + a− a+ + a2− ,
4m2 ω 2 +
~2
hn| a2+ a2− + a+ a− a+ a− + a+ a− a− a+ + a− a+ a+ a− + a− a+ a− a+ + a2− a2+ |ni.
2
2
4m ω
(Note that only terms with equal numbers of raising and lowering operators will survive). Using Eq. 2.67,
hx4 i =
hx4 i =
h p
~2
hn| a2+
n(n − 1) |n − 2i + a+ a− (n |ni) + a+ a− (n + 1) |ni
2
2
4m ω
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
249
p
i
+a− a+ (n |ni) + a− a+ (n + 1) |ni + a2−
(n + 1)(n + 2) |n + 2i
=
hp
p
~2
hn|
n(n
−
1)
n(n
−
1)
|ni
+ n (n |ni) + (n + 1) (n |ni)
4m2 ω 2
p
i
p
+n (n + 1) |ni + (n + 1) (n + 1) |ni + (n + 1)(n + 2)
(n + 1)(n + 2) |ni
~2 n(n − 1) + n2 + (n + 1)n + n(n + 1) + (n + 1)2 + (n + 1)(n + 2)
4m2 ω 2
2
2
~
~
2
2
2
2
2
2
(n − n + n + n + n + n + n + n + 2n + 1 + n + 3n + 2) =
(6n2 + 6n + 3).
=
2mω
2mω
~2
3 ~2 ω 2
mω 4
2
·
3(2n
(2n2 + 2n + 1).
+
2n
+
1)
=
−
Er1 = − 2 ·
8c
4m2 ω 2
32 mc2
=
Problem 7.18
Quoting the Laplacian in spherical coordinates (Eq. 4.13), we have, for states with no dependence on θ or φ:
~2 d
2 d
2
2 2
r
.
p = −~ ∇ = − 2
r dr
dr
Question: Is it Hermitian?
Using integration by parts (twice), and test functions f (r) and g(r):
Z ∞
Z ∞
1 d
d
dg
dg
hf |p2 gi = −~2
f 2
r2
4πr2 dr = −4π~2
f
r2
dr
r dr
dr
dr
dr
0
0
Z ∞
dg ∞
df dg
dr
−
r2
= −4π~2 r2 f
dr 0
dr dr
0
Z ∞
∞
∞
dg
df
d
df
= −4π~2 r2 f
− r2 g
+
r2
g dr
dr 0
dr 0
dr
dr
0
∞
dg
df
= −4π~2 r2 f
− r2 g
+ hp2 f |gi.
dr
dr 0
The boundary term at infinity vanishes for functions f (r) and g(r) that go to zero exponentially; the boundary
term at zero is killed by the factor r2 , as long as the functions (and their derivatives) are finite. So
hf |p2 gi = hp2 f |gi,
and hence p2 is Hermitian.
As for p4 , first note that
1 d
d
1 d
∇2 e−kr = 2
r2 e−kr = 2
−kr2 e−kr
r dr
dr
r dr
k
2k
= − 2 2re−kr − kr2 e−kr = k 2 −
e−kr .
r
r
So
∇4 e−kr = ∇2
k2 −
2k
r
1 −kr
e−kr = k 2 ∇2 e−kr − 2k∇2
e
.
r
250
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
Now, for any functions f and g,
∇2 (f g) = ∇ · ∇(f g) = ∇ · (f ∇g + g∇f ) = 2∇f · ∇g + f ∇2 g + g∇2 f,
so (using ∇2 (1/r) = −4π δ 3 (r))
1 2 −kr 1 −kr
1
1
2
−kr
−kr 2
∇
e
·∇ e
= 2∇
+ ∇ e
+e
∇
r
r
r
r
1
1
2k
= 2 − 2 r̂ · −ke−kr r̂ +
k2 −
e−kr + e−kr −4πδ 3 (r)
r
r
r
2
2k
k
k 2 −kr
2k
= 2 e−kr + e−kr − 2 e−kr − 4πδ 3 (r) =
e
− 4πδ 3 (r).
r
r
r
r
Therefore
2
2k
k −kr
e−kr − 2k
e
∇4 e−kr = k 2 k 2 −
− 4πδ 3 (r)
r
r
3
4k
e−kr + 8πkδ 3 (r).
= k4 −
r
Then
he
−jr
4 −kr
|∇ e
4k 3
−kr
3
i= e
k −
e
+ 8πkδ (r) d3 r
r
Z ∞
4k 3
4
= 4π
e−(j+k)r r2 dr + 8πk
k −
r
0
Z ∞
Z ∞
3
2 −(j+k)r
2
−(j+k)r
= 4πk k
r e
dr − 4k
re
dr + 2
0
0
1
2
− 4k 2
+2
= 4πk k 3
(j + k)3
(j + k)2
3
k − 2k 2 (j + k) + (j + k)3
= 8πk
(j + k)3
8πk
=
k 3 − 2k 2 j − 2k 3 + j 3 + 3j 2 k + 3jk 2 + k 3
3
(j + k)
8πjk
=
j 2 + 3jk + k 2 = h∇4 e−jr |e−kr i.
3
(j + k)
Z
−jr
4
(The final step follows immediately, because the penultimate expression is symmetric under j ↔ k.) Thus
p4 = (~4 /4m4 )∇4 is hermitian for functions of the form ψn00 . [Actually, I’ve only proved it for functions of the
form e−kr , whereas ψn00 = fn (r)e−r/na (the exponential is multiplied by a polynomial of degree n − 1). But
remember, the potential problems come from boundary terms at r → 0, and higher powers of r are only going
to make such terms go to zero even more rapidly. It’s e−kr with no factor of r that’s going to cause trouble if
anything does. So we’re safe.]
Problem 7.19
(a)
[L · S, Lx ] = [Lx Sx + Ly Sy + Lz Sz , Lx ] = Sx [Lx , Lx ] + Sy [Ly , Lx ] + Sz [Lz , Lx ]
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
251
= Sx (0) + Sy (−i~Lz ) + Sz (i~Ly ) = i~(Ly Sz − Lz Sy ) = i~(L × S)x .
Same goes for the other two components, so [L · S, L] = i~(L × S).
(b) [L · S, S] is identical, only with L ↔ S: [L · S, S] = i~(S × L).
(c) [L · S, J] = [L · S, L] + [L · S, S] = i~(L × S + S × L) = 0.
(d) L2 commutes with all components of L (and S) , so
(e) Likewise,
L · S, L2 = 0.
L · S, S 2 = 0.
(f ) L · S, J 2 = L · S, L2 + L · S, S 2 + 2 [L · S, L · S] = 0 + 0 + 0 =⇒ L · S, J 2 = 0.
Problem 7.20
With the plus sign, j = ` + 1/2 (` = j − 1/2) :
1
Equation 7.67 ⇒ Eso
=
=
(En )2
1
1
Efs
= Er1 + Eso
=
2
Eq. 7.58 ⇒ Er1 = −
(En )2
2mc2
4n
−3 .
j
(En )2 n j(j + 1) − (j − 12 )(j + 21 ) − 34
mc2
(j − 21 )j(j + 12 )
(En )2 n(j 2 + j − j 2 + 41 − 34 )
(En )2
n
=
.
1
1
2
2
mc
mc j(j + 21 )
(j − 2 )j(j + 2 )
4n
2n
−
+3+
2mc
j
j(j + 12 )
(En )2
2n
1
(En )2
4n
=
3+
1−2 j+
=
3−
.
2mc2
2
2mc2
j(j + 12 )
j + 21
With the minus sign, j = ` − 1/2 (` = j + 1/2) :
(En )2 n
1
Equation 7.67 ⇒ Eso
=
2
mc
Eq. 6.57 ⇒ Er1 = −
(En )2
2mc2
4n
−3 .
j+1
j(j + 1) − (j + 12 )(j + 23 ) − 34
(j + 12 )(j + 1)(j + 32 )
(En )2 n(j 2 + j − j 2 − 2j − 34 − 43 )
(En )2
−n
=
.
1
3
2
mc
mc2 (j + 1)(j + 12 )
(j + 2 )(j + 1)(j + 2 )
(En )2
4n
2n
(En )2
2n
1
1
Efs
=
−
+
3
−
=
3
−
1
+
2
j
+
2mc2
j+1
2mc2
2
(j + 1)(j + 21 )
(j + 1)(j + 12 )
4n
(En )2
4n
(En )2
1
3
−
.
For
both
signs,
then,
E
=
3−
. QED
=
fs
1
2
2
2mc
2mc
j+2
j + 21
=
252
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
Problem 7.21
1 1
5
36 2π~c
−
= − E1 ⇒ λ = −
; E1 = −13.6 eV;
λ
9 4
36
5 E1
36 (2π)(1.97 × 10−11 × 106 eV · cm)
~c = 1.97 × 10−11 MeV·cm; λ =
= 6.55 × 10−5 cm= 655 nm.
5
(13.6 eV)
c
4n
3.00 × 108 m/s
(En )2
1
14
ν= =
4.58
×
10
Hz.
Equation
7.68
⇒
E
3
−
=
=
:
fs
λ
6.55 × 10−7 m
2mc2
j + 12
2π~c
E30 − E20 = hν =
= E1
For n = 2: ` = 0 or ` = 1, so j = 1/2 or 3/2. Thus n = 2 splits into two levels :
j = 1/2 : E21 =
j = 3/2 :
(E2 )2
2mc2
8
1
3−
(E2 )2
E21 =
2
2mc
=−
8
3−
2
5
5 (E2 )2
=−
2
2 mc
2
=−
2
5
1
(E1 )2
(13.6 eV)2
=
−
= −5.66×10−5 eV.
4
mc2
32 (.511 × 106 eV)
1
1 (E2 )2
= − (3.62 × 10−4 eV) = −1.13 × 10−5 eV.
2
2 mc
32
For n = 3: ` = 0, 1 or 2, so j = 1/2, 3/2 or 5/2. Thus n = 3 splits into three levels :
(E3 )2
9 1 (E1 )2
1
12
9 (E3 )2
1
=−
= − (3.62×10−4 eV) = −2.01×10−5 eV.
j = 1/2 : E3 =
3−
=−
2mc2
1
2 mc2
2 92
mc2
18
(E3 )2
3 (E3 )2
1
12
1
j = 3/2 : E3 =
=
−
= − (3.62 × 10−4 eV) = −0.67 × 10−5 eV.
3
−
2mc2
2
2 mc2
54
(E3 )2
1
12
1 (E3 )2
j = 5/2 : E31 =
=−
(3.62 × 10−4 eV) = −0.22 × 10−5 eV.
3
−
=−
2
2mc
3
2 mc2
162
0
E3
j= 5/2
j= 3/2
j=1/2
0
E2
1 2 3
4
5
6
j= 3/2
j= 1/2
There are six transitions here; their energies are (E30 + E31 ) − (E20 + E21 ) = (E30 − E20 ) + ∆E, where
∆E ≡ E31 − E21 . Let β ≡ (E1 )2 /mc2 = 3.62 × 10−4 eV. Then:
1
3
( → ):
2
2
(
3
3
→ ):
2
2
(
5
3
→ ):
2
2
(
1
1
→ ):
2
2
1
7
∆E =
− −
β=−
β = −8.80 × 10−6 eV.
32
288
1
1
11
− −
β=
β = 4.61 × 10−6 eV.
∆E =
−
54
32
864
1
1
65
∆E =
−
+
β=
β = 9.08 × 10−6 eV.
162
32
2592
5
1
29
∆E =
−
β=
β = 36.45 × 10−6 eV.
32
18
288
1
−
18
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
3
1
( → ):
2
2
1
5
( → ):
2
2
253
1
5
119
∆E =
−
+
β=
β = 49.86 × 10−6 eV.
54
32
864
1
5
389
∆E =
−
+
β=
β = 54.33 × 10−6 eV.
162
32
2592
Conclusion: There are six lines; one of them ( 12 → 32 ) has a frequency less than the unperturbed line, the
other five have (slightly) higher frequencies. In order they are: 23 → 32 ; 52 → 23 ; 12 → 21 ; 32 → 12 ; 52 → 12 . The
frequency spacings are:
ν2 − ν1
ν3 − ν2
ν4 − ν3
ν5 − ν4
ν6 − ν5
=
=
=
=
=
(∆E2 − ∆E1 )/2π~
(∆E3 − ∆E2 )/2π~
(∆E4 − ∆E3 )/2π~
(∆E5 − ∆E4 )/2π~
(∆E6 − ∆E5 )/2π~
=
=
=
=
=
3.23 × 109 Hz
1.08 × 109 Hz
6.60 × 109 Hz
3.23 × 109 Hz
1.08 × 109 Hz
Problem 7.22
s
s
"
2 #
1
α
1
α2
1
1−
1−
= (j + ) −
≈ j+
.
1
2
2 j+2
2
2(j + 12 )
α
α
α
q
=
≈
2
1
1
α
α2
2
n − j + 2 + j + 2 − 2(j+ 1 )
n − 2(j+
1
n − (j + 12 ) +
j + 21 − α2
2
2)
α
α
α2
i≈
= h
1
+
.
2
n
2n(j + 12 )
n 1− α 1
1
j+
2
2
− α2 =
1
j+
2
α
j + 12
2
2n(j+ 2 )



1 + 
α
q
2 −1/2
−1/2
α2
α2
 
≈ 1+ 2 1+

n
n(j + 21 )
2
2
n − (j + 12 ) +
j + 12 − α
α2
α4
−n
3
1 α2
α2
3 α4
=
1
−
+
+
≈1−
1
+
+
.
2 n2
8 n4
2n2
2n4 j + 12
4
n(j + 12 )
3
3
α2
α4
−n
α2 mc2
α2
n
2
+
−
Enj ≈ mc 1 − 2 + 4
−1 =−
1+ 2
2n
2n
4
2n2
n
4
j + 12
j + 21
3
13.6 eV
α2
n
−
=−
1+ 2
, confirming Eq. 7.69.
1
2
n
n
4
j+2
Problem 7.23
Equation 7.60 ⇒ B =
B=
=
e
1
L.
4π0 mc2 r3
Say L = ~, r = a; then
1
e~
4π0 mc2 a3
(1.60 × 10−19 C)(1.05 × 10−34 J · s)
2
3 = 12 T.
4π 8.9 × 10−12 C2 /N · m2 (9.1 × 10−31 kg) (3 × 108 m/s) (0.53 × 10−10 m)
So a “strong” Zeeman field is Bext 10 T, and a “weak” one is Bext 10 T. Incidentally, the earth’s field
(10−4 T) is definitely weak.
254
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
Problem 7.24
For n = 2, ` = 0 (j = 1/2) or ` = 1 (j = 1/2 or 3/2). The eight states are:
|1i = |2 0 21 12 i


|2i = |2 0 21 − 12 i
(1/2)(3/2) + (3/4)
3/2
gJ = 1 +
=1+
= 2.

2(1/2)(3/2)
3/2
|3i = |2 1 21 12 i


|4i = |2 1 21 − 12 i
(1/2)(3/2) − (1)(2) + (3/4)
−1/2
gJ = 1 +
=1+
= 2/3.

2(1/2)(3/2)
3/2
α2 2 3
5 2
13.6 eV
−
1+
= −3.4 eV 1 + α .
In these four cases, Enj = −
4
4 1 4
16
|5i = |2 1 32 32 i
|6i = |2 1 23 12 i










(3/2)(5/2) − (1)(2) + (3/4)
5/2
gJ = 1 +
=1+
= 4/3.

2(3/2)(5/2)
15/2
|7i = |2 1 23 − 12 i 






3
3 
|8i = |2 1 2 − 2 i
α2 2 3
1
In these four cases, Enj = −3.4 eV 1 +
−
= −3.4 eV 1 + α2 .
4 2 4
16
5 2
E1 = −3.4 eV 1 + 16
α + µB Bext .
5 2
E2 = −3.4 eV 1 + 16
α − µB Bext .
5 2
E3 = −3.4 eV 1 + 16
α + 31 µB Bext .
5 2
E4 = −3.4 eV 1 + 16
α − 31 µB Bext .
The energies are:
1 2
E5 = −3.4 eV 1 + 16
α + 2µB Bext .
1 2
E6 = −3.4 eV 1 + 16
α + 32 µB Bext .
1 2
E7 = −3.4 eV 1 + 16
α − 32 µB Bext .
1 2
E8 = −3.4 eV 1 + 16
α − 2µB Bext .
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
E
255
µBBext
5 (slope 2)
6 (slope 2/3)
-3.4 (1+_2/16) eV
7 (slope -2/3)
8 (slope -2)
1 (slope 1)
-3.4 (1+5_2/16) eV
3 (slope 1/3)
4 (slope -1/3)
2 (slope -1)
Problem 7.25
The Wigner-Eckart theorem tells us that


1
1
` 1 `0
` 1 `0

 √2 Cm−1 m0 − √2 Cm 0 m0 

i
` 1 `0
` 1 `0
hn0 `0 m0 |V|n ` mi = √i2 Cm−1
hn0 `0 ||V ||n `i = Mhn0 `0 ||V ||n `i,
0 + √ Cm 1 m0
m
2




` 1 `0
Cm
0 m0
where M stands for the coefficient in the curly brackets. By the same token,
hn0 `0 m0 |W|n ` mi = Mhn0 `0 ||W ||n `i, with the same coefficients M. So the two vectors are proportional.
Problem 7.26
1
0
Efs
= hn`m` ms |(Hr0 + Hso
)|n`m` ms i = −
e2
En2
4n
~2 m` ms
+
−
3
.
2mc2 ` + 1/2
8π0 m2 c2 `(` + 1/2)(` + 1)n3 a3
2E1
2En2
E1
α2
=
−
−
= 4 (13.6 eV). (Problem 7.14.)
2
2
4
mc
mc
n
n "
2 2 # 2
Now
2 2
2 2
2 3
e
~
e
~
(me
)
m
e
e2


=
=
= α2 (13.6 eV).

 8π m2 c2 a3
2 · 4π0 m2 c2 (4π0 ~2 )3
2~2 4π0
4π0 ~c
0





1
Efs
=
13.6 eV 2
1
3
m` ms
13.6 eV 2 3
`(` + 1) − m` ms
α
−
+
+
=
α
−
. QED
n3
(` + 1/2) 4n `(` + 1/2)(` + 1)
n3
4n `(` + 1/2)(l + 1)
256
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
Problem 7.27
The Bohr energy is the same for all of them: E2 = −13.6 eV/22 = −3.4 eV. The Zeeman contribution is the
1
second term in Eq. 7.83: µB Bext (m` +2ms ). The fine structure is given by Eq. 7.86: Efs
= (13.6 eV/8)α2 {· · · } =
2
(1.7eV)α {· · · }. In the
table below I record the 8 states, the value of (m` + 2ms ), the value of {· · · } ≡
`(` + 1) − m` ms
3
−
, and (in the last column) the total energy, −3.4 eV [1−(α2 /2){· · · }]+(m` +2ms )µB Bext .
8
`(` + 1/2)(` + 1)
State = |n`m` ms i
|1i = |2 0 0 12 i
|2i = |2 0 0 − 12 i
|3i = |2 1 1 12 i
|4i = |2 1 − 1 − 12 i
|5i = |2 1 0 12 i
|6i = |2 1 0 − 12 i
|7i = |2 1 1 − 12 i
|8i = |2 1 − 1 12 i
(m` + 2ms )
1
−1
2
−2
1
−1
0
0
{· · · }
−5/8
−5/8
−1/8
−1/8
−7/24
−7/24
−11/24
−11/24
Total Energy
-3.4 eV [1 + (5/16)α2 ] + µB Bext
-3.4 eV [1 + (5/16)α2 ] − µB Bext
-3.4 eV [1 + (1/16)α2 ] + 2µB Bext
-3.4 eV [1 + (1/16)α2 ] − 2µB Bext
-3.4 eV [1 + (7/48)α2 ] + µB Bext
-3.4 eV [1 + (7/48)α2 ] − µB Bext
-3.4 eV [1 + (11/48)α2 ]
-3.4 eV [1 + (11/48)α2 ]
Ignoring fine structure there are five distinct levels—corresponding to the possible values of (m` + 2ms ):
2
(d = 1); 1
(d = 2); 0
(d = 2); −1
(d = 2); −2
(d = 1).
Problem 7.28
e
e
Bext · hL + 2Si =
Bext 2ms ~ = 2ms µB Bext (same as the Zeeman term in Eq. 7.83,
2m
2m 13.6 eV
α2
3
(since j = 1/2). So the total energy is
1
+
n
−
Equation 7.69 ⇒ Enj = −
n2
n2
4
Equation 7.74 ⇒ Ez1 =
with ml = 0).
13.6 eV
α2
3
E=−
+ 2ms µB Bext .
1+ 2 n−
n2
n
4
13.6 eV 2
3
13.6 eV 2 3
α
n
−
=
α
−
1
, which is the same as
n4
4
n3
4n
Eq. 7.86, with the term in square brackets set equal to 1. QED
1
Fine structure is the α2 term: Efs
= −
Problem 7.29
E12
8
E1
α2
=
3
−
;
=
−
(Problem 7.14), so
32mc2
j + 1/2
mc2
2
E1 α2
8
13.6 eV 2
8
8
1
Efs = −
3−
=
α 3−
=γ 3−
.
32
2
j + 1/2
64
j + 1/2
j + 1/2
1
Equation 7.68 ⇒ Efs
=
E22
2mc2
3−
8
j + 1/2
8
For j = 1/2 (ψ1 , ψ2 , ψ6 , ψ8 ), Hfs1 = γ(3 − 8) = −5γ. For j = 3/2 (ψ3 , ψ4 , ψ5 , ψ7 ), Hfs1 = γ(3 − ) = −γ.
2
This confirms all the γ terms in −W (p. 309). Meanwhile, Hz0 = (e/2m)Bext (Lz + 2Sz ) (Eq. 7.73); ψ1 , ψ2 , ψ3 , ψ4
are eigenstates of Lz and Sz ; for these there are only diagonal elements:
hHz0 i =
e~
Bext (ml + 2ms ) = (ml + 2ms )β;
2m
hHz0 i11 = β;
hHz0 i22 = −β;
hHz0 i33 = 2β;
hHz0 i44 = −2β.
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
This confirms the upper left corner of −W. Finally:
q

hHz0 i55 = (2/3)β,
(Lz + 2Sz )|ψ5 i = +~ 23 |1 0i| 12 12 i 



q
hHz0 i66 = (1/3)β,


(Lz + 2Sz )|ψ6 i = −~ 13 |1 0i| 12 12 i 
hHz0 i77 = −(2/3)β,
q
so
hHz0 i88 = −(1/3)β, √

(Lz + 2Sz )|ψ7 i = −~ 23 |1 0i| 12 − 12 i 


q
hHz0 i56 = hHz0 i65 = −(√2/3)β,


(Lz + 2Sz )|ψ8 i = −~ 13 |1 0i| 12 − 12 i 
hHz0 i78 = hHz0 i87 = −( 2/3)β,
which confirms the remaining elements.
Problem 7.30
There are eighteen n = 3 states (in general, 2n2 ).
WEAK FIELD
3
1
13.6 eV
α2
3
α2
1
−
−
Equation 7.69 ⇒ E3j = −
1+
= −1.51 eV 1 +
.
9
9 j + 1/2 4
3 j + 1/2 4
Equation 7.79 ⇒ Ez1 = gJ mj µB Bext .
State |3 ` j mj i
gJ (Eq. 6.75)
1
3
1
1
j+1/2 − 4
` = 0, j = 1/2
|3 0 21 12 i
2
1/4
` = 0, j = 1/2
|3 0 21 − 12 i
2
1/4
` = 1, j = 1/2
|3 1 21 12 i
2/3
1/4
` = 1, j = 1/2
|3 1 21 − 12 i
2/3
1/4
` = 1, j = 3/2
|3 1 23 32 i
4/3
1/12
` = 1, j = 3/2
|3 1 23 12 i
4/3
1/12
` = 1, j = 3/2
|3 1 23 − 12 i
4/3
1/12
` = 1, j = 3/2
|3 1 23 − 32 i
4/3
1/12
` = 2, j = 3/2
|3 2 23 32 i
4/5
1/12
` = 2, j = 3/2
|3 2 23 12 i
4/5
1/12
` = 2, j = 3/2
|3 2 23 − 12 i
4/5
1/12
` = 2, j = 3/2
|3 2 23 − 32 i
4/5
1/12
` = 2, j = 5/2
|3 2 25 52 i
6/5
1/36
` = 2, j = 5/2
|3 2 25 32 i
6/5
1/36
` = 2, j = 5/2
|3 2 25 12 i
6/5
1/36
` = 2, j = 5/2
|3 2 25 − 12 i
6/5
1/36
` = 2, j = 5/2
|3 2 25 − 32 i
6/5
1/36
` = 2, j = 5/2
|3 2 25 − 52 i
6/5
1/36
Total Energy
2
−1.51 eV 1 + α4 + µB Bext
2
−1.51 eV 1 + α4 − µB Bext
2
−1.51 eV 1 + α4 + 13 µB Bext
2
−1.51 eV 1 + α4 − 13 µB Bext
2
−1.51 eV 1 + α12 + 2µB Bext
2
−1.51 eV 1 + α12 + 23 µB Bext
2
−1.51 eV 1 + α12 − 23 µB Bext
2
−1.51 eV 1 + α12 − 2µB Bext
2
−1.51 eV 1 + α12 + 65 µB Bext
2
−1.51 eV 1 + α12 + 25 µB Bext
2
−1.51 eV 1 + α12 − 25 µB Bext
2
−1.51 eV 1 + α12 − 65 µB Bext
2
−1.51 eV 1 + α36 + 3µB Bext
2
−1.51 eV 1 + α36 + 95 µB Bext
2
−1.51 eV 1 + α36 + 35 µB Bext
2
−1.51 eV 1 + α36 − 35 µB Bext
2
−1.51 eV 1 + α36 − 95 µB Bext
2
−1.51 eV 1 + α36 − 3µB Bext
257
258
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
STRONG FIELD
Equation 7.83 ⇒ −1.51 eV + (m` + 2ms )µB Bext ;
13.6 eV 2
Equation 7.86 ⇒
α
27
1
`(` + 1) − m` ms
α2
`(` + 1) − m` ms
1
−
= −1.51 eV
−
.
4
`(` + 1/2)(` + 1)
3
`(` + 1/2)(` + 1) 4
1
Etot = −1.51 eV(1 + α A) + (m` + 2ms )µB Bext , where A ≡
3
2
`(` + 1) − m` ms
1
−
`(` + 1/2)(` + 1) 4
.
These terms are given in the table below:
State |n ` m` ms i
(m` + 2ms )
A
`=0
|3 0 0 12 i
Total Energy
1
1/4
`=0
|3 0 0 − 12 i
−1
1/4
`=1
|3 1 1 12 i
2
1/12
`=1
|3 1 − 1 − 12 i
−2
1/12
`=1
|3 1 0 12 i
1
5/36
`=1
|3 1 0 − 12 i
−1
5/36
`=1
|3 1 − 1 12 i
0
7/36
`=1
|3 1 1 − 12 i
0
7/36
`=2
|3 2 2 12 i
3
1/36
`=2
|3 2 − 2 − 12 i
−3
1/36
`=2
|3 2 1 12 i
2
7/180
`=2
|3 2 − 1 − 12 i
−2
7/180
`=2
|3 2 0 12 i
1
1/20
`=2
|3 2 0 − 12 i
−1
1/20
`=2
|3 2 − 1 12 i
0
11/180
`=2
|3 2 1 − 12 i
0
11/180
`=2
|3 2 − 2 12 i
−1
13/180
`=2
|3 2 2 − 12 i
1
13/180
2
−1.51 eV 1 + α4 + µB Bext
2
−1.51 eV 1 + α4 − µB Bext
2
−1.51 eV 1 + α12 + 2µB Bext
2
−1.51 eV 1 + α12 − 2µB Bext
2
−1.51 eV 1 + 5α
36 + µB Bext
5α2
−1.51 eV 1 + 36 − µB Bext
2
−1.51 eV 1 + 7α
36 2
−1.51 eV 1 + 7α
36
2
−1.51 eV 1 + α36 + 3µB Bext
2
−1.51 eV 1 + α36 − 3µB Bext
2
−1.51 eV 1 + 7α
180 + 2µB Bext
7α2
−1.51 eV 1 + 180 − 2µB Bext
2
−1.51 eV 1 + α20 + µB Bext
2
−1.51 eV 1 + α20 − µB Bext
2
−1.51 eV 1 + 11α
180 2
−1.51 eV 1 + 11α
180 2
− µB Bext
−1.51 eV 1 + 13α
180
13α2
−1.51 eV 1 + 180 + µB Bext
INTERMEDIATE FIELD
As in the book, I’ll use the basis |n ` j mj i (same as for weak field); then the fine structure matrix elements
are diagonal: Eq. 7.68 ⇒
1
Efs
=
γ≡
E32
2mc2
3−
12
j + 1/2
E12
=
54mc2
1−
4
j + 1/2
E1 α2
=−
108
4
1−
j + 1/2
4
= 3γ 1 −
j + 1/2
13.6 eV 2
1
1
1
α . For j = 1/2, Efs
= −9γ; for j = 3/2, Efs
= −3γ; for j = 5/2, Efs
= −γ.
324
,
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
259
The Zeeman Hamiltonian is Eq. 7.73: Hz0 = ~1 (Lz + 2Sz )µB Bext . The first eight states (` = 0 and ` = 1) are
the same as before (p. 309), so the β terms in W are unchanged; recording just the non-zero blocks of −W:
√
(9γ − β), (9γ + β), (3γ − 2β), (3γ + 2β),
2
(3γ√− 32 β)
3 β
2
(9γ − 13 β)
3 β
√
!
,
2
(3γ√+ 32 β)
3 β
2
(9γ + 31 β)
3 β
!
.
The other 10 states (` = 2) must first be decomposed into eigenstates of Lz and Sz :
| 25 52 i
= |2 2i| 12 21 i
=⇒ (γ − 3β)
| 52 − 52 i = |2 − 2i| 12 − 12 i =⇒ (γ + 3β)
| 52 32 i =
| 23 32 i =
| 52 12 i =
| 23 12 i =
q

q
1
|2 2i| 12 − 12 i + 45 |2 1i| 21 12 i 
q5
q
4
1
1
5 |2 2i| 2 − 2 i −
q
1
1 1 
5 |2 1i| 2 2 i
| 32 − 12 i =
| 25 − 32 i =
| 32 − 32 i =
2
(γ − 59 β)
5β
2
β
(3γ
− 65 β)
5
=⇒
6
(γ √
− 53 β)
5 β
6
(3γ − 52 β)
5 β
√

q
2
|2 1i| 12 − 12 i + 35 |2 0i| 21 12 i 
q5
q
| 52 − 12 i =
=⇒
3
1
1
5 |2 1i| 2 − 2 i −
2
1 1 
5 |2 0i| 2 2 i
q
q
1
1
2
5 |2 0i| 2 − 2 i −
q
√

3
|2 0i| 12 − 12 i + 25 |2 − 1i| 12 12 i 
q5
q
1 1 
3
5 |2 − 1i| 2 2 i
q
=⇒

4
|2 − 1i| 12 − 12 i + 15 |2 − 2i| 21 21 i 
q5
q
1
1
1
5 |2 − 1i| 2 − 2 i −
4
1 1 
5 |2 − 2i| 2 2 i
!
6
(γ √
+ 53 β)
5 β
6
(3γ + 25 β)
5 β
=⇒
!
2
(γ + 59 β)
5β
2
β
(3γ
+ 65 β)
5
[Sample Calculation: For the last two, letting Q ≡ ~1 (Lz + 2Sz ), we have
q
q
Q| 25 − 23 i = −2 45 |2 − 1i| 12 − 12 i − 15 |2 − 2i| 12 12 i;
q
q
Q| 32 − 23 i = −2 15 |2 − 1i| 12 − 12 i + 45 |2 − 2i| 12 12 i.
h 52 − 23 |Q| 52 − 32 i = (−2) 45 − 15 = − 95 ; h 32 − 32 |Q| 32 − 32 i = (−2) 15 − 45 = − 56 ;
q q
q q
h 52 − 23 |Q| 32 − 32 i = −2 45 15 + 15 45 = − 45 + 25 = − 52 = h 23 − 32 |Q| 52 − 23 i.]
So the 18 × 18 matrix −W splits into six 1 × 1 blocks and six 2 × 2 blocks. We need the eigenvalues of the
2 × 2 blocks. This means solving 3 characteristic equations (the other 3 are obtained trivially by changing the
sign of β):
2
1
2
3γ − β − λ
9γ − β − λ − β 2 = 0 =⇒ λ2 + λ(β − 12γ) + γ(27γ − 7β) = 0.
3
3
9
9
6
4 2
33
2
2
2
γ− β−λ
3γ − β − λ − β = 0 =⇒ λ + λ(3β − 4γ) + 3γ − γβ + 2β
= 0.
5
5
25
5
3
2
6
11
γ− β−λ
3γ − β − λ − β 2 = 0 =⇒ λ2 + λ(β − 4γ) + γ 3γ − β = 0.
5
5
25
5
The solutions are:
260
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
p
λ = −β/2 + 6γ ± q
(β/2)2 + βγ + 9γ 2
λ = −3β/2 + 2γ ± (β/2)2 + 35 βγ + γ 2
q
λ = −β/2 + 2γ ± (β/2)2 + 15 βγ + γ 2
1 = E3 − 9γ + β
2 = E3 − 3γ + 2β
3 = E3 − γ + 3β
p
4 = E3 − 6γ + β/2 + p9γ 2 + βγ + β 2 /4
5 = E3 − 6γ + β/2 − q9γ 2 + βγ + β 2 /4
⇒
γ 2 + 53 βγ + β 2 /4
6 = E3 − 2γ + 3β/2 +
q
7 = E3 − 2γ + 3β/2 − γ 2 + 53 βγ + β 2 /4
q
8 = E3 − 2γ + β/2 + γ 2 + 15 βγ + β 2 /4
q
9 = E3 − 2γ + β/2 − γ 2 + 15 βγ + β 2 /4
eV 2
(The other 9 ’s are the same, but with β → −β.) Here γ = 13.6
324 α , and β = µB Bext .
In the weak-field limit (β γ):
4 ≈ E3 − 6γ + β/2 + 3γ
p
2
1 + β/9γ ≈ E3 − 6γ + β/2 + 3γ(1 + β/18γ) = E3 − 3γ + β.
3
1
5 ≈ E3 − 6γ + β/2 − 3γ(1 + β/18γ) = E3 − 9γ + β.
3
9
6 ≈ E3 − 2γ + 3β/2 + γ(1 + 3β/10γ) = E3 − γ + β.
5
6
7 ≈ E3 − 2γ + 3β/2 − γ(1 + 3β/10γ) = E3 − 3γ + β.
5
3
8 ≈ E3 − 2γ + β/2 + γ(1 + β/10γ) = E3 − γ + β.
5
2
9 ≈ E3 − 2γ + β/2 − γ(1 + β/10γ) = E3 − 3γ + β.
5
Noting that γ = −(E3 /36)α2 = 1.5136eV α2 , we see that the weak field energies are recovered as in the first table.
In the strong-field limit (β γ):
p
4 ≈ E3 − 6γ + β/2 + β/2 1 + 4γ/β ≈ E3 − 6γ + β/2 + β/2(1 + 2γ/β) = E3 − 5γ + β.
5 ≈ E3 − 6γ + β/2 − β/2(1 + 2γ/β) = E3 − 7γ.
7
6 ≈ E3 − 2γ + 3β/2 + β/2(1 + 6γ/5β) = E3 − γ + 2β.
5
7 ≈ E3 − 2γ + 3β/2 − β/2(1 + 6γ/5β) = E3 −
13
γ + β.
5
9
8 ≈ E3 − 2γ + β/2 + β/2(1 + 2γ/5β) = E3 − γ + β.
5
9 ≈ E3 − 2γ + β/2 − β/2(1 + 2γ/5β) = E3 −
11
γ.
5
Again, these reproduce the strong-field results in the second table.
In the figure below each line is labeled by the level number and (in parentheses) the starting and ending
slope; for each line there is a corresponding one starting from the same point but sloping down.
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
261
3(3)
E
6(9/5<> 2)
8(3/5<>1)
2(2)
L3
7(6/5<>1)
4(2/3<>1)
L 3 <a
9(2/5<>0)
L 3 <3a
1(1)
5(1/3<>0)
L 3 <9a
Problem 7.31
R
I ≡ (a · r̂)(b · r̂) sin θ dθ dφ
R
= (ax sin θ cos φ + ay sin θ sin φ + az cos θ)(bx sin θ cos φ + by sin θ sin φ + bz cos θ) sin θ dθ dφ.
Z 2π
But
Z 2π
sin φ dφ =
0
Z
I=
0
Z 2π
0
Z π
0
sin φ cos φ dφ = 0, so only three terms survive :
0
(ax bx sin2 θ cos2 φ + ay by sin2 θ sin2 φ + az bz cos2 θ) sin θ dθ dφ.
But
I=
Z 2π
cos φ dφ =
sin2 φ dφ =
Z 2π
0
cos2 φ dφ = π,
Z 2π
dφ = 2π, so
0
π(ax bx + ay by ) sin2 θ + 2πaz bz cos2 θ sin θ dθ.
Z π
But
0
4
sin θ dθ = ,
3
3
Z π
cos2 θ sin θ dθ =
0
2
,
3
4
2
4π
4π
so I = π(ax bx + ay by ) + 2πaz bz =
(ax bx + ay by + az bz ) =
(a · b). QED
3
3
3
3
[Alternatively, noting that I has to be a scalar bilinear in a and b, we know immediately that I = A(a·b), where
R
A is some constant (same for all a and b). To determine A, pick a = b = k̂; then I = A = cos2 θ sin θ dθ dφ =
4π/3.]
√
For states with ` = 0, the wave function is independent of θ and φ (Y00 = 1/ 4π), so
Z ∞
Z
3(Sp · r̂)(Se · r̂) − Sp · Se
1
2 2
=
|ψ(r)| r dr
[3(Sp · r̂)(Se · r̂) − Sp · Se ] sin θ dθ dφ.
r3
r3
0
262
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
R
The first angular integral is 3(4π/3)(Sp · Se ) = 4π(Sp · Se ), while the second is −(Sp · Se ) sin θ dθ dφ =
−4π(Sp · Se ), so the two cancel, and the result is zero. QED [Actually, there is a little sleight-of-hand here,
since for ` = 0, ψ → constant as r → 0, and hence the radial integral diverges logarithmically at the origin.
Technically, the first term in Eq. 7.90 is the field outside an infinitesimal sphere; the delta-function gives the
field inside. For this reason it is correct to do the angular integral first (getting zero) and not worry about the
radial integral.]
Problem 7.32
g
From Eq. 7.93 we see that ∆E ∝
; we want reduced mass in a, but not in mp me (which come from
mp me a3
Eq. 7.89); the notation in Eq. 7.97 obscures this point.
(a) g and mp are unchanged; me → mµ = 207me , and a → aµ .
From Eq. 4.72, a ∝ 1/m, so
a
207
mµ (reduced)
mµ mp
1
207
207
=
=
=
·
=
=
= 186.
(9.11×10−31 )
aµ
me
mµ + mp me
1 + 207(me /mp )
1.11
1 + 207
−27
1.67×10
∆E = (5.88 × 10
−6
)
3
eV) (1/207) (186) = 0.183 eV.
a
mp (reduced)
m2e
1
1
=
=
·
= .
ap
me
me + me me
2
3
2
1.67 × 10−27
1
∆E = (5.88 × 10−6 eV)
= 4.82 ×10−4 eV.
5.59
9.11 × 10−31
2
(b) g : 5.59 → 2;
mp → me ;
a
mm (reduced)
me mµ
1
207
.
=
=
·
=
am
me
me + mµ me
208
3
2
1.67 × 10−27
207
−6
∆E = (5.88 × 10 )
= 1.84×10−5 eV.
5.59
(207)(9.11 × 10−31 )
208
(c) g : 5.59 → 2;
mp → mµ ;
Problem 7.33
Use perturbation theory:
e2
1 1
H0 = −
−
,
4π0 b
r
for
0 < r < b. ∆E = hψ|H 0 |ψi,
with
ψ≡√
1
πa3
e−r/a .
Z b
Z b
Z b
e2 1
1 1 −2r/a 2
e2
1
2 −2r/a
−2r/a
4π
−
e
r
dr
=
−
r
e
dr
−
re
dr
4π0 πa3
b
r
π0 a3 b 0
0
0
b
2
2
a 2 −2r/a
a
2r
a
2r
1
e2
−2r/a
−2r/a
−
r
e
+
a
e
−
−
1
−
e
−
−
1
=−
π0 a3 b
2
2
a
2
a
0
e2
a 2 −2b/a a3 −2b/a
2b
a2 −2b/a
2b
a3
a2
=−
− b e
+ e
−
−1 − e
−
−1 +
−
π0 a3
2b
4b
a
4
a
4b
4
2
2
3
2
2
e
ab a
a
ab a
a a
=−
e−2b/a −
−
−
+
+
+
−1
π0 a3
2
2
4b
2
4
4 b
e2
a2
a
a2 a
e2
a
a −2b/a
−2b/a
=−
e
−
+
1
+
−
1
=
1
−
+
1
+
e
.
π0 a3
4
b
4 b
4π0 a
b
b
∆E = −
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
263
Let +2b/a = (very small). Then the term in square brackets is:
2
2
2
3
1−
+ 1+
1−+
−
+ ···
2
6
2
2
3
2
2
2
1 + − − 2+
+ −
−
+ ( )3 + · · · =
+ ( )3 + ( )4 · · ·
=
1 − + 2
6
3
6
To leading order, then, ∆E =
E = E1 = −
∆E
e2
=
E
4π0
e2 1 4b2
.
4π0 a 6a2
me4
;
2(4π0 )2 ~2
−
a=
4π0 ~2
;
me2
so Ea = −
e2
.
2(4π0 )
2
4 b
2(4π0 ) 2b2
−
=
.
e2
3a2
3 a
Putting in a = 5 × 10−11 m:
∆E
4
=−
E
3
10−15
5 × 10−11
By contrast,
2
=−
16
× 10−10 ≈ −5 × 10−10 .
3
fine structure:
∆E/E ≈ α2 = (1/137)2 = 5 × 10−5 ,
hyperfine structure: ∆E/E ≈ (me /mp )α2 = (1/1800)(1/137)2 = 3 × 10−8 .
So the correction for the finite size of the nucleus is much smaller (about 1% of hyperfine).
Problem 7.34
(a) H̃0 |ψn0 i = H0 |ψn0 i + PD H 0 PD |ψn0 i. But PD |ψn0 i = |ψa0 ihψa0 |ψn0 i + |ψb0 ihψb0 |ψn0 i = 0, because eigenstates of
H0 belonging to different eigenvalues are orthogonal. And H0 |ψn0 i = En0 |ψn0 i, so H̃0 |ψn0 i = En0 |ψn0 i. X
H̃0 |ψ 0 i = H0 |ψ 0 i + PD H 0 PD |ψ 0 i. But H0 |ψ 0 i = E 0 |ψ 0 i, and PD |ψ 0 i = |ψ 0 i, so
= E 0 α|ψa0 i + β|ψb0 i + |ψa0 ihψa0 | + |ψb0 ihψb0 | H 0 α|ψa0 i + β|ψb0 i
= E 0 α|ψa0 i + β|ψb0 i + α Waa |ψa0 i + Wba |ψb0 i + β Wab |ψa0 i + Wbb |ψb0 i
= αE 0 + αWaa + βWab |ψa0 i + βE 0 + αWba + βWbb |ψb0 i
= (αE 0 + αE 1 )|ψa0 i + (βE 0 + βE 1 )|ψb0 i = (E 0 + E 1 )|ψ 0 i.
X
1
1
from Equation 7.33.
In the last line I used Equations 7.27 and 7.29; E 1 can be either E+
or E−
(b) Using nondegenerate perturbation theory, to second order (Equations 7.9 and 7.15), for ψ+ :
1
0
0
E+ = (E 0 + E+
) + hψ+
|H̃ 0 |ψ+
i+
0
0 2
X |hψm
|H̃ 0 |ψ+
i|
m6=+
1 − E0
E 0 + E+
m
But
0
0
0
0
0
0
hψ+
|H̃ 0 |ψ+
i = hψ+
|H 0 |ψ+
i − hψ+
|PD H 0 PD |ψ+
i = 0,
.
264
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
0
0
0
0
because PD |ψ+
i = |ψ+
i and hψ+
|PD = hψ+
|, while
0
0
0
0
0
0
hψm
|H̃ 0 |ψ+
i = hψm
|H 0 |ψ+
i − hψm
|PD H 0 PD |ψ+
i=
(
0
0
hψm
|H 0 |ψ+
i, (m 6= ±)
0,
(m = ±)
0
(note that PD |ψm
i = 0 if m 6= ±). So
1
E+ = E 0 + E+
+
0
0 2
X |hψm
|H 0 |ψ+
i|
1 − E0 .
E 0 + E+
m
m6=±
(Change the subscripts + to − for E− .) Typically one drops the E 1 term in the denominator, since it is
really a third-order correction.
Problem 7.35
(a)
 
 

 
 

0 0 0
1 0 0
1
0
1 0 0
PD = 0 1 0 0 + 1 0 1 0 = 0 0 0 + 0 1 0 = 0 1 0 .
0 0 0
0 0 0
0
0
0 0 0

 


 
 

0 0
1 0 0
0 a b
1 0 0
0 0
0 a0
H̃0 = H0 + PD H0 PD = 0 0  + 0 1 0 a∗ 0 c  0 1 0 = 0 0  + a∗ 0 0
0 0 0
0 0 0
b∗ c∗ 0
0 0 0
0 0 0
0 0 0


a 0
= a∗ 0 
0 0 0

 
 

0 0 b
0 a 0
0 a b
H̃0 = H0 − PD H0 PD = a∗ 0 c  − a∗ 0 0 =  0 0 c  .
b∗ c∗ 0
0 0 0
b∗ c ∗ 0
(b) Eigenvalues of H̃0 :
−λ a
0
a∗ − λ 0
= ( − λ)2 (0 − λ) − (0 − λ)a∗ a = 0 ⇒ λ1 = + |a|, λ2 = − |a|, λ3 = 0 .
0
0
0 −λ
The spectrum is nondegenerate (provided that a 6= 0 and 0 6= ± |a|). To get the eigenvectors:



 
 
r
a 0
x
x
 x + ay = ( + |a|)x 
a∗
∗
∗
λ1 a 0  y  = ( + |a|) y  ⇒ a x + y = ( + |a|)y ⇒ y =
x, z = 0.


a
0 0 0
z
z
0 z = ( + |a|)z



 
 
r
a 0
x
x
 x + ay = ( − |a|)x 
a∗
∗
∗
λ2 a 0  y  = ( − |a|) y  ⇒ a x + y = ( − |a|)y ⇒ y = −
x, z = 0.


a
0 0 0
z
z
0 z = ( − |a|)z



 
 
a 0
x
x
 x + ay = 0 x 
λ3 a∗ 0  y  = 0 y  ⇒ a∗ x + y = 0 y ⇒ x = y = 0.


0 0 0
z
z
0 z = 0 z
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY


1
1 p ∗ 
So the (normalized) eigenvectors are |ψ1 i = √
a /a .
2
0
265


1
1  p ∗ 
|ψ2 i = √
− a /a .
2
0
 
0
|ψ3 i = 0 .
1
(c) The “good” states are the eigenstates of H̃0 : |ψ1 i, |ψ2 i, and |ψ3 i from part (b). To first order, the energies
are the eigenvalues, ± |a| and 0 .
Problem 7.36
(a) In terms of the one-dimensional harmonic oscillator states {ψn (x)}, the unperturbed ground state is
|0i = ψ0 (x)ψ0 (y)ψ0 (z).
E01 = h0|H 0 |0i = hψ0 (x)ψ0 (y)ψ0 (z)|λx2 yz|ψ0 (x)ψ0 (y)ψ0 (z)i = λhx2 i0 hyi0 hzi0 .
But
hyi0 = hzi0 = 0.
So there is no change, in first order.
(b) The (triply degenerate) first excited states are

 |1i = ψ0 (x)ψ0 (y)ψ1 (z)
|2i = ψ0 (x)ψ1 (y)ψ0 (z)

|3i = ψ1 (x)ψ0 (y)ψ0 (z)
In this basis the perturbation matrix is
Wij = hi|H 0 |ji,
i = 1, 2, 3.
h1|H 0 |1i = hψ0 (x)ψ0 (y)ψ1 (z)|λx2 yz|ψ0 (x)ψ0 (y)ψ1 (z)i = λhx2 i0 hyi0 hzi1 = 0,
h2|H 0 |2i = hψ0 (x)ψ1 (y)ψ0 (z)|λx2 yz|ψ0 (x)ψ1 (y)ψ0 (z)i = λhx2 i0 hyi1 hzi0 = 0,
h3|H 0 |3i = hψ1 (x)ψ0 (y)ψ0 (z)|λx2 yz|ψ1 (x)ψ0 (y)ψ0 (z)i = λhx2 i1 hyi0 hzi0 = 0,
h1|H 0 |2i = hψ0 (x)ψ0 (y)ψ1 (z)|λx2 yz|ψ0 (x)ψ1 (y)ψ0 (z)i = λhx2 i0 h0|y|1ih1|z|0i
2
~
~
2
=λ
|h0|x|1i| = λ
[using Problems 2.11 and 3.39].
2mω
2mω
h1|H 0 |3i = hψ0 (x)ψ0 (y)ψ1 (z)|λx2 yz|ψ1 (x)ψ0 (y)ψ0 (z)i = λh0|x2 |1ihyi0 h1|z|0i = 0,
h2|H 0 |3i = hψ0 (x)ψ1 (y)ψ0 (z)|λx2 yz|ψ1 (x)ψ0 (y)ψ0 (z)i = λh0|x2 |1ih1|y|0i0 hzi0 = 0.


0 a0
W = a 0 0 ,
0 0 0
Eigenvalues of W :
where
a≡λ
~
2mω
2
.
2
−E a 0
~
a −E 0 = −E 3 + Ea2 = 0 ⇒ E = {0, ±a} = 0, ±λ
.
2mω
0 0 −E
266
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
Problem 7.37
(a) The first term is the nucleus/nucleus interaction, the second is the interaction between the nucleus of
atom 2 and the electron in atom 1, the third is between nucleus 1 and electron 2, and the last term is the
interaction between the electrons.
x x 2
1
x −1
1 1
1−
+
=
=
1+
+ ... ,
R−x
R
R
R
R
R
so
(
"
2 #)
x x 2 x x 2 x1 − x2
x1 − x2
1 e2
1
2
2
1
0 ∼
+
− 1−
+
+ 1+
1− 1+
+
H =
4π0 R
R
R
R
R
R
R
1 e2
e2 x1 x2
2x1 x2
≈
=−
. X
−
2
4π0 R
R
2π0 R3
(b) Expanding Eq. 7.106:
1
1
e2
p2+ + p2− + k x2+ + x2− −
x2+ − x2−
3
2m
2
4π0 R
1
1
e2
=
p21 + p22 + k x21 + x22 −
(2x1 x2 ) = H 0 + H 0
2m
2
4π0 R3
H =
(Eqs. 7.103 and 7.105).
(c)
r
ω± =
k
m
1∓
e2
2π0 R3 k
1/2
"
∼
= ω0
1
1∓
2
e2
2π0 R3 mω02
1
−
8
e2
2π0 R3 mω02
2
#
+ ... .
2
1
e2
1
e2
1
~ω0 1 −
−
+
2
2 2π0 R3 mω02
8 2π0 R3 mω02
2 e2
1
e2
1
1+
−
− ~ω0
2 2π0 R3 mω02
8 2π0 R3 mω02
2
2 2
1
e2
1 ~
e
1
~ω0
−
=
−
. X
=
2
4
2π0 R3 mω02
8 m2 ω03 2π0
R6
∆V ∼
=
(d) In first order:
E01 = h0|H 0 |0i = −
e2
e2
hψ
(x
)ψ
(x
)|x
x
|ψ
(x
)ψ
(x
)i
=
−
hxi0 hxi0 = 0.
0
1
0
2
1
2
0
1
0
2
2π0 R3
2π0 R3
In second order:
E02 =
∞
X
|hψn |H 0 |ψ0 i|2
. Here |ψ0 i = |0i|0i, |ψn i = |n1 i|n2 i, so
E0 − En
2 X
∞ X
∞
e2
|hn1 |x1 |0i|2 |hn2 |x2 |0i|2
=
[use Problem 3.39]
3
2π0 R
E0,0 − En1 ,n2
n1 =1 n2 =1
2
e2
|h1|x|0i|2 |h1|x|0i|2
=
[zero unless n1 = n2 = 1]
2π0 R3
( 12 ~ω0 + 12 ~ω0 ) − ( 32 ~ω0 + 32 ~ω0 )
2 2
2 2
e2
e
1
~
~
1
−
. X
=
=− 2 3
2π0 R3
2~ω0
2mω0
8m ω0 2π0
R6
n=1
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
267
[There is an interesting fraud in this well-known problem. If you expand H 0 to order 1/R5 , the extra term
has a nonzero expectation value in the ground state of H 0 , so there is a non-zero first-order perturbation,
and the dominant contribution goes like 1/R5 , not 1/R6 (as desired). The model gets the power “right” in
three dimensions (where the expectation value is zero), but not in one. See A. C. Ipsen and K. Splittorff,
Am. J. Phys. 83, 150 (2015).]
Problem 7.38
(a) Let the unperturbed Hamiltonian be H(λ0 ), for some fixed value λ0 . Now tweak λ to λ0 + dλ. The
perturbing Hamiltonian is H 0 = H(λ0 + dλ) − H(λ0 ) = (∂H/∂λ) dλ (derivative evaluated at λ0 ).
The change in energy is given by Eq. 7.9:
dEn = En1 = hψn0 |H 0 |ψn0 i = hψn |
∂H
|ψn i dλ (all evaluated at λ0 );
∂λ
so
∂En
∂H
= hψn |
|ψn i.
∂λ
∂λ
[Note: Even though we used perturbation theory, the result is exact, since all we needed (to calculate the
derivative) was the infinitesimal change in En .]
(b) En = (n + 12 )~ω;
H=−
1
~2 d2
+ mω 2 x2 .
2m dx2
2
(i)
1
∂En
= (n + )~;
∂ω
2
V =
1
mω 2 x2 ,
2
∂H
= mωx2 ;
∂ω
1
so F-H ⇒ (n + )~ = hn|mωx2 |ni. But
2
1
1
1
so hV i = hn| mω 2 x2 |ni = ω(n + )~;
2
2
2
hV i = 21 (n + 12 )~ω.
(ii)
1
∂En
= (n + )ω;
∂~
2
∂H
~ d2
2
=−
=
∂~
m dx2
~
1
2
so F-H ⇒ (n + )ω = hn|T |ni,
2
~
or
~2 d 2
2
−
= T;
2m dx2
~
hT i = 12 (n + 12 )~ω.
(iii)
∂En
= 0;
∂m
∂H
~2 d2
1
1
=
+ ω 2 x2 = −
∂m
2m2 dx2
2
m
1
1
So F-H ⇒ 0 = − hT i + hV i,
m
m
Problems 2.12 and 3.37.
or
hT i = hV i.
−
~2 d2
2m dx2
+
1
m
1
mω 2 x2
2
=−
1
1
T + V.
m
m
These results are consistent with what we found in
268
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
Problem 7.39
(a)
a − λ 0
V
0 a − λ V
= (a − λ)2 (c − λ) − 2V V ∗ (a − λ) = 0
∗
∗
V
V
c − λ
s
2
a + c
a − c
⇒
+ 2|V |2 .
λ1 = a , λ± =
±
2
2
(b)
s
8|V |2
a + c
a − c
a − c
4|V |2
1+
≈
λ± =
±
±
1
+
2
(a − c )2
2
2
(a − c )2
a + c
a − c
2|V |2
=
±
+
.
2
2
(a − c )
a + c
2
λ+ ≈ a +
λ 1 = a ,
2|V |2
,
a − c
λ− ≈ c −
2|V |2
.
a − c
(c) Naively applying nondegenerate perturbation theory to the states (1,0,0), (0,1,0), and (0,0,1):
0
E1 = a + H11
+
X |H 0 |2
n1
n6=1
0
E2 = a + H22
+
0
E3 = c + H33
+
+ · · · = a +
0
a − En
X |H 0 |2
n2
a − En0
n6=2
X |H 0 |2
n3
c − En0
n6=3
0 2
0 2
|H21
|H31
0
|V |2
|
|
+
= a + +
.
a − a
a − c
0 a − c
+ · · · = a +
0 2
0 2
|H12
|
|V |2
|H32
|
0
+
= a + +
.
a − a
a − c
0 a − c
+ · · · = c +
0 2
0 2
|H13
|
|H23
|
|V |2
+
= c − 2
.
c − a
c − a
a − c
The nondegenerate state comes out right (E3 = λ− ), but the other two are at best ambiguous (the 0/0
terms warning us, perhaps, that something is amiss).
Problem 7.40
(a) In Equation 7.10, let ψn → ψ ≡ αψa + βψb :
(H 0 − E 0 )ψ 1 = −(H 0 − E 1 )ψ 0 .
0
Now take the inner product with ψm
, where m 6= a, b:
0 0
0
0
0
0
Em
hψm
|ψ 1 i − E 0 hψm
|ψ 1 i = −hψm
|H 0 |ψ 0 i + E 1
hψ
m |ψ i,
so
0
hψm
|ψ 1 i =
0
0
0
X αH 0 + βH 0
hψm
|H 0 |ψ 0 i
αHma
+ βHmb
ma
1
mb 0
=
,
and
(Equation
7.13)
ψ
=
ψm .
0
0
0
E 0 − Em
E 0 − Em
E 0 − Em
m6=a,b
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
269
(b) Similarly, write Equation 7.8 for ψn → ψ:
H 0 ψ2 + H 0 ψ1 = E 0 ψ2 + E 1 ψ1 + E 2 ψ0
and take the inner product with ψa0 and ψb0 :
0 2
0
0 1
0 2
1
0 1
2
0 0
E 0
hψ
E 0
hψ
a |ψ i + hψa |H |ψ i = a |ψ i + E hψa |ψ i + E hψa |ψ i.
As before (Equation 7.11) we might as well choose hψa0 |ψ 1 i = 0. And hψa0 |ψ 0 i = α. So
hψa0 |H 0 |ψ 1 i = E 2 α, and similarly hψb0 |H 0 |ψ 1 i = E 2 β.
Putting in the result of part (a):
X αH 0 + βH 0
ma
0
mb
Hbm
= E2β
0
E 0 − Em
X αH 0 + βH 0
ma
0
mb
Ham
= E 2 α,
0
E 0 − Em
m6=a,b
m6=a,b
This has the form of an eigenvalue equation:
0
0
0
X
1
α
|Hma
|2 Ham
Hmb
2 α
=
E
.
0
0
0
0
β
β
Hbm
Hma
|Hmb
|2
E 0 − Em
m6=a,b
{z
}
|
W2
(c) In the case of Problem 7.39 the matrix is
1
W =
a − c
2
Its (normalized) eigenvectors are
1
1
√
,
±1
2
|V |2 |V |2
.
|V |2 |V |2
so the “good” states are
0
ψ±
=
ψa0 ± ψb0
√
,
2
and the second-order correction to the energies are the eigenvalues
2
E+
=2
|V |2
,
a − c
2
E−
= 0.
This agrees precisely with the result in Problem 7.39(b), obtained by Taylor expansion of the exact
energies.
Problem 7.41
(a)
−
~2 d2 ψ
d2 ψ
=
Eψ
⇒
= −k 2 ψ ⇒ ψ = Aeikx ; ψ(x + L) = ψ(x) ⇒ eikL = 1
2m dx2
dx2
1
⇒ kL = 2nπ (n = 0, ±1, ±2, . . .); ψn0 (x) = √ ei2πnx/L . X
L
270
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
(b)
Z
x
V0 L i2π(n−m)x/L i2πx/L
e
+ e−i2πx/L dx
e−i2πmx/L cos 2π
ei2πnx/L dx =
e
L
2L 0
0
Z Lh
i
V0
=
ei2π(n−m+1)x/L + ei2πx(n−m−1)/L dx.
2L 0
0
Hmn
=
V0
L
Z L
But, for any nonzero integer N,
Z L
e
i2πN x/L
0
ei2πN x/L
dx =
i2πN/L
L
=
0
ei2πN − 1
1−1
=
= 0,
i2πN/L
i2πN/L
while for N = 0 the integral is (obviously) L, so
0
Hmn
=
V0
(δm,n+1 + δm,n−1 ) .
2
(c)
0
0
0
[W ]1,1 = H11
= 0, [W ]1,−1 = H10 −1 = 0, [W ]−1,1 = H−1
1 = 0, [W ]−1,−1 = H−1 −1 = 0,
so
W=
00
,
00
and (Equation 7.33) there are no first-order corrections to the energies.
(d)
2 0
0
0 2
0 2
H1n
Hn1
|H10
|
|H21
|
V0
1
1
V2
=
+
=
+
= 00 ,
0
0
0
0
0
0
0
0
0
E − En
E1 − E0
E1 − E2
2
E1 − 0 E1 − 4E1
6E1
n6=1, −1 1
2
0
0
X H1n
H10
H00 −1
Hn0 −1
V0
1
V02
=
=
=
,
[W 2 ]1,−1 =
E10 − En0
E10 − E00
2
E10 − 0
4E10
n6=1, −1
2 0
0
0
0
X H−1
H−1
1
V0
V02
n Hn 1
0 H0 1
2
[W ]−1,1 =
=
=
=
,
0
0
0
0
E1 − En0
E1 − E0
2
E1 − 0
4E10
n6=1, −1
2 0
0
2
0
2
0
X H−1
|H−1
|H−2
V0
1
V2
1
0|
−1 |
n Hn −1
=
+
=
+
= 00 .
[W 2 ]−1,−1 =
0
0
0
0
0
0
0
0
0
E1 − En
E1 − E0
E1 − E2
2
E1 − 0 E1 − 4E1
6E1
[W 2 ]1,1 =
X
n6=1, −1
so
V2
W = 00
12E1
2
2 3
.
3 2
23
The eigenvalues of
are given by
32
(2 − λ)
3
= (2 − λ)2 − 9 ⇒ λ± = 2 ± 3; λ+ = 5, λ− = −1,
3
(2 − λ)
so the degeneracy lifts at second order. The eigenvectors are given by
23
x
x
=λ
⇒ 2x + 3y = λx ⇒ 2x + 3y = λx.
32
y
y
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
271
1
;
1
1
1
λ− = −1 ⇒ 2x + 3y = −x ⇒ y = −x ⇒ v− = √
,
2 −1
1
λ+ = 5 ⇒ 2x + 3y = 5x ⇒ y = x ⇒ v+ = √
2
so the “good” states are
r
2
2πx
cos
,
L
L
r
1
2
1 1 i2πx/L
2πx
−i2πx/L
= i
e
−e
ψ− = √ (ψ1 − ψ−1 ) = √ √
sin
.
L
L
2
2 L
1 1 i2πx/L
1
e
ψ+ = √ (ψ1 + ψ−1 ) = √ √
+ e−i2πx/L =
2
2 L
(e) The energy, to second order, is
0
1
E =E +E +E
2
= E10 + 0 +
V02
12E10
(2 ± 3), where E10 =
~2 k 2
2π 2 ~2
=
,
2m
mL2
so the perturbed energies are
E+ =
2π 2 ~2
V 2 mL2
+5 0 2 2,
2
mL
24π ~
E− =
2π 2 ~2
V 2 mL2
− 0 2 2.
2
mL
24π ~
Problem 7.42
(a)
∂En
4me3
4
=−
= En ;
∂e
32π 2 20 ~2 (jmax + ` + 1)2
e
∂H
2e 1
=−
. So the F-H theorem says:
∂e
4π0 r
"
2 2 #
e
1
8π0
8π0 E1
8π0
m
e2 m 1
4
e
1
1
En = −
, or
= − 2 En = − 2 2 = − 2 − 2
=
.
2
e
2π0 r
r
e
e n
e
2~
4π0
n
4π0 ~2 n2
But
4π0 ~2
= a (by Eq. 4.72), so
me2
1
1
= 2 .
r
n a
(Agrees with Eq. 7.56.)
(b)
∂En
2me4
2En
∂H
~2
=
=
−
;
=
(2` + 1); so F-H says
∂`
32π 2 20 ~2 (jmax + ` + 1)3
n
∂`
2mr2
~2 (2` + 1) 1
1
4mEn
4mE1
2En
=
, or
=−
=− 3
.
−
n
2m
r2
r2
n(2` + 1)~2
n (2` + 1)~2
4mE1
2
1
1
But −
=
,
so
= 3
. (Agrees with Eq. 7.57.)
~2
a2
r2
n (` + 21 )a2
272
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
Problem 7.43
e2
4π0
1
u.
r
2 2
2m m
e
1
2mEn
1
= 2
= 2 2 . So
−
~2
~ 2~2 4π0
n2
a n
2m
`(` + 1) 2mEn
Equation 4.53 ⇒ u =
−
− 2
2
2
r
~
~
00
me2
1
= (Eq. 4.72), and
4π0 ~2
a
2
`(` + 1)
1
00
−
u.
F u =
+
r2
ar n2 a2
Z
Z
2
2
1
1
s 00
s `(` + 1)
∴ (ur u ) dr = ur
−
+ 2 2 u dr = `(` + 1)hrs−2 i − hrs−1 i + 2 2 hrs i
2
r
ar n a
a
n a
Z
Z
Z
d
=−
(urs )u0 dr = − (u0 rs u0 ) dr − s (urs−1 u0 ) dr.
dr
Z
Z
Z
Z
d
(urs )u dr = − (u0 rs u) dr − s urs−1 u dr ⇒
Lemma 1: (urs u0 ) dr = −
dr
Z
Z
s
2 (urs u0 ) dr = −shrs−1 i, or
(urs u0 ) dr = − hrs−1 i.
2
Z
Z
Z
Z
d
s+1 0
0 s 0
00 s+1 0
0
(r u ) dr = −(s + 1) (u r u ) dr − (u0 rs+1 u00 ) dr.
Lemma 2: (u r u ) dr = − u
dr
Z
Z
Z
Z
2
2 (u00 rs+1 u0 ) dr = −(s + 1) (u0 rs u0 ) dr, or:
(u0 rs u0 ) dr = −
(u00 rs+1 u0 ) dr.
s+1
Lemma 3: Use F in Lemma 2, and exploit Lemma 1:
But
Z
Z `(` + 1)
2
1
2
−
+
(u r u ) dr = −
(urs+1 u0 ) dr
s+1
r2
ar n2 a2
Z
Z
Z
2
1
2
s−1 0
s 0
s+1 0
`(` + 1) (ur u ) dr −
(ur u ) dr + 2 2 (ur u ) dr
= −
s+1
a
n a
2
2
s s−1
1
s − 1 s−2
s+1 s
= −
hr i −
− hr i + 2 2 −
hr i
`(` + 1) −
s+1
2
a
2
n a
2
s−1
2
s
1
= `(` + 1)
hrs−2 i −
hrs−1 i + 2 2 hrs i.
s+1
a s+1
n a
0 s 0
Plug Lemmas 1 and 3 into :
2
1
`(` + 1)hrs−2 i − hrs−1 i + 2 2 hrs i
a n a
s−1
2
s
1
s(s − 1) s−2
= −`(` + 1)
hrs−2 i +
hrs−1 i − 2 2 hrs i +
hr i.
s+1
a s+1
n a
2
2
2
s
s−1
s(s − 1)
s
s−1
hr i −
1+
hr i + `(` + 1) 1 +
−
hrs−2 i = 0.
n2 a 2
a
s+1
s+1
2
|
|
{z
}
{z
}
2s
s+1
2s+1
s+1
2(s + 1) s
2
(s2 − 1)
s−1
2
hr i − (2s + 1)hr i + 2s ` + ` −
hrs−2 i = 0, or, finally,
n2 a2
a
4
(s + 1) s
sa2 2
s−1
hr
i
−
a(2s
+
1)hr
i
+
(4` + 4` + 1 −s2 )hrs−2 i = 0. QED
{z
}
n2
4 |
(2`+1)2
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
273
Problem 7.44
(a)
1
1
1
+0=0⇒
= 2 .
r
r
n a
1
1
2
1
2
a hri − 3ah1i +
(2` + 1)2 − 1 a2
= 0 ⇒ 2 hri = 3a − `(` + 1)a2 2 = 2 3n2 − `(` + 1) .
2
n
4
r
n
n a
n
a 2
hri =
3n − `(` + 1) .
2
1
h1i − a
n2
2
a2 3 2
1
3 2
a 2
2
hr
hr
i
−
5ahri
+
(2`
+
1)
−
4
a
=
0
⇒
i
=
5a
3n
−
`(`
+
1)
−
(2` + 1)2 − 4
n2
2
n2
2
2
a2 a2 3 2
2
hr
i
=
15n
−
5`(`
+
1)
−
4`(`
+
1)
−
1
+
4
=
15n2 − 9`(` + 1) + 3
2
n
2
2
2 2 2 3a
n a
=
5n2 − 3`(` + 1) + 1 ;
hr2 i =
5n2 − 3`(` + 1) + 1 .
2
2
4 3
3
hr i − 7ahr2 i +
(2` + 1)2 − 9 a2 hri = 0 =⇒
2
n
4
3
n2 a2 2
a
4 3
hr i = 7a
5n − 3`(` + 1) + 1 − [4`(` + 1) − 8] a2 3n2 − `(` + 1)
2
n
2
4
2
a3 =
35n4 − 21`(` + 1)n2 + 7n2 − [3`(` + 1) − 6] 3n2 − `(` + 1)
2
a3 35n4 − 21`(` + 1)n2 + 7n2 − 9`(` + 1)n2 + 3`2 (` + 1)2 + 18n2 − 6`(` + 1)
=
23
a =
35n4 + 25n2 − 30`(` + 1)n2 + 3`2 (` + 1)2 − 6`(` + 1) .
2
hr3 i =
n2 a3 35n4 + 25n2 − 30`(` + 1)n2 + 3`2 (` + 1)2 − 6`(` + 1) .
8
(b)
0+a
1
r2
−
1
(2` + 1)2 − 1 a2
4
1
r3
1
r3
=0⇒
1
r2
= a`(` + 1)
1
.
r3
(c)
a`(` + 1)
1
r3
=
1
⇒
(` + 1/2)n3 a2
=
1
.
`(` + 1/2)(` + 1)n3 a3
Agrees with Eq. 7.66.
Notice a curious fact about these results: for a given n,
hri decreases with increasing `, whereas h1/ri is constant;
hr2 i decreases with increasing `, but so does h1/r2 i;
hr3 i decreases with increasing `, and so does h1/r3 i.
Off hand, you might suppose that if hrs i decreases, h1/rs i should increase, but this is not the case. What
happens (as Paul Muzikar points out) is that the electron density is more sharply peaked as ` increases; for small
` it is both closer in and farther out than it is for large `.
274
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
Problem 7.45
(a)
|1 0 0i = √
1
πa3
−r/a
e
(Eq. 4.80),
Es1 = h1 0 0|H 0 |1 0 0i = eEext
Z
e−2r/a (r cos θ)r2 sin θ dr dθ dφ.
π
Z π
cos θ sin θ dθ =
But the θ integral is zero:
1
πa3
0
sin2 θ
= 0. So Es1 = 0. QED
2 0

r −r/2a
1 1 

√
1
−
e
|1i
=
ψ
=
2
0
0


2a

2πa 2a


1 1


re−r/2a sin θeiφ
 |2i = ψ2 1 1 = − √
2
8a
πa
(b) From Problem 4.13:
1
1

 |3i = ψ2 1 0 = √
re−r/2a cos θ

2

4a

2πa


1 1


re−r/2a sin θe−iφ
 |4i = ψ2 1−1 = √
πa 8a2

Z π


cos θ sin θ dθ = 0
h1|Hs0 |1i = {. . .}



0
Z π



2
0

h2|Hs |2i = {. . .}
sin θ cos θ sin θ dθ = 0 



Z0 π



2
0

h3|Hs |3i = {. . .}
cos θ cos θ sin θ dθ = 0 



Z0 π



2
0
sin θ cos θ sin θ dθ = 0 
h4|Hs |4i = {. . .}



of Hs0 are zero

Z0 2π
 All matrix elements

0
except h1|Hs |3i and h3|Hs0 |1i
h1|Hs0 |2i = {. . .}
eiφ dφ = 0
(which are complex conjugates,

Z0 2π



so only one needs to be evaluated).
−iφ

0

h1|Hs |4i = {. . .}
e
dφ = 0



0

Z 2π




h2|Hs0 |3i = {. . .}
e−iφ dφ = 0



0

Z 2π



−2iφ
0

h2|Hs |4i = {. . .}
e
dφ = 0



0

Z 2π



−iφ
0


h3|Hs |4i = {. . .}
e
dφ = 0
0
1 1 1
1
√
2
2πa 2a
Z π2πa 4a
Z r −r/2a −r/2a
e
re
cos θ(r cos θ)r2 sin θ dr dθ dφ
1−
2a
Z ∞
eEext
r −r/a 4
2
=
(2π)
cos
θ
sin
θ
dθ
1
−
e
r dr
2πa8a3 Z
2a 0
0
Z ∞
∞
eEext 2
1
eEext
1
4 −r/a
5 −r/a
5
6
=
r
e
dr
−
r
e
dr
=
4!a
−
5!a
8a4 3
2a 0
12a4
2a
0
eEext
5
5
=
24a 1 −
= eaEext (−3) = −3aeEext .
12a4
2
h1|Hs0 |3i = eEext √


0 0 1 0
0 0 0 0

W = −3aeEext 
1 0 0 0.
0 0 0 0
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
275
We need the eigenvalues of this matrix. The characteristic equation is:
−λ 0 1 0
0 −λ 0
−λ 0 0
0 −λ 0 0
= −λ 0 −λ 0 + 1 0 0 = −λ(−λ)3 + (−λ2 ) = λ2 (λ2 − 1) = 0.
1 0 −λ 0
0 0 −λ
0 0 −λ
0 0 0 −λ
The eigenvalues are 0, 0, 1, and −1, so the perturbed energies are
E2 , E2 , E2 + 3aeEext , E2 − 3aeEext .
Three levels.
 
 
0
0
0
1
 

(c) The eigenvectors with eigenvalue 0 are |2i = 
 0  and |4i =  0 ; the eigenvectors with eigenvalues ±1
1
0


1
1 
0 

. So the “good” states are ψ2 1 1 , ψ2 1−1 , √1 (ψ2 0 0 + ψ2 1 0 ), √1 (ψ2 0 0 − ψ2 1 0 ).
are |±i ≡ √ 

±1
2
2
2
0
hpe i4 = −e
1 1
πa 64a4
Z 2π
But
Z
h
i
r2 e−r/a sin2 θ r sin θ cos φî + r sin θ sin φĵ + r cos θk̂ r2 sin θ dr dθ dφ.
Z 2π
cos φ dφ =
0
Z π
sin φ dφ = 0,
0
π
sin3 θ cos θ dθ =
0
hpe i4 = 0.
Likewise
sin4 θ
= 0,
4 0
so
hpe i2 = 0.
Z
1
hpe i± = − e (ψ1 ± ψ3 )2 (r)r2 sin θ dr dθ dφ
2
Z h
i2
1 1 1
r r
=− e
1−
±
cos θ e−r/a r(sin θ cos φ î + sin θ sin φ ĵ + cos θ k̂)r2 sin θ dr dθ dφ
2
2 2πa 4a Z
2a
2a
h
i2
e k̂ 1
r r
=−
2π
1
−
±
cos
θ
r3 e−r/a cos θ sin θ dr dθ.
2 2πa 4a2
2a
2a
Rπ
Rπ
But 0 cos θ sin θ dθ = 0 cos3 θ sin θ dθ = 0, so only the cross-term survives:
Z 1
r e
1−
r cos θ r3 e−r/a cos θ sin θ dr dθ
hpe i± = − 3 k̂ ±
8a
a
2a Z
Z
π
∞
e e 2
r 4 −r/a
1
2
5
6
=∓
k̂
cos θ sin θ dθ
1−
r e
dr = ∓
k̂
4!a − 5!a
4
8a
2a
8a4
3
2a
0
0
1
5
5
= ∓ek̂
24a 1 −
= ±3aek̂.
12a4
2
Problem 7.46
(a) The nine states are:
276
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY

` = 0 : |3 0 0i = R30 Y00




` = 1 : |3 1 1i = R31 Y11




|3 1 0i = R31 Y10




|3 1 − 1i = R31 Y1−1

` = 2 : |3 2 2i = R32 Y22


|3 2 1i = R32 Y21




|3 2 0i = R32 Y20




|3 2 − 1i = R32 Y2−1



|3 2 − 2i = R32 Y2−2
Hs0 contains no φ dependence, so the φ integral will be:
hn ` m|Hs0 |n0 `0 m0 i = {· · · }
Z 2π
0
e−imφ eim φ dφ,
which is zero unless
m0 = m.
0
Rπ
2
For diagonal elements: hn ` m|Hs0 |n ` mi = {· · · } 0 [P`m (cos θ)] cos θ sin θ dθ. But (p. 136 in the text)
P`m is a polynomial (even or odd) in cos θ, multiplied (if m is odd) by sin θ. Since sin2 θ = 1 − cos2 θ,
2
[P`m (cos θ)] is a polynomial in even powers of cos θ. So the θ integral is of the form
Z π
π
(cos θ)2j+1 sin θ dθ = −
0
(cos θ)2j+2
= 0. All diagonal elements are zero.
(2j + 2) 0
There remain just 4 elements to calculate:
m = m0 = 0 : h3 0 0|Hs0 |3 1 0i, h3 0 0|Hs0 |3 2 0i, h3 1 0|Hs0 |3 2 0i; m = m0 = ±1 : h3 1 ± 1|Hs0 |3 2 ± 1i.
Z
Z
h3 0 0|Hs0 |3 1 0i = eEext R30 R31 r3 dr Y00 Y10 cos θ sin θ dθ dφ. From Table 4.7 :
Z
1
8
1 1
2
√ 3/2
R30 R31 r dr = √
3/2
a
a
a
27
27 6
3
Z 2r
2r2
1−
+
3a 27a2
r −r/3a 3
e−r/3a 1 −
re
r dr.
6a
Let x ≡ 2r/3a:
5 Z ∞ 24
3a
x2 x 4 −x
√
1−x+
1−
x e dx
6
4
35 2a4 2
0
Z ∞
a
5
5
1
a
5
5
1
= √
1 − x + x2 − x3 x4 e−x dx = √
4! − 5! + 6! − 7!
4
12
24
4
12
24
2 2 0
2 2
√
= −9 2a.
r
√
√
√
Z
Z
Z π
1
3
3
32
3
Y00 Y10 cos θ sin θ dθ dφ = √
cos θ cos θ sin θ dθ dφ =
2π
cos2 θ sin θ dθ =
=
.
4π
2
3
3
4π 4π
0
√ !
√
√
3
h3 0 0|Hs0 |3 1 0i = eEext (−9 2a)
= −3 6aeEext .
3
Z
Z
0
3
h3 0 0|Hs |3 2 0i = eEext R30 R32 r dr Y00 Y20 cos θ sin θ dθ dφ.
Z
Z
R30 R31 r3 dr =
1
Y00 Y20 cos θ sin θ dθ dφ = √
4π
r
5
16π
Z
(3 cos2 θ − 1) cos θ sin θ dθ dφ = 0.
h3 0 0 |Hs0 |3 2 0i = 0.
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
h3 1 0|Hs0 |3 2 0i = eEext
Z
Z
Z
R31 R32 r3 dr
Z
277
Y10 Y20 cos θ sin θ dθ dφ.
Z r −r/3a 2 −r/3a 3
1 1 4
1 1
8
√ 3/2
√
1
−
re
r e
r dr
R31 R32 r dr =
6a
27 6 a a 81 30 a3/2 a2
√
7 Z ∞ 24
x 6 −x
a
3a
1
9 5
= √
1−
x e dx = √
6! − 7! = −
a.
4
4
2
38 5a6 2
24 5
0
3
r
r
Z
3
5
cos θ(3 cos2 θ − 1) cos θ sin θ dθ dφ
4π 16π
√ √
π
Z π
15
15
2
3
1
4
2
5
3
(3 cos θ − cos θ) sin θ dθ =
=√ .
=
2π
− cos θ + cos θ
8π
4
5
3
15
0
0
√ !
√
9 5
2
√
h3 1 0|Hs0 |3 2 0i = eEext −
a
= −3 3aeEext .
2
15
Z
Z
∗
h3 1 ± 1|Hs0 |3 2 ± 1i = eEext R31 R32 r3 dr
Y1±1 Y2±1 cos θ sin θ dθ dφ.
Y10 Y20 sin θ cos θ dθ dφ =
Z
!
r !Z
3
15
∓
sin θe∓iφ sin θ cos θe±iφ cos θ sin θ dθ dφ
∓
8π
8π
√
π
Z π
3 5
cos3 θ cos5 θ
3√
=
5 −
2π
+
cos2 θ(1 − cos2 θ) sin θ dθ =
8π
4
3
5
0
0
1
= √ .
5
√ !
1
9
5
0
√
h3 1 ± 1|Hs |3 2 ± 1i = eEext −9
a
= − aeEext .
2
2
5
∗
Y1±1 Y2±1 cos θ sin θ dθ dφ =
r
Thus the matrix representing Hs0 is (all empty boxes are zero; all numbers multiplied by −aeEext ):
300 310 320 311 321 31-1 32-1 322 32-2
300
336
310 336
320
333
333
311
321
9/2
9/2
31-1
32-1
322
32-2
9/2
9/2
278
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
(b) The perturbing matrix (below) breaks into a 3×3 block, two 2 × 2 blocks, and two 1 × 1 blocks, so we can
work out the eigenvalues in each block separately.
√
√


−λ
20
2 0
√
√0
√
3 × 3 : 3 3 2 0 1;
2 −λ 1 = −λ3 + λ + 2λ = −λ(λ2 − 3) = 0;
0 1 0
0 1 −λ
√
λ = 0, ± 3 ⇒ E11 = 0, E21 = 9aeEext , E31 = −9aeEext .
9 01
−λ 1
2×2:
= λ2 − 1 = 0 ⇒ λ = ±1.
;
1 −λ
2 10
9
9
aeEext , E51 = − aeEext . From the other 2×2 we get E61 = E41 , E71 = E51 , and from the 1×1’s we
2
2
0
(degeneracy 3)
(9/2)aeEext
(degeneracy 2)
−(9/2)aeEext (degeneracy 2)
get E81 = E91 = 0. Thus the perturbations to the energy (E3 ) are:
9aeEext
(degeneracy 1)
−9aeEext
(degeneracy 1)
E41 =
Problem 7.47
1
µ0 gd e2
hSd · Se i; Eq. 7.95 ⇒ Sd · Se = (S 2 − Se2 − Sd2 ).
3πmd me a3
2
Electron has spin 12 , so Se2 = 12 23 ~2 = 34 ~2 ; deuteron has spin 1, so Sd2 = 1(2)~2 = 2~2 .
1
3 2
2
2
Total spin could be 23 [in which case S 2 = 32 25 ~2 = 15
4 ~ ] or 2 [in which case S = 4 ~ ]. Thus

 1 15 2 3 2
 2 4 ~ − 4 ~ − 2~2 = 21 ~2 
3 2
µ0 gd e2 ~2
;
the
difference
is
~
,
so
∆E
=
.
hSd · Se i =

1 3 2 3 2
2
2πmd me a3
2
= −~2
2 4 ~ − 4 ~ − 2~
1
Equation 7.93 ⇒ Ehf
=
1
1
2gd e2 ~2
2gd ~4
3 gd mp
⇒
µ
=
,
so
∆E
=
=
=
∆Ehydrogen (Eq. 7.97).
0
2
2
2
3
2
2
4
c
0 c
4π0 md me c a
md me c a 2 gpmd
c
ch
2 gp md
4 5.59
Now, λ = =
, so λd =
λh , and since md = 2mp , λd =
(21 cm) = 92 cm.
ν
∆E
3 gd mp
3 1.71
But
µ0 0 =
Problem 7.48
(a) The potential energy of the electron (charge −e) at (x, y, z) due to q’s at x = ±d alone is:
"
#
eq
1
1
p
V =−
+p
. Expanding (with d x, y, z) :
4π0
(x + d)2 + y 2 + z 2
(x − d)2 + y 2 + z 2
1
p
= (x2 ± 2dx + d2 + y 2 + z 2 )−1/2 = (d2 ± 2dx + r2 )−1/2 =
(x ± d)2 + y 2 + z 2
1
x
r2
3 4x2
1
x
1
2
2
1∓ − 2 +
=
1 ∓ + 2 (3x − r ) .
≈
d
d 2d
8 d2
d
d 2d
1
d
1±
2x r2
+ 2
d
d
−1/2
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
279
eq
x
2eq
x
1
1
eq
2
2
2
2
V =−
(3x2 − r2 )
1 − + 2 (3x − r ) + 1 + + 2 (3x − r ) = −
−
4π0 d
d 2d
d 2d
4π0 d 4π0 d3
e q
= 2βd2 + 3βx2 − βr2 , where β ≡ −
.
4π0 d3
Thus with all six charges in place
H 0 = 2(β1 d21 + β2 d22 + β3 d23 ) + 3(β1 x2 + β2 y 2 + β3 z 2 ) − r2 (β1 + β2 + β3 ). QED
Z
1
(b) h1 0 0|H 0 |1 0 0i =
e−2r/a H 0 r2 sin θ dr dθ dφ
πa3
Z
Z
(β1 + β2 + β3 )
3
e−2r/a (β1 x2 + β2 y 2 + β3 z 2 )r2 sin θdr dθ dφ −
r2 e−2r/a r2 sin θ dr dθ dφ.
= V0 + 3
πa
πa3
Z
I1 ≡
r2 e−2r/a r2 sin θ dr dθ dφ = 4π
Z ∞
0
Z
I2 ≡
e
−2r/a
2
2
2
a
r4 e−2r/a dr = 4π4!( )5 = 3πa5 .
2
2
(β1 x + β2 y + β3 z )r sin θ dr dθ dφ
Z
r4 e−2r/a (β1 sin2 θ cos2 φ + β2 sin2 θ sin2 φ + β3 cos2 θ) sin θ dr dθ dφ.
Z 2π
Z 2π
Z 2π
2
2
But
cos φ dφ =
sin φ dφ = π,
dφ = 2π. So
0
0
0
Z ∞
Z π
=
r4 e−2r/a dr
π(β1 + β2 ) sin2 θ + 2πβ3 cos2 θ sin θ dθ.
0
0
Z π
Z π
2
4
3
cos2 θ sin θ dθ = . So
But
sin θ dθ = ,
3
3
0
0
a 5 4π
4π
= 4!
(β1 + β2 ) +
β3 = πa5 (β1 + β2 + β3 ).
2
3
3
=
(β1 + β2 + β3 )
3
πa5 (β1 + β2 + β3 ) −
3πa5 = V0 .
πa3
πa3


|2 0 0i = R20 Y00 





|2 1 1i = R21 Y11
(functional forms in Problem 4.13).
(c) The four states are
|2 1 − 1i = R21 Y1−1 





|2 1 0i = R21 Y10
Diagonal elements: hn ` m|H 0 |n ` mi = V0 + 3 β1 hx2 i + β2 hy 2 i + β3 hz 2 i − (β1 + β2 + β3 )hr2 i.
h1 0 0|H 0 |1 0 0i = V0 +
For |2 0 0i, hx2 i = hy 2 i = hz 2 i = 31 hr2 i (Y00 does not depend on φ, θ; this state has spherical symmetry),
so h2 0 0|H 0 |2 0 0i = V0 .
(I could have used the same argument in (b).)
n2 a 2 2
From Problem 7.44(a), hr2 i =
5n − 3`(` + 1) + 1 , so for n = 2, ` = 1 : hr2 i = 30a2 . Moreover,
2
Z 2π
Z 2π
2
2
since hx i = {. . .}
cos φ dφ = {. . .}
sin2 φ dφ = hy 2 i, and hx2 i + hy 2 i + hz 2 i = hr2 i, it follows that
0
0
1
1
hx i = hy i = (hr2 i − hz 2 i) = 15a2 − hz 2 i. So all we need to calculate is hz 2 i.
2
2
Z
1
1
h2 1 0|z 2 |2 1 0i =
r2 e−r/a cos2 θ(r2 cos2 θ)r2 sin θ dr dθ dφ
2πa 16a4
2
2
280
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
=
1
16a5
Z ∞
r6 e−r/a dr
0
Z π
cos4 θ sin θ dθ =
0
1
2
6!a7 = 18a2 ;
16a5
5
hx2 i = hy 2 i = 15a2 − 9a2 = 6a2 .
h2 1 0|H 0 |2 1 0i = V0 + 3(6a2 β1 + 6a2 β2 + 18a2 β3 ) − 30a2 (β1 + β2 + β3 )
= V0 − 12a2 (β1 + β2 + β3 ) + 36a2 β3 .
Z
1 1
r2 e−r/a sin2 θ(r2 cos2 θ)r2 sin θ dr dθ dφ
πa 64a4
Z π
Z ∞
1
1
2 2
2
2
6 −r/a
7
(1
−
cos
θ)
cos
θ
sin
θ
dθ
=
=
r
e
dr
6!a
−
= 6a2 ;
5
32a5 0
32a
3
5
0
h2 1 ± 1|z 2 |2 1 ± 1i =
hx2 i = hy 2 i = 15a2 − 3a2 = 12a2 .
h2 1 ± 1|H 0 |2 1 ± 1i = V0 + 3(12a2 β1 + 12a2 β2 + 6a2 β3 ) − 30a2 (β1 + β2 + β3 )
= V0 + 6a2 (β1 + β2 + β3 ) − 18a2 β3 .
We need h2 0 0|H 0 |2 1 0i, h2 0 0|H 0 |2 1 ± 1i, h2 1 0|H 0 |2 1 ± 1i, and h2 1 − 1|H 0 |2 1 1i.
Off-diagonal elements:
Now hn ` m|V0 |n0 `0 m0 i = 0, by orthogonality, and hn ` m|r2 |n0 `0 m0 i = 0, by orthogonality of Y`m , so
all we need are the matrix elements of x2 and y 2 (h|z 2 |i = −h|x2 |i − h|y 2 |i). For h2 0 0|x2 |2 1 ± 1i and
R 2π
R 2π
R 2π
h2 1 0|x2 |2 1 ± 1i the φ integral is 0 cos2 φe±iφ dφ = 0 cos3 φ dφ ± i 0 cos2 φ sin φ dφ = 0, and the
same goes for y 2 . So h2 0 0|H 0 |2 1 ± 1i = h2 1 0|H 0 |2 1 ± 1i = 0.
Rπ
π
For h2 0 0|x2 |2 1 0i and h2 0 0|y 2 |2 1 0i the θ integral is 0 cos θ(sin2 θ) sin θ dθ = sin4 θ/4 0 = 0,
0
h2 0 0|H |2 1 0i = 0.
so
Finally:
Z
1 1
r2 e−r/a sin2 θe2iφ (r2 sin2 θ cos2 φ)r2 sin θ dr dθ dφ
πa 64a4
Z π
Z ∞
Z 2π
1
5
6 −r/a
= −
r
e
dr
sin
θdθ
e2iφ cos2 φ dφ
64πa5 0
0
0
|
{z
}|
{z
}|
{z
}
h2 1 − 1|x2 |2 1 1i = −
6!a7
16/15
π/2
1
16 π
6!a7
= −6a2 .
= −
5
64πa
15 2
For y 2 , the φ integral is
R 2π
0
e2iφ sin2 φ dφ = −π/2, so h2 1 − 1|y 2 |2 1 1i = 6a2 , and h2 1 − 1|z 2 |2 1 1i = 0.
h2 1 − 1|H 0 |2 1 1i = 3 β1 (−6a2 ) + β2 (6a2 ) = −18a2 (β1 − β2 ).
The perturbation matrix is:
2 0 0
2 1 0
2 1 1
2 1 -1
2 0 0
V0
0
0
0
2 1 0
0
V0 − 12a2 (β1 + β2 ) + 24a2 β3
0
0
2 1 1
0
0
V0 + 6a2 (β1 + β2 ) − 12a2 β3
−18a2 (β1 − β2 )
2 1 -1
0
0
−18a2 (β1 − β2 )
V0 + 6a2 (β1 + β2 ) − 12a2 β3
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
The 2 × 2 block has the form
or
A
B
B
A
281
; its characteristic equation is (A − λ)2 − B 2 = 0, so A − λ = ±B,
2
2
2
λ = A ∓ B = V0 + 6a (β1 + β2 ) − 12a β3 ± 18a (β1 − β2 ) =
V0 + 24a2 β1 − 12a2 β2 − 12a2 β3 ,
V0 − 12a2 β1 + 24a2 β2 − 12a2 β3 .
1 = V 0
= V0 − 12a2 (β1 + β2 − 2β3 )
The first-order corrections to the energy (E2 ) are therefore: 2
3 = V0 − 12a2 (−2β1 + β2 + β3 )
4 = V0 − 12a2 (β1 − 2β2 + β3 )
(i) If β1 = β2 = β3 , then 1 = 2 = 3 = 4 = V0 : one level (still 4-fold degenerate).
(ii) If β1 = β2 6= β3 , then 1 = V0 , 2 = V0 − 24a2 (β1 − β3 ), 3 = 4 = V0 + 12a2 (β1 − β3 ): three levels
(one remains doubly degenerate).
(iii) If all three β’s are different, there are four levels (no remaining degeneracy).
Problem 7.49
We first calculate the mechanical angular momentum:
1
q 2
r B0 − r(r · B0 ) .
Lmechanical = L − q(r × A) = L − q r × − r × B0
=L−
2
2
The Hamiltonian (Problem 4.72) is
H=
so
p2
q2 2 2
+ qϕ − B · (γo L + γS) +
r B0 − (r · B0 )2 ,
2m
8m
∂H
q2 2
r B0 − (r · B0 )z ,
= −(γo Lz + γSz ) +
∂B0
4m
and since γo = q 2 /2m,
h
i
∂H
q 2
r B0 − (r · B0 )z − γSz = −γo Lmechanicalz − γSz = −µz .
= −γo Lz −
∂B0
2
Therefore, from the Feynman-Hellman theorem,
∂En
= −hψn |µz |ψn i. X
∂B0
Problem 7.50
(a) At first order in perturbation theory
E 1 = −B0 (γo hLz i + γhSz i) +
Z
e2 2 X 2
B0
hxi + yi2 i
8m
i=1
282
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
and at second order
E 2 = B02
X |hψn |(γo Lz + γSz )|ψ0 i|2
E0 − En
n6=0
.
(I’ve ignored the second term in H 0 , when computing E 2 , because it will add higher powers of B0 .) Now,
since ψ0 is an eigenstate of Lz and Sz with eigenvalue 0, we are left with only one contribution:
∆E =
2
e2 2 X 2
B0
hxi + yi2 i.
8m
i=2
The ground state is spherically symmetric, so hx2i i = hyi2 i = hr2 i/3, where hr2 i is the expectation value of
r2 in the ground state of hydrogen (Problem 4.15, only with Z = 2): hr2 i = 3a2 /Z 2 :
∆E =
e2 2 a 2
e2 2 2
B0 hr i =
B
.
6m
2m 0 Z
χ=−
N ∂2E
N e2 a 2
µ0
= − µ0
.
2
V
∂B0
V
m Z
(b) Combining these results, we get
Now
N
=
V
atoms
volume
=
1
mass/atom
mass ρ
=
,
volume
mHe
χ=−
so
ρ µ0 e2 a 2
.
mHe m
Z
With Z = 2 this yields a numerical value of −6.2 × 10−10 , but if you use the screened value for Z of 1.69
(see Equation 8.34) it rises to −8.7 × 10−10 .
Problem 7.51
(a) (i) Equation 7.10:
(H 0 − E00 )ψ01 = −(H 0 − E01 )ψ00 .
~2 2
e2 1
~2
2
2
H =−
∇ −
=−
∇ +
, since
2m
4π0 r
2m
ar
~2
.
E00 = −
2ma2
H 0 = eEext r cos θ; E01 = 0 (Problem 7.45(a)).
1
ψ00 = √
e−r/a ; ψ01 = f (r)e−r/a cos θ.
πa3
0
a=
4π0 ~2
.
me2
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
283
Equation 4.13 ⇒
f e−r/a d cos θ d
d
−r/a
2 d
+
f
e
r
sin
θ
(cos
θ)
r2 dr
dr
r2 sin θ dθ
dθ
−r/a
cos θ d
1
fe
d =
r2 f 0 − f e−r/a + 2
− sin2 θ
r2 dr
a
r sin θ dθ
1
2 cos θ −r/a
cos θ
2 0
1
0
−r/a
−r/a
2
00
2r f − f e
−
fe
+ r f − f + 2f e
=
r2
a
a
a
r2
2
1
1
1
1
= cos θe−r/a f 00 − f 0 + 2 f + 2 f 0 − f
− 2f 2 .
a
a
a
r
r
∇2 ψ01 =
Plug this into Eq. 7.10:
~2
1
11
1
1
1
1
2
1
−
e−r/a ,
cos θe−r/a f 00 − f 0 + 2 f + 2 f 0 − f
− 2f 2 + 2f
− f 2 = −eEext r cos θ √
2m
a
a
a
r
r
ar
a
πa3
1
2
1
2meEext
4γ
√
r,
f 00 − f 0 + 2f 0 − 2f 2 =
r=
2
3
a
r
r
a
~ πa
f (r) = A + Br + Cr2 ,
Now let
so
f 0 = B + 2Cr
and f 00 = 2C.
where
γ≡
meEext
√ .
2~2 πa
Then
2
2
2
4γ
2C − (B + 2Cr) + (B + 2Cr) − 2 (A + Br + Cr2 ) =
r.
a
r
r
a
Collecting like powers of r:
r−2 :
A = 0.
r−1 :
2B − 2B = 0
(automatic).
0
r :
2C − 2B/a + 4C − 2C = 0 ⇒ B = 2aC.
1
−4C/a = 4γ/a ⇒ C = −γ.
r :
Evidently the function suggested does satisfy Eq. 7.10, with the coefficients A = 0, B = −2aγ, C = −γ;
the first-order correction to the wave function is
ψ01 = −γr(r + 2a)e−r/a cos θ.
(ii) Equation 7.14 says, in this case:
Z
1 meEext
√
E02 = hψ00 |H 0 |ψ01 i = − √
eE
e−r/a (r cos θ)r(r + 2a)e−r/a cos θ r2 sin θ dr dθ dφ
ext
πa3 2~2 πa
Z ∞
Z π
m(eEext )2
4
−2r/a
= −
2π
r
(r
+
2a)e
dr
cos2 θ sin θ dθ
2πa2 ~2
0
0
2 a 5 cos3 θ π
eEext
a 6
5!
+ 2a 4!
−
= −m
a~
2
2
3
0
2 2 2
eEext
27 6 2
3eEext a
= −m
a
= −m
.
a~
8
3
2~
284
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
(b) (i) This is the same as (a) [note that E01 = 0, as before, since ψ00 is spherically symmetric, so hcos θi = 0]
except for the r-dependence of H 0 . So Eq. ⇒
2mep
2β
1 1
1
1
mep
00
0
√
√
f + 2f
= − 2 , where β ≡
.
−
− 2f 2 = −
2
2
3
r
a
r
r
r
4π0 ~ πa
4π0 ~2 πa3
The solution this time it obvious: f (r) = β (constant). [For the general solution we would add the general
solution to the homogeneous equation (right side set equal to zero), but this would simply reproduce the
unperturbed ground state, ψ00 , which we exclude—see p. 282.] So
ψ01 = βe−r/a cos θ.
(ii) The electric dipole moment of the electron is
hpe i = h−er cos θi = −ehψ00 +ψ01 |r cos θ|ψ00 +ψ01 i = −e hψ00 |r cos θ|ψ00 i + 2hψ00 |r cos θ|ψ01 i + hψ01 |r cos θ|ψ01 i .
But the first term is zero, and the third is higher order, so
Z
1
β e−r/a (r cos θ)e−r/a cos θ r2 sin θ dr dθ dφ
hpe i = −2e √
πa3
Z ∞
Z π
mep
a 4
2
me2 p
3 −2r/a
2
= −2e
2π
3!
r
e
dr
cos
θ
sin
θ
dθ
=
−
2
3
2
3
4π0 ~ πa
~
πa
2
3
0
0
0
4 2
2
me p
3a
2
me pa
=−
=−
= −p.
0 ~2 πa3
8
3
4π0 ~2
Evidently the dipole moment associated with the perturbation of the electron cloud cancels the dipole
moment of the nucleus, and the total dipole moment of the atom is zero.
(iii) The first-order correction is zero (as noted in (i)). The second-order correction is
Z
cos θ
mep
−r/a
√
e
e−r/a cos θ r2 sin θ dr dθ dφ
r2
4π0 ~2 πa3
Z ∞
Z π
a 2
(ep)2
(ep)2
−2r/a
2
2π
= −m
e
dr
cos θ sin θ dθ = −2m
(4π0 )2 ~2 πa3
(4π0 )2 ~2 a3 2
3
0
0
4
me4
p2
4 p 2
E1 .
=
−
=
2
2
2
2
3
2(4π0 ) ~
e a
3 ea
1
E02 = hψ00 |H 0 |ψ01 i = √
πa3
ep
−
4π0
Problem 7.52
(a) The Schrödinger equation separates into two one-dimensional harmonic oscillators, and the solutions can
be written as products, ψ(x, y) = ψm (x)ψn (y), with energies Emn = (m + n + 1)~ω. The ground state is
m = n = 0, and (Equation 2.60)
r
ψ0 (x, y) =
mω − mω (x2 +y2 )
e 2~
, E0 = ~ω.
π~
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
285
The first excited states are m = 0, n = 1 and m = 1, n = 0; using Equation 2.63:
r
2 mω − mω (x2 +y2 )
y e 2~
,
π
~
r
2 mω − mω (x2 +y2 )
x e 2~
.
π
~
ψ1a (x, y) = ψ0 (x)ψ1 (y) =
ψ1b (x, y) = ψ1 (x)ψ0 (y) =
Any orthonormal linear combination of these would also do; the energy is E1 = 2~ω.
(b) For the ground state we use nondegenerate perturbation theory (Equation 7.9):
mω
E01 = hψ0 |H 0 |ψ0 i =
π~
qB0
−
2m
Z
mω
2
2
mω
2
2
e− 2~ (x +y ) (xpy − ypx )e− 2~ (x +y ) dx dy.
But
(xpy − ypx )e
2
2
− mω
2~ (x +y )
2
2
mω
∂
∂
−y
e− 2~ (x +y )
= −i~ x
∂y
∂x
mω mω 2 2 i
h mω mω 2 2
y e− 2~ (x +y ) − y −
x e− 2~ (x +y ) = 0,
= −i~ x −
~
~
so E01 = 0. For the first excited state(s) we need degenerate perturbation theory (Equation 7.33):
mω 2 qB Z
2
2
2
2
mω
mω
0
−
ye− 2~ (x +y ) (xpy − ypx )ye− 2~ (x +y ) dx dy.
π
~
2m
2
Waa = hψ1a |H 0 |ψ1a i =
But
2
2
mω
∂
∂
(xpy − ypx )y e
= −i~ x
−y
y e− 2~ (x +y )
∂y
∂x
h mω mω 2 2 i
2
2
mω
mω 2 − mω (x2 +y2 )
= −i~ x 1 −
y e 2~
− y2 −
x e− 2~ (x +y ) = −i~ xe− 2~ (x +y ) ,
~
~
2
2
− mω
2~ (x +y )
so
Z
2
2
mω
2 mω 2
qB0
Waa =
−
(−i~) x y e− ~ (x +y ) dx dy = 0,
π
~
2m
Z
2
2
mω
2 mω 2
qB0
b
0 a
Wba = hψ1 |H |ψ1 i =
−
(−i~) x2 e− ~ (x +y ) dx dy
π
~
2m
! r
!
r
mω 2 qB0
~
π~
π~
qB0 ~
=i
=i
.
π~
2mω mω
mω
2m
(Wab is the complex conjugate of Wba ; Wbb the same as Waa with x ↔ y, so it too is zero.) From Equation
7.33:
qB0 ~
1
E± = ±|Wab | = ±
.
2m
286
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
Problem 7.53
(a) The first-order correction to the nth energy is
En1 = ψn0 H 0 ψn0
Z a
2
ψn0 (x) δ(x − x0 )
=λ
0
2
= λ ψn0 (x0 )
n π x 2λ
0
=
.
sin2
a
L
Here small means that
Vmn
1.
0 − E0
Em
n
The matrix element |Vmn | = (2 λ/a) |sin(m π x0 /a) sin(n π x0 /a)| is of order 2 λ/a and the spacing between
energy levels is of order π 2 ~2 /2 m a2 so that we need to have
λ
a π 2 ~2
.
2 2 m a2
(b) To compute the second-order correction we first need
0
Vmn = ψm
H 0 ψn0
Z
0
∗
=λ
ψm (x) δ(x − x0 ) ψn0 (x) dx
=
m π x n π x 2λ
0
0
sin
sin
.
a
a
a
Then the second-order correction to the allowed energies is then
2
En2 =
X
m6=n
|Vmn |
0
En0 − Em
2
X sin2 m π x0 sin2 n π x0
(2 λ/a)
a
a
= 2 2
π ~ /(2 m a2 )
n2 − m2
m6=n
X sin2 m π x0 8 m λ2
2 n π x0
a
= 2 2 sin
.
π ~
a
n2 − m2
m6=n
(c) We construct a piecewise solution. The time-independent Schrödinger equation for 0 < x < x0 or for
x0 < x < a reads
−
~2 00
ψ =Eψ
2m
for which the general solution is ψ(x) = A sin(k x) + B cos(k x) where k =
piecewise solution is
A sin(k x) + B cos(k x) 0 < x < x0
ψ(x) =
.
C sin(k x) + D cos(k x) x0 < x < a
p
2 m E/~2 so that our
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
287
The boundary condition at x = 0 gives B = 0 and the boundary condition at x = a gives D = −C tan(k a)
so with these two boundary conditions satisfied
A sin(k x)
0 < x < x0
ψ(x) =
.
C 0 sin [k (a − x)] x0 < x < a
where (unimportantly) we’ve defined a new constant C 0 = −C/ cos(k a). Now we look at the boundary
conditions at x = x0 . Continuity of the wave function requires
A sin(k x0 ) = C 0 sin [k (a − x0 )]
F
and the condition on the derivative (Equation 2.128) is
2mλ
ψ(x0 )
~2
2mλ
−C 0 k cos [k (a − x0 )] − A k cos(k x0 ) =
A sin(k x0 )
~2
2mλ
A k cos(k x0 ) +
A sin(k x0 ) = −C 0 k cos [k (a − x0 )]
~2
∆ψ 0 =
Dividing Equation by Equation F gives
2mλ
= −k cot [k (a − x0 )]
~2
u cot(p u) + Λ = −u cot [u (1 − p)]
k cot(k x0 ) +
u cos(p u) sin [u (1 − p)] + Λ sin(p u) sin [u (1 − p)] = −u cos [u (1 − p)] sin(p u)
which simplifies to
u sin(u) + Λ sin(p u) sin [u (1 − p)] = 0 .
FF
Now, to calculate the corrections to the energy from this expression note that
r
r
2 m a2 E
2 m a2 (En0 + En1 + . . .)
u=
=
~2
~2
s
r
2 m a2 En0
m a2
=
+
E1 + . . .
2
~
2 En0 ~2 n
m a2 1
E + ...
n π ~2 n
= u0 + δu + . . .
= nπ +
Plugging into Equation FF and keeping terms of order δu gives
(u0 + δu) sin(u0 + δu) + Λ sin(p u0 ) sin [u0 (1 − p)] ≈ 0
where we can ignore the δu in the second term because the Λ ensures that the whole term is already order
δ. Expanding this




u0 sin(u0 ) + u0 cos(u0 ) + sin(u0 ) δu + Λ sin(p u0 ) ≈ 0
| {z }
| {z }
0
(−1)n
288
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
and
δu =
m a2 1
n Λ
E = −(−1)
sin(p u0 ) sin [u0 (1 − p)]
n π ~2 n
u0
2
n nπ~ 2maλ
sin(p u0 ) [sin u0 cos(p u0 )
En1 = − (−1)
2
m a n π ~2
= − sin(p u0 ) cos u0 ]
2λ
2
=
|sin(p u0 )|
a
which is precisely the result obtained above in part (a) since p u0 = n π x0 /a.
(d) In this case we want negative-energy solutions and we write
r
r
2 m E a2
2 m |E| a2
=i
≡ iv.
u=
2
~
~2
Plugging this into Equation FF and noting that sin(i z) = i sinh z we have
v sinh v + Λ sinh(p v) sinh [(1 − p) v] = 0 .
For the case p = 1/2,
v sinh v + Λ sinh2
v
2
=0
v
1
v = − Λ tanh
.
2
2
If a is huge so that the boundaries don’t matter, then tanh(v/2) = tanh(κ a/2) ≈ 1 and
v≈−
Λ
2
and then
E=−
~2 v 2
~2 Λ 2
m λ2
≈
−
=
−
,
2 m a2
8 m a2
2 ~2
which is precisely the energy of the delta-function well.
(e) For the case of p = 1/2,
v
1
v = |Λ| tanh
2
2
as shown above. From the graph (below) it’s clear that the functions on the right- and left-hand sides of
this equation intersect at most once. In order to intersect at all the slope of the function on the right-hand
side must be greater than 1 at the origin.
v
d 1
|Λ| tanh
>1
dv 2
2 0
|Λ|
>1
4
as claimed for p = 1/2.
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
In[! ]:=
Abs[Λ]
Λ = - 12; Plotv,
2
v
Tanh , {v, 0, 10}, AxesLabel → {"v", None}
2
10
8
6
Out[! ]=
4
2
2
4
6
8
As a function of p, solutions for v exist between
1
2
s
1−
|Λ| − 4
|Λ|
!
1
<p<
2
s
1+
|Λ| − 4
|Λ|
!
.
These values are shown with gridlines in the following plot.
10
v
289
290
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
Part e
Transcendental equation (LHS) for negative-energy solutions:
In[! ]:=
f[Λ_, p_, v_] := v Sinh[v] + Λ Sinh[p v] Sinh[(1 - p) v]
This will solve the transcendental equation. A starting guess of 4 works for this range of Λ values
In[! ]:=
solution[λ_, P_] := v /. FindRoot[f[λ, P, v], {v, 4}]
These are the minimum and maximum values of p for which there should be solutions for a given Λ
In[! ]:=
p1[λ_] :=
1
2
1-
- 4 + Abs[λ]
Abs[λ]
; p2[λ_] :=
1
2
1+
- 4 + Abs[λ]
Abs[λ]
;
Here is a plot of the solutions. Vertical grid lines mark the boundaries of the regions for which we
expect solutions and match up perfectly
In[! ]:=
Plot[
{solution[- 4.1, P], solution[- 5, P], solution[- 10, P]}, {P, 0, 1},
PlotRange → {0, 5},
GridLines →
{{p1[- 41 / 10], p2[- 41 / 10], p1[- 5], p2[- 5], p1[- 10], p2[- 10]}, None},
AxesLabel → {"p", "v"}]
Out[! ]=
(f ) Note that (for p = 1/2), for negative energy solutions to exist we must have
|Λ| >
1
= 4.
1/2 (1 − 1/2)
So we need to switch between solving the two transcendental equations at this point.
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
Part (f)
291
Transcendental equation (LHS) for positive-energy solutions:
In[! ]:=
g[Λ_, p_, u_] := u Sin[u] + Λ Sin[p u] Sin[(1 - p) u]
This solves the appropriate transcendental equation and returns u (or i v) depending on whether it is a
positive- (or negative-) energy solution
In[! ]:=
solutionF[λ_] := If[Abs[λ] > 4, I solution[λ, 1 / 2],
u /. FindRoot[g[λ, 1 / 2, u], {u, 4}]]
This is the wave function for a given value of u = k (for p = 1/2)
In[! ]:=
ψ[k_, x_] := If[x < 1 / 2, Sin[k x], Sin[k (1 - x)]]
This is a table with the wave function for each value of Λ we want to plot
In[! ]:=
ψ[x_] = Table[ψ[solutionF[λ], x], {λ, {0, - 2, - 3, - 3.5, - 4.5, - 5, - 10}}];
This computes the norm for each ψ
In[! ]:=
norm = Sqrt[Integrate[Abs[ψ[x]] ^ 2, {x, 0, 1}]];
And the normalized wave function
In[! ]:=
ψ[x_] = ψ[x] / norm;
Here it is...
In[! ]:=
Plot[Abs[ψ[x]], {x, 0, 1}, AxesLabel → {"x/a", "ψ"}]
ψ
2.0
1.5
Out[! ]=
1.0
0.5
0.2
0.4
0.6
0.8
1.0
x/a
Problem 7.54
(a) hH 0 i = 2 Re
X |hψ 0 |H 0 |ψ 0 i|2
m
m6=n
n
0
En0 − Em
. The sum is manifestly real (so we can drop the “Re”), and in fact it is
precisely En2 (Equation 7.15): hH 0 i = 2En2 .
Explanation: The second-order correction to the energy is the term of order λ2 in
hψn |H0 + λH 0 |ψn i
hψn |H0 |ψn i
hψn |H 0 |ψn i
=
+λ
,
hψn |ψn i
hψn |ψn i
hψn |ψn i
where ψn is the exact nth eigenstate. We just calculated the second term; we need to evaluate the first
292
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
term, to order λ2 .
hψn |H0 |ψn i = h(ψn0 + λψ01 + λ2 ψn2 + · · · )|H0 |(ψn0 + λψ01 + λ2 ψn2 + · · · )i
= hψn0 |H0 |ψn0 i + λ hψn0 |H0 |ψn1 i + hψn1 |H0 |ψn0 i
+ λ2 hψn0 |H0 |ψn2 i + hψn1 |H0 |ψn1 i + hψn2 |H0 |ψn0 i + · · ·
≈ En0 + λ En0 hψn0 |ψn1 i + hψn1 |ψn0 i + λ2 En0 hψn0 |ψn2 i + hψn2 |ψn0 i + hψn1 |H0 |ψn1 i
= En0 + λ2 hψn1 |H0 |ψn1 i.
(As always, we choose hψni |ψn0 i = 0—see comments after Equation 7.11.) Similarly,
hψn |ψn i = hψn0 |ψn0 i + λ hψn0 |ψn1 i + hψn1 |ψn0 i + λ2 hψn0 |ψn2 i + hψn1 |ψn1 i + hψn2 |ψn0 i + · · ·
≈ 1 + λ2 hψn1 |ψn1 i.
So
−1 0
hψn |H0 |ψn i 0
≈ En + λ2 hψn1 |H0 |ψn1 i 1 − λ2 hψn1 |ψn1 i
= En + λ2 hψn1 |H0 |ψn1 i 1 + λ2 hψn1 |ψn1 i
hψn |ψn i
= En0 + λ2 hψn1 |H0 |ψn1 i − En0 hψn1 |ψn1 i = En0 + λ2 hψn1 |(H0 − En0 )|ψn1 i.
Finally, using Equation 7.13:
X
hψn1 |(H0 − En0 )|ψn1 i = hψn1 |(H0 − En0 )
m6=n
∗
=−
XX
p6=n m6=n
0
X
Hmn
0
1
0
0
|ψ
i
=
−hψ
|
Hmn
|ψm
i
m
n
0 )
(En0 − Em
m6=n
0
X |H 0 |2
Hpn
mn
0
0
Hmn
hψp0 |ψm
i=−
= −En2
0
0
0
En − Ep
En0 − Em
m6=n
(Equation 7.15). So the second-order correction the nth energy is −En2 + 2En2 = En2 , as it should be.
(b) To first order, the expectation value of pe in the nth state is
hpe i = 2Re
X hψ 0 |qx|ψ 0 ihψ 0 |(−qEext x)|ψ 0 i
n
m6=n
m
m
n
0
En0 − Em
so
α = −2q 2
= −2q 2 Eext
X |hψ 0 |x|ψ 0 i|2
n
m6=n
X |hψ 0 |x|ψ 0 i|2
n
m6=n
m
0
En0 − Em
m
0
En0 − Em
,
. X
For the one-dimensional oscillator (Equation 3.114)
r
r
r
√
~ √
~ √
~
0
0
0
0
hψn |x|ψm i =
m δn,m−1 + n δm,n−1 ⇒ hψ0 |x|ψm i =
m δ0,m−1 =
δm,1 .
2mω
2mω
2mω
(don’t confuse the mass m, in the factor out front, with the index m), so
α = −2q 2
~/2mω
q2
=
.
(~ω/2) − (3~ω/2)
mω 2
Classically, the electric force qEext balances the spring force kd, so pe = qd = q 2 Eext /k, and hence
α = q 2 /k = q 2 /(mω 2 ), in agreement with the quantum result.
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
293
(c)


0 
1
0
0
3
X hψn0 |x|ψm
X hψ 0 |x|ψ 0 ihψ 0 |x3 |ψ 0 i
ihψ
|
−
κx
|ψ
i
κ
m
n
n
m
m
n 
6
=−
hxi1 = 2Re 
Re 
1
1
3~ω
(n − m)
(n
+
)~ω
−
(m
+
)~ω
2
2
m6=n
m6=n


r
√
√
X ( m δn,m−1 + n δm,n−1 )
κ
~
0
=−
Re 
hψm
|x3 |ψn0 i
3~ω 2mω
(n − m)
m6=n
r
√
√
~
κ
0
0
Re − n + 1hψn+1
|x3 |ψn0 i + nhψn−1
|x3 |ψn0 i .
=−
3~ω 2mω

Now, from (Equation 2.70),
3
~
2mω
~
2mω
x =
=
3/2
3/2
(a+ + a− )3
(a3+ + a2+ a− + a+ a− a+ + a+ a2− + a− a2+ + a− a+ a− + a2− a+ + a3− ).
We need
3
~
2mω
~
2mω
hn + 1|x |ni =
hn − 1|x3 |ni =
3/2
3/2
hn + 1| a2+ a− + a+ a− a+ + a− a2+ |ni,
hn − 1| a+ a2− + a− a+ a− + a2− a+ |ni.
(In the first line we need two a+ ’s and one a− , to raise |ni up to |n + 1i; in the second line we need two
a− ’s and one a+ , to lower |ni to |n − 1i.) Using Equation 2.67 repeatedly:
√
√ √
√
a2+ a− |ni = n a2+ |n − 1i = n n a+ |ni = n n + 1 |n + 1i,
√
√
√
√
a+ a− a+ |ni = n + 1 a+ a− |n + 1i = n + 1 n + 1 a+ |ni = (n + 1) n + 1 |n + 1i,
√
√
√
√
a− a2+ |ni = n + 1 a− a+ |n + 1i = n + 1 n + 2 a− |n + 2i = (n + 2) n + 1 |n + 1i,
so
√
√
√
hn + 1| a2+ a− + a+ a− a+ + a− a2+ |ni = hn + 1| n n + 1 + (n + 1) n + 1 + (n + 2) n + 1 |n + 1i
√
= 3(n + 1) n + 1.
In the same way,
√ √
√
n n − 1 a+ |n − 2i = (n − 1) n |n − 1i,
√
√ √
√
a− a+ a− |ni = n a− a+ |n − 1i = n n a− |ni = n n |n − 1i,
√
√
√
√
a2− a+ |ni = n + 1 a2− |n + 1i = n + 1 n + 1 a− |ni = (n + 1) n |n − 1i,
a+ a2− |ni =
√
n a+ a− |n − 1i =
so
√
√
√ hn − 1| a+ a2− + a− a+ a− + a2− a+ |ni = hn − 1| (n − 1) n + n n + (n + 1) n |n − 1i
√
= 3n n.
294
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
Thus
3
hn + 1|x |ni =
~
2mω
3/2
√
3(n + 1) n + 1,
3
hn − 1|x |ni =
~
2mω
3/2
√
3n n.
Finally,
r
3/2
√
√
√
√ κ
~
~
hxi = −
− n + 1 3(n + 1) n + 1 + n 3n n
3~ω 2mω 2mω
2
κ
~κ
~
=−
−(n + 1)2 + n2 = (2n + 1) 2 3 .
~ω 2mω
4m ω
1
Problem 7.55
In the absence of the perturbation, the ground state is
1
ψ0 (x) = √
L
and the remaining stationary states come in degenerate pairs,
r
2
2nπx
ψn,e (x) =
cos
L
L
r
2
2nπx
ψn,o (x) =
cos
.
L
L
The unperturbed energies are
En0 =
2 n2 π 2 ~2
.
m L2
We now calculate the first-order energy corrections due to the perturbation. The correction to the ground state
is
Z L
α
2
1
E0 =
[ψ0 (x)] [−α δ(x)] ψ0 (x) dx = − .
L
0
The degenerate states are already the good states (calculate the matrix W if you wish) since they are eigenstates
of a reflection about x = 0 with distinct eigenvalues (±1), a symmetry of the Hamiltonian. The corrections to
their energies are then
1
En,e
=
Z L
2
[ψn (x)] [−α δ(x)] ψn (x) dx = −
0
1
En,o
=
Z L
2α
L
2
[ψn (x)] [−α δ(x)] ψn (x) dx = 0 .
0
These results are shown on the next page fro the first seven states. The sequence of node numbers is 0, 2, 2, 4, 4, 6, 6, . . ..
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
���� �������������
n
E
Nodes
295
Parity
2.0
1.5
0
0
-α
L
Even
1.0
0.5
0.2
0.4
0.6
0.8
1.0
0.2
0.4
0.6
0.8
1.0
0.2
0.4
0.6
0.8
1.0
0.2
0.4
0.6
0.8
1.0
0.2
0.4
0.6
0.8
1.0
0.2
0.4
0.6
0.8
1.0
0.2
0.4
0.6
0.8
1.0
1.5
1.0
1
2 2
- 2 α + 2 π2 ℏ
L
L m
0.5
2
Even
-0.5
-1.0
-1.5
1.5
1.0
1
0.5
2 π2 ℏ2
2
L2 m
Odd
-0.5
-1.0
-1.5
1.5
1.0
2
2 2
- 2 α + 8 π2 ℏ
L
L m
0.5
4
Even
-0.5
-1.0
-1.5
1.5
1.0
2
0.5
8 π2 ℏ2
4
L2 m
Odd
-0.5
-1.0
-1.5
1.5
1.0
3
- 2α +
L
18 π 2 ℏ 2
L2 m
0.5
6
Even
-0.5
-1.0
-1.5
1.5
1.0
3
18 π 2 ℏ 2
L2 m
0.5
6
Odd
-0.5
-1.0
-1.5
Problem 7.56
(a) The first order correction to the ground state is
D
E
e2
e2
a
ψ0 =
ψ0
E01 = ψ0
ψ0
4 π 0 r
4 π 0 a
r
296
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
where r = |r1 − r2 |. The matrix element evaluates to 5/4; see the solution for Problem 5.15 for the details.
Therefore
E01 =
5 e2
5
= |E1 |
2 8 π0 a
2
where |E1 | = 13.6 eV.
(b) Now we turn to the degenerate first-excited states. The states are
1
ψ± (r1 , r2 ) = √ [ψ100 (r1 ) ψ200 (r2 ) ± ψ200 (r1 ) ψ100 (r2 )] .
2
Of course we need to worry about whether to use degenerate perturbation theory. It turns out its unnecessary. The Hamiltonian is invariant under the interchange of r1 and r2 and these two states are even and
odd under that interchange so they won’t be mixed by the perturbation. We may therefore compute
e2
ψ±
ψ±
4 π 0 r
Z Z 1
e2
∗
∗
=
ψ100
(r1 ) ψ200
(r2 )
ψ100 (r1 ) ψ200 (r2 )
2
4 π 0 r
e2
∗
∗
± ψ100
(r1 ) ψ200
(r2 )
ψ200 (r1 ) ψ100 (r2 )
4 π 0 r
e2
∗
∗
ψ100 (r1 ) ψ200 (r2 )
± ψ200
(r1 ) ψ100
(r2 )
4 π 0 r
e2
∗
∗
+ψ200
(r1 ) ψ100
(r2 )
ψ200 (r1 ) ψ100 (r2 ) d3 r1 d3 r2 .
4 π 0 r
1
E±
=
The orbitals are real so we can drop the complex conjugation. Additionally, we can interchange r1 and r2
in the third and fourth terms to see that they are equal to the first two integrals and therefore
1
E±
=
1
(K ± J) ,
2
with the integrals as given in the problem statement.
Now we turn to the business of actually evaluating these two integrals.
Z
K=2
2
|ψ200 (r2 )|
e2
2
|ψ100 (r1 )|
dr1 dr2
4 π 0 r
|
{z
}
Z
≡V1
where
Z
e2
8 −4 r1 /a
a
dr1
e
4 π 0 a
π a3
|r1 − r2 |
Z
Z
16 ∞ −4 r1 /a π
a sin θ1
p
= 2 |E1 | 3
e
dθ1 r12 dr1 ,
2
2
a 0
r1 + r2 − 2 r1 r2 cos θ1
0
|
{z
}
V1 =
Z1
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
297
having already done the trivial 2 π integral. Next we compute (setting u = cos θ)
Z 1
a
p
du
Z1 =
2
2
r1 + r2 − 2 r1 r2 u
−1
q
1
a
=−
r12 + r22 − 2 r1 , r2 u
r1 r2
−1
a
=
(r1 + r2 − |r1 − r2 |)
r r
1 2
2 a/r1 r1 > r2
=
.
2 a/r2 r1 < r2
Returning with this result to the expression for V1 we have
Z r2
Z ∞
a 2
32
e−4 r1 /a
r1 dr1 +
e−4 r1 /a a r1 dr1
V1 = 2 |E1 | 3
a
r2
0
r2
" Z
#
Z ∞
4
r
/a
2
32 a 3 a
2 −z
−z
z e dz + 4
z e dz
= 2 |E1 | 3
a 4
r2 0
4 r2 /a
( "
#
2 !
a
8 r2
4 r2
= |E1 |
2− 2+
+
e−4 r2 /a
r2
a
a
4 r2
= +4 1 +
e−4 r2 /a
a
a
a
= 2 |E1 |
− 2+
e−4 r2 /a .
r2
r2
Now since V1 and ψ200 depend only on r2 , not θ2 , φ2 we have
Z ∞
2
K = 8π
|ψ200 (r2 )| V1 (r2 ) r22 dr2
0
Z ∞
1 r2 2 −2 r2 /a a
a
−4 r2 /a
= 8 π 2 |E1 |
1
−
−
2
+
e
e
r22 dr2
π a3
a
r2
r2
0
Z ∞
16
r2 2 −2 r2 /a
a r2 1 −
e
dr2
= 3 |E1 |
a
a
0
Z ∞
r2 2 −6 r2 /a
=−
2 r22 + a r2 1 −
e
dr2
a
0
Now we’ll perform a variable substitution where z = 2 r2 /a in the first integral and z = 6 r2 /a in the
second integral. We then have
Z ∞ 2 a z
z 2 −z a
16
1−
e
dz
K = 3 |E1 |
a
2
2
2
0
Z ∞
1 2 2 1 2
z 2 −z a
=−
a z + a z
1−
e
dz
18
6
6
6
0
Z ∞
16
14 3
1 4
= 2 |E1 |
z − 2 z2 +
z −
z
e−z dz
9
27
486
0
16
42
4
= 2 |E1 |
−4+
−
9
9
81
298
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
and
K=
136
|E1 | .
81
Next we compute
Z
Z
e2
J = 2 ψ200 (r2 ) ψ100 (r2 ) ψ200 (r1 ) ψ100 (r1 )
d3 r1 d3 r2 .
4 π 0 r
{z
}
|
≡V2
Here
√ Z
r1 −3 r1 /a
a
e2 2 2 1−
e
dr1
3
4 π 0 a π a
a
|r1 − r2 |
√
Z ∞
r1 −3 r1 /a
2 2
1
−
2
π
e
Z1 (r1 ) r12 dr1
= 2 |E1 |
π a3
a
0
V2 =
where the 2 π is from the φ integral and where the function Z1 is the same one that appeared in the
computation of K. Then
√ Z r2
8 2
a r12 r1 −3 r1 /a
V2 = 2 |E1 | 3
e
dr1
1−
a
r2
a
0
Z ∞
r1 −3 r1 /a
=+
a r1 1 −
e
dr1
a
r2
"
√
Z 3 r2 /a 3 2 a z
8 2
z −z a
dz
= 2 |E1 | 3
1−
e
a
9 r2
3
3
0
#
Z ∞
z −z a
a2 z 1−
e
dz
=+
3
3
3r2 /a 3
with the variable change z = 3 r1 /a. Then
√ "
Z 3 r2 /a 8 2
a
z3
a
2
V2 = 2 |E1 |
z −
e−z dz
9
3 r2 0
3
3
#
Z ∞
z2
a
e−z dz
=+
z−
3
3
3r2 /a
√ 8 2
a
6 r2
9 r2
= 2 |E1 |
2− 2+
+ 22 e−3 r2 /a
9
3 r2
a
a
6 r2
9 r22
9 r23
3 r1
=− 2− 2+
+ 2 + 3
e−3 r2 /a
+ 1+
e−3 r2 /a
a
a
a
a
6 r2
9 r2
1
=−
2+
+ 22 e−3 r2 /a
3
a
a
√ 3 r2
16 2
1+
e−3 r2 /a .
= |E1 |
27
a
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
299
We now plug this back into our expression for J. Once again the angular integration is trivial and
Z
J = 8π
ψ200 (r2 ) ψ100 (r2 ) V2 (r2 ) r22 dr2
√
√ Z
16 2 2 2 ∞ r2 −3 r2 /a
3 r2
= 8 π |E1 |
1−
e
1+
e−3 r2 /a r22 dr2
27 π a3 0
a
a
Z ∞
z −z 2
1 a3
z
512
1+
e z dz
|E1 | 3
1−
=
27
a
6
6
2
0
Z ∞
64
1 4
z3
=
|E1 |
z2 +
−
z
e−z dz
729
3
12
0
64
|E1 | (2 + 2 − 2)
=
729
and
J=
128
|E1 | .
729
So the energies are
64
2969
1
68
E1 =
E1 = −55.39 eV
E 0 + (K + J) = 4 E1 + 4 E2 − E1 −
|{z} 81
2
729
729
E1 /4
for the symmetric spatial state (parahelium) and
68
64
3097
1
E1 =
E1 = −57.78 eV
E 0 + (K − J) = 4 E1 + 4 E2 − E1 +
|{z} 81
2
729
729
E1 /4
for the antisymmetric spatial state (orthohelium). Looking at Figure 5.1, this calculation correctly predicts
that orthohelium is lower in energy, roughly by the correct amount.
Problem 7.57
(a) Setting q = 0 in Equation 6.14:
Then, for q 6= 0,
H0 =
−
~2 d2 un0
+ V (x) uno = En0 un0 , so
2m dx2
H0 =
p̂2
+ V (x).
2m
~q
~2 q 2
p̂ +
.
m
2m
(b) From Equations 7.9 and 7.15:
Enq ≈ En0 + hun0 |H 0 |un0 i +
X |hum0 |H 0 |un0 i|2
m6=n
Now,
hum0 |H 0 |un0 i =
~q
~2 q 2
hum0) |p̂|un0 i +
δmn ,
m
2m
En0 − Em0
.
so
~q
~2 q 2
Enq ≈ En0 + hun0) |p̂|un0 i +
+
m
2m
~q
m
2 X
m6=n
|hum0 |p̂|un0 i|2
.
En0 − Em0
300
CHAPTER 7. TIME-INDEPENDENT PERTURBATION THEORY
(c) From Problem 2.1(b), the solutions un0 can be chosen to be real. Then
Z a
Z
d
i~ a d
i~ 2
hun0 | p̂ |un0 i =
un0 (x) −i~
un0 (x) dx = −
[un0 (x)] dx = −
un0 (a)2 − un0 (0)2 = 0.
dx
2 0 dx
2
0
CHAPTER 8. THE VARIATIONAL PRINCIPLE
301
Chapter 8
The Variational Principle
Problem 8.1
(a)
2
Z ∞
hV i = 2αA
−2bx2
xe
2
dx = 2αA
0
hHi =
~2 b
α
+√
.
2m
2πb
~2
hHimin =
2m
2
1
− e−2bx
4b
∞
0
αA2
α
=
=
2b
2b
r
2b
α
=√
.
π
2bπ
α m
~2
1 α
∂hHi
=
− √ b−3/2 = 0 =⇒ b3/2 = √
; b=
∂b
2m 2 2π
2π ~2
mα
√
2π~2
2/3
√
2π~2
mα
α
+√
2π
!1/3
α2/3 ~2/3
= 1/3
m (2π)1/3
1
+1
2
3
=
2
mα
√
2π~2
2/3
.
α2 ~2
2πm
1/3
3α~4
4m2
1/3
.
(b)
2
Z ∞
hV i = 2αA
4 −2bx2
x e
0
~2 b
3α
hHi =
+
.
2m 16b2
hHimin =
~2
2m
3αm
4~2
3
dx = 2αA
8(2b)2
2
r
3α
π
=
2b
16b2
r
π
2b
r
3α
2b
=
.
π
16b2
∂hHi
~2
3α
3αm
=
− 3 = 0 =⇒ b3 =
; b=
∂b
2m 8b
4~2
1/3
+
3α
16
4~2
3αm
2/3
=
α1/3 ~4/3 1/3 −1/3
3 4
m2/3
Problem 8.2
r
1
π
2b3
2
2 π
Normalize: 1 = 2|A|
dx = 2|A| 3 = 3 |A| . A =
.
2
2
2
(x + b )
4b
2b
π
0
Z
∞
~2
1
d2
1
Kinetic Energy: hT i = −
|A|2
dx.
2 + b2 ) dx2
2 + b2 )
2m
(x
(x
−∞
2
Z ∞
3αm
4~2
1/3
1 1
+
2 4
.
=
3
4
.
302
CHAPTER 8. THE VARIATIONAL PRINCIPLE
But
d2
dx2
1
(x2 + b2 )
d
=
dx
−2x
(x2 + b2 )2
=
−2
4x
2(3x2 − b2 )
+
2x
=
, so
(x2 + b2 )2
(x2 + b2 )3
(x2 + b2 )3
Z ∞
Z ∞
Z ∞
~2 2b3
4~2 b3
1
(3x2 − b2 )
1
2
hT i = −
dx = −
dx − 4b
dx
4
3
2m π
(x2 + b2 )4
πm
(x2 + b2 )3
(x2 + b2 )4
0
0
0
4~2 b3
~2
3π
2 5π
= −
−
4b
=
.
3
πm
16b5
32b7
4mb2
Potential Energy: hV i =
hHi =
1
~2
+ mω 2 b2 .
4mb2
2
~2
hHimin =
4m
√
1
mω 2 |A|2 2
2
Z ∞
x2
0
(x2 + b2 )2
dx = mω 2
2b3 π
1
= mω 2 b2 .
π 4b
2
∂hHi
~2
~2
1 ~
+ mω 2 b = 0 =⇒ b4 =
=⇒ b2 = √
=−
.
3
∂b
2mb
2m2 ω 2
2 mω
1 ~
2mω 1
+ mω 2 √
= ~ω
~
2
2 mω
√
1
2
+ √
4
2 2
√
!
=
1
2
~ω = 0.707 ~ω > ~ω. X
2
2
Problem 8.3


A(x + a/2), (−a/2 < x < 0),
ψ(x) = A(a/2 − x), (0 < x < a/2),


0,
(otherwise).
1 = |A|2 2
Z a/2 0
a/2
2
3
a
1 a
− x dx = −2|A|2
−x
2
3 2
0

A,
(−a/2 < x < 0),
dψ 
= −A, (0 < x < a/2),

dx

0,
(otherwise).
=
2 2 a 3
a3 2
|A|
|A| ;
=
3
2
12
r
A=
12
a3
(as before).
d2 ψ
a
a
=
Aδ
x
+
−
2Aδ(x)
+
Aδ
x
−
.
dx2
2
2
Z h ~2
a
a i
~2
~2 2 a
ψ Aδ x +
− 2Aδ(x) + Aδ x −
dx =
2Aψ(0) =
A
2m
2
2
2m
m
2
~2 a 12
~2
=
=6
(as before).
2m a3
ma2
Z
a 2
α
~2
α
hV i = −α |ψ|2 δ(x) dx = −α|ψ(0)|2 = −αA2
= −3 . hHi = hT i + hV i = 6
−3 .
2
a
ma2
a
hT i = −
∂
~2
α
~2
hHi = −12
.
+
3
=
0
⇒
a
=
4
∂a
ma3
a2
mα
hHimin = 6
mα mα2
~2 mα 2
− 3α
= 2
2
m 4~
4~2
~
3 3
−
8 4
= −
3mα2
8~2
> −
mα2
. X
2~2
CHAPTER 8. THE VARIATIONAL PRINCIPLE
303
Problem 8.4
(a) Follow the proof in §8.1: ψ =
∞
X
∞
X
cn ψn , where ψ1 is the ground state. Since hψ1 |ψi = 0, we have:
n=1
cn hψ1 |ψn i = c1 = 0; the coefficient of the ground state is zero. So
n=1
hHi =
∞
X
En |cn |2 ≥ Efe
n=2
∞
X
|cn |2 = Efe , since En ≥ Efe for all n except 1.
n=2
(b)
1 = |A|
2
Z ∞
2 −2bx2
x e
−∞
1
dx = |A| 2
8b
2
r
r
π
2b
2
=⇒ |A| = 4b
.
2b
π
Z ∞
2 2 d
~2
2
−bx2
|A|
xe−bx
xe
dx
hT i = −
2m
dx2
−∞
2
2
2
d −bx2
d2 −bx2 2 −bx2
xe
=
e
−
2bx
e
= −2bxe−bx − 4bxe−bx + 4b2 x3 e−bx
2
dx
dx
r
r
r
r Z ∞
2
−2bx2
~
2b
1
π
π
2~2 b 2b
2 3
2
2 4
hT i = −
4b
2
2 −6b
+ 4b
−6bx + 4b x e
dx = −
2
2m
π
m
π
8b 2b
32b
2b
0
2
2
4~ b
3 3
3~ b
=−
.
− +
=
m
4 8
2m
r
r
Z ∞
1
3
3mω 2
1
2b
π
2
2
2 −2bx2 2
2
x e
x dx = mω 4b
2
=
.
hV i = mω |A|
2
2
π 32b2 2b
8b
−∞
∂hHi
3~2
3mω 2
m2 ω 2
mω
2
=
−
.
=
0
=⇒
b
=
=⇒ b =
∂b
2m
8b2
4~2
2~
3~2 mω 3mω 2 2~
3 3
3
hHimin =
+
= ~ω
+
= ~ω.
2m 2~
8 mω
4 4
2
hHi =
3~2 b 3mω 2
+
;
2m
8b
This is exact, since the trial wave function is in the form of the true first excited state.
Problem 8.5
The trial function has to vanish at x = 0 and (of course) go to zero at ∞. I’ll choose
2
ψ(x) = A x e−λ x .
We first normalize the wave function:
Z ∞
Z ∞
Z ∞
2
2
1
2
1=
|ψ| dx = A2
x2 e−2 λ x dx = A2 √
u2 e−u du
8 λ3 | 0
0
0
{z
}
√
π/4
304
CHAPTER 8. THE VARIATIONAL PRINCIPLE
(where u =
√
2 λ x) so that
r
2 λ3
A2 = 8
.
π
Now
2
2
dψ
= A e−λ x − 2 λ A x2 e−λ x
dx
and exploiting the hermiticity of p̂ we can then compute the kinetic energy as
Z ∞
Z ∞
1
p̂2
2
|p̂ ψ| dx
ψ dx =
ψ∗
2m
2m 0
0
Z ∞
2
2
1
2 2
1 − 2 λ x2 e−2 λ x dx
=
A ~
2m
0
r
Z ∞
2
3
2
2
~
2λ
1
√
=
8
1 − u2 e−u du
2m
π
2λ 0
|
{z
}
3
√
π/8
2
=
3 ~
λ
.
2 m
The potential energy is
Z ∞
Z ∞
2
ψ ∗ (x) V (x) ψ(x) = A2 m g
x3 e−2 λ x dx
0
0
r
Z ∞
3
2
2λ
1
mg
u3 e−u du
=8
2
π
(2 λ) | 0
{z
}
1/2
=√
2
mg.
πλ
Therefore the expectation value of the energy is
E(λ) =
3 ~2
2
λ
+√
mg.
2 m
πλ
We still need to minimize the energy to find the tightest upper bound.
r dE
3 ~2
2
1
1
=
+ mg
−
=0
dλ
2m
π
2 λ3/2
so that
2 m4 g 2
λ=
9 π ~4
1/3
and evaluating E using this result for λ we get
Egs ≤ 3
3
2π
1/3 m g 2 ~2
2
1/3
This is fairly close to the exact result.
≈ 2.345
m g 2 ~2
2
1/3
.
CHAPTER 8. THE VARIATIONAL PRINCIPLE
305
Problem 8.6
0
(a) Use the unperturbed ground state (ψgs
) as the trial wave function. The variational principle says
0
0
0
0
0
0
0
0
0
0
hψgs |H|ψgs i ≥ Egs . But H = H 0 + H 0 , so hψgs
|H|ψgs
i = hψgs
|H 0 |ψgs
i + hψgs
|H 0 |ψgs
i. But hψgs
|H 0 |ψgs
i=
0
0
0 0
Egs (the unperturbed ground state energy), and hψgs |H |ψgs i is precisely the first order correction to the
0
1
ground state energy (Eq. 7.9), so Egs
+ Egs
≥ Egs . QED
2
2
(b) The second order correction (Egs
) is Egs
=
X |hψ 0 |H 0 |ψgs i|2
m
0 − E0
Egs
m
. But the numerator is clearly positive,
m6=gs
0
0
2
and the denominator is always negative (since Egs
< Em
for all m), so Egs
is negative.
Problem 8.7
He+ is a hydrogenic ion (see Problem 4.19); its ground state energy is (2)2 (−13.6 eV), or −54.4 eV. It takes
79.0 − 54.4 = 24.6 eV to remove one electron.
Problem 8.8
I’ll do the general case of a nucleus with Z0 protons. Ignoring electron-electron repulsion altogether gives
ψ0 =
Z03 −Z0 (r1 +r2 )/a
e
,
πa3
(generalizing Eq. 8.18)
and the energy is 2Z02 E1 . hVee i goes like 1/a (Eqs. 8.21 and 8.26), so the generalization of Eq. 8.26 is hVee i =
− 45 Z0 E1 , and the generalization of Eq. 8.27 is hHi = (2Z02 − 54 Z0 )E1 .
If we include shielding, the only change is that (Z − 2) in Eqs. 8.29, 8.30, and 8.33 is replaced by (Z − Z0 ).
Thus Eq. 8.33 generalizes to
5
5
2
2
hHi = 2Z − 4Z(Z − Z0 ) − Z E1 = −2Z + 4ZZ0 − Z E1 .
4
4
5
∂hHi
5
= −4Z + 4Z0 −
E1 = 0 =⇒ Z = Z0 − .
∂Z
4
16
"
2
#
5
5
5
5
hHimin = −2 Z0 −
+ 4 Z0 −
Z0 −
Z0 −
E1
16
16
4
16
5
25
5
5
25
2
2
= −2Z0 + Z0 −
+ 4Z0 − Z0 − Z0 +
E1
4
128
4
4
64
5
25
(16Z0 − 5)2
= 2Z02 − Z0 +
E1 =
E1 ,
4
128
128
generalizing Eq. 8.35. The first term is the naive estimate ignoring electron-electron repulsion altogether; the
second term is hVee i in the unscreened state, and the third term is the effect of screening.
Z0 = 1 (H− ): Z = 1 −
5
11
=
= 0.688. The effective nuclear charge is less than 1, as expected.
16
16
306
CHAPTER 8. THE VARIATIONAL PRINCIPLE
hHimin =
112
121
E1 =
E1 = −12.9 eV.
128
128
27
272
729
5
=
= 1.69 (as before); hHimin =
E1 =
E1 = −77.5 eV.
16
16
128
128
43
1849
5
432
Z0 = 3 (Li+ ): Z = 3 −
=
= 2.69 (somewhat less than 3); hHimin =
E1 =
E1 = −196 eV.
16
16
128
128
Z0 = 2 (He): Z = 2 −
Problem 8.9
Z
1
1
1
1
ψ0 (r1 )i = ahψ0 (r2 )
ψ0 (r2 )i = a 3
D = ahψ0 (r1 )
e−2r2 /a d3 r
r2
r1
πa
r1
Z
Z ∞ Z π
√
√
2
2
1
2π
−a
−a
r 2 +R2 −2rR cos θ 1 2
r 2 +R2 −2rR cos θ
e
e
r
=
r
sin
θ
dr
dθ
dφ
=
sin
θ
dθ
dr.
πa2
r
πa2 0
0
Z r+R
a
a i
1
a h −2(r+R)/a r+R+
[. . .] =
e−2y/a y dy = −
e
− e−2|r−R|/a |r − R| +
rR |r−R|
2rR
2
2
Z
2 a −2R/a ∞ −2r/a a
D= 2 −
e
dr
e
r+R+
a
2R
2
0
#
Z R
Z ∞
a
a
−2R/a
2R/a
2r/a
−2r/a
−e
dr − e
dr
e
R−r+
e
r−R+
2
2
0
R
a a
a 2 1
a a 2r/a R
−2R/a
e
+ R+
e
− e−2R/a R +
=−
aR
2
2
2
2
2
0
∞
R
a 2
a 2
a a −2r/a ∞
2r
2r
+e−2R/a
− e2R/a −R +
e2r/a
−1
− e
− e2R/a
e−2r/a − − 1
2
a
2
2
2
a
R
0
R
2
2
2
2
2
2
2
2
2
2
1
aR a
aR a
a
a 2R a
aR a
a 2R a
a
aR a
=−
+
+
+
+
+
−
+
−
+
−
−
−
e−2R/a
+ −
aR
4
2
4
2
4
4
2
4
4 a
4
2
4
4 a
4
i
1 h −2R/a 2
a a −2R/a
=−
e
a + aR + −a2 =⇒ D =
− 1+
e
(confirms Eq. 8.48).
aR
R
R
Z
1
1
1
X = ahψ0 (r1 )
ψ0 (r2 )i = a 3
e−r1 /a e−r2 /a d3 r
r1
πa
r1
Z π √
Z
Z ∞
√
2π
1
−r/a − r 2 +R2 −2rR cos θ/a 1 2
−r/a
− r 2 +R2 −2rR cos θ/a
e
e
r
sin
θ
dr
dθ
dφ
=
re
e
sin
θ
dθ
dr.
=
πa2
r
πa2 0
0
i
a h −(r+R)/a
[. . .] = −
e
(r + R + a) − e−|r−R|/a (|r − R| + a)
rR
Z
2
a −R/a ∞ −2r/a
X= 2 −
e
e
(r + R + a)dr
a
R
0
#
Z R
Z ∞
−R/a
R/a
−2r/a
−e
(R − r + a)dr − e
e
(r − R + a)dr
=−
2
aR
e−R/a
0
R
a 2
a R2
−R/a
+ (R + a)
−e
(R + a)R −
2
2
2
CHAPTER 8. THE VARIATIONAL PRINCIPLE
307
∞
2
∞
a
2r
R/a a
−2r/a
−2r/a
−e
e
−e (−R + a) − e
− −1
2
2
a
R
R
2
2
2
2
2
2
2
a
aR
a
R
aR
a
a
2R
a
−R/a
2
=−
e
+
+
− R − aR +
+
−
−
−
aR
4
2
2
2
2
2
4 a
4
2
2 −R/a
aR R
R
=−
e
−
−
=⇒ X = e−R/a 1 +
(confirms Eq. 8.49).
aR
2
2
a
R/a
Problem 8.10
There are two changes: (1) the 2 in Eq. 8.39 changes sign . . . which amounts to changing the sign of I in Eq. 8.44;
(2) the last term in Eq. 8.45 changes sign . . . which amounts to reversing the sign of X. Thus Eq. 8.50 becomes
D−X
E1 , and hence Eq. 8.52 becomes
hHi = 1 + 2
1−I
Etot
D−X
2
1/x − (1 + 1/x) e−2x − (1 + x)e−x
2a
−1−2
= −1 + − 2
=
−E1
R
1−I
x
1 − (1 + x + x2 /3)e−x
2
−x
−2x
2 1 − (1 + x + x /3)e − 1 + (x + 1)e
+ (x + x2 )e−x
= −1 +
x
1 − (1 + x + x2 /3)e−x
"
#
2 (1 + x)e−2x + 23 x2 − 1 e−x
= −1 +
.
x
1 − (1 + x + x2 /3)e−x
F (x) =
The graph (with plus sign for comparison) has no minimum, and remains above −1, indicating that the energy
is greater than for the proton and atom dissociated. Hence, no evidence of bonding here.
F(x)
-0.5
-0.6
(+)
(<)
-0.7
-0.8
-0.9
2
4
6
8
x
-1.1
Problem 8.11
According to Mathematica, the minimum occurs at x = 2.493, and at this point F 00 = 0.1257.
r
E1 00
1 −(0.1257)E1
2
00
.
mω = V = − 2 F , so ω =
a
a
m
mp mp
1
Here m is the reduced mass of the proton: m =
= mp .
mp + mp
2
s
3 × 108 m/s
(0.1257)(13.6 eV)
ω=
= 3.42 × 1014 /s.
(0.529 × 10−10 m) (938 × 106 eV)/2
308
CHAPTER 8. THE VARIATIONAL PRINCIPLE
1
1
~ω = (6.58 × 10−16 eV · s)(3.42 × 1014 /s) = 0.113 eV (ground state vibrational energy).
2
2
Mathematica says that at the minimum F = −1.1297, so the binding energy is (0.1297)(13.6 eV) = 1.76 eV.
Since this is substantially greater than the vibrational energy, it stays bound. The highest vibrational energy is
1.76
1
given by (n + 21 )~ω = 1.76 eV, so n =
− = 7.29. I estimate eight bound vibrational states (including
0.226 2
n = 0).
Problem 8.12
I’ll drop the (irrelevant) normalization factors:
ψ(r1 , r2 ) = ψB (r1 )ψA (r2 ) − ψA (r1 )ψB (r2 )
= [ψ0 (r1 ) + ψ0 (r10 )] [ψ0 (r2 ) − ψ0 (r20 )] − [ψ0 (r1 ) − ψ0 (r10 )] [ψ0 (r2 ) + ψ0 (r20 )]
0
0
0 (((
0
(((
=(
ψ0(
(r(
ψ0(
(r(
1 )ψ0 (r2 ) − ψ0 (r1 )ψ0 (r2 ) + ψ0 (r1 )ψ0 (r2 ) − (
1 )ψ0 (r2 )
0
0
0 (((
0
(((
−(
ψ0(
(r(
ψ0(
(r(
1 )ψ0 (r2 ) − ψ0 (r1 )ψ0 (r2 ) + ψ0 (r1 )ψ0 (r2 ) + (
1 )ψ0 (r2 )
= −2 [ψ0 (r1 )ψ0 (r20 ) − ψ0 (r10 )ψ0 (r2 )] . X
Problem 8.13
−
e2
4π0 r1
e2
4π0 a
a
ψ0 (r1 ) hψ0 (r20 ) | ψ0 (r20 )i
ψ0 (r1 )
r1
±
a
a
0
0
0
0
+ ψ0 (r1 )
ψ0 (r1 ) hψ0 (r2 ) | ψ0 (r2 )i ± 2 ψ0 (r1 )
ψ0 (r1 ) hψ0 (r2 ) | ψ0 (r2 )i
r1
r1
2 e
1
=−
[(1)(1) + (D)(1) ± 2(X)(I)]
4π0 a 2(1 ± I 2 )
2 1
e
1 + D ± 2I X
=−
. X
2 4π0 a
1 ± I2
=−
A2±
Problem 8.14
(a) The φ1 integral is trivial:
2π
Φ(r2 ) =
πa2
Z
1
e−2r1 /a p 2
r12 dr1 sin θ1 dθ1 .
2
r1 + r2 − 2r1 r2 cos θ1
Now do the θ1 integral:
Z π
q
π
sin θ1
−1
p
dθ
=
r12 + r22 − 2r1 r2 cos θ1
1
2
2
r1 r2
0
r1 + r2 − 2r1 r2 cos θ1
0
r + r − |r − r |
p
p
1
1
2
1
2
=
(r1 + r2 )2 − (r1 − r2 )2 =
r1 r2
r1 r2
(
2/r2 , if r1 < r2 ,
=
2/r1 if r1 > r2 .
CHAPTER 8. THE VARIATIONAL PRINCIPLE
309
Finally, the r1 integral:
Z r2
Z ∞
1
4
−2r1 /a 2
−2r1 /a
e
Φ(r2 ) = 2
r1 dr1 +
e
r1 dr1
a
r2 0
r2
)
( r2 2
r ∞
a −2r1 /a r1
4
1
a3 −2r1 /a
r1 2
1
+ − e
+1
2 +1
= 2
− e
2
+2
a
r2
4
a
a
4
a
0
r2
a
a
=
− 1+
e−2r2 /a . X
r2
r2
(b) Again, the φ2 integral is trivial:
Z
√
2π
a
a
−2 R2 +r22 −2Rr2 cosθ2 /a
−2r2 /a
e
D2 =
−
1
+
e
r22 dr2 sin θ2 dθ2 .
πa3
r2
r2
p
4Rr2 2
R2 + r22 − 2Rr2 cos θ2 /a, so du =
a u sin θ2 dθ2 . Then
P
Z π
Z u2
√
u2
a2
a2
−2 R2 +r22 −2Rr2 cosθ2 /a
Iθ ≡
e
sin θ2 dθ2 =
ue−u du = −
(u + 1)e−u .
4Rr2 u1
4Rr2
u1
0
For the θ2 integral, let u ≡ 2
Now, letting ρ ≡ 2R/a and x ≡ r2 /R,
q
2
2
u2 =
R2 + r22 + 2Rr2 = (R + r2 ) = ρ(1 + x),
a
a
Then
Iθ =
u1 =
2
a
q
2
R2 + r22 − 2Rr2 = |R − r2 | = ρ|1 − x|.
a
o
1 n
[ρ|1 − x| + 1]e−ρ|1−x| − [ρ(1 + x) + 1]e−ρ(1+x) .
ρ2 x
Finally, the r2 integral becomes
Z
o
1 ∞n
D2 =
[ρ|1 − x| + 1]e−ρ|1−x| − [ρ(1 + x) + 1]e−ρ(1+x) 2 − (ρx + 2) e−ρx dx
4 0
Z
e−ρ 1
{[ρ(1 − x) + 1]eρx } 2 − (ρx + 2) e−ρx dx
=
4 0
Z
eρ ∞ +
[ρ(x − 1) + 1]e−ρx 2 − (ρx + 2) e−ρx dx
4 1
Z
e−ρ ∞ −
[ρ(1 + x) + 1]e−ρx 2 − (ρx + 2) e−ρx dx
4 0
Z 1
Z 1
−ρ
e
2
[(ρ + 1) − ρx]eρx dx −
[2(ρ + 1) + (ρ − 1)ρx − (ρx)2 ] dx
=
4
0
Z0 ∞
Z ∞
eρ
−ρx
+
2
[(1 − ρ) + ρx]e
dx −
2(1 − ρ) + ρx(3 − ρ) + (ρx)2 e−2ρx dx
4
1
1
Z ∞
Z ∞
e−ρ
−ρx
−
2
[(1 + ρ) + ρx]e
dx −
2(1 + ρ) + (3 + ρ)ρx + (ρx)2 e−2ρx dx
4
0
0
−ρ
e
2
3
1 2
eρ 4 −ρ
3 2 −2ρ
ρ
=
(2e − ρ − 2) − 2 + ρ + ρ
+
e −
+
e
4
ρ
2
6
4 ρ
4 ρ
−ρ
e
2
5 2
2 1
16 ρ2
−
2 1+
−
+
= −
11 + 3ρ +
+
e−ρ .
4
ρ
4 ρ
ρ 8
ρ
3
310
CHAPTER 8. THE VARIATIONAL PRINCIPLE
Or, puitting back ρ ≡ 2R/a:
a
D2 =
−
R
Problem 8.15
11 3R
a
R2
+
+ + 2
8
4a
R 6a
e−2R/a .
X
Problem 8.15 NPN
�������
DInt[ρ_] := 1 / ρ - (1 + 1 / ρ) Exp[- 2 ρ];
XInt[ρ_] :=8.15
(1 + ρ)
Exp[- ρ];
Problem
NPN
�������
�������
IInt[ρ_] := Exp[- ρ] 1 + ρ + 1
3 ρ^2 ;
D2Int[ρ_]
/ ρ+ 2 ρ]
XInt[ρ_]:=
:= 1(1
ρ)Exp[Exp[- ρ];
ρ 1 + 11
�������
DInt[ρ_] := 1 / ρ - (1 + 1 / ρ) Exp[- 2 ρ];
������� �������
IInt[ρ_]:=
:= 1Exp[1 +2ρρ]
+ 1253 ρ8
^ 2- ;23
X2Int[ρ_]
/ 5 ρ]
Exp[-
������� �������
�������
8ρ + 3
4 ρ^2 + 1
4 ρ - 3 ρ^2 - 1
6 ρ^3 ;
3 ρ^3
6 ρ IInt[ρ]
EulerGamma
+ +Log[ρ]
4 ρ] D2Int[ρ_]
:= 1 / ^
ρ 2- Exp[2 ρ] ρ 1
11 8 ρ ++ 3IInt[4 ρ ^ 2 ρ]
+ 1^ 26ExpIntegralEi[ρ^3 ;
�������
X2Int[ρ_] := 1 / 5 Exp[- 2 ρ] 25
2 IInt[ρ] IInt[- ρ] ExpIntegralEi[- 2 ρ]
8 - 23
4 ρ - 3 ρ^2 - 1
3 ρ^3
+
6 ρ :=
IInt[ρ]
^ 2- EulerGamma
+ Log[ρ] + IInt[- ρ] ^ 2 ExpIntegralEi[- 4 ρ] H[pm_, ρ_]
-2 1
1/ρ +
2 IInt[ρ] IInt[- ρ] ExpIntegralEi[- 2 ρ]
2 DInt[ρ] - D2Int[ρ] + pm 2 IInt[ρ] XInt[ρ] - X2Int[ρ]
�������
�������
+
�������
1 + pm IInt[ρ] ^ 2
H[pm_, ρ_] := - 2 1 - 1 / ρ +
Plot {H[1, ρ], H[- 1, ρ]}, {ρ, 0, 6},
2 DInt[ρ] - D2Int[ρ] + pm 2 IInt[ρ] XInt[ρ] - X2Int[ρ]
PlotRange → {- 2.5, 0}, AxesLabel →
������� Plot {H[1, ρ], H[- 1, ρ]}, {ρ, 0, 6},
"R/a", "
<H>
E1
1 + pm IInt[ρ] ^ 2
" ,
<H>
PlotRange
{- 2.5, 0},{Black,
AxesLabel
→ "R/a",GridLines
"
" , → {None, {- 2}}
PlotStyle
→ →{{Black},
Dashed}},
E1
<H>
PlotStyle → {{Black}, {Black, Dashed}}, GridLines → {None, {- 2}}
E1 < H >
E1
1
2
1
3
2
4
3
4
5
5
R/a
6
6
R/a
-0.5
-0.5
������� -1.0
������� -1.0
-1.5
-1.5
-2.0 -2.0
-2.5
�������
�������
�������
1 + pm XInt[ρ] IInt[ρ]
Kinetic[pm_, ρ_] := - 2 1 - 2 1 + pm XInt[ρ] IInt[ρ]
1 + pm IInt[ρ] ^ 2
�������
IonPotential[pm_, ρ_] := - 4
Kinetic[pm_, ρ_] := - 2 1 - 2
IonPotential[pm_, ρ_] := - 4
��������
��������
-2.5
1 + pm IInt[ρ] ^ 2
1 + DInt[ρ] + pm 2 XInt[ρ] IInt[ρ]
1 + DInt[ρ]
pm 2 XInt[ρ]
IInt[ρ]
1 + pm+IInt[ρ]
^2
ElectronPotential[pm_, ρ_] :=
ElectronPotential[pm_, ρ_] :=
1 + pm IInt[ρ]
^2
2 D2Int[ρ]
2 pm X2Int[ρ]
+
2
1 + pm
IInt[ρ]
2 pm X2Int[ρ]
1 + pm
IInt[ρ]2
2 D2Int[ρ]
1 + pm IInt[ρ]2
+
1 + pm IInt[ρ]2
2
Problem8.15(NPN)Notebook.nb
CHAPTER 8. THE VARIATIONAL
PRINCIPLE
311
2
Problem8.15(NPN)Notebook.nb
2 ��������
Problem8.15(NPN)Notebook.nb
Plot[{Kinetic[1, ρ] , Kinetic[- 1, ρ]},
{ρ, 0, 6}, PlotStyle → {{Black}, {Black, Dashed}}]
��������
��������
Plot[{Kinetic[1, ρ] , Kinetic[- 1, ρ]},
Plot[{Kinetic[1, ρ] , Kinetic[- 1, ρ]},
{ρ, 0, 6}, PlotStyle → {{Black}, {Black, Dashed}}]
3.5
{ρ, 0, 6}, PlotStyle → {{Black}, {Black, Dashed}}]
3.5
3.5
3.0
�������� 3.0
3.0
��������
��������
2.5
2.5
2.5
2.0
2.0
2.0
1
2
1
1
��������
��������
��������
2
2
3
3
3
4
4
4
5
5
5
6
6
6
Plot[{IonPotential[1, ρ] , IonPotential[- 1, ρ]},
Plot[{IonPotential[1, ρ] , IonPotential[- 1, ρ]},
Plot[{IonPotential[1,
ρ] ,
1, ρ]}, Dashed}}]
{ρ,0,
0,6},
6},PlotStyle
PlotStyle
→ IonPotential[{{Black},
{Black,
{ρ,
→ {{Black},
{Black,
Dashed}}]
{ρ, 0, 6}, PlotStyle → {{Black}, {Black, Dashed}}]
-5
-5
-5
-6
-6
-6
��������
��������
��������
-7
-7
-7
-8
-8
-8
-9
-9
-9
1
1
2
2
1
��������
��������
��������
3
3
2
4
4
3
5
5
4
6
6
5
6
Plot[{ElectronPotential[1, ρ], ElectronPotential[- 1, ρ]},
Plot[{ElectronPotential[1, ρ], ElectronPotential[- 1, ρ]},
{ρ, 0, 6}, PlotStyle → {{Black}, {Black, Dashed}}]
{ρ, 0, 6}, PlotStyle → {{Black}, {Black, Dashed}}]
Plot[{ElectronPotential[1, ρ], ElectronPotential[- 1, ρ]},
{ρ, 0, 6}, PlotStyle → {{Black}, {Black, Dashed}}]
1.0
1.0
��������
��������
1.0
0.5
0.5
0.5
��������
1
1
2
2
3
3
4
4
5
5
6
6
-0.5
-0.5
1
2
3
4
-0.5
Problem 8.16
(a) First we normalize the function
Z a
1=
2
|ψ(x)| dx = A2
0
= A2 a5
Z a
2
x2 (a − x) dx
0
Z 1
0
2
u2 (1 − u) du =
A2 a5
30
5
6
312
CHAPTER 8. THE VARIATIONAL PRINCIPLE
so that A2 = 30/a5 . The expectation value of the Hamiltonian is


Z a
 ~2 d 2 ψ 

hHi =
ψ ∗ (x) 
− 2 m dx2  dx
0
|{z}
2A
2
=
~ A
m
2
Z a
x (a − x) dx
0
30 ~2 3
a
m a5
~2
.
=5
m a2
Z 1
u (1 − u) du
=
0
Therefore
Egs ≤
10 π 2 ~2
.
π 2 2 m a2
which is approximately 1.3% larger than the exact answer.
(b) Again we start by normalizing
Z a
Z a
Z 1
2
2p
2p
1=
|ψ| dx = A2
x2 p (a − x) dx = A2 a4 p+1
|u (1 − u)| du .
0
0
0
I’ll leave the integral alone for a second. The expectation value of the Hamiltonian is
Z a
~2 d2 ψ
1
ψ∗ −
dx
hHi =
2m 0
2 m dx2
∗
Z a
~2
dψ
dψ
=
dx
2m 0
dx
dx
Z
~2 2 a 2
2
2 p−2
A
=
p (a − 2 x) [x (a − x)]
dx
2m
0
Z
~2 4 p−1 2 1
2
2 p−2
a
A
(1 − 2 u) [u (1 − u)]
du
= p2
2m
0
R1
2
2 p−2
(1 − 2 u) [u (1 − u)]
du
~2
0
= p2
.
R1
2p
2 m a2
|u (1 − u)| du
0
Now note that the integral in the numerator is
Z 1
Z 1
d
1
2 p−1
2
2 p−2
(1 − 2 u)
[u (1 − u)]
du
(1 − 2 u) [u (1 − u)]
du =
2
p
−
1
du
0
0
Z 1
1
2 p−1
=
0+2
[u (1 − u)]
du
2p − 1
0
where in the second line we’ve integrated by parts. Therefore
R1
2 p−1
[u (1 − u)]
du
2 p2
~2
0
hHi =
R
1
2p
2 p − 1 2 m a2
|u (1 − u)| du
0
CHAPTER 8. THE VARIATIONAL PRINCIPLE
313
These integrals can be looked up in a table (for example Schaum’s 18.24) to give
hHi =
2 p (4 p + 1) ~2
.
2 p − 1 2 m a2
(8.1)
We can then minimize this
d hHi
=0
dp
~2 16 p2 − 16 p − 2
= 0.
2
2 m a2
(2 p − 1)
The positive√root
of the quadratic (the other root would give a solution that diverges at 0 and a) is
pmin = 2 + 6 /4. Plugging this into Equation 8.1 gives
√
5 + 2 6 π 2 ~2
.
Egs ≤
π2
2 m a2
This bound is approximately 0.3% greater than the exact value.
Problem 8.17
(a)
Z
Z a/2
r
2
⇒ A=
.
dx = |A|
a
2
a
−a/2
Z
Z
~2
d2 ψ
~2 π 2
π 2 ~2
hT i = −
ψ 2 dx =
ψ 2 dx =
.
2m
dx
2m a
2ma2
Z
Z
Z
πx 1
1
2 a/2 2
mω 2 a 3 π/2 2
hV i = mω 2 x2 ψ 2 dx = mω 2
x cos2
dx =
y cos2 y dy
2
2
a −a/2
a
a
π
−π/2
3 2
π/2
2
2 2
2 2
mω a y
y
1
y cos 2y
mω a
π
=
+
−
sin
2y
+
=
−
1
.
π3
6
4
8
4
4π 2
6
−π/2
π 2 ~2
mω 2 a2 π 2
∂hHi
π 2 ~2
mω 2 a π 2
hHi =
+
−1 ;
=−
+
−1 =0 ⇒
2ma2
4π 2
6
∂a
ma3
2π 2
6
1=
2
2
|ψ| dx = |A|
cos
2
πx r
a=π
~
mω
2a
2
π 2 /6 − 1
1/4
.
s
r
π 2 ~2 mω π 2 /6 − 1 mω 2 π 2
2
2 ~
+
−1 π
hHimin =
2mπ 2 ~
2
4π 2
6
mω π 2 /6 − 1
r
1
π2
1
1
− 2 = ~ω(1.136) > ~ω. X
= ~ω
2
3
2
2
[We do not need to worry about the kink at ±a/2. It is true that d2 ψ/dx2 has delta functions there, but
since ψ(±a/2) = 0 no “extra” contribution to T comes from these points.]
314
CHAPTER 8. THE VARIATIONAL PRINCIPLE
(b) Because this trial function is odd, it is orthogonal to the ground state, so by Problem 8.4 hHi will give
an upper bound to the first excited state.
Z
Z a
πx 1
dx = |B|2 a ⇒ B = √ .
1=
|ψ|2 dx = |B|2
sin2
a
a
−a
2 2
2 Z
2
2 2 Z
π ~
~
d ψ
~
π
ψ 2 dx =
hT i = −
ψ 2 dx =
.
2m
dx
2m a
2ma2
Z
Z a
Z
πx 1
1
mω 2 a 3 π 2 2
1
hV i = mω 2 x2 ψ 2 dx = mω 2
x2 sin2
dx =
y sin y dy
2
2
a −a
a
2a π
−π
2
π
mω 2 a2 2π 2
mω 2 a2 y 3
y
1
y cos 2y
=
=
−
−
sin 2y −
−1 .
2π 3
6
4
8
4
4π 2
3
−π
π 2 ~2
mω 2 a2 2π 2
mω 2 a 2π 2
∂hHi
π 2 ~2
hHi =
+
+
−1 ;
=−
−1 =0 ⇒
2ma2
4π 2
3
∂a
ma3
2π 2
3
r
a=π
~
mω
2
2π 2 /3 − 1
1/4
.
s
r
π 2 ~2 mω 2π 2 /3 − 1 mω 2 2π 2
2
2 ~
+
−1 π
hHimin =
2mπ 2 ~
2
4π 2
3
mω 2π 2 /3 − 1
r
1
4π 2
1
3
= ~ω
− 2 = ~ω(3.341) > ~ω. X
2
3
2
2
Problem 8.18
We will need the following integral repeatedly:
Z ∞
Γ
1
xk
dx = 2l−k−1
l
2
2
2b
(x + b )
0
k+1
2
Γ 2l−k−1
2
.
Γ(l)
(a)
s
|A|2 Γ 21 Γ 4n−1
b4n−1 Γ(2n)
2
.
1=
dx = 4n−1
|ψ| dx = 2|A|
⇒A=
2n
b
Γ(2n)
Γ 12 Γ 4n−1
(x2 + b2 )
−∞
0
2
"
#
Z ∞
Z
~2
d2 ψ
~2 2 ∞
1
d
−2nx
hT i = −
ψ
dx = −
A
dx
2
2 n
2
2 n+1
2m −∞ dx2
2m
−∞ (x + b ) dx (x + b )
"
#
Z
n~2 2 ∞
1
1
2(n + 1)x2
=
A
dx
n+1 −
n+2
2
2 n
m
(x2 + b2 )
(x2 + b2 )
−∞ (x + b )
"Z
#
Z ∞
∞
2n~2 2
x2
1
=
A
2n+1 dx − 2(n + 1)
2n+2 dx
m
(x2 + b2 )
(x2 + b2 )
0
0
"
#
2n~2 b4n−1 Γ(2n)
1 Γ 12 Γ 4n+1
2(n + 1) Γ 32 Γ 4n+1
~2 n(4n − 1)
2
2
=
−
=
.
m Γ 21 Γ 4n−1
2b4n+1 Γ(2n + 1)
2b4n−1
Γ(2n + 2)
4mb2 (2n + 1)
2
Z ∞
2
2
Z ∞
1
CHAPTER 8. THE VARIATIONAL PRINCIPLE
Z ∞
x2
1
2
2
ψ x dx = mω 2A
2n dx
2
(x2 + b2 )
−∞
0
b4n−1 Γ(2n)
1 Γ 32 Γ 4n−3
mω 2 b2
2
= mω 2
=
.
1
4n−1 2b4n−3
Γ(2n)
2(4n − 3)
Γ 2 Γ
2
1
hV i = mω 2
2
Z ∞
315
2 2
~2 n(4n − 1)
mω 2 b2
∂hHi
~2 n(4n − 1)
mω 2 b
+
;
=
−
+
=0 ⇒
4mb2 (2n + 1)
(4n − 3)
∂b
2mb3 (2n + 1)
(4n − 3)
r
1/4
~ n(4n − 1)(4n − 3)
.
b =
mω
2(2n + 1)
s
s
~2 n(4n − 1) mω
2(2n + 1)
mω 2
~
n(4n − 1)(4n − 3)
hHimin =
+
4m (2n + 1) ~
n(4n − 1)(4n − 3) 2(4n − 3) mω
2(2n + 1)
s
r
1
1
2n(4n − 1)
8n2 − 2n
1
= ~ω
= ~ω
> ~ω. X
2
2
(2n + 1)(4n − 3)
2
8n − 2n − 3
2
hHi =
(b)
s
2 Γ 3 Γ 4n−3
|B|
b4n−3 Γ(2n)
2
2
.
dx = 4n−3
1 = 2|B|2
⇒B=
2n
b
Γ(2n)
Γ 32 Γ 4n−3
(x2 + b2 )
0
2
#
"
Z
d
2nx2
~2 2 ∞
x
1
−
dx
hT i = −
B
n+1
2
2 n
2
2 n
2m
(x2 + b2 )
−∞ (x + b ) dx (x + b )
#
"
Z
~2 B 2 ∞
4nx
4n(n + 1)x3
x
−2nx
=−
−
dx
n+1 +
n+2
2m −∞ (x2 + b2 )n (x2 + b2 )n+1
(x2 + b2 )
(x2 + b2 )
" Z
#
Z ∞
∞
4n~2 B 2
x2
x4
=
3
2n+1 dx − 2(n + 1)
2n+2 dx
2m
(x2 + b2 )
(x2 + b2 )
0
0
"
#
2n~2 b4n−3 Γ(2n)
3 Γ 32 Γ 4n−1
2(n + 1) Γ 25 Γ 4n−1
3~2 n(4n − 3)
2
2
=
−
=
.
3
4n−3
4n−1
4n−1
m Γ 2 Γ
2b
Γ(2n + 1)
2b
Γ(2n + 2)
4mb2 (2n + 1)
2
Z ∞
1
hV i = mω 2 2B 2
2
2
x2
Z ∞
0
x4
4n−3
1
2 Γ
2 b Γ(2n) dx
=
mω
2n
3
4n−3
4n−5
2
2
2
2b
Γ 2 Γ
(x + b )
2
5
2
Γ 4n−5
3 mω 2 b2
2
=
.
Γ(2n)
2 (4n − 5)
3~ n(4n − 3) 3 mω 2 b2
∂hHi
3~2 n(4n − 3)
3mω 2 b
+
;
=−
+
=0 ⇒
2
3
4mb (2n + 1)
2 (4n − 5)
∂b
2mb (2n + 1)
(4n − 5)
r
1/4
~ n(4n − 3)(4n − 5)
b=
.
mω
2(2n + 1)
s
s
3~2 n(4n − 3) mω
2(2n + 1)
3 mω 2
~
n(4n − 3)(4n − 5)
hHimin =
+
4m (2n + 1) ~
n(4n − 3)(4n − 5) 2 (4n − 5) mω
2(2n + 1)
s
r
3
2n(4n − 3)
3
8n2 − 6n
3
= ~ω
= ~ω
> ~ω. X
2
(2n + 1)(4n − 5)
2
8n2 − 6n − 5
2
hHi =
316
CHAPTER 8. THE VARIATIONAL PRINCIPLE
(c) As n → ∞, ψ becomes more and more “gaussian”. In the figures I have plotted the trial wave functions
for n = 2, n = 3, and n = 4, as well as the exact states (heavy line). Even for n = 2 the fit is pretty good,
so it is hard to see the improvement, but the successive curves do move perceptably toward the correct
result.
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
1
2
r
Analytically, for large n, b ≈
2 n
2
x +b
4
3
=b
~
mω
2n
n · 4n · 4n
2 · 2n
x2
1+ 2
b
1/4
r
=
n
≈b
2n
1
2
→
b2n emωx /2~ .
4
3
2n~
, so
mω
mωx2
1+
2~n
n
2
Meanwhile, using Stirling’s approximation in the form Γ(z + 1) ≈ z z e−z :
2n−1
p
b4n−1 Γ(2n)
b4n−1 (2n − 1)2n−1 e−(2n−1)
b4n−1 1
2n − 1
√
≈ √ √
2n − 3/2.
A =
3
1 ≈
1
2n−3/2
π 2n − 3
π
e 2n − 2
Γ 2 Γ 2n − 2
e−(2n−3/2)
2
1 1 − 2n
3
3
1
1
1
But
1
+
≈1+
−
=1+
;
≈
1
−
3
2n
4n
4n 2n
4n
1 − 4n
2n−1 n 2
2 √
2n − 1
1
1
so
≈
1
+
→ e1/4 = e.
3
4n
1 + 1/4n
2n − 2
r
1/4
b4n−1 √ √
2n 4n−1
2n
= √
e 2n =
b
⇒ A≈
b2n−1/2 . So
π
π
πe
1/4
1/4 2n
2n
mω 1/4 −mωx2 /2~ mω 1/4 −mωx2 /2~
2n−1/2 1 −mωx2 /2~
ψ ≈
b
e
=
e
=
e
,
π
b2n
π
2n~
π~
2
which is precisely the ground state of the harmonic oscillator (Eq. 2.60). So it’s no accident that we get
the exact energies, in the limit n → ∞.
Problem 8.19
1 = |A|2
Z
2
e−2br r2 sin θ dr dθ dφ = 4π|A|2
Z ∞
2
r2 e−2br dr = |A|2
0
e2
hV i = −
|A|2 4π
4π0
Z ∞
e
0
e2
r dr = −
r
4π0
−2br 2 1 2
2b
π
3/2
π 3/2
2b
⇒ A=
r
1
e2
2b
4π = −
2
.
4b
4π0
π
2b
π
3/4
.
CHAPTER 8. THE VARIATIONAL PRINCIPLE
~2
hT i = −
|A|2
2m
Z
2
317
2
e−br (∇2 e−br ) r2 sin θ dr dθ dφ
−2b
2
2
2
2
1 d
d
1 d −2br3 e−br = 2 3r2 − 2br4 e−br .
But (∇2 e−br ) = 2
r2 e−br = 2
r dr
dr
r dr
r
r
r 3/2
3/2 Z ∞
−~2 2b
~2
π
π
2b
1
3
2
4 −2br 2
=
(4π)(−2b)
(3r − 2br )e
dr =
πb4
3
− 2b
2
2m π
m
π
8b 2b
32b
2b
0
~2
2b
3
3
3~2 b
=
4πb
−
=
.
m
π
8b 16b
2m
r
e2
2b
3~2 b
−
2
;
hHi =
2m
4π0
π
∂hHi
3~2
e2
=
−
∂b
2m
4π0
r
2 1
√ =0
π b
⇒
√
e2
b=
4π0
r
2 2m
.
π 3~2
r 2
r
2 2
e2
2 4m2
e2
2
e2
2 2m
e
m
4
8
−
2
=
−
4π0
π 9~4
4π0
π 4π0
π 3~2
4π0
~2 3π 3π
2 2
m
8
e
8
= − 2
=
E1 = −11.5 eV.
2~
4π0
3π
3π
hHimin =
3~2
2m
Problem 8.20
Adopting the variational wave function
−br 2
ψ(r, θ, φ) = Ae
2
Z ∞
1 = |A|
−2br 2 2
e
Z
r dr
|Y10 |2 dΩ = |A|2
0
2
e
hV i = −
|A|2
4π0
0
e−2br
2
e2
1 2
r dr = −
r
4π0
1
8b
r
4
2
3
Ae−br cos θ,
4π
3 1/4
8b
π
(1) ⇒ A = 2
.
2b
π
!
r
8b3 1
e2
8b
=−
.
π
4b
4π0 π
r
1 ∂
1
∂
∂ψ
2 ∂ψ
r
+
sin
θ
r2 ∂r
∂r
r2 sin θ ∂θ
∂θ
(
)
r
2
h
i
3
cos θ d 2
e−br ∂
2
−br 2
=
A
r (−2br)e
+ 2
(− sin θ)
4π
r2 dr
r sin θ ∂θ
(
)
r
−br 2
2
3
2b cos θ 2 −br2
e
=
A −
3r e
− 2br4 e−br − 2
(2 sin θ cos θ)
4π
r2
r sin θ
r
3 −br2 cos θ
=
Ae
2b2 r4 − 3br2 − 1 .
2
π
r
∇2 ψ =
Z ∞
r
Y10 (θ, φ) =
318
CHAPTER 8. THE VARIATIONAL PRINCIPLE
r r
Z
Z
2
2 cos θ
~2
3
3
~2
∗ 2
3
hT i = −
ψ ∇ ψd r = −
2b2 r4 − 3br2 − 1 r2 sin θ dr dθ dφ
|A|2 e−br cos θe−br
2
2m
2m π 4π
r
Z ∞
Z π
2
~2 3
|A|2 2π
e−2br 2b2 r4 − 3br2 − 1 dr
=−
cos2 θ sin θ dθ
2m 2π
0
0
√
√
√
2
1
1
1
3~2 2
2
− 3b 2 π √
−
|A| 2b 12 π √
π √
=−
5
3
2m
3
(2 2b)
(2 2b)
(2 2b)
r
√ 2
2
2
2
24b
6b
~
~
π
π 11
11b~
√
− √
−1 =
= − |A|2 √
|A|2
=
.
m
m
2b 16
2m
2 2b (2 2b)4
(2 2b)2
11 b~2
e2
hHi =
−
2m
4π0
r
8b
;
π
∂hHi
11~2
e2
=
−
∂b
2m
4π0
11~2 e4 2 4m2
e2
−
hHimin =
2m 16π 2 20 π 121~4
4π0
=
r
8 e2
π 4π0
r
r
2
=0
πb
2 2m
=−
π 11~2
⇒
4
11π
√
e2
b=
4π0
r
2 2m
.
π 11~2
me4
~2 (4π0 )2
32
E2 = 0.926E2 .
11π
Since E2 is negative this is indeed an upper bound, within about 7% of the exact answer.
Problem 8.21
Let ψ = √
1
πb3
e−r/b (same as hydrogen, but with a → b adjustable). From Eq. 4.218, we have hT i = −E1 =
~2
~2
for
hydrogen,
so
in
this
case
hT
i
=
.
2ma2
2mb2
Z ∞
Z
−µr
e2 4π
e2 4 ∞ −(µ+2/b)r
1
e2
1
e2 4
2
−2r/b e
hV i = −
r
dr
=
−
=
−
.
e
e
r
dr
=
−
2
4π0 πb3 0
r
4π0 b3 0
4π0 b3 (µ + 2/b)2
4π0 b(1 + µb
2 )
~2
e2
1
−
.
2
2
2mb
4π0 b(1 + µb
2 )
~2
e2
µ
~2
e2 (1 + 3µb/2)
∂hHi
1
=− 3 +
+
=− 3 +
=0
2
2
3
∂b
mb
4π0 b (1 + µb/2)
b(1 + µb/2)
mb
4π0 b2 (1 + µb/2)3
~2 4π0
(1 + 3µb/2)
(1 + 3µb/2)
=b
, or b
= a.
2
3
m
e
(1 + µb/2)
(1 + µb/2)3
hHi =
⇒
This determines b, but unfortunately it’s a cubic equation. So we use the fact that µ is small to obtain a suitable
approximate solution. If µ = 0, then b = a (of course), so µa 1 =⇒ µb 1 too. We’ll expand in powers of
µb:
"
2 #
3µb
3µb
µb
9
6
3
2
2
2
a≈b 1+
1−
+6
≈ b 1 − (µb) + (µb) = b 1 − (µb) .
2
2
2
4
4
4
Since the 43 (µb)2 term is already a second-order correction, we can replace b by a:
a
3
2
b≈ (µa)
.
≈
a
1
+
4
1 − 34 (µb)2
CHAPTER 8. THE VARIATIONAL PRINCIPLE
319
~2
e2
1
−
2
2
3
3
2
2
2
4π
0 a 1 + (µa)
2ma 1 + 4 (µa)
1 + 12 (µa)
4
µa 2 3
~2
3
e2 1
µa
2
2
1 − 2 (µa) −
≈
1 − (µa)
+3
1−2
2ma2
4
4π0 a
4
2
2
3
3
3
3
= −E1 1 − (µa)2 + 2E1 1 − µa + (µa)2 − (µa)2 = E1 1 − 2(µa) + (µa)2 .
2
4
4
2
hHimin =
Problem 8.22
(a)
H=
1
λ=
2
Ea h
h Eb
; det(H − λ) = (Ea − λ)(Eb − λ) − h2 = 0 =⇒ λ2 − λ(Ea + Eb ) + Ea Eb − h2 = 0.
q
h
i
p
2
2
2
Ea + Eb ± Ea + 2Ea Eb + Eb − 4Ea Eb + 4h ⇒ E± = 12 Ea + Eb ± (Ea − Eb )2 + 4h2 .
(b) Zeroth order: Ea0 = Ea , Eb0 = Eb . First order: Ea1 = hψa |H 0 |ψa i = 0, Eb1 = hψb |H 0 |ψb i = 0. Second
order:
|hψb |H 0 |ψa i|2
h2
|hψa |H 0 |ψb i|2
h2
Ea2 =
=−
;
Eb2 =
=
;
Ea − Eb
Eb − Ea
Eb − Ea
Eb − Ea
E− ≈ Ea −
h2
;
(Eb − Ea )
E+ ≈ Eb +
h2
.
(Eb − Ea )
(c)
hHi = hcos φ ψa + sin φ ψb |(H 0 + H 0 )| cos φ ψa + sin φ ψb i
= cos2 φ hψa |H 0 |ψa i + sin2 φ hψb |H 0 |ψb i + sin φ cos φ hψb |H 0 |ψa i + sin φ cos φ hψa |H 0 |ψb i
= Ea cos2 φ + Eb sin2 φ + 2h sin φ cos φ.
∂hHi
= −Ea 2 cos φ sin φ + Eb 2 sin φ cos φ + 2h(cos2 φ − sin2 φ) = (Eb − Ea ) sin 2φ + 2h cos 2φ = 0.
∂φ
tan 2φ = −
2h
= −
Eb − Ea
sin2 2φ(1 + 2 ) = 2 ;
or
cos 2φ = √
∓1
1 + 2
where
≡
sin 2φ = √
2h
.
Eb − Ea
±
;
1 + 2
(sign dictated by tan 2φ =
sin 2φ
p
1 − sin2 2φ
= −;
sin2 2φ = 2 (1 − sin2 2φ);
cos2 2φ = 1 − sin2 2φ = 1 −
2
1
=
;
1 + 2
1 + 2
sin 2φ
= −).
cos 2φ
1
1
1
1
1
1
2
√
√
cos φ = (1 + cos 2φ) =
1∓
; sin φ = (1 − cos 2φ) =
1±
.
2
2
2
2
1 + 2
1 + 2
1
1
1
1
1
(Eb − Ea + 2h)
√
+ Eb 1 ± √
± h√
=
Ea + Eb ±
hHimin = Ea 1 ∓ √
2
2
2
1 + 2
1 + 2
1 + 2
1 + 2
2
320
CHAPTER 8. THE VARIATIONAL PRINCIPLE
2h
(Eb − Ea ) + 2h (Eb −Ea )
p
(E − Ea )2 + 4h2
(Eb − Ea + 2h)
√
q
p b
=
=
=
(Eb − Ea )2 + 4h2 ,
4h2
1 + 2
(Eb − Ea )2 + 4h2
1+
2
But
So
(Eb −Ea )
i
p
1h
Ea + Eb ± (Eb − Ea )2 + 4h2
we want the minus sign (+ is maximum)
2
i
h
p
= 12 Ea + Eb − (Eb − Ea )2 + 4h2 .
hHimin =
h
i
q
2
(d) If h is small, the exact result (a) can be expanded: E± = 12 (Ea + Eb ) ± (Eb − Ea ) 1 + (Eb4h
−Ea )2 .
1
1
2h2
2h2
=⇒ E± ≈
=
Ea + Eb ± (Eb − Ea ) 1 +
Ea + Eb ± (Eb − Ea ) ±
,
2
(Eb − Ea )2
2
(Eb − Ea )
h2
h2
, E− ≈ Ea −
,
(Eb − Ea )
(Eb − Ea )
confirming the perturbation theory results in (b). The variational principle (c) gets the ground state (E− )
exactly right—not too surprising since the trial wave function Eq. 8.75 is almost the most general state
(there could be a relative phase factor eiθ ).
so
E+ ≈ Eb +
Problem 8.23
For the electron,
γ= −e/m, so E± =
±eBz ~/2m (Eq. 4.161). For consistency with Problem 8.22, Eb > Ea ,
eBz ~
eBz ~
1
0
so χb = χ+ =
, χa = χ− =
, Eb = E+ =
, Ea = E− = −
.
0
1
2m
2m
(a)
01
1
eBx ~
eBx ~
0
01
0 1
=
= 0;
10
1
0
m 2
2m
01
0 1
eBx ~
eBx ~
eBx ~
1
0
10
1 0
hχb |H 0 |χb i =
;
= 0; hχb |H 0 |χa i =
=
10
1 0
0
1
2m
2m
2m
01
0
eBx ~
eBx ~
eBx ~
eBx ~
1
0
01
0 1
hχa |H |χb i =
=
=
. So h =
,
1
0
0
1
2m
2m
2m
2m
and the conditions of Problem 8.22 are met.
hχa |H 0 |χa i =
(b) From Problem 8.22(b),
h2
eBz ~ (eBx ~/2m)2
e~
Egs ≈ Ea −
=−
−
= −
(Eb − Ea )
2m
(eBz ~/m)
2m
Bx2
Bz +
.
2Bz
h
i
p
(c) From Problem 8.22(c), Egs = 12 Ea + Eb − (Eb − Ea )2 + 4h2 (it’s actually the exact ground state).
s
2
2
1
eBz ~
eBx ~
e~ p 2
Egs = −
+4
= −
Bz + Bx2
2
m
2m
2m
(which was obvious from the start, since the square root is simply the magnitude of the total field).
CHAPTER 8. THE VARIATIONAL PRINCIPLE
321
Problem 8.24
(a)
1
1
1
r1 = √ (u + v); r2 = √ (u − v); r12 + r22 = (u2 + 2u · v + v 2 + u2 − 2u · v + v 2 ) = u2 + v 2 .
2
2
2
2
∂2f
∂2f
∂2f
∂2f
∂2f
∂ f
+
+
+
+
+
.
(∇21 + ∇22 )f (r1 , r2 ) =
∂x21
∂y12
∂z12
∂x22
∂y22
∂z22
∂f
∂f ∂ux
∂f ∂vx
1
∂f
∂f
∂f
∂f ∂ux
∂f ∂vx
1
∂f
∂f
=
+
=√
+
;
=
+
=√
−
.
∂x1
∂ux ∂x1 ∂vx ∂x1
∂vx
∂x2
∂ux ∂x2 ∂vx ∂x2
∂vx
2 ∂ux
2 ∂ux
2
1 ∂
∂f
∂f
1
∂ f ∂ux
∂ 2 f ∂vx
∂ 2 f ∂ux
∂ 2 f ∂vx
∂2f
=√
+
=√
+
+
+ 2
∂x21
∂vx
∂ux ∂vx ∂x1
∂vx ∂ux ∂x1
∂vx ∂x1
2 ∂x1 ∂ux
2 ∂u2x ∂x1
2
1 ∂ f
∂2f
∂2f
=
+2
+ 2 ;
2
2 ∂ux
∂ux ∂vx
∂vx
2
∂2f
1 ∂
∂ f ∂ux
∂f
∂f
1
∂ 2 f ∂vx
∂ 2 f ∂ux
∂ 2 f ∂vx
=√
−
=√
+
−
− 2
∂x22
∂vx
∂ux ∂vx ∂x2
∂vx ∂ux ∂x2
∂vx ∂x2
2 ∂x2 ∂ux
2 ∂u2x ∂x2
2
∂2f
1 ∂ f
∂2f
−
2
=
+
.
2 ∂u2x
∂ux ∂vx
∂vx2
2
2
∂ f
∂2f
∂ f
∂2f
So
+
=
+
, and likewise for y and z: ∇21 + ∇22 = ∇2u + ∇2v .
∂x21
∂x22
∂u2x
∂vx2
1
λ
~2
(∇2u + ∇2v ) + mω 2 (u2 + v 2 ) − mω 2 2v 2
H = −
2m
2
4
2
2
1
~
1
1
~
∇2u + mω 2 u2 + −
∇2v + mω 2 v 2 − λmω 2 v 2 . QED
= −
2m
2
2m
2
2
√
√
(b) The energy is 23 ~ω (for the u part) and 23 ~ω 1 − λ (for the v part): Egs = 23 ~ω 1 + 1 − λ .
(c) The ground state for a one-dimensional oscillator is
ψ0 (x) =
mω 1/4
π~
2
e−mωx /2~
(Eq. 2.60).
So, for a 3-D oscillator, the ground state is ψ0 (r) =
ψ(r1 , r2 ) =
mω 3/2
π~
mω
2
mω 3/4 −mωr 2 /2~
e
, and for two particles
π~
2
e− 2~ (r1 +r2 ) . (This is the analog to Eq. 8.18.)
3
3
~ω + ~ω + hVee i = 3~ω + hVee i (the analog to Eq. 8.20).
2
2
mω 3 Z
2
2
mω
λ
hVee i = − mω 2
e− ~ (r1 +r2 ) (r1 − r2 )2 d3 r1 d3 r2 (the analog to Eq. 8.21).
| {z }
4
π~
hHi =
r12 −2r1 ·r2 +r22
322
CHAPTER 8. THE VARIATIONAL PRINCIPLE
The r1 · r2 term integrates to zero, by symmetry, and the r22 term is the same as the r12 term, so
3 Z
2
2
mω
λ
2 mω
hVee i = − mω
2 e− ~ (r1 +r2 ) r12 d3 r1 d3 r2
4
π~
Z ∞
Z ∞
mω 3
2
2
λ
e−mωr2 /~ r22 dr2
e−mωr1 /~ r14 dr1
(4π)2
= − mω 2
2
π~
0
0
"
#" #
r
2 r
π~
π~
3
8m4 ω 5 1 ~
3
~
= − λ~ω.
= −λ
3
π~
4 mω mω
8 mω
mω
4
3
λ
hHi = 3~ω − λ~ω = 3~ω 1 −
.
4
4
The variational principle says this must exceed the exact ground-state energy (b); let’s check it:
√
√
λ
λ √
λ2
λ
3 3~ω 1 −
> ~ω 1 + 1 − λ ⇔ 2 − > 1 + 1 − λ ⇔ 1 − > 1 − λ ⇔ 1 − λ +
> 1 − λ.
4
2
2
2
4
It checks. In fact, expanding the exact answer in powers of λ, Egs ≈ 32 ~ω(1 + 1 − 12 λ) = 3~ω 1 − λ4 ,
we recover the variational result.
Problem 8.25
Z
2 3
3
|ψ| d r1 d r2 = |A|
1=
Z
2
ψ12 d3 r1
Z
ψ22 d3 r2 + 2
Z
3
ψ1 ψ2 d r 1
Z
3
ψ1 ψ2 d r2 +
Z
ψ22 d3 r1
Z
ψ12 d3 r2
= |A|2 (1 + 2S 2 + 1),
where
p
Z
3
S≡
ψ1 (r)ψ2 (r) d r =
A2 =
h
1
6
2 1 + (y/x)
(Z1 Z2 )3
πa3
Z
e
−(Z1 +Z2 )r/a
4 y 3
2a3
y 3
4πr dr = 3
.
=
a 2
(Z1 + Z2 )3
x
2
i.
~2
e2
1
1
e2
1
(∇21 + ∇22 ) −
+
+
,
2m
4π0 r1
r2
4π0 |r1 − r2 |
~2
e2
Z1
Z2
Hψ = A −
(∇21 + ∇22 ) −
+
ψ1 (r1 )ψ2 (r2 )
2m
4π0 r1
r2
~2
e2
Z1
Z2
+ −
(∇21 + ∇22 ) −
+
ψ2 (r1 )ψ1 (r2 )
2m
4π0 r1
r2
2
e
Z1 − 1 Z2 − 1
Z2 − 1 Z1 − 1
+A
+
ψ1 (r1 )ψ2 (r2 ) +
+
ψ2 (r1 )ψ1 (r2 ) + Vee ψ,
4π0
r1
r2
r1
r2
H=−
where Vee ≡
e2
1
.
4π0 |r1 − r2 |
CHAPTER 8. THE VARIATIONAL PRINCIPLE
323
The term in first curly brackets is (Z12 + Z22 )E1 ψ1 (r1 )ψ2 (r2 ) + (Z22 + Z12 )E1 ψ2 (r1 )ψ1 (r2 ), so
Hψ = (Z12 + Z22 )E1 ψ
e2
Z1 − 1 Z2 − 1
Z2 − 1 Z1 − 1
+A
+
ψ1 (r1 )ψ2 (r2 ) +
+
ψ2 (r1 )ψ1 (r2 ) + Vee ψ
4π0
r1
r2
r1
r2
2 e
hHi = (Z12 + Z22 )E1 + hVee i + A2
4π0
Z1 − 1 Z2 − 1
Z2 − 1 Z1 − 1
ψ1 (r1 )ψ2 (r2 )i +
ψ2 (r1 )ψ1 (r2 )i
.
× hψ1 (r1 )ψ2 (r2 ) + ψ2 (r1 )ψ1 (r2 )
+
+
r1
r2
r1
r2
1
1
ψ1 (r1 )i + (Z2 − 1)hψ2 (r2 )
ψ2 (r2 )i
r1
r2
1
+ (Z2 − 1)hψ1 (r1 )
ψ2 (r1 )ihψ2 (r2 )|ψ1 (r2 )i
r1
1
1
+ (Z1 − 1)hψ1 (r1 )|ψ2 (r1 )ihψ2 (r2 )
ψ1 (r2 )i + (Z1 − 1)hψ2 (r1 )
ψ1 (r1 )ihψ1 (r2 )|ψ2 (r2 )i
r2
r1
1
1
+ (Z2 − 1)hψ2 (r1 )|ψ1 (r1 )ihψ1 (r2 )
ψ2 (r2 )i + (Z2 − 1)hψ2 (r1 )
ψ2 (r1 )i
r2
r1
1
ψ1 (r2 )i
+ (Z1 − 1)hψ1 (r2 )
r2
1
1
1
1
= 2(Z1 − 1)
+ 2(Z2 − 1)
+ 2(Z1 − 1)hψ1 |ψ2 ihψ1
ψ2 i + 2(Z2 − 1)hψ1 |ψ2 ihψ1
ψ2 i.
r 1
r 2
r
r
1
1
Z1
1
Z2
But
= hψ1 (r)
ψ1 (r)i =
;
=
, so hHi = (Z12 + Z22 )E1
r 1
r
a
r 2
a
2 1
1
1
e
2
ψ2 i + hVee i.
2 (Z1 − 1)Z1 + (Z2 − 1)Z2 + (Z1 + Z2 − 2)hψ1 |ψ2 ihψ1
+A
4π0
a
a
r
= (Z1 − 1)hψ1 (r1 )
3
hψ1 |ψ2 i = S = (y/x) , so
p
2
Z
(Z1 Z2 )3
1
a
y3
y3
−(Z1 +Z2 )r/a
hψ1
r
dr
=
ψ2 i =
4π
e
=
.
r
πa3
2a3 Z1 + Z2
2ax2
And
2 y 3 y 3 1
e
2 2
hHi = (x2 − y 2 )E1 + A2
Z1 + Z22 − (Z1 + Z2 ) + (x − 2)
+ hVee i
2
4π0 a
x 2x2
1
1
y6
1
= (x2 − y 2 )E1 − 4E1 A2 x2 − y 2 − x + (x − 2) 5 + hVee i.
2
2
2
x
e2
1
hψ
ψi
4π0
|r1 − r2 |
2 e
1
=
A2 hψ1 (r1 )ψ2 (r2 ) + ψ2 (r1 )ψ1 (r2 )
ψ1 (r1 )ψ2 (r2 ) + ψ2 (r1 )ψ1 (r2 )i
4π0
|r1 − r2 |
2 e
1
1
2
A 2hψ1 (r1 )ψ2 (r2 )
ψ1 (r1 )ψ2 (r2 )i + 2hψ1 (r1 )ψ2 (r2 )
ψ2 (r1 )ψ1 (r2 )i
=
4π0
|r1 − r2 |
|r1 − r2 |
2 e
A2 (B + C), where
= 2
4π0
hVee i =
324
CHAPTER 8. THE VARIATIONAL PRINCIPLE
B ≡ hψ1 (r1 )ψ2 (r2 )
1
ψ1 (r1 )ψ2 (r2 )i;
|r1 − r2 |
C ≡ hψ1 (r1 )ψ2 (r2 )
1
ψ2 (r1 )ψ1 (r2 )i.
|r1 − r2 |
Z
1
Z13 Z23
e−2Z1 r1 /a e−2Z2 r2 /a
d3 r1 d3 r2 . As on page 334, the r2 integral is
(πa3 )2
|r1 − r2 |
Z
1
d3 r2
e−2Z2 r2 /a p 2
2
r1 + r2 − 2r1 r2 cos θ2
πa3
Z2 r1
2
−2Z2 r1 /a
= 3
1− 1+
e
(Eq. 8.25, but with a →
a).
Z2 r1
a
Z2
Z ∞
Z13 Z23 (πa3 )
Z2 r1
−2Z1 r1 /a 1
−2Z2 r1 /a
4π
e
B=
1− 1+
e
r12 dr1
(πa3 )2 Z23
r1
a
0
Z Z2 2 −2(Z1 +Z2 )r1 /a
4Z 3 ∞
r1 e−2Z1 r1 /a − r1 e−2(Z1 +Z2 )r1 /a −
dr1
= 31
r1 e
a
a
0
"
2
3 #
2 Z13
4Z13
Z2
a
1
Z2
a
a
1
=
= 3
−
2
−
−
−
a
2Z1
2(Z1 + Z2 )
a
2(Z1 + Z2 )
a Z12
(Z1 + Z2 )2
(Z1 + Z2 )3
Z1 Z2
y2
Z1 Z2
y2
1+
1
+
=
=
.
a(Z1 + Z2 )
(Z1 + Z2 )2
4ax
4x2
Z
Z 3Z 3
1
e−Z1 r1 /a e−Z2 r2 /a e−Z2 r1 /a e−Z1 r2 /a
d3 r1 d3 r2
C = 1 3 22
(πa )
|r1 − r2 |
Z
1
(Z1 Z2 )3
e−(Z1 +Z2 )(r1 +r2 )/a
d3 r1 d3 r2 .
=
(πa3 )2
|r1 − r2 |
B=
4
a. Comparing Eqs. 8.21 and 8.26, we see that the
The integral is the same as in Eq. 8.21, only with a → Z1 +Z
2
integral itself was
πa3
8
2
5 2 5
(Z1 Z2 )3 5π 2
45 a5
20 (Z1 Z2 )3
5 y6
π a . So C =
=
=
.
3
2
5
5
256
(πa ) 256 (Z1 + Z2 )
a (Z1 + Z2 )
16a x5
2 2 e
y2
y2
5y 4
y
5 y6
y2
2
2
hVee i = 2
1+ 2 +
1+ 2 + 4 .
A
= 2A (−2E1 )
4π0
4ax
4x
16a x5
4x
4x
4x
6
2
1 2
1
y
y2
y2
5y 4
2
1 2
2
2
x − y − x + (x − 2) 5 −
1+ 2 + 4
hHi = E1 x − y −
2
[1 + (y/x)6 ]
2
2
x
[1 + (y/x)6 ] 4x
4x
4x
1
1
1 y6
y4
E1
y6
y2
5y 6
(x2 − y 2 )(x6 + y 6 ) − 2x6 x2 − y 2 − x +
+
= 6
− 5+
+
6
4
3
(x + y )
2
2
2x
x
4x 16x
16x5
E1
1 6 2 1 8
1 5 2 1 3 4 5 6
8
2 6
8
6 2
7
2 6
6
= 6
x
+
x
y
−
x
y
−
y
−
2x
+
x
y
+
2x
−
x
y
+
2xy
−
x
y
−
x
y
−
xy
(x + y 6 )
2
2
2
8
8
E1
1
1
1
11
1
=
−x8 + 2x7 + x6 y 2 − x5 y 2 − x3 y 4 + xy 6 − y 8 .
(x6 + y 6 )
2
2
8
8
2
5
4a
=
Mathematica finds the minimum of hHi at x = 1.32245, y = 1.08505, corresponding to Z1 = 1.0392, Z2 =
0.2832. At this point, hHimin = 1.0266E1 = −13.962 eV, which is less than −13.6 eV—but not by much!
CHAPTER 8. THE VARIATIONAL PRINCIPLE
325
Problem 8.26
The calculation is the same as before, but with me → mr , the reduced mass of the muon:
mµ md
mµ 2mp
mµ
=
=
.
mµ + md
mµ + 2mp
1 + mµ /2mp
mr =
From Problem 7.32, mµ = 207me , so
1+
mµ
207 (9.11 × 10−31 )
=1+
= 1.056;
2mp
2 (1.67 × 10−27 )
mr =
207me
= 196me .
1.056
This shrinks the muonic “Bohr radius” down by a factor of nearly 200. Equation 8.50 is still valid, but now E1
is the ground state energy of the muonic “atom.” The potential energy associated with the deuteron-deuteron
repulsion is the same as Vpp (Eq. 8.51), and since the product aE1 is independent of mass, it doesn’t matter
whether we write it using the electron values or the muon values. Thus Eq. 8.52 still holds, and the entire
molecule shrinks by that same factor of 196. The equilibrium separation for the electron case was 2.493a
(Problem 8.11), so for muons
R=
2.493
(0.529 × 10−10 m) = 6.73 × 10−13 m.
196
Problem 8.27
(a)
~2 ∂ 2 ψ ∂ 2 ψ
−
+
= Eψ. Let ψ(x, y) = X(x)Y (y).
2m ∂x2
∂y 2
d2 Y
2mE
1 d2 X
1 d2 Y
2mE
d2 X
+
X
=
−
XY
;
+
=− 2 .
2
2
2
2
2
dx
dy
~
X dx
Y dy
~
2
d2 X
d
Y
2mE
= −kx2 X;
= −ky2 Y, with kx2 + ky2 =
. The general solution to the y equation is
dx2
dy 2
~2
nπ
π
Y (y) = A cos ky y + B sin ky y; the boundary conditions Y (±a) = 0 yield ky =
with minimum
.
2a
2a
[Note that ky2 has to be positive, or you cannot meet the boundary conditions at all.] So
~2
π2
2
E ≥
kx + 2 . For a traveling wave kx2 has to be positive. Conclusion: Any solution with E <
2m
4a
Y
π 2 ~2
will be a bound state.
8ma2
(b) Here’s a graph of ψ(x, y):
��������
a = 0.9
��������
0.9
��������
g[x_, y_] :=
Piecewise
��������
��������
Cos π x
2 + Cos π y
2
ⅇ ^ (- a), Abs[x] < 1 && Abs[y] < 1 ,
Cos π x
2
ⅇ ^ - a Abs[y] , Abs[x] < 1 && Abs[y] > 1 ,
Cos π y
2
ⅇ ^ - a Abs[x] , Abs[x] > 1 && Abs[y] < 1
Plot3D[g[x, y], {x, - 4, 4}, {y, - 4, 4}, PlotRange -> {0, .85}]
��������
a = 0.9
��������
0.9
��������
g[x_, y_] :=
Piecewise
326
��������
��������
��������
��������
Cos π x
2
ⅇ ^ (- a), Abs[x] < 1 && Abs[y] < 1 ,
2
ⅇ ^ - a Abs[y] , Abs[x] < 1 && Abs[y] > 1 ,
Cos π y
2
ⅇ ^ - a Abs[x] , Abs[x] > 1 && Abs[y] < 1
a = 0.9
CHAPTER 8. THE VARIATIONAL PRINCIPLE
Plot3D[g[x,
y], {x, - 4, 4}, {y, - 4, 4}, PlotRange -> {0, .85}]
0.9
g[x_, y_] :=
Piecewise
��������
2 + Cos π y
Cos π x
Cos π x
2 + Cos π y
2
ⅇ ^ (- a), Abs[x] < 1 && Abs[y] < 1 ,
Cos π x
2
ⅇ ^ - a Abs[y] , Abs[x] < 1 && Abs[y] > 1 ,
Cos π y
2
ⅇ ^ - a Abs[x] , Abs[x] > 1 && Abs[y] < 1
Plot3D[g[x, y], {x, - 4, 4}, {y, - 4, 4}, PlotRange -> {0, .85}]
��������
��������
��������
j[x_] := g[x, .5]
It has roof-lines along the edges of the central square, as you can see by plotting g[x, 0.5] across the kink:
��������
��������
Plot[j[x], {x, .8, 1.2}]
j[x_] := g[x, .5]
0.40
��������
Plot[j[x], {x, .8, 1.2}]
0.40
0.35
��������
0.35
�������� 0.30
0.30
0.25
0.9
0.25
1.0
0.9
1.0
1.1
1.1
1.2
1.2
y
a
II
I
a
x
To normalize, we integrate |ψ(x, y)|2 over regions I and II (in the figure), and multiply by 8.
Z ∞ Z a
πy e−2αx/a dx dy
2a
x=a y=0
(
)
∞) (
πy πy 1 a
a
1
= A2 − e−2αx/a
sin
cos
+ y
2α
2(π/2a)
2a
2a
2
a
0
n a
o
2
1
2
−2α
2a
−2α
=A
e
0+ a =A
e
.
2α
2
4α
III = A2
cos2
CHAPTER 8. THE VARIATIONAL PRINCIPLE
327
Z a Z a h
πx πy i2
cos
+ cos
e−2α dx dy
2a
2a
x=0 y=0
Z a
Z a
Z a
Z a
πx πy 1 2 −2α
2 πx
= A e
dx
dy + 2
cos
dx
cos
dy
2
cos
2
2a
2a
2a
0
0
0
0


#
"
 a πx a 2 
a2
2a
8
2 −2α
=A e
= A2 e−2α 1 + 2 .
a+
sin
 2

π
2a
2
π
II =
1 2
A
2
0
Normalizing:
1 = 8(II + III ) = 8A2 a2 e−2α
1
1
+
4α 2
so
A2 =
1+
8
π2
8
1
= 4A2 a2 e−2α 1 + 2 +
,
π
2α
e2α
1 .
4a2 1 + π82 + 2α
2
~
Next, ignoring the roof-lines for the moment, we calculate hHia = −8 2m
(JI + JII ), where
Z ∞ Z a h
πy πy i ∂2
i
∂2 h
cos
+
cos
e−αx/a
e−αx/a dx dy
2
2
2a
∂x
∂y
2a
x=a y=0
2
2
2
π
α
1
π
α2 −
e−2α .
−
III = A2
=
a
2a
4α
4
JII = A2
2
Z a Z a h
πx πy i
πy i
πx ∂
∂2 h
cos
+ cos
e−α
+
+
cos
e−α dx dy
cos
2a
2a
∂x2
∂y 2
2a
2a
x=0 y=0
π2
π2
= − 2 II = −A2 1 +
e−2α .
4a
8
JI =
1 2
A
2
So far, then,
~2
π2
1
π2
−A2 1 +
e−2α + A2
α2 −
e−2α
2m
8
4α
4
2
2
~
π
=
A2 e−2α 8 + π 2 − 2α +
.
2m
2α
hHia = −8
2
~
Now the roof-lines, along the four sides of the central square: hHib = −8 2m
JIII , where
Z a+ Z a
∂2
JIII =
ψ(x, y)
ψ(x, y) dx dy
∂x2
x=a− y=0
(
)
Z a
πy i
h
πx i
∂ h
∂
=A
ψ(a, y)
cos
e−αx/a
−
cos
e−α
dy
∂x
2a
∂x
2a
y=0
x=a
x=a
Z a h
πy in α
πy π
o
π
= A2
cos
e−α − cos
e−α +
sin
e−α dy
2a
a
2a
2a
2
y=0
Z ah
i
π
πy
πy
α
1
cos
− α cos2
dy = A2 e−2α 1 −
,
= A2 e−2α
a
2
2a
2a
2
0
328
CHAPTER 8. THE VARIATIONAL PRINCIPLE
so
hHib = −8
~2 2 −2α ~2 2 −2α
α
=
A e
1−
A e
4(α − 2).
2m
2
2m
Putting it all together,
~2 2 −2α
π2
~2 2 −2α
2
A e
+
A e
8 + π − 2α +
(4α − 8)
hHi = hHia + hHib =
2m
2α
2m
"
#
π2
π 2 + 2α
+ 2α
π2
~2
~2 2 −2α
A e
π2 +
+ 2α =
=
1
2m
2α
2m 4a2 1 + π82 + 2α
~2 π 2
1 − (α/4)
=
−
.
ma2 8
1 + (8/π 2 ) + (1/2α)
Minimizing:
dhHi
=0
dα
⇒
d
dα
4−α
1 + (8/π 2 ) + (1/2α)
= 0;
4−α
−1
−
1 + (8/π 2 ) + (1/2α) (1 + (8/π 2 ) + (1/2α))2
1 + (8/π 2 ) + (1/2α) −
α=
−1 ±
4−α
= 0;
2α2
−1
2α2
= 0;
8
α2 1 + 2 + α − 2 = 0;
π
p
p
−1 + 9 + (8/π)2
1 + 8[1 + (8/π 2 )]
=
= 0.81053
2[1 + (8/π 2 )]
2[1 + (8/π 2 )]
(note that α must be positive). Putting this value of α into the expression for hHi, we get
hHimin = 0.905221
π 2 ~2
~2
=
1.2337
, so E0 is definitely less than Ethreshold .
8 ma2
ma2
But Ethreshold =
Problem 8.28
(a) To normalize the variational state
Z
2
1 = |ψβ (r)| d3 r
Z ∞
= A2 4 π
e−2 β r/r0 r2 dr
0
2
= A 4π
= A2
~2
.
ma2
π r03
β3
r0
2β
3 Z ∞
0
e−z z 2 dz
CHAPTER 8. THE VARIATIONAL PRINCIPLE
and
s
A=
β3
.
π r03
(b) Now we calculate the expectation value of the Hamiltonian:
Z Z
~2 1
e−r/r0
2 ∂ψβ
hHi =
ψβ∗ (r) −
r
−
r
V
ψ
(r)
r2 dr dΩ
0 0
β
2 µ r2
∂r
r
Z ∞
2
~ β 2 r0
−β r/r0
e−β r/r0
−
β
e
= 4 π A2
r2 dr
2
2
µ
r
r
0
0
Z
−(2 β+1) r/r0
= − r0 V0
r dr .
e
In the first integral we let z = 2 β r/r0 and in the second integral we let z = (2 β + 1) r/r0 to get
"
#
Z ∞
Z ∞
1 2
r03 V0
4 β 3 ~2 r0
−z
−z
z− z
e dz −
z e dz
hHi = 3
2
r0
4µβ 0
4
(2 β + 1) 0
#
"
1
r03 V0
4 β 3 ~2 r0
1−
−
= 3
2
r0
4µβ
2
(2 β + 1)
=
~2 β 2
4 β3
~2 γ
−
2
2
2 µ r0 (2 β + 1) 2 µ r02
which can be put in the desired form:
"
#
~2
4β γ
2
E(β) =
β 1−
.
2
2 µ r02
(2 β + 1)
(c) ] Finding the value of β that minimizes the energy:
dE(β)
=0
dβ
#
"
~2
12 β 2 γ
4 β3 γ
2β −
+2
2
3 2 =0
2
2 µ r0
(2 β + 1)
(2 β + 1)
3
(2 β + 1) − 6 β (2 β + 1) γ + 8 β 2 γ = 0
3
(2 β + 1) − 2 β (3 + 2 β) γ = 0 .
Solving this for γ,
3
γ=
(2 β + 1)
,
2 β (3 + 2 β)
and plugging back into our expression for the energy we have
"
#
3
~2
(2 β + 1)
4β
2
Emin =
β 1−
2
2 µ r02
2β
(2 β + 1)
"
#
3
4
β
(2
β
+
1)
~2
β2 1 −
=
2
2 µ r02
(2 β + 1) 2 β (3 + 2 β)
329
330
CHAPTER 8. THE VARIATIONAL PRINCIPLE
which simplifies to
Emin =
~2 β 2 1 − 2 β
.
2 µ r02 3 + 2 β
(d) From our plot we see that Emin < 0 for β > 1/2. We can turn this into a value for γ; there will be a
negative-energy (or bound) state when
3
γ>
(2 · 1/2 + 1)
=2
2 · 1/2 (3 + 2 · 1/2)
which in turn means
2 µ0 r02
V0 > 2
~2
or
2
m π c2
~2
= 42 MeV .
V0 > 2 = 2
µr0
mp c2
So we know there is a bound state for V0 > 42 MeV but a better variational wave function could show a
bound state for a smaller value of V0 .
2 μ r0 2
ℏ2
Emin
0.1
0.2
0.3
0.4
0.5
0.6
0.7
β
-0.02
-0.04
-0.06
Problem 8.29
(a) Let ψ1 be the true ground state (with energy E1 ) of the Hamiltonian with potential V1 and use it as the
variational wave function for the Hamiltonian with potential V2 . The true ground state energy of this
CHAPTER 8. THE VARIATIONAL PRINCIPLE
331
second Hamiltonian then satisfies
Z ∞
~2
∗
E2 ≤
ψ1 (x) −
+ V2 (x) ψ1 (x) dx
2m
−∞
Z ∞
~2
ψ1∗ (x) −
=
+ V1 (x) ψ1 (x) dx
2m
−∞
Z ∞
=+
ψ1∗ (x) [V2 (x) − V1 (x)] ψ1 (x) dx
−∞
Z ∞
2
|ψ1 (x)| [V2 (x) − V1 (x)] dx .
= E1 +
−∞
Now since V2 (x) < V1 (x) for all x we have
Z ∞
2
E2 ≤ E1 −
|ψ1 (x)| |V2 (x) − V1 (x)| dx ≤ E1
−∞
since the remaining integral is clearly non-negative in this expression. That means that if the ground state
of E1 is negative (meaning it has a bound state), then so is the ground state of E2 , and it has a bound
state as well.
(b) A finite potential well of arbitrarily small width a and arbitrarily small depth V0 has a bound state, as
shown in Section 2.6. Now for any potential, one can place a finite square well inside it as long as you
make it narrow and shallow enough, and since that finite well has a bound state, so does the potential of
interest by part (a).
(c) The theorem of course generalizes to any dimension. Just replace dx by dd r, nothing else changes. The
corollary though only extends to two dimensions because a finite well of arbitrarily small depth and width
binds a state in two dimensions, but not in three.
Problem 8.30
(a)
P
P
c∗m Hmn cn
∂E
n Hjn cn
nm
P
P
=
−
cj = 0,
∗
2
∂cj
( n |cn |2 )2
n |cn |
or, multiplying by
2
n |cn | ,
P
X
Hjn cn − E cj = 0. X
n
(b)
∂E
=
∂cj
P ∗
P
cm Hmj
c∗m Hmn cn ∗
m
nm
P
P
−
cj = 0,
2
( n |cn |2 )2
n |cn |
Taking the complex conjugate of both sides:
X
⇒
X
c∗m Hmj = E c∗j .
m
∗
c∗m Hmj
= E ∗ cj .
m
∗
But H is Hermitian, so Hmj
= Hjm , and E ∗ = E, and hence
X
m
Hjm cm − E cj = 0. X
332
CHAPTER
Sloped infinite well (Problem
8.30) 8. THE VARIATIONAL PRINCIPLE
Variational
(c) The lowest bound
on thecalculation
ground state energy is 39.982 ~2 /ma2 . See Mathematica output below.
�������
a = ℏ = m = 1;
�������
V[x_] := 100 x;
�������
Psi[n_, x_] := Sqrt 2
�������
f[n_, p_] = FullSimplify[Integrate[Psi[n, x] V[x] Psi[p, x], {x, 0, a}]];
�������
fD[n_] = FullSimplify[Integrate[Psi[n, x] V[x] Psi[n, x], {x, 0, a}]];
�������
Num = 10;
�������
H = 1. Table If p ⩵ n,
�������
E0 = Last[Eigenvalues[H]]
a Sin[n Pi x / a];
n ^ 2 π2
2 m ℏ2
+ fD[n], f[n, p] , {n, 1, Num}, {p, 1, Num}
�������
39.982
��������
c = Normalize[Last[Eigenvectors[H]]];
��������
c = c
��������
ψ[x_] = Sum[c[[i]] Psi[i, x], {i, 1, Num}];
��������
plot1 = Plot[ψ[x], {x, 0, 1}];
��������
Show[plot1]
Sign[c[[1]]];
1.5
��������
1.0
0.5
0.2
0.4
0.6
0.8
1.0
// Chop;
CHAPTER 9. THE WKB APPROXIMATION
333
Chapter 9
The WKB Approximation
Problem 9.1
Z a
with n = 1, 2, 3, . . . and
p(x) =
p
2m[E − V (x)]
√
a
√
p(x) dx = nπ~,
(Eq. 9.17).
0
Z a
Here
p
E(E − V0 ) =
4
2m
0
⇒ E + E − V0 + 2
Square again:
p
a
+
2mE
p(x) dx =
2
2m(E − V0 )
nπ~
a
2
2
=
= 4En0 ;
2m
a √
2
E+
p
E − V0 = nπ~
p
2 E(E − V0 ) = (4En0 − 2E + V0 ).
2
4E(E − V0 ) = 4E 2 − 4EV0 = 16En0 + 4E 2 + V02 − 16EEn0 + 8En0 V0 − 4EV0
2
⇒ 16EEn0 = 16En0 + 8En0 V0 + V02 ⇒ En = En0 +
V0
V2
+ 0 0.
2
16En
V0
Perturbation theory gave En = En0 + ; the extra term goes to zero for very small V0 , or (since En0 ∼ n2 ), for
2
large n.
Problem 9.2
(a)
dψ
i
= f 0 eif /~ ;
dx
~
d2 ψ
i
=
dx2
~
i 0 2 if /~
i 00
1 0 2 if /~
00 if /~
f e
+ (f ) e
=
f − 2 (f ) e
.
~
~
~
p2
i 00
1 0 2 if /~
p2
d2 ψ
=
−
ψ
=⇒
f
−
(f
)
e
= − 2 eif /~ =⇒ i~f 00 − (f 0 )2 + p2 = 0. QED
2
2
2
dx
~
~
~
~
(b) f 0 = f00 + ~f10 + ~2 f20 + · · · =⇒ (f 0 )2 = (f00 + ~f10 + ~2 f20 + · · · )2 = (f00 )2 + 2~f00 f10 + ~2 [2f00 f20 + (f10 )2 ] + · · ·
f 00 = f000 + ~f100 + ~2 f200 + · · · .
~0 : (f00 )2 = p2 ;
i~(f000 + ~f100 + ~2 f200 ) − (f00 )2 − 2~f00 f10 − ~2 [2f00 f20 + (f10 )2 ] + p2 + · · · = 0.
~1 : if000 = 2f00 f10 ;
~2 : if100 = 2f00 f20 + (f10 )2 ;
...
334
CHAPTER 9. THE WKB APPROXIMATION
Z
df0
i ±p0
df1
i f000
i d
i
(c)
=
= ±p =⇒ f0 = ± p(x)dx + constant ;
=
=
ln p =⇒ f1 = ln p + const.
dx
dx
2 f00
2 ±p
2 dx
2
Z
Z
i
i
i
if
p dx p−1/2 eiK/~
= exp
± p(x) dx + ~ ln p + K
= exp ±
ψ = exp
~
~
2
~
Z
C
i
p dx . QED
= √ exp ±
p
~
Problem 9.3
γ=
1
~
Z
|p(x)| dx =
1
~
Z 2a p
2a p
2m(V0 − E) dx =
2m(V0 − E).
~
0
T ≈ e−4a
√
2m(V0 −E)/~
.
From Problem 2.33, the exact answer is
T =
1
V2
1 + 4E(V00−E) sinh2 γ
.
Now, the WKB approximation assumes the tunneling probability is small (p. 359)—which is to say that γ is
large. In this case, sinh γ = 12 (eγ − e−γ ) ≈ 12 eγ , and sinh2 γ ≈ 41 e2γ , and the exact result reduces to
16E(V0 − E)
1
T ≈
≈
e−2γ .
2
V02
V
2γ
0
1 + 16E(V0 −E) e
The coefficient in { } is of order 1; the dominant dependence on E is in the exponential factor. In this sense
T ≈ e−2γ (the WKB result).
Problem 9.4
I take the masses from Thornton and Rex, Modern Physics, Appendix 8. They are all atomic masses, but the
electron masses subtract out in the calculation of E. All masses are in atomic units (u): 1 u = 931 MeV/c2 .
The mass of He4 is 4.002602 u, and that of the α-particle is 3727 MeV/c2 .
U238 : Z = 92, A = 238, m = 238.050784 u → Th234 : m = 234.043593 u.
r1 = (1.07 × 10−15 m)(238)1/3 = 6.63 × 10−15 m.
E = (238.050784 − 234.043593 − 4.002602)(931) MeV = 4.27 MeV.
r
r
2E
(2)(4.27)
V =
=
× 3 × 108 m/s = 1.44 × 107 m/s.
m
3727
p
90
− 1.485 90(6.63) = 86.19 − 36.28 = 49.9.
γ = 1.980 √
4.27
τ=
(2)(6.63 × 10−15 ) 98.8
7.46 × 1021
21
e
s
=
7.46
×
10
s
=
yr = 2.4 × 1014 yrs.
1.44 × 107
3.15 × 107
Po212 : Z = 84, A = 212, m = 211.988842 u → Pb208 : m = 207.976627 u.
CHAPTER 9. THE WKB APPROXIMATION
335
r1 = (1.07 × 10−15 m)(212)1/3 = 6.38 × 10−15 m.
E = (211.988842 − 207.976627 − 4.002602)(931) MeV = 8.95 MeV.
r
r
2E
(2)(8.95)
=
× 3 × 108 m/s = 2.08 × 107 m/s.
V =
m
3727
p
82
− 1.485 82(6.38) = 54.37 − 33.97 = 20.4.
γ = 1.980 √
8.95
(2)(6.38 × 10−15 ) 40.8
e
s = 3.2 × 10−4 s.
2.08 × 107
These results are way off—but note the extraordinary sensitivity to nuclear masses: a tiny change in E produces
enormous changes in τ .
√
Much more impressive results are obtained when you plot the logarithm of lifetimes against 1/ E, as in
Figure 9.6. Thanks to David Rubin for pointing this out. Some experimental values are listed below (all energies
in MeV):
A
E
τ
238 4.198 4.468 × 109 yr
A
E
τ
236 4.494 2.342 × 107 yr
224 7.488 0.79 s
5
234 4.775 2.455 × 10 yr
Protactinium (Z = 91): 222 8.540 2.9 ms
Uranium (Z = 92):
232 5.320
68.9 yr
220 9.650 0.78 µs
230 5.888
20.8 day
218 9.614 0.12 ms
228 6.680
9.1 min
226 7.570
0.35 s
A
E
τ
A
E
τ
226 4.784 1600 yr
232 4.012 1.405 × 1010 yr
224 5.685 3.66 day
4
Thorium (Z = 90): 230 4.687 7.538 × 10 yr
Radium (Z = 88):
222 6.559 38 s
228 5.423
1.912 yr
220 7.455 18 ms
226 6.337
30.57 min
218 8.389 25.6 µs
τ=
Problem 9.5
From Equation 9.23, the tunneling probability is (approximately)
Z
p
1 a
−2γ
T =e
, where γ =
|p(x)| dx, and p(x) = 2m(E − V (x)) .
~ 0
An electron in the lower band, with energy E, has to tunnel through the barrier shown here:
V(x)
Eg
E+E g-eE 0 x
E
a
x
336
CHAPTER 9. THE WKB APPROXIMATION
|p(x)| =
q
q
q
2m[E − (E + Eg − eE0 x)] =
−2m(Eg − eE0 x) = 2m(Eg − eE0 x) .
Then (letting u ≡ eE0 x/Eg )
Z s
2mEg a
eE0 x
2m(Eg − eE0 x) dx =
1−
dx
~
Eg
0
0
p
Z
3/2 √
2mEg 1 √
Eg
2 Eg
2m
=
1 − u du =
,
eE0
~
3 eE0 ~
0
1
γ=
~
Z aq
p
and the probability of tunneling is
3/2 √
4 Eg
T = exp −
3 eE0
2m
~
!
.
Problem 9.6
(a) V (x) = mgx.
(b)
−
~2 d2 ψ
d2 ψ
2m2 g
+ mgxψ = Eψ =⇒
=
2
2
2m dx
dx
~2
x−
E
mg
. Let y ≡ x −
E
, and α ≡
mg
2m2 g
~2
1/3
.
E
d2 ψ
d2 ψ
3
=
α
yψ.
Let
z
≡
αy
=
α(x
−
),
so
= zψ. This is the Airy equation (Eq. 9.37), and
dy 2
mg
dz 2
the general solution is ψ = aAi(z) + bBi(z). However, Bi(z) blows up for large z, so b = 0 (to make ψ
h
i
E
normalizable). Hence ψ(x) = aAi α(x − mg
) .
Then
(c) Since V (x) = ∞ for x < 0, we require ψ(0) = 0; hence Ai [α(−E/mg)] = 0. Now, the zeros of Ai are
an (n = 1, 2, 3, . . .). Abramowitz and Stegun list a1 = −2.338, a2 = −4.088, a3 = −5.521, a4 = −6.787,
2 1/3
αEn
mg
~
etc. Here −
= an , or En = −
an = −mg
an , or En = −( 12 mg 2 ~2 )1/3 an . In this case
mg
α
2m2 g
1
1
2 2
2 2
−34
J·s)2 = 5.34 × 10−68 J3 ; ( 12 mg 2 ~2 )1/3 = 3.77 × 10−23 J.
2 mg ~ = 2 (0.1 kg)(9.8 m/s ) (1.055 × 10
E1 = 8.81 × 10−23 J,
E2 = 1.54 × 10−22 J,
E3 = 2.08 × 10−22 J,
E4 = 2.56 × 10−22 J.
(d)
dV
1
i = hmgxi = hV i, so hT i = hV i.
dx
2
3
2
2En
But hT i + hV i = hHi = En , so
hV i = En , or hV i = En . But hV i = mghxi, so hxi =
.
2
3
3mg
1/3 1/3
1
1
2 2
−31
2
−34 2
For the electron,
mg ~
=
(9.11 × 10 )(9.8) (1.055 × 10 )
= 7.87 × 10−33 J.
2
2
2hT i = hx
dV
i (Eq. 3.113);
dX
here
dV
= mg,
dx
E1 = 1.84 × 10−32 J = 1.15 × 10−13 eV.
hxi =
so
hx
2(1.84 × 10−32 )
= 1.37 × 10−3 = 1.37 mm.
3(9.11 × 10−31 )(9.8)
CHAPTER 9. THE WKB APPROXIMATION
337
Problem 9.7
(a)
V(x)
E
mgx
x2
Z x2
Eq. 9.48 =⇒
0
Z x2
p(x) dx =
x
p
1
p(x) dx = (n − )π~, where p(x) = 2m(E − mgx) and E = mgx2 =⇒ x2 = E/mg.
4
√
0
x2
√
2
E − mgx dx = 2m −
(E − mgx)3/2
3mg
0
0
r
h
i 2r 2 1
2 2 1
=−
E 3/2 .
(E − mgx2 )3/2 − E 3/2 =
3 mg
3 mg
2m
Z x2 p
1
1
√
(2E)3/2 = (n − )π~,
4
3 mg
or
En =
1 2 1/3
2
2 2
.
8 π mg ~ (n − 4 )
9
(b)
9 2 2 2
π mg ~
8
1/3
9 2
=
π (0.1)(9.8)2 (1.055 × 10−34 )2
8
1/3
= 1.0588 × 10−22 J.
2/3
3
= 8.74 × 10−23 J,
4
2/3
7
E2 = (1.0588 × 10−22 )
= 1.54 × 10−22 J,
4
2/3
11
−22
E3 = (1.0588 × 10 )
= 2.08 × 10−22 J,
4
2/3
15
E4 = (1.0588 × 10−22 )
= 2.56 × 10−22 J.
4
E1 = (1.0588 × 10−22 )
These are in very close agreement with the exact results (Problem 9.6(c)). In fact, they agree precisely
(to 3 significant digits), except for E1 (for which the exact result was 8.81 × 10−23 J).
(c) From Problem 9.6(d),
2En
hxi =
,
3mg
n=
so
2 (1.0588 × 10−22 )
1=
3
(0.1)(9.8)
1
+ (1.388 × 1022 )3/2 = 1.64 × 1033 .
4
1
n−
4
2/3
2/3
1
, or n −
= 1.388 × 1022 .
4
338
CHAPTER 9. THE WKB APPROXIMATION
Problem 9.8
Z x2
p(x) dx =
x1
1
n−
2
s
π~;
1
2
2
2m E − mω x ;
2
p(x) =
1
x2 = −x1 =
ω
r
2E
.
m
q
x2
Z x2 q
Z x2 r
1
2E
2 − x2 dx = mω x x2 − x2 + x2 sin−1 (x/x )
2 dx = 2mω
n−
−
x
x
π~ = mω
2
2
2
2
2
mω 2
0
−x2
0
= mωx22 sin−1 (1) =
π
πE
2E
π
=
mωx22 = mω
.
2
2
2
mω
ω
En = n − 12 ~ω (n = 1, 2, 3, . . .)
Since the WKB numbering starts with n = 1, whereas for oscillator states we traditionally start with n = 0,
letting n → n + 1 converts this to the usual formula En = (n + 21 )~ω. In this case the WKB approximation
yields the exact results.
Problem 9.9
(a)
1
mω 2 x22 = En =
2
1
n+
~ω
2
r
(counting n = 0, 1, 2, . . .);
x2 =
(2n + 1)~
.
mω
(b)
Vlin (x) =
1
1
mω 2 x22 + (mω 2 x2 )(x − x2 ) =⇒ Vlin (x2 + d) = mω 2 x22 + mω 2 x2 d.
2
2
1
mω 2 (x2 + d)2 − 21 mω 2 x22 − mω 2 x2 d
V (x2 + d) − Vlin (x2 + d)
= 2
1
2 2
V (x2 )
2 mω x2
=
x22 + 2x2 d + d2 − x22 − 2x2 d
=
x22
d
x2
2
= 0.01.
d = 0.1 x2 .
(c)
2m
α=
mω 2 x2
~2
1/3
(Eq. 9.35),
2m2 ω 2 (2n + 1)2 ~2
≥ (50)3 ;
~2
m2 ω 2
nmin = 125.
or
so
0.1 x2
(2n + 1)2 ≥
2m2 ω 2
x2
~2
1/3
2m2 ω 2 4
≥ 5 =⇒
x2
~2
(50)3
= 62500;
2
2n + 1 ≥ 250;
1/3
n≥
≥ 50.
249
= 124.5.
2
However, as we saw in Problems 9.7 and 9.8, WKB may be valid at much smaller n.
CHAPTER 9. THE WKB APPROXIMATION
339
Problem 9.10
overlap 1
overlap 2
sP
s
s
WKB
WKB
E
Nonclassical
Classical
x
0
patching region
Shift origin to the turning point.

R0
0
0
1
1


De− ~ x |p(x )| dx
(x < 0)
p
|p(x)|
i
ψWKB =
Rx
0
0
i
1 h i R x p(x0 ) dx0


Be ~ 0
+ Ce− ~ 0 p(x ) dx (x > 0)
p
p(x)
Linearized potential in the patching region:
0
V (x) ≈ E+V (0)x.
0
Note : V (0) is negative.
2mV 0 (0)
d2 ψp
=
xψp = −α3 xψp , where α ≡
dx2
~2
2m|V 0 (0)|
~2
ψp (x) = aAi(−αx) + bBi(−αx). (Note change of sign, as compared with Eq. 9.38).
√
p
p
p
√
p(x) = 2m[E − E − V 0 (0)x] = −2mV 0 (0)x = 2m|V 0 (0)|x = α3 ~2 x = ~α3/2 x.
Overlap region 1 (x < 0):
Z 0
|p(x0 )| dx0 = ~α3/2
x
Z 0
√
−x0 dx0 = ~α3/2
x
ψWKB ≈
2
− (−x0 )3/2
3
0
=
x
2 3/2
2
~α (−x)3/2 = ~(−αx)3/2 .
3
3
3/2
2
1
De− 3 (−αx) . For large positive argument (−αx 1) :
~1/2 α3/4 (−x)1/4
3/2
3/2
2
2
1
1
ψp ≈ a √
e− 3 (−αx) + b √
e 3 (−αx) . Comparing ⇒ a = 2D
1/4
1/4
2 π(−αx)
π(−αx)
r
π
;
α~
b = 0.
Overlap region 2 (x > 0):
x
2 0 3/2
2
(x )
= ~(αx)3/2 .
3
3
0
0
0
h
i
3/2
3/2
2
2
1
ψWKB ≈ 1/2 3/4 1/4 Bei 3 (αx) + Ce−i 3 (αx)
. For large negative argument (−αx −1) :
~ α x
Z x
p(x0 ) dx0 = ~α3/2
Z x√
x0 dx0 = ~α3/2
1/3
.
340
CHAPTER 9. THE WKB APPROXIMATION
i
1
2
1 h iπ/4 i 2 (αx)3/2
π
a
−iπ/4 −i 23 (αx)3/2
3/2
3
√
sin
e
e
−
e
e
(remember : b = 0).
(αx)
+
=
3
4
π(αx)1/4
π(αx)1/4 2i
r
r
a α~ iπ/4
a α~ −iπ/4
Comparing the two: B =
e
, C=−
e
.
2i
π
2i
π
ψp (x) ≈ a √
Inserting the expression for a from overlap region 1 : B = −ieiπ/4 D; C = ie−iπ/4 D. For x > 0, then,
Z x
i
Rx
0
0
i
π
−iD h i R x p(x0 ) dx0 +i π
2D
1
π
4 − e− ~ 0 p(x ) dx −i 4
ψWKB = p
e~ 0
=p
sin
p(x0 ) dx0 +
.
~ 0
4
p(x)
p(x)
Finally, switching the origin back to x1 :


Rx
1
D
1 |p(x0 )| dx0
−~


x
p


,
(x
<
x
)
;
e
1


|p(x)| Z
ψWKB (x) =
x
2D
1
π



sin
p(x0 )dx0 +
, (x > x1 ). 

p
~
4
p(x)
x1
QED
Problem 9.11
At x1 , we have an upward-sloping turning point. Follow the method in the book. Shifting origin to x1 :

i
R0
0
0
i
1 h − i R 0 p(x0 ) dx0


Ae ~ x
+ Be ~ x p(x ) dx
(x < 0)
p
p(x)
h
i
ψWKB (x) =
R
R
x
x
0
0
0
0
1
1
1


Ce ~ 0 |p(x )| dx + De− ~ 0 |p(x )| dx (x > 0)
p
|p(x)|
In overlap region 2, Eq. 9.40 becomes ψWKB ≈
1
~1/2 α3/4 x1/4
h
2
Ce 3 (αx)
3/2
2
3/2
+ De− 3 (αx)
i
,
r
r
π
π
whereas Eq. 9.41 is unchanged. Comparing them =⇒ a = 2D
, b=C
.
α~
α~
h
i
1
−i 23 (−αx)3/2
i 23 (−αx)3/2
In overlap region 1, Eq. 9.44 becomes ψWKB ≈ 1/2 3/4
Ae
+
Be
,
~ α (−x)1/4
and Eq. 9.45 (with b 6= 0) generalizes to
a
2
π
b
2
π
3/2
3/2
ψp (x) ≈ √
sin (−αx) +
+√
cos (−αx) +
3
4
3
4
π(−αx)1/4
π(−αx)1/4
h
i
1
i 23 (−αx)3/2 iπ/4
−i 23 (−αx)3/2 −iπ/4
= √
e
+
(ia
+
b)e
e
. Comparing them =⇒
(−ia
+
b)e
2 π(−αx)1/4
r
r
~α ia + b −iπ/4
~α −ia + b iπ/4
A=
e
; B=
e
. Putting in the expressions above for a and b :
π
2
π
2
C
C
−iπ/4
A=
+ iD e
; B=
− iD eiπ/4 .
2
2
These are the connection formulas relating A, B, C, and D, at x1 .
CHAPTER 9. THE WKB APPROXIMATION
341
At x2 , we have a downward-sloping turning point, and follow the method of Problem 9.10. First rewrite the
middle expression in Eq. 9.53:
ψWKB = p
Let γ ≡
R x2
x1
1
|p(x)|
h
1
2 |p(x0 )| dx0 + 1
~
Rx
Ce ~ x1
Rx
x2
|p(x0 )| dx0
1
−~
+ De
2 |p(x0 )| dx0 − 1
~
x1
Rx
Rx
x2
|p(x0 )| dx0
i
.
|p(x)| dx/~, as before (Eq. 9.23), and let C 0 ≡ De−γ , D0 ≡ Ceγ . Then (shifting the origin to x2 ):

h
i
R0
R0
0
0
0
0
1
1
1


C 0 e ~ x |p(x )| dx + D0 e− ~ x |p(x )| dx , (x < 0);
p
|p(x)|
ψWKB =
R
 p 1 F e ~i 0x p(x0 ) dx0 ,

(x > 0).

p(x)
In the patching region ψp (x) = aAi(−αx) + bBi(−αx), where α ≡
Z 0
In overlap region 1 (x < 0):
|p(x0 )| dx0 =
x
2m|V 0 (0)|
~2
1/3
;
√
p(x) = ~α3/2 x.
2
~(−αx)3/2 , so
3

r
h
i 
1
π 0

0 23 (−αx)3/2
0 − 23 (−αx)3/2


ψWKB ≈ 1/2 3/4
Ce
+D e
a = 2

D
~ α (−x)1/4
~α
r
Comparing
=⇒
3/2
3/2
2
2
a
b
π 0



e− 3 (−αx) + √
e 3 (−αx) 
ψp ≈ √
b =
C
2 π(−αx)1/4
π(−αx)1/4
~α
Z x
3/2
2
2
1
In overlap region 2 (x > 0):
p(x0 ) dx0 = ~(αx)3/2 =⇒ ψWKB ≈ 1/2 3/4 1/4 F ei 3 (αx) .
3
~ α x
0
2
π
2
π
a
b
ψp ≈ √
sin (αx)3/2 +
+√
cos (αx)3/2 +
3
4
3
4
π(αx)1/4
π(αx)1/4
i
h
3/2
3/2
2
2
1
−i π
iπ
4 ei 3 (αx)
4 e−i 3 (αx)
+
(ia
+
b)e
. Comparing =⇒ (ia + b) = 0;
= √
(−ia
+
b)e
2 π(αx)1/4
r
r
r
r
~α −ia + b iπ/4
~α iπ/4
π −iπ/4
π −iπ/4
F =
e
=b
e
. b=
e
F; a = i
e
F.
π
2
π
~α
~α
r
r
~α
1 ~α
i
i
0
−iπ/4
0
C =
b=e
F, D =
a = e−iπ/4 F. D = eγ e−iπ/4 F ; C = e−γ e−iπ/4 F.
π
2
π
2
2
These are the connection formulas at x2 . Putting them into the equation for A:
−γ
C
i −γ −iπ/4
e
A=
+ iD e−iπ/4 =
e e
F + ieγ e−iπ/4 F e−iπ/4 =
+ eγ F.
2
4
4
T =
F
A
2
=
1
e−2γ
=
−γ
2.
(eγ + e 4 )2
[1 + (e−2γ /4)]
If γ 1, the denominator is essentially 1, and we recover T = e−2γ
(Eq. 9.23).
342
CHAPTER 9. THE WKB APPROXIMATION
Problem 9.12
In the plot below the blue and red curves make up the piecewise WKB solution, and the black dashed line is
the exact result. I chose the width of the patching region to be 0.3 x2 , centered at x2 .
$Assumptions = m > 0 && En > 0 && ω > 0;
Analytical Integral
p[x_] := Sqrt 2 m En - V[x]
1
V[x_] :=
2
2
x2 =
;
m ω2 x2 ;
En
m ω
;
Numerical Method
n = 3;
En =
2n -
1
ℏ ω;
2
ℏ = m = ω = 1;
ψ1[x_] :=
ψ3[x_] :=
α =
2m
ℏ2
2
p[x]
Abs[p[x]]
4π
1
NIntegrate[p[xp], {xp, x, x2}] +
ℏ
1
V '[x2]
ψ2[x_] :=
Sin
1/3
Exp -
1
π
4
NIntegrate[Abs[p[xp]], {xp, x2, x}]
ℏ
;
AiryAi α x - x2
αℏ
ψWKB[x_] := If[x < .85 x2, ψ1[x], If[x < 1.15 x2, ψ2[x], ψ3[x]]];
1
norm =
;
NIntegrate[ψWKB[x] ^ 2, {x, 0, 2 x2}]
ψexact[n_, x_] :=
2
mω
πℏ
1/4
1
2n n !
HermiteH n,
mω
ℏ
x Exp -
1
mω
2
ℏ
2
x
2
Prob9P12.nb
CHAPTER 9. THE WKB APPROXIMATION
343
Show[
Plot[norm ψ1[x], {x, 0, .75 x2}, PlotRange → {{0, 1.5 x2}, {- 1, 1}},
Ticks → {{{x2, "x2 "}}, None}, PlotStyle → Thickness[0.01]],
Plot[norm ψ3[x], {x, 1.15 x2, 1.5 x2}, PlotRange → All,
PlotStyle → Thickness[0.01]],
Plot[norm ψ2[x], {x, .75 x2, 1.15 x2}, PlotRange → {{0, 2 x2}, {- 2, 2}},
PlotStyle → {Red, Thickness[0.01]}],
Plot[ψexact[2 n - 1, x], {x, 0, 1.5 x2}, PlotStyle → {Black, Dashed}]
]
x2
Problem 9.13
Z x2 r
Z x2 p
√
1
α
ν
Equation 9.52 ⇒ n −
2m(E − αx ) dx = 2 2mE
1 − xν dx;
π~ = 2
2
E
0
0
1/ν
1/ν
α ν
zE
E
1 1 −1
z ≡ x , so x =
; dx =
z ν dz. Then
E
α
α
ν
E = αxν2 . Let
1/ν Z 1
1/ν
√
√
√
1
E
E
1
1 Γ(1/ν)Γ(3/2)
1
π~ = 2 2mE
n−
z ν −1 1 − z dz = 2 2mE
2
α
ν 0
α
ν Γ( ν1 + 23 )
√
1/ν
1/ν
√
Γ( ν1 + 1) 21 π √
Γ( ν1 + 1)
E
E
=
.
= 2 2mE
2πmE
1
3
α
α
Γ( ν + 2 )
Γ( ν1 + 32 )
E
1
1
ν+2
(n − 1 )π~ 1/ν Γ( ν1 + 32 )
= √ 2
α
;
Γ( ν1 + 1)
2πm
En =
1
n−
2
r
( 2ν )
π Γ( ν1 + 32 ) ν+2
~
α.
2mα Γ( ν1 + 1)
r
r
1
π Γ(2)
1
2α
For ν = 2: En =
n−
~
α = (n − )~
.
2
2mα Γ(3/2)
2
m
For a harmonic oscillator, with α = 12 mω 2 , En = n − 21 ~ω (n = 1, 2, 3, . . .). X
344
CHAPTER 9. THE WKB APPROXIMATION
Problem 9.14
Z x2 s
~2 a 2
1
~2 a2
2
2
V (x) = −
sech (ax). Eq. 9.52 =⇒ n −
π~ = 2
2m E +
sech (ax) dx
m
2
m
0
Z x2 r
√
mE
= 2 2~a
sech2 (ax) + 2 2 dx.
~
a
0
√
~2 a2
1
mE
sech2 (ax2 ) defines x2 . Let b ≡ − 2 2 , z ≡ sech2 (ax), so that x = sech−1 z, and hence
m
~ a
a
Z z2 √
√
1
1
1
−1
1
1
1
z−b
√ dz = −
√
√
dz. Then n −
dz.
dx =
π = 2 2a −
√ √
a
2a z 1 − z
2
2a
z 1−z 2 z
z
1−z
z1
(
)
r
x = 0 =⇒ z1 = sech2 (0) = 1
√ Z 1 1 z−b
1
Limits :
.
n−
π= 2
dz.
mE
2
1−z
x = x2 =⇒ z2 = sech2 (ax2 ) = − 2 2 = b
b z
~ a
r
1 z−b
1
(z − b)
1
b
= p
=p
− p
.
z 1−z
z (1 − z)(z − b)
(1 − z)(z − b) z (1 − z)(z − b)
E=−
"
#
Z 1
√ Z 1
1
1
1
p
p
n−
π = 2
dz − b
dz
2
2
(1 − z)(z − b)
b
b z −b + (1 + b)z − z
(
r
) 1
√
1−z √
−1 (1 + b)z − 2b
−1
= 2 −2 tan
− b sin
z−b
z(1 − b)
b
i √ √
√
√ h
π √ π √ π
−1
−1
−1
= 2 −2 tan (0) + 2 tan (∞) − b sin (1) + b sin−1 (−1) = 2 0 + 2 − b − b
2
2
2
√
√
√
√
(n − 12 )
1
1
√
= 1 − b;
b=1− √
n−
.
= 2π(1 − b);
2
2
2
√
√
Since the left side is positive, the right side must also be: (n − 21 ) < 2, n < 12 + 2 = 0.5 + 1.414 = 1.914.
So the only possible n is 1; there is only one bound state (which is correct—see Problem 2.52).
√
For n = 1,
1
b=1− √ ;
2 2
1
1
9
1
b=1− √ + = − √ ;
8
2 8
2
The exact answer (Problem 2.52(c)) is −0.5
~2 a2
E1 = −
m
9
1
−√
8
2
~2 a2
. Not bad.
m
Problem 9.15
Z r0 p
1
n−
π~ =
2m [E − V0 ln(r/a)] dr; E = V0 ln(r0 /a) defines r0 .
4
0
Z r0 p
Z r0 p
p
√
= 2m
V0 ln(r0 /a) − V0 ln(r/a) dr = 2mV0
ln(r0 /r) dr.
0
0
Let x ≡ ln(r0 /r), so ex = r0 /r, or r = r0 e−x =⇒ dr = −r0 e−x dx.
Z x2
p
√ −x
1
r = 0 =⇒ x1 = ∞
n−
π~ = 2mV0 (−r0 )
xe dx. Limits :
.
r = r0 =⇒ x2 = 0
4
x1
= −0.418
~2 a2
.
m
CHAPTER 9. THE WKB APPROXIMATION
345
√
Z ∞
p
p
p
√ −x
π
1
π~ = 2mV0 r0
xe dx = 2mV0 r0 Γ(3/2) = 2mV0 r0
.
n−
4
2
0
r
r
r
2π
2π
2π
1
~
1
1
~
r0 =
~ n−
⇒
En = V0 ln
n−
= V0 ln n −
+ V0 ln
.
mV0
4
a mV0
4
4
a mV0
3
1
n + 3/4
En+1 −En = V0 ln n +
−V0 ln n −
= V0 ln
, which is indeed independent of m (and a).
4
4
n − 1/4
Problem 9.16
Z r2 s
Z r2 r
√
1
e2 1
~2 `(` + 1)
A
B
0
n −
dr = −2mE
π~ =
2m E +
−
−1 + − 2 dr,
2
2
4π
r
2m
r
r
r
0
r1
r1
e2 1
~2 `(` + 1)
and B ≡ −
4π0 E
2m
E
Z r2 √ 2
√
1
−r + Ar − B
π~ = −2mE
dr.
n0 −
2
r
r1
where A ≡ −
are positive constants, since E is negative.
Let r1 and r2 be the roots of the polynomial in the numerator: − r2 + Ar − B = (r − r1 )(r2 − r).
Z r2 p
√
√
(r − r1 )(r2 − r)
√ 2
1
π √
0
n −
π~ = −2mE
dr = −2mE ( r2 − r1 ) .
2
r
2
r1
√
√
1
2 n0 −
~ = −2mE (r2 + r1 − 2 r1 r2 ) . But − r2 + Ar − B = −r2 + (r1 + r2 )r − r1 r2
2
=⇒ r1 + r2 = A; r1 r2 = B. Therefore
!
r
√
2
2 `(` + 1)
√
√
1
e
~
1
2 n0 −
~ = −2mE A − 2 B = −2mE −
−2 −
2
4π0 E
2m
E
r
p
2m
e2
−
− 2~ `(` + 1).
=−
4π0
E
r
2
1 p
(e2 /4π0 )2
e
2m
E
0
= 2~ n − + `(` + 1) ; −
=
−
−
h
i2 .
p
4π0
E
2
2m
4~2 n0 − 21 + `(` + 1)
−(m/2~2 )(e2 /4π0 )2
−13.6 eV
En0 ` = h
i2 = 2 .
p
p
n0 − 21 + `(` + 1)
n0 − 21 + `(` + 1)
r
p
1
1
1
1
1
`(` + 1) = ` 1 + = ` 1 +
− 2 + ... ≈ ` + −
,
`
2` 8`
2 8`
so
h
n0 − (1/2) +
p
i h
i
p
`(` + 1) = n − ` − (1/2) + `(` + 1) ≈ [n − ` − (1/2) + ` + (1/2) − (1/8`)] = n − (1/8`),
and hence
En` ≈
which is the Bohr formula.
−13.6 eV
2 ≈
[n − (1/8`)]
−13.6 eV
,
n2
346
CHAPTER 9. THE WKB APPROXIMATION
Problem 9.17
(a)
(i) ψWKB (x) = p
(ii) ψWKB (x) = p
D
|p(x)|
1
p(x)
(iii) ψWKB (x) = p
e
1
−~
Rx
x2
|p(x0 )| dx0
(x > x2 );
i
h i Rx
Rx
0
0
0
0
i
2
2
Be ~ x p(x ) dx + Ce− ~ x p(x ) dx (x1 < x < x2 );
1
|p(x)|
h
1
1 |p(x0 )| dx0
Rx
Fe~ x
1
Rx
+ Ge− ~ x
1 |p(x0 )| dx0
Z x2
1
π
2D
0
0
Equation 9.47 =⇒ (ii) ψWKB = p
sin
p(x )dx +
~ x
4
p(x)
i
(0 < x < x1 ).
(x1 < x < x2 ).
To effect the join at x1 , first rewrite (ii):
Z x2
Z x
Z
1
1 x
π
1
π
2D
2D
sin
p(x0 ) dx0 −
p(x0 ) dx0 +
sin
p(x0 ) dx0 − θ −
= −p
,
(ii) ψWKB = p
~ x1
~ x1
4
~ x1
4
p(x)
p(x)
where θ is defined in Eq. 9.59. Now shift the origin to x1 :


h 1 R0
i
R0
0
0
0
0
1
1


~ x |p(x )| dx + Ge− ~ x |p(x )| dx
p


F
e
(x
<
0)


|p(x)|
Z
.
ψWKB =
x
1
2D
π



(x > 0) 
p(x0 ) dx0 − θ −
sin
 −p

~ 0
4
p(x)
Following Problem 9.10: ψp (x) = aAi(−αx)+bBi(−αx), with α ≡
Z 0
Overlap region 1 (x < 0):
x
|p(x0 )| dx0 =
2m|V 0 (0)|
~2
1/3
√
; p(x) = ~α3/2 x.
2
~(−αx)3/2 .
3

h 2
i
3/2
1
− 23 (−αx)3/2
r
r

3 (−αx)
ψWKB ≈ 1/2 3/4
F
e
+
Ge

π
π
~ α (−x)1/4
=⇒ a = 2G
; b=F
.
3/2
3/2
2
2
b
a

− 3 (−αx)
~α
~α
3 (−αx)

√
e
+
e
ψp
≈ √
2 π(−αx)1/4
π(−αx)1/4
Z x
2
Overlap region 2 (x > 0):
p(x0 ) dx0 = ~(αx)3/2 .
3
0
2D
π
2
=⇒ ψWKB ≈ − 1/2 3/4 1/4 sin (αx)3/2 − θ −
,
3
4
~ α x
a
2
π
b
2
π
3/2
3/2
ψp ≈ √
sin (αx) +
+√
cos (αx) +
.
3
4
3
4
π(αx)1/4
π(αx)1/4
i
3/2
2
−2D 1 h i 2 (αx)3/2 −iθ −iπ/4
Equating the two expressions: 1/2 3/4
e 3
e e
− e−i 3 (αx) eiθ eiπ/4
2i
~ α
h
i bh 2
i
3/2
3/2
3/2
2
2
1
a i 2 (αx)3/2 iπ/4
= √ 1/4
e 3
e
− e−i 3 (αx) e−iπ/4 +
ei 3 (αx) eiπ/4 + e−i 3 (αx) e−iπ/4
2i
2
πα


r
r
π −iθ −iπ/4
π −iθ 

iπ/4


 −2D

e e
= (a + ib)e
, or (a + ib) = 2D
ie
α~
α~
r
r
=⇒
π iθ iπ/4
π iθ 


 2D

e e
= (−a + ib)e−iπ/4 ,
or (a − ib) = −2D
ie 
α~
α~
CHAPTER 9. THE WKB APPROXIMATION
347


r
r
π
π


−iθ
iθ


 2a = 2D
i(e
− e ) ⇒ a = 2D
sin θ, 
α~
α~
r
r
=⇒
π
π




 2ib = 2D
i(e−iθ + eiθ ) ⇒ b = 2D
cos θ. 
α~
α~
Combining these with the results from overlap region 1 =⇒
r
r
r
r
π
π
π
π
2G
= 2D
sin θ, or G = D sin θ; F
= 2D
cos θ,
α~
α~
α~
α~
Putting these into (iii) : ψWKB (x) = p
D
|p(x)|
or
F = 2D cos θ.
i
h
Rx
Rx
0
0
0
0
1
1
1
1
2 cos θe ~ x |p(x )| dx + sin θe− ~ x |p(x )| dx (0 < x < x1 ).
(b)
1
R x1
0
0
1
R x1
0
0
Odd(−) case: (iii) =⇒ ψ(0) = 0 ⇒ 2 cos θe ~ 0 |p(x )| dx + sin θe− ~ 0 |p(x )| dx = 0.
Z
1 x1
1
|p(x0 )| dx0 = φ, with φ defined by Eq. 8.60. So sin θe−φ/2 = −2 cos θeφ/2 , or tan θ = −2eφ .
~ 0
2
i
d|p(x)| h
D
1
φ/2
−φ/2
2
cos
θe
+
sin
θe
Even(+) case: (iii) =⇒ ψ 0 (0) = 0 ⇒ −
2 (|p(x)|)3/2 dx 0
R x1
R x1
0
0
0
0
1
1
1
1
D
2 cos θe ~ 0 |p(x )| dx − |p(0)| + sin θe− ~ 0 |p(x )| dx
|p(0)|
= 0.
+p
~
~
|p(x)|
Now
√ 1
d|p(x)|
d p
1
dV
dV
d|p(x)|
2m[V (x) − E] = 2m √
=
, and
= 0, so
= 0.
dx
dx
2 V − E dx
dx 0
dx 0
2 cos θeφ/2 = sin θe−φ/2 , or tan θ = 2eφ . Combining the two results: tan θ = ±2eφ . QED
(c)
sin n + 21 π + (−1)n cos cos 1
=
tan θ = tan
π+ =
=−
≈− .
1
n+1
(−1)
sin sin cos n + 2 π + 1
1
1
1
1
1
π ≈ ∓ e−φ , so θ ≈ n +
π ∓ e−φ . QED
So − ≈ ±2eφ , or ≈ ∓ e−φ , or θ − n +
2
2
2
2
2
1
n+
2
[Note: Since θ (Eq. 9.59) is positive, n must be a non-negative integer: n = 0, 1, 2, . . .. This is like
harmonic oscillator (conventional) numbering, since it starts with n = 0.]
V(x)
-a
(d)
x1
a
x2
x
348
CHAPTER 9. THE WKB APPROXIMATION
Z x2 s
1
2
2
2m E − mω (x − a) dx. Let z = x − a (shifts the origin to a).
2
x1
s
Z
2 z2
1
1
=
2m E − mω 2 z 2 dz, where E = mω 2 z22 .
~ 0
2
2
q
z2
Z z2 q
mω 2 −1
mω
π mω 2
2
−1
2
2
2
2
2
=
z2 − z dz =
z z2 − z + z2 sin (z/z2 )
z2 sin (1) =
z ,
= mω
~
~
~
2 ~ 2
0
0
1
θ =
~
=
π mω 2E
πE
=
.
2 ~ mω 2
~ω
Putting this into Eq. 9.62 yields
πE
≈
~ω
n+
1
2
1
π ∓ e−φ ,
2
or En± ≈
n+
1
2
~ω ∓
~ω −φ
e . QED
2π
(e)
+
−
1 Ψ(x, t) = √ ψn+ e−iEn t/~ + ψn− e−iEn t/~ =⇒
2
i
−
+
−
+
1h +2
|ψn | + |ψn− |2 + ψn+ ψn− ei(En −En )t/~ + e−i(En −En )t/~ .
|Ψ(x, t)|2 =
2
En− − En+
1 ~ω
ω
(Note that the wave functions (i), (ii), (iii) are real ). But
≈ 2 e−φ = e−φ , so
~
~ 2π
π
1 + 2
ω −φ
|Ψ(x, t)|2 =
ψ (x) + ψn− (x)2 + ψn+ (x)ψn− (x) cos
e t .
2 n
π
2π 2 φ
2π
e . QED
=
It oscillates back and forth, with period τ =
−φ
(ω/π) e
ω
(f )
Z x1 s
Z x1 r
1
2√
mω 2
2
2
2m mω (x − a) − E dx =
2mE
(x − a)2 − 1 dx.
2
~
2E
0
0


r
r
r
m


m
2E 1
x = 0 =⇒ z =
ωa ≡ z0
Let z ≡
ω(a − x), so dx = −
dz. Limits:
.
2E


2E
m ω
x = x1 =⇒ radicand = 0 =⇒ z = 1
1
φ=2
~
r
Z
Z
i z0
p
2√
2E 1 z0 p 2
4E z0 p 2
4E 1 h p 2
φ =
2mE
z − 1 dz =
z − 1 dz =
z z − 1 − ln(z + z 2 − 1)
~
m ω 1
~ω 1
~ω 2
1
q
q
2E
=
z0 z02 − 1 − ln z0 + z02 − 1 ,
~ω
r
where z0 = aω
m
.
2E
V (0) = 21 mω 2 a2 ,
so
V (0) E ⇒
m 2 2
ω a E ⇒ aω
2
In that case
2E 2
2E 2
2E 2 2 m
mωa2
φ≈
z0 − ln(2z0 ) ≈
z0 =
a ω
=
.
~ω
~ω
~ω
2E
~
This, together with Eq. 9.65, gives us the period of oscillation in a double well.
r
m
1,
2E
or z0 1.
CHAPTER 9. THE WKB APPROXIMATION
349
Problem 9.18
(a) En ≈
n2 π 2 ~2
.
2m(2a)2
With n = 1,
E1 =
π 2 ~2
.
8ma2
(b)
V(x)
-a
a
x
V0
E1
V(x)
x0
a
-a
E
E1
x
tunneling
(c)
Z
1 x0
V0 − E1
|p(x)| dx. αx0 = V0 − E1 ⇒ x0 =
.
~
α
p a
p(x) = 2m [E − V (x)]; V (x) = −αx, E = E1 − V0 .
p
√
√
√
√
= 2m(E1 − V0 + αx) = 2mα x − x0 ; |p(x)| = 2mα x0 − x.
√
√
x0
Z x0
√
2mα
2
2 2mα
1√
γ =
2mα
x0 − x dx =
− (x0 − x)3/2
=
(x0 − a)3/2 .
~
~
3
3
~
a
a
γ =
Now x0 − a = (V0 − E1 − aα)/α, and αa ~2 /ma2 ≈ E1 V0 , so we can drop E1 and αa. Then
p
√
3/2
8mV03
2 2mα V0
γ≈
=
.
3 ~
α
3α~
Equation 9.29
⇒ τ=
4a 2γ
e ,
v
where
τ=
1
π 2 ~2
π 2 ~2
mv 2 ≈
⇒ v2 =
,
2
2
8ma
4m2 a2
4a
8ma2 2γ
2ma e2γ =
e .
π~
π~
or
v=
π~
.
2ma
So
350
CHAPTER 9. THE WKB APPROXIMATION
(d)
2
(8) 9.1 × 10−31 10−10
τ =
e2γ = 2 × 10−16 e2γ ;
π (1.05 × 10−34 )
q
3
(8) (9.1 × 10−31 ) (20 × 1.6 × 10−19 )
γ =
= 4.4 × 104 ;
(3) (1.6 × 10−19 ) (7 × 106 ) (1.05 × 10−34 )
τ = 2 × 10−16 × 1038,000 s = 1038,000 yr.
4
e2γ = e8.8×10 = 10log e
8.8×104
= 1038,000 .
Seconds, years . . . it hardly matters; nor is the factor out front significant. This is a huge number—the
age of the universe is about 1010 years. In any event, this is clearly not something to worry about.
Problem 9.19
Equation 9.23 ⇒ the tunneling probability: T = e−2γ , where
Z
p
1 x0 p
γ =
2m(V − E) dx. Here V (x) = mgx, E = 0, x0 = R2 + (h/2)2 − h/2 (half the diagonal).
~
√ 0
Z x0
x
2m √
2m p
m p 2 3/2 0
3/2
=
mg
2g x
=
2g x0 .
x1/2 dx =
~
~
3
3~
0
0
I estimate:
h = 10 cm, R = 3 cm, m = 300 gm; let g = 9.8 m/s2 .
γ=
Then x0 =
√
9 + 25 − 5 = 0.83 cm, and
p
(2)(0.3)
(2)(9.8) (0.0083)3/2 = 6.4 × 1030 .
−34
(3)(1.05 × 10 )
Frequency of “attempts”: say f = v/2R. We want the product of the number of attempts (f t) and the
probability of toppling at each attempt (T ), to be 1:
t
v −2γ
e
=1
2R
Estimating the thermal velocity: 21 mv 2 = 21 kB T
p
is the temperature, 300 K) ⇒ v = kB T /m.
r
t = 2R
⇒
t=
2R 2γ
e .
v
(I’m done with the tunneling probability; from now on T
s
30
30
m 2γ
0.3
12.8×1030
8
log e 13×10
e = 2(0.03)
e
=
5
×
10
10
= (5 × 108 ) × 105.6×10 s
kB T
(1.4 × 10−23 )(300)
30
= 16 × 105.6×10
yr.
Don’t hold your breath.
[Actually, to tunnel all the way through the classically forbidden region, the center of mass must not only
rise from 0 to x0 , but also drop back to 0. This doubles γ, and makes the final exponent 1 × 1031 .]
CHAPTER 9. THE WKB APPROXIMATION
351
Problem 9.20
Z
p
1 x
1
(a) φ(x) =
p(x0 ) dx0 ; φ0 (x) = p(x); p(x) = 2m(E − V (x)). But V (x) = 0 for |x| ≥ a, so
~ 0
~
√
0
p(±a) = 2mE = ~k; φ (±a) = k; p0 (±a) = 0 (the latter follows because p(x) is constant for |x| ≥ a (so
it’s derivative is zero in that region), and by assumption V (x) is smooth).
Boundary conditions at −a:
h
i
1
Ae−ika + Beika = p
C+ eiφ(−a) + C− e−iφ(−a) ,
p(−a)
h
i
i
C+ eiφ(−a) − C− e−iφ(−a) φ0 (−a).
ik Ae−ika − Beika = p
p(−a)
p
(There’s another!term in the second equation, where 1/ p(x) is differentiated, but since
d
1
1
1
p
p0 (x), and p0 (−a) = 0, it contributes nothing.) Meanwhile, using φ0 (−a) = k,
=−
dx
2 (p(x))3/2
p(x)
the second equation simplifies to
h
i
1
Ae−ika − Beika = p
C+ eiφ(−a) − C− e−iφ(−a) .
p(−a)
Subtracting this from the first equation, we find:
1
Beika = p
C− e−iφ(−a) .
p(−a)
Boundary conditions at a:
i
1 h
Ceika = p
C+ eiφ(a) + C− e−iφ(a) ,
p(a)
h
i
i
C+ eiφ(a) − C− e−iφ(a) φ0 (a).
ikCeika = p
p(a)
(Again, the other term in the second equation is zero.) Thus
i
1 h
Ceika = p
eiφ(a)
− ^C
e−iφ(a)
.
PlotC+
41 Cos[x]
2 ,−
{x,
0, 1.5}, PlotRange
→ {0, 2}
p(a)
2.0
Subtracting this from the first equation yields
1
0= p
C− e−iφ(a)
p(a)
1.5
1.0
0.5
⇒
C− = 0.
Combining the results of the boundary conditions at −a and at +a, B = 0, and hence R = 0.
(b)
0.0
Plot
0.2
0.4
0.6
0.8
1.0
Sech[x] 2 , {x, - 4, 4}, PlotRange → {0, 1}
1.0
0.8
0.6
0.4
0.2
-4
-2
0
2
4
352
CHAPTER 9. THE WKB APPROXIMATION
1
sech(iz) =
, so p(iy) =
cos(z)
s
2m E −
V0
. Here’s the graph:
cos2 (y/a)
��������
Plot
4- 1
Cos[x] ^ 2 , {x, 0, 1.5}, PlotRange → {0, 2}
2.0
1.5
�������� 1.0
0.5
0.0
��������
0.2
Plot
0.4
0.6
0.8
1.0
Sech[x] 2 , {x, - 4, 4}, PlotRange → {0, 1}
1.0
Z s
Z u0 r
2 y0
V0
1
2a p
p(iy) dy =
2m E −
2mV0
β2 −
du,
dy =
2 (y/a)
2u
~
cos
~
cos
0.8
0
0
0
p
(where β ≡ E/V0 and u ≡ y/a)
0.6
Z u0 p 2
2a p
β cos2 u − 1
=
2mV��������
du. Now let z ≡ sin u, so dz = cos u du :
0
~
cos u
0
0.4
p
Z z0
Z z0 p 2
2
2
p
β (1 − z ) − 1
z0 − z 2
2a p
2a
=
dz =
dz, where z02 ≡ 1 − (1/β 2 ).
2mV0
2mV0 β
2
2
0.2
~
1
−
z
~
1
−
z
0
0
q
π
1
π
Mathematica says this integral is
1 − 1 − z02 =
1−
2
β
-4
-2
0
2
42
!
r
p
2a p
πa p
E
π
πa √
=
2mV0 (β − 1) =
2mV0
2mE − 2mV0 . X
−1 =
~
2
~
V0
~
2
9P18.nb
2
λ=
~
Z y0
Here’s the graph of R(E):
�������
Plot ⅇ1-
x
, {x, 1, 10}, PlotRange → {0, 1.1}, AxesOrigin → {1, 0}
1.0
0.8
0.6
�������
0.4
0.2
0
(c) Here’s the plot of V (x):
2
4
6
8
10
0.8
0.6
0.4
0.2
CHAPTER 9. THE WKB APPROXIMATION
0
��������
Plot
2
1
1 + x2
353
4
6
8
10
, {x, - 5, 5}, PlotRange → {0, 1}
1.0
0.8
0.6
��������
0.4
0.2
9P18.nb
-4
0
-2
2
3
4
In the notation of Part (b):
��������
Z y0 s
2m E −
Z u0 r
2a p
V0
1
dy =
du
2mV0
β2 −
2
1 − (y/a)
~
1 − u2
0
0
Z u0 p 2
u − u2
2aβ p
√0
=
2mV0
du, where u20 = 1 − (1/β 2 );
~
1 − u2
0
Z u0 p
1 − (u/u0 )2
2a p
2aβ p
√
2mV0 u0
du =
2m(E − V0 )E(u0 |u−2
=
0 ),
2
~
~
1
−
u
0
2
λ=
~
ClearAll
where E(x|m) is the elliptic integral of the second kind (Arfken, third edition, Equation (5.132b)). Here’s the
�������� ClearAll
plot of R:
1
��������
w[x_] := 1 -
��������
L[x_] :=
��������
Plot ⅇ-L[x] , {x, 1, 10}
x
w[x]
1 - w[x]
EllipticE ArcSin
w[x] ,
1
w[x]
1.0
0.8
0.6
��������
0.4
0.2
2
4
6
8
10
354
CHAPTER 10. SCATTERING
Chapter 10
Scattering
Problem 10.1
(a)
q1
b
e
q
r
q2
Conservation of energy:
E=
1
m(ṙ2 + r2 φ̇2 ) + V (r),
2
Conservation of angular momentum: J = mr2 φ̇.
ṙ2 +
So
where
φ̇ =
V (r) =
q1 q2 1
.
4π r
J
.
mr2
J2
2
= (E − V ). We want r as a function of φ (not t).
m2 r2
m
Also, let u ≡ 1/r.
Then
2
1 du J 2
J du
J du
J2
2
− 2
u =−
. Then: −
+ 2 u2 = (E − V ), or
u
dφ m
m dφ
m dφ
m
m
r
2
du
2m
du
2m
du
du
2
= 2 (E −V )−u ;
=
(E − V ) − u2 ; dφ = q
=p
, where
dφ
J
dφ
J2
2m
I(u)
2
(E − V ) − u
ṙ =
dr
dr du dφ
=
=
dt
du dφ dt
J2
2m
(E − V ) − u2 . Now, the particle q1 starts out at r = ∞ (u = 0), φ = 0, and the point
J2
Z umax
du
√ . It now swings through an equal angle Φ
of closest approach is rmin (umax ), Φ : Φ =
I
0
I(u) ≡
CHAPTER 10. SCATTERING
355
Z umax
on the way out, so
Φ + Φ + θ = π,
θ = π − 2Φ.
or
θ =π−2
0
du
p
.
I(u)
So far this is general ; now we put in the specific potential:
2m q1 q2
2mE
− 2
u − u2 = (u2 − u)(u − u1 ), where u1 and u2 are the two roots.
J2
J 4π0
p
(Since du/dφ = I(u), umax is one of the roots; setting u2 > u1 , umax = u2 .)
u2
Z u2
du
−2u + u1 + u2
p
θ =π−2
= π + 2 sin−1
u2 − u1
(u2 − u)(u − u1 )
0
0
u
+
u
1
2
= π + 2 sin−1 (−1) − sin−1
u2 − u1
π
u1 + u2
u1 + u2
−1
−1
= π + 2 − − sin
= −2 sin
.
2
u2 − u1
u2 − u1
I(u) =
Now J = mvb, E = 12 mv 2 , where v is the incoming velocity, so J 2 = m2 b2 (2E/m) = 2mb2 E, and hence
2m/J 2 = 1/b2 E. So
1
1 1 q1 q2
A
1
q1 q2
I(u) = 2 − 2
, so
− I(u) = u2 + 2 u − 2 .
u − u2 . Let A ≡
b
b
E 4π0
4π0 E
b
b


s
"
#
r
2
2
A
1
1
A
A
4
A
2b 
To get the roots: u2 + 2 u − 2 = 0 =⇒ u =
− 2±
+ 2 = 2 −1 ± 1 +
.
b
b
2
b
b4
b
2b
A




s
s
2
2
2b 
A
2b 
u1 + u2
−1
A
, u1 = 2 −1 − 1 +
;
=q
.
Thus u2 = 2 −1 + 1 +
2b
A
2b
A
u2 − u1
2
1 + (2b/A)


2
1
θ
2b
1
1
 , or q
θ = 2 sin−1  q
= sin
; 1+
=
;
2
2
A
2
2
sin (θ/2)
1 + (2b/A)
1 + (2b/A)
2b
A
2
=
cos2 (θ/2)
1 − sin2 (θ/2)
=
;
sin2 (θ/2)
sin2 (θ/2)
2b
= cot(θ/2),
A
or
b=
q1 q2
cot(θ/2).
8π0 E
(b)
D(θ) =
db
q1 q2
b
db
. Here
=
sin θ dθ
dθ
8π0 E
−
1
2 sin2 (θ/2)
.
q1 q2 cos(θ/2) q1 q2
1
1
=
=
2
2 sin(θ/2) cos(θ/2) 8π0 E sin(θ/2) 8π0 E 2 sin (θ/2)
q1 q2
16π0 E sin2 (θ/2)
2
.
(c)
Z
σ=
D(θ) sin θ dθ dφ = 2π
q1 q2
16π0 E
2 Z π
0
sin θ
dθ.
sin4 (θ/2)
This integral does not converge,
for near θ = 0 (and again near π) we have sin θ ≈ θ, sin(θ/2) ≈ θ/2, so
R
the integral goes like 16 0 θ−3 dθ = − 8θ−2 0 → ∞.
356
CHAPTER 10. SCATTERING
Problem 10.2
r
e
x
Two dimensions:
eikr
ψ(r, θ) ≈ A eikx + f (θ) √ .
r
One dimension:
ψ(x) ≈ A eikx + f (x/|x|)e−ikx .
Problem 10.3
Multiply Eq. 10.32 by P`0 (cos θ) sin θ dθ and integrate from 0 to π, exploiting the orthogonality of the Legendre polynomials (Eq. 4.34)—which, with the change of variables x ≡ cos θ, says
Z π
2
P` (cos θ)P`0 (cos θ) sin θ dθ =
δ``0 .
2` + 1
0
The delta function collapses the sum, and we get
h
i
0
(1)
2i` j`0 (ka) + ika`0 h`0 (ka) = 0,
and hence (dropping the primes)
a` = −
j` (ka)
(1)
. QED
ikh` (ka)
Problem 10.4
Keeping only the ` = 0 terms, Eq. 10.29 says that in the exterior region:
h
i
sin(kr)
eikr
sin(kr)
eikr
(1)
ψ ≈ A j0 (kr) + ika0 h0 (kr) P0 (cos θ) = A
+ ika0 −i
=A
+ a0
kr
kr
kr
r
(r > a).
In the internal region Eq. 10.18 (with n` eliminated because it blows up at the origin) yields
ψ(r) ≈ bj0 (kr) = b
sin(kr)
kr
(r < a).
The boundary conditions hold independently for each `, as you can check by keeping the summation over ` and
exploiting the orthogonality of the Legendre polynomials:
CHAPTER 10. SCATTERING
357
sin ka
eika
sin ka
A
+ a0
=b
.
ka
a
ka
(1) ψ continuous at r = a:
(2) ψ 0 discontinuous at r = a: Integrating the radial equation across the delta function gives
Z 2
Z Z
~2
~2
d u
~2 `(` + 1)
2mα
dr+
u
dr
=
E
u
dr
⇒
−
−
αδ(r
−
a)
+
∆u0 +αu(a) = 0, or ∆u0 = 2 u(a).
2m
dr2
2m r2
2m
~
Now
u = rR,
so u0 = R+rR0 ;
∆u0 = ∆R+a∆R0 = a∆R0 =
2mα
2mα
β
aR(a), or ∆ψ 0 = 2 ψ(a) = ψ(a).
~2
~
a
((
A A b β sin(ka)
b
(ika
(ke
(
(
sin ka = b
k cos(ka) + a0 ik 2 eika − 2 (sin(ka)
k
cos(ka)
+
.
+
a
−
(
0
2
ka
ka (
ka
ka
a
ka
β
sin(ka) .
The indicated terms cancel (by (1)), leaving A cos(ka) + ia0 keika = b cos(ka) +
ka
β Using (1) to eliminate b: A cos(ka) + ia0 keika = cot(ka) +
sin(ka) + a0 keika A.
ka
+ ia keika = cos(ka)
β sin(ka) + a k cot(ka)eika + β a0 eika .
cos(ka)
0
0
+
ka
a
β
cos(ka)
1
β
ia0 keika 1 + i cot(ka) + i
sin(ka). But ka 1, so sin(ka) ≈ ka, and cot(ka) =
≈
.
=
ka
ka
sin(ka)
ka
i
i
i
ia0 k(1 + ika) 1 +
(1 + β) = β; ia0 k 1 +
(1 + β) + ika − 1 − β ≈ ia0 k
(1 + β) = β.
ka
ka
ka
aβ
a0 = −
.
1+β
aβ
Equation 10.25 ⇒ f (θ) ≈ a0 = −
.
1+β
Equation 10.27
⇒ σ = 4πD = 4π
aβ
1+β
2
Equation 10.14 ⇒ D = |f | =
2
.
Problem 10.5
(a) In the region to the left
ψ(x) = Aeikx + Be−ikx
(x ≤ −a).
In the region −a < x < 0, the Schrödinger equation gives
−
where k 0 =
p
~2 d2 ψ
− V0 ψ = Eψ
2m dx2
2m(E + V0 )/~.
⇒
d2 ψ
= −(k 0 )2 ψ
dx2
The general solution is
ψ = C sin(k 0 x) + D cos(k 0 x)
But ψ(0) = 0 implies D = 0, so
ψ(x) = C sin(k 0 x)
(−a ≤ x ≤ 0).
aβ
1+β
2
.
358
CHAPTER 10. SCATTERING
The continuity of ψ(x) and ψ 0 (x) at x = −a says
Ae−ika + Beika = −C sin(k 0 a),
ikAe−ika − ikBeika = k 0 C cos(k 0 a).
Divide and solve for B:
ikAe−ika − ikBeika
= −k 0 cot(k 0 a),
Ae−ika + Beika
ikAe−ika − ikBeika = −Ae−ika k 0 cot(k 0 a) − Beika k 0 cot(k 0 a),
Beika [−ik + k 0 cot(k 0 a)] = Ae−ika [−ik − k 0 cot(k 0 a)] .
k − ik 0 cot(k 0 a)
B = Ae−2ika
.
k + ik 0 cot(k 0 a)
(b)
2
|B| = |A|
2
k − ik 0 cot(k 0 a)
k + ik 0 cot(k 0 a)
·
= |A|2 . X
k + ik 0 cot(k 0 a)
k − ik 0 cot(k 0 a)
(c) From part (a) the wave function for x < −a is
ψ(x) = Aeikx + Ae−2ika
k − ik 0 cot(k 0 a) −ikx
e
.
k + ik 0 cot(k 0 a)
But by definition of the phase shift (Eq. 10.40)
h
i
ψ(x) = A eikx − ei(2δ−kx) .
so
k − ik 0 cot(k 0 a)
= −e2iδ .
k + ik 0 cot(k 0 a)
p
√
This is exact. For a very deep well, E V0 , k = 2mE/~ 2m(E + V0 )/~ = k 0 , so
0
0
−2ika −ik cot(k a)
e
δ = −ka.
= −e2iδ ; e−2ika = e2iδ ;
ik 0 cot(k 0 a)
e−2ika
Problem 10.6
1 iδ`
e sin δ` ,
k
From Eq. 10.46,
a` =
But (Eq. 10.19)
h` (x) = j` (x) + in` (x),
and Eq. 10.33,
(1)
eiδ` sin δ` = i
a` = i
j` (ka)
(1)
kh` (ka)
,
it follows that
j` (ka)
eiδ` sin δ` = i (1)
.
h` (ka)
so
j` (ka)
1
1 − i(n/j)
(n/j) + i
=i
=i
=
,
j` (ka) + in` (ka)
1 + i(n/j)
1 + (n/j)2
1 + (n/j)2
(writing (n/j) as shorthand for n` (ka)/j` (ka)). Equating the real and imaginary parts:
cos δ` sin δ` =
(n/j)
;
1 + (n/j)2
sin2 δ` =
1
.
1 + (n/j)2
Dividing the second by the first, I conclude that
tan δ` =
1
,
(n/j)
or
δ` = tan−1
j` (ka)
.
n` (ka)
CHAPTER 10. SCATTERING
359
Problem 10.7
r > a : u(r) = A sin(kr + δ);
r < a : u(r) = B sin kr + D cos kr = B sin kr,
because u(0) = 0 =⇒ D = 0.
Continuity at r = a =⇒ B sin(ka) = A sin(ka + δ) =⇒ B = A
sin(ka + δ)
sin(ka + δ)
. So u(r) = A
sin kr.
sin(ka)
sin(ka)
From Problem 10.4,
β
du
sin(ka + δ)
β
= u(a) ⇒ Ak cos(ka + δ) − A
∆
k cos(ka) = A sin(ka + δ).
dr r=a
a
sin(ka)
a
cos(ka + δ) −
β
sin(ka + δ)
cos(ka) =
sin(ka + δ),
sin(ka)
ka
sin(ka) cos(ka + δ) − sin(ka + δ) cos(ka) =
sin(ka − ka − δ) =
− sin δ = β
β
sin(ka) [sin(ka) cos δ + cos(ka) sin δ] ,
ka
sin2 (ka)
[cot δ + cot(ka)] .
ka
ka
cot δ = − cot(ka) +
.
β sin2 (ka)
sin2 (ka)
[cos δ + cot(ka) sin δ] ;
ka
cot δ = − cot(ka) −
β
sin(ka + δ) sin(ka),
ka
ka
;
β sin2 (ka)
−1 = β
Problem 10.8
eikr
1 1 ikr
1
=⇒ ∇G = −
∇e + eikr ∇
=⇒
4πr
4π r
r
1
1
1
1
2 ∇
· (∇eikr ) + ∇2 (eikr ) + eikr ∇2
.
∇2 G = ∇ · (∇G) = −
4π
r
r
r
G=−
1 d
∇2 eikr = ik∇ · (eikr r̂) = ik 2 (r2 eikr ),
r dr
ik
2
1
2 ikr
ikr
2 ikr
ikr
2
∇ e = 2 (2re + ikr e ) = ike
+ ik ; and ∇
= −4πδ 3 (r)
r
r
r
But ∇
1
1
= − 2 r̂;
r
r
∇(eikr ) = ikeikr r̂;
(see reference in footnote 13).
1
1
1
2
So ∇2 G = −
2 − 2 r̂ · ikeikr r̂ + ikeikr
+ ik − 4πeikr δ 3 (r) .
4π
r
r
r
eikr δ 3 (r) = δ 3 (r), so
2ik 2ik k 2
eikr
1 ikr
2
3
∇ G = δ (r) −
e
− 2 + 2 −
= δ 3 (r) + k 2
= δ 3 (r) − k 2 G.
4π
r
r
r
4πr
But
Therefore
(∇2 + k 2 )G = δ 3 (r). QED
360
CHAPTER 10. SCATTERING
Problem 10.9
ψ=√
1
e
−r/a
√
e2
~2 1
V =−
=−
4π0 r
ma r
;
(Eq. 4.72);
k=i
−2mE
i
= .
~
a
πa3
In this case there is no “incoming” wave, and ψ0 (r) = 0. Our problem is to show that
Z ik|r−r0 |
m
e
−
V (r0 )ψ(r0 ) d3 r0 = ψ(r).
2
2π~
|r − r0 |
We proceed to evaluate the left side (call it I):
Z −|r−r0 |/a
e
m ~2
1
1 −r0 /a 3
√
I= −
−
e
d r0
2
3
2π~
ma
|r − r0 | r0
πa
√
Z − r2 +r2 −2rr0 cos θ/a −r0 /a
0
e
e
1
1
√
p
r02 sin θ dr0 dθ dφ.
=
2
2
2πa πa3
r + r0 − 2rr0 cos θ r0
(I have set the z0 axis along the—fixed—direction r, for convenience.) Doing the φ integral (2π):
√
#
"Z
Z ∞
π − r 2 +r02 −2rr0 cos θ/a
1
e
p
I= √
sin θ dθ dr0 . The θ integral is
r0 e−r0 /a
r2 + r02 − 2rr0 cos θ
a πa3 0
0
Z π
0
√2 2
i
a h −(r+r0 )/a
a −√r2 +r02 −2rr0 cos θ/a π
e− r +r0 −2rr0 cos θ/a
−|r−r0 |/a
p
=
−
e
−
e
.
e
sin
θ
dθ
=
−
rr0
rr0
0
r2 + r02 − 2rr0 cos θ
Z ∞
h
i
1
I=− √
e−r0 /a e−(r0 +r)/a − e−|r0 −r|/a dr0
r πa3 0
Z ∞
Z r
Z ∞
1
−r/a
−2r0 /a
−r/a
r/a
−2r0 /a
e
e
dr0 − e
=− √
dr0 − e
e
dr0
r πa3
0
0
r
a
∞i
1 h −r/a a e
− e−r/a (r) − er/a − e−2r0 /a
=− √
2
2
r
r πa3
h
i
1
a −r/a
a
1
=− √
e
− re−r/a − er/a e−2r/a = √
e−r/a = ψ(r). QED
2
r πa3 2
πa3
Problem 10.10
For the potential in Eq. 10.81, Eq. 10.88 =⇒
2m
f (θ) = − 2 V0
~ κ
Z a
r sin(κr) dr = −
0
where (Eq. 10.89) κ = 2k sin(θ/2).
2mV0
~2 κ
a
1
r
2mV0
sin(κr)
−
cos(κr)
= − 2 3 [sin(κa) − κa cos(κa)] ,
κ2
κ
~ κ
0
For low-energy scattering (ka 1):
1
1
(κa)3 ; cos(κa) ≈ 1 − (κa)2 ; so
3!
2
2mV0
1
1
2 mV0 a3
3
3
f (θ) ≈ − 2 3 κa − (κa) − κa + (κa) = −
,
~ κ
6
2
3 ~2
sin(κa) ≈ κa −
in agreement with Eq. 10.82.
CHAPTER 10. SCATTERING
361
Problem 10.11
Z
i
1 ∞ h −(µ−iκ)r
e
− e−(µ+iκ)r dr
2i 0
0
−(µ−iκ)r
∞
1 e
1
κ
e−(µ+iκ)r
1
1
1 µ + iκ − µ + iκ
=
=
= 2
.
−
−
=
2i −(µ − iκ) −(µ + iκ) 0
2i µ − iκ µ + iκ
2i
µ2 + κ2
µ + κ2
sin(κr) =
So
1 iκr
e − e−iκr ,
2i
f (θ) = −
Z ∞
so
e−µr sin(κr) dr =
2mβ
2mβ
κ
=− 2 2
. QED
~2 κ µ2 + κ2
~ (µ + κ2 )
Problem 10.12
2
2mβ
1
, where Eq. 10.89 ⇒ κ = 2k sin(θ/2).
~2
(µ2 + κ2 )2
2
Z
Z π
2mβ
1
1
σ = D(θ) sin θ dθ dφ = 2π
h
i2 2 sin(θ/2) cos(θ/2) dθ.
~2
µ4 0
2
1 + (2k/µ) sin2 (θ/2)
Equation 10.91 =⇒ D(θ) = |f (θ)|2 =
2k
µ
µ
sin(θ/2) ≡ x, so 2 sin(θ/2) = x, and cos(θ/2) dθ = dx. Then
µ
k
k
2
Z
x
1
2mβ
1 µ 2
x
θ = 0 =⇒ x = x0 = 0,
dx.
The
limits
are
σ = 2π
2 2
θ = π =⇒ x = x1 = 2k/µ.
~2
µ4 k
x0 (1 + x )
Let
So
2
2k/µ
2
2mβ
2mβ
1
1
1
1
1
=
π
−
1
−
~2
(µk)2
2 (1 + x2 ) 0
~2
(µk)2
1 + (2k/µ)2
2
2
2mβ
4(k/µ)2
4mβ
1
2mE
1
1
= π
. But k 2 =
, so
=
π
~2
(µk)2 1 + 4k 2 /µ2
~2
µ2 µ2 + 4k 2
~2
σ = 2π
σ=π
4mβ
µ~
2
1
.
(µ~)2 + 8mE
Problem 10.13
(a)
m
V (r) = αδ(r − a). Eq. 10.80 =⇒ f = −
2π~2
2mα
f = − 2 a2 ;
~
2
D = |f | =
2mα 2
a
~2
Z
m
V (r) d r = −
α4π
2π~2
3
2
;
σ = 4πD = π
4mα 2
a
~2
Z ∞
δ(r − a)r2 dr.
0
2
.
(b)
2m
Eq. 10.88 =⇒ f = − 2 α
~ κ
Z ∞
rδ(r − a) sin(κr) dr = −
0
2mα
a sin(κa)
~2 κ
(κ = 2k sin(θ/2)).
362
CHAPTER 10. SCATTERING
(c) Note first that (b) reduces to (a) in the low-energy regime (ka 1 =⇒ κa 1). Since Problem 10.4
was also for low energy, what we must confirm is that Problem 10.4 reproduces (a) in the regime for
which the Born approximation holds. Inspection shows that the answer to Problem 10.4 does reduce to
f = −2mαa2 /~2 when β 1, which is to say when f /a 1. This is the appropriate condition, since
(Eq. 10.12) f /a is a measure of the relative size of the scattered wave, in the interaction region.
Problem 10.14
F=
1 q1 q2
r̂;
4π0 r2
F⊥ =
1 q1 q2
cos φ;
4π0 r2
b
cos φ = ,
r
so
F⊥ =
1 q1 q2 b
;
4π0 r3
dt =
dx
.
v
F
q
q1
r
b
x
q2
Z
I⊥ =
F⊥ dt =
1 q1 q2 b
4π0 v
Z ∞
Z ∞
dx
=2
2 + b2 )3/2
(x
−∞
0
Z ∞
dx
−∞
(x2 + b2 )3/2
. But
∞
2
dx
2x
= 2,
= √
b
(x2 + b2 )3/2
b2 x 2 + b2 0
q1 q2 1
I⊥
q1 q2
1
=
tan θ =
=
.
mv
4π0 b( 21 mv 2 )
4π0 bE
q1 q2
1
b=
=
4π0 E tan θ
r
q1 q2
8π0 E
−1
θ = tan
so
I⊥ =
1 2q1 q2
.
4π0 bv
q1 q2
.
4π0 bE
(2 cot θ).
The exact answer is the same, only with cot(θ/2) in place of 2 cot θ. So I must show that cot(θ/2) ≈ 2 cot θ,
for small θ (that’s the regime in which the impulse approximation should work). Well:
cot(θ/2) =
cos(θ/2)
1
2
≈
= ,
sin(θ/2)
θ/2
θ
for small θ,
while 2 cot θ = 2
cos θ
1
≈ 2 . So it works.
sin θ
θ
Problem 10.15
First let’s set up the general formalism. From Eq. 10.101:
Z
Z
Z
ψ(r) = ψ0 (r) + g(r − r0 )V (r0 )ψ0 (r0 ) d3 r0 + g(r − r0 )V (r0 )
g(r0 − r1 )V (r1 )ψ0 (r1 ) d3 r1 d3 r0 + · · ·
Put in ψ0 (r) = Aeikz ,
g(r) = −
m eikr
:
2π~2 r
CHAPTER 10. SCATTERING
363
Z ik|r−r0 |
e
mA
V (r0 )eikz0 d3 r0
ψ(r) = Ae −
2
2π~
|r − r0 |
Z ik|r0 −r1 |
m 2 Z eik|r−r0 |
e
ikz1 3
+
A
V
(r
)
V
(r
)e
d
r
d3 r0 .
0
1
1
2π~2
|r − r0 |
|r0 − r1 |
ikz
In the scattering region r r0 , Eq. 10.73 =⇒
eik|r−r0 |
eikr −ik·r0
≈
e
,
|r − r0 |
r
with
k ≡ kr̂,
so
Z
m eikr
ψ(r) = A eikz −
e−ik·r0 V (r0 )eikz0 d3 r0
2π~2 r
Z ik|r0 −r1 |
m 2 eikr Z
e
−ik·r0
ikz1 3
3
e
V (r0 )
V (r1 )e
d r1 d r0
+
2π~2
r
|r0 − r1 |
f (θ, φ) = −
m
2π~2
Z
0
ei(k −k)·r V (r) d3 r +
Z ik|r−r0 |
m 2 Z
e
3
ikz0 3
−ik·r
d
r
V
(r
)e
e
V
(r)
0 d r.
0
2π~2
|r − r0 |
I simplified the subscripts, since there is no longer any possible ambiguity. For low-energy scattering we drop
the exponentials (see p. 393):
m
f (θ, φ) ≈ −
2π~2
Z
Z
m 2 Z
1
3
V (r) d r +
V (r)
V (r0 ) d r0 d3 r.
2π~2
|r − r0 |
3
Now apply this to the potential in Eq. 10.81:
Z
1
V (r0 ) d3 r0 = V0
|r − r0 |
Z a
0
1
r2 sin θ0 dr0 dθ0 dφ0 .
|r − r0 | 0
Orient the z0 axis along r, so |r − r0 | = r2 + r02 − 2rr0 cos θ0 .
Z
Z π
1
V (r0 ) d3 r0 = V0 2π
|r − r0 |
Z a
r02
Z π
0
1
1
p
sin θ0 dθ0 =
2
2
rr0
r + r0 − 2rr0 cos θ0
0
0
q
1
p
r2 + r02 − 2rr0 cos θ0
sin θ0 dθ0 dr0 . But
π
r2 + r02 − 2rr0 cos θ0
=
0
1
[(r0 + r) − |r0 − r|] =
rr0
2/r, r0 < r;
2/r0 , r0 > r.
Here r < a (from the “outer” integral), so
Z
Z r
3
Z a
1
1r
1 2
1 2
1
3
2
2
2
V (r0 ) d r0 = 4πV0
r dr0 +
r0 dr0 = 4πV0
+ (a − r ) = 2πV0 a − r .
|r − r0 |
r 0 0
r 3
2
3
r
Z
Z
V (r)
Z a
1
1
a3
1 a5
32 2 2 5
V (r0 ) d3 r0 d3 r = V0 (2πV0 )4π
a2 − r2 r2 dr = 8π 2 V02 a2 −
=
π V0 a .
|r − r0 |
3
3
3
5
15
0
f (θ) = −
m
4 3 m 2 32 2 2 5
V
πa +
π V0 a = −
0
2π~2 3
2π~2
15
2mV0 a3
3~2
4 mV0 a2
1−
.
5
~2
364
CHAPTER 10. SCATTERING
Problem 10.16
Z
d2
1
2
√
+
k
eisx g(s) ds (analog to Eq. 10.54).
G(x)
=
δ(x)
(analog
to
Eq.
10.52).
G(x)
=
dx2
2π
2
Z
Z
1
1
1
d
2
2
2
isx
√
+
k
G
=
eisx ds =⇒ g(s) = √
(−s
+
k
)g(s)e
ds
=
δ(x)
=
.
2
dx
2π
2π
2π(k 2 − s2 )
Z ∞
1
eisx
G(x) =
ds. Skirt the poles as in Fig. 10.10. For x > 0, close above:
2
2π −∞ k − s2
isx I isx 1
eikx
e
1
1
e
G(x) = −
= −i
ds = − 2πi
. For x < 0, close below:
2π
s+k s−k
2π
s + k s=k
2k
isx I isx e−ikx
e
1
1
e
1
= −i
ds =
2πi
.
G(x) = +
2π
s−k s+k
2π
s − k s=−k
2k
In either case, then, G(x) = −
Z
ψ(x) =
i ik|x|
e
.
2k
(Analog to Eq. 10.65.)
i 2m
2m
G(x − x0 ) 2 V (x0 )ψ(x0 ) dx0 = −
~
2k ~2
Z
eik|x−x0 | V (x0 )ψ(x0 ) dx0 ,
plus any solution ψ0 (x) to the homogeneous Schrödinger equation:
2
d
2
+ k ψ0 (x) = 0. So:
dx2
ψ(x) = ψ0 (x) −
im
~2 k
Z ∞
eik|x−x0 | V (x0 )ψ(x0 ) dx0 .
−∞
Problem 10.17
For the Born approximation let ψ0 (x) = Aeikx , and ψ(x) ≈ Aeikx .
Z
im ∞ ik|x−x0 |
ikx
ikx0
ψ(x) ≈ A e − 2
e
V (x0 )e
dx0
~ k −∞
Z x
Z
im
im ∞ ik(x0 −x)
ik(x−x0 )
ikx0
ikx0
ikx
e
V (x0 )e
e
V (x0 )e
=A e − 2
dx0 − 2
dx0 .
~ k −∞
~ k x
Z x
Z ∞
im
im
ψ(x) = A eikx − 2 eikx
V (x0 ) dx0 − 2 e−ikx
e2ikx0 V (x0 ) dx0 .
~ k
~ k
−∞
x
Now assume V (x) is localized; for large positive x, the third term is zero, and
Z
im ∞
ikx
ψ(x) = Ae
1− 2
V (x0 ) dx0 . This is the transmitted wave.
~ k −∞
For large negative x the middle term is zero:
Z
im −ikx ∞ 2ikx0
ikx
ψ(x) = A e − 2 e
e
V (x0 )dx0 .
~ k
−∞
CHAPTER 10. SCATTERING
365
Evidently the first term is the incident wave and the second the reflected wave:
R=
m 2 Z ∞
~2 k
2
e2ikx V (x) dx .
−∞
If you try in the same spirit to calculate the transmission coefficient, you get
2
Z
2
m 2 Z ∞
im ∞
T = 1− 2
V (x)dx = 1 + 2
V (x)dx ,
~ k −∞
~ k
−∞
which is nonsense (greater than 1). The first Born approximation gets R right, but all you can say to this order
is T ≈ 1 (you would do better using T = 1 − R).
Problem 10.18
Z ∞
Delta function:
V (x) = −αδ(x).
e2ikx V (x) dx = −α,
so
R=
−∞
mα 2
~2 k
,
or, in terms of energy (k 2 = 2mE/~2 ):
R=
m2 α2
mα2
= 2 ;
2
2mE~
2~ E
The exact answer (Eq. 2.144) is
1
mα2
1 + 2~
2E
Finite square well:
mα2
.
2~2 E
T =1−R= 1−
V (x) =
≈1−
mα2
,
2~2 E
−V0 (−a < x < a)
0
(otherwise)
so they agree provided E mα2
.
2~2
.
a
V0 e2ika − e−2ika
V0
e2ikx
=−
= − sin(2ka).
e
V (x) dx = −V0
e
dx = −V0
2ik −a
k
2i
k
−∞
−a
2
2
2a √
m 2 V0
V0
sin(2ka)
.
sin
So R =
T
=
1
−
2mE
.
~2 k
k
2E
~
Z ∞
Z a
2ikx
2ikx
If E V0 , the exact answer (Eq. 2.172) becomes
2
2
V0
2a √
V0
2a √
−1
T ≈1+
sin
2mE
=⇒ T ≈ 1 −
sin
2mE
,
2E
~
2E
~
so they agree provided E V0 .
Problem 10.19
The Legendre polynomials satisfy Pl (1) = 1 (see footnote 39, p. 124), so Eq. 10.47 ⇒
∞
f (0) =
1X
(2l + 1)eiδl sin δl . Therefore
k
∞
Im[f (0)] =
l=0
and hence (Eq. 10.48):
σ=
4π
Im[f (0)]. QED
k
1X
(2l + 1) sin2 δl ,
k
l=0
366
CHAPTER 10. SCATTERING
Problem 10.20
Using Eq. 10.88 and integration by parts:
Z
Z
2mA ∞ d
2m ∞
1 −µr2
−µr 2
rAe
sin(κr) dr = − 2
f (θ) = − 2
− e
sin(κr) dr
~ κ 0
~ κ 0 dr
2µ
Z ∞
∞
2mA
−µr 2 d
−µr 2
e
−
e
sin(κr)
[sin(κr)]
dr
=
2µ~2 κ
dr
0
0
√
Z ∞
2
π −κ2 /4µ
mA
mA
−µr
=
0
−
κ
e
e
cos(κr)
dr
=
−
√
µ~2 κ
µ~2 2 µ
0
√
2
mA π
= − 2 3/2 e−κ /4µ , where κ = 2k sin(θ/2) (Eq. 10.89).
2~ µ
From Eq. 10.14, then,
dσ
πm2 A2 −κ2 /2µ
e
=
,
dΩ
4~4 µ3
and hence
Z
2
2
dσ
πm2 A2
σ =
dΩ =
e−4k sin (θ/2)/2µ sin θ dθ dφ
dΩ
4~4 µ3
Z
π 2 m2 A2 π −2k2 sin2 (θ/2)/µ
e
sin θ dθ; write sin θ = 2 sin(θ/2) cos(θ/2)
=
2~4 µ3 0
Z
Z
π 2 m2 A2 1 −2k2 x2 /µ
2π 2 m2 A2 1 −2k2 x2 /µ
=
e
2x 2 dx =
xe
dx
2~4 µ3 0
~4 µ3
0
1
π 2 m2 A2 −2k2 /µ
2π 2 m2 A2 h µ −2k2 x2 /µ i
=
−
e
=
−
e
−
1
~4 µ3
4k 2
2~4 µ2 k 2
0
π 2 m 2 A2 −2k2 /µ
1
−
e
.
=
2~4 µ2 k 2
Z
and let x ≡ sin(θ/2)
Problem 10.21
(a) In the first Born approximation (Equation 10.79):
Z
Z
0
m
m 2π~2 b X
i(k0 −k)·r0
3
e
V
(r
)
d
r
=
−
ei(k −k)·r0 δ 3 (r0 − ri ) d3 r0
0
0
2
2
2π~
2π~
m
i
X
X
0
i(k −k)·ri
−iq·ri
= −b
e
= −b
e
;
f (θ, φ) ≈ −
i
i
X
2
dσ
= |f |2 = b2
e−iq·ri . X
dΩ
i
(b)
X
i
e
−iq·ri
=
X
l,m,n
e
−i(qx la+qy ma+qz na)
=
N
−1
X
l=0
e
−iqx la
N
−1
X
m=0
e
−iqy ma
N
−1
X
n=0
e−iqz na .
CHAPTER 10. SCATTERING
367
Now, letting u ≡ exp(−iqx a),
N
−1
X
e
−iqx la
=
N
−1
X
l=0
l=0
e−iqx aN/2 eiqx aN/2 − e−iqx aN/2
1 − e−iqx aN
1 − uN
=
=
u =
1−u
1 − e−iqx a
e−iqx a/2 eiqx a/2 − e−iqx a/2
l
= e−iqx a(N −1)/2
sin(qx aN/2)
,
sin(qx a/2)
so
dσ
sin2 (qx aN/2) sin2 (qy aN/2) sin2 (qz aN/2)
. X
= b2
dΩ
sin2 (qx a/2) sin2 (qy a/2) sin2 (qz a/2)
(c) Here’s the graph:
1
Sin N x
f[N_, x_] :=
�������
Plot[{f[1, x], f[5, x], f[10, x]}, {x, 0, 4 π}, PlotRange → {0, 10}]
N
2
Sin x
2
2
�������
10
8
6
�������
4
2
0
2
4
6
8
10
12
(d) The vectors k, k0 and G form an isosceles triangle (Figure 10.11), and
sin(θ/2) =
πp2
G/2
=
l + m2 + n2 ,
k
ka
so, in terms of the wavelength λ = 2π/k,
θ = 2 arcsin
λp2
2
2
l +m +n .
2a
The smallest (nonzero) angles occur when (l2 + m2 + n2 ) = 1, 2, or 3, corresponding to angles
π/3, π/2, and 2π/3.
Problem 10.22
~2
(a) −
2m
∂2
1 ∂
1 ∂2
+
+
ψ(r, θ) + V (r)ψ(r, θ) = Eψ(r, θ). Let ψ(r, θ) = R(r) Θ(θ) :
∂r2
r ∂r r2 ∂θ2
2
~2
d R 1 dR
R d2 Θ
Θ
+
+ 2 2 + V (r) RΘ = E RΘ. Multiply by r2 and divide by RΘ:
−
2m
dr2
r dr
r dθ
368
CHAPTER 10. SCATTERING
r dR
~2 r2 d2 R
~2 1 d2 Θ
+
+ r2 V (r) = r2 E. The Θ term must be constant:
−
−
2m R dr2
R dr
2m Θ dθ2
1 d2 Θ
= −j 2
Θ dθ2
⇒
d2 Θ
= −j 2 Θ
dθ2
⇒
Θ(θ) = Aeijθ ,
where A is a constant (which we might as well absorb into R), and j could be positive or negative (it has to be
an integer, since Θ(θ + 2π) = Θ(θ)). This leaves the radial equation,
dR
~2 2
~2
d2 R
r2 2 + r
+
j R + r2 V (r)R = r2 E R.
−
2m
dr
dr
2m
Let
1 d2 u
1 du 3 u
dR
1 du 1 u
d2 R
u
=√
− 3/2
=√
−
,
+
. Then
r R(r), so R = √ ,
2
3/2
dr
2r
dr
dr
4 r5/2
r
r dr
r dr2
r
~2
1 du 3 u
u
u
1 d2 u
1
1 du 1 u
~2 2 1 u
√
√
√ + V (r) √ = E √ ;
−
−
+
+
−
+
j
2
3/2
5/2
3/2
2m
dr
4r
r
2r
2m r2 r
r dr
r dr
r
r
r
u(r) ≡
√
u
ψ(r, θ) = √ eijθ ,
r
−
~2 d2 u
~2 (j 2 − 1/4)
+
V
(r)
+
u = Eu.
2m dr2
2m
r2
√
~2 d2 u
d2 u
= Eu ⇒
= −k 2 u where k ≡ 2mE
. The general solution is u(r) = Aeikr + Be−ikr , but
~
2
2
2m dr
dr
eikr
for an outgoing wave we want only the first term: R(r) = A √ . Notice that this asymptotic form is the same
r
for all j, so when we form the general linear combination of separable solutions (from Part (a)), at large r the
eikr
radial term will factor out, leaving some function of θ: ψ ≈ Af (θ) √ . X
r
2
2
2
2
2
~
∂
~
∂
~
∇2 ψ = −
+ 2 Aeikx = −
(−k 2 )Aeikx = E ψ. X
−
2m
2m ∂x2
∂y
2m
(b) −
(c)
−
~2 d2 u
~2 (j 2 − 1/4)
+
u = Eu
2m dr2
2m
r2
⇒
d2 u j 2 − 1/4)
−
u = −k 2 u.
dr2
r2
This is a form of Bessel’s
√ equation; in Boas’ notation (her Equation 16.1), a = 1/2, b = k, c = 1, p = j, and
the solution is u(r) = r Zj (kr), where Z is a Bessel/Neumann/Hankel function of order j. The outgoing one
√ (1)
(1)
is H (1) , so u(r) = rHj (kr), the separable solution is ψ(r, θ) = Hj (kr)eijθ , and the general solution is the
linear combination
∞
X
(1)
ψout (r, θ) = A
cj Hj (kr) eijθ .
−∞
Combining this with the incident wave (Aeikx ) we have
ψ(r, θ) = A e
ikx
+
∞
X
!
(1)
cj Hj (kr) eijθ
.
−∞
(d) At large r:
(
ψ(r, θ) ∼ A e
r
ikx
+
2 −iπ/4
e
π
∞
X
−∞
!
j ijθ
cj (−i) e
eikr
√
kr
)
.
CHAPTER 10. SCATTERING
369
Comparing Equation (10.109), we read off
r
f (θ) =
∞
2 −iπ/4 X
e
cj (−i)j eijθ . X
πk
−∞
(e) Following the argument leading to Equation (10.14), the probability that an incident particle (traveling at
speed v) passes through an infinitesimal length (differential impact parameter) db is
|A|2 |f |2
(v dt)r dθ ⇒ db = |f |2 dθ ⇒ D(θ) = |f (θ)|2 . X
r



Z 2π
Z 2π
Z 2π
X
X
0
0
2 
|f (θ)|2 dθ =
D(θ) dθ =
B=
(−i)j cj eijθ   (i)j c∗j 0 e−ij θ  dθ
πk
0
0
0
0
j
dP = |ψincident |2 da = |A|2 (v dt)db = |ψscattered |2 da =
j
=
2 X j 0 −j ∗
(i)
cj cj 0
πk 0
Z 2π
j,j
0
ei(j−j )θ dθ =
0
∞
2 X j 0 −j ∗
4 X
(i)
cj cj 0 (2πδj,j 0 ) =
|cj |2 . X
πk 0
k j=−∞
j,j
(f)
ψ(a, θ) = A

∞
 X

∞
X
(i)j Jj (ka)eijθ +
(1)
cj Hj (ka)eijθ
∞ X
=A

j=−∞
j=−∞


(1)
ij Jj + cj Hj
eijθ = 0.
j=−∞
0
Each coefficient must vanish (prove it, if you like, by multiplying by e−ij θ and integrating over θ from 0 to 2π):
(1)
i Jj (ka) + cj Hj (ka) = 0
j
(1)
To create the plot, note that Hj
Jj
(1)
Hj
⇒
cj = −
ij Jj (ka)
(1)
⇒
Hj (ka)
∞
4 X
Jj (ka)
B=
k j=−∞ H (1) (ka)
= Jj + iNj , while Jj and Nj (the Neumann function) are real, so
−1
= 1 + (Nj /Jj )2
. Then
100
2 -1
b[x_] :=
�������
Plot[b[x], {x, 0, 2}, PlotRange → {0, 15}]
x j=-100
1+
BesselY[j, x]
�������
BesselJ[j, x]
14
12
10
8
�������
6
4
2
0.0
��������
N[b[100]]
��������
4.07415
.
j
2
4
2
0.5
1.0
1.5
2.0
370
CHAPTER 10. SCATTERING
Problem 10.23
(a) The center of mass and relative coordinates are R = (r1 + r2 )/2 and r = r2 − r1 . Therefore interchanging
the two particles leaves R untouched, but switches the sign of r. Since the wave function must be
symmetric under interchange, this means ψr (−r) = ψr (r). X
(b) Because the potential is spherically symmetric, if ψr (r) is a solution to Equation 10.118 with energy Er , so
too is ψr (−r). We can therefore construct a symmetric solution from the sum: ψr (r) + ψr (−r). Applying
this to Equation 10.12 gives
ψ(r, θ) = A e
ikz
+e
−ikz
eikr
+ [f (θ) + f (π − θ)]
r
,
using the fact that in polar coordinates r → −r is achieved by (r, θ, φ) → (r, π − θ, φ + π). We can then
read off the scattering amplitude: fB (θ) = f (θ) + f (π − θ). X
(c) From Equation 10.25,
fB (θ) =
∞
X
(2` + 1)a` [P` (cos θ) + P` (cos(π − θ))]
`=0
∞
X
X
=
(2` + 1)a` [P` (cos θ) + P` (− cos θ)] =
2(2` + 1)a` P` (cos θ),
`=0
` even
since P` (−x) = (−1)` P` (x) (Equation 4.28).
(d) In this case the wave function must be odd: ψr (−r) = −ψr (r). We construct antisymmetric solutions,
ψr (r) − ψr (−r), and by the same reasoning we get fF (θ) = f (θ) − f (π − θ). This time the partial waves
include only odd values of `.
(e) fF (π/2) = f (π/2) − f (π − π/2) = f (π/2) − f (π/2) = 0. X
1
1
1
fB/F = A
±
. But sin2 (θ/2) = [1 − cos θ], and
2
2
2
sin (θ/2) sin ((θ − π/2)/2)
1
1
1
1
2
hence sin ((θ − π/2)/2) = [1 − cos(θ − π/2)] = [1 + cos θ], so fB/F = 2A
±
=
2
1 − cos θ 1 + cos θ
2
(1 + cos θ) ± (1 − cos θ)
1
cos θ
2A
. fB = 4A
, fF = 4A
.
1 − cos2 θ
sin2 θ
sin2 θ
A
(f ) f (θ) =
;
2
sin (θ/2)
dσ
dΩ
B
= (4A)2
1
sin4 θ
;
dσ
dΩ
F
= (4A)2
cos2 θ
sin4 θ
In the plot below the dashed line is (dσ/dΩ)B , and the solid line is (dσ/dΩ)F .
.
CHAPTER 10. SCATTERING
371
x, 0,
π
�������
LogPlot
Sin[x] -4 ,
,
2
PlotRange → .001, 105 , PlotStyle -> Dashing[{0.01, 0.01}]
�������
LogPlot
Cos[x] 2 Sin[x] -4 ,
��������
Show[%9, %5]
x, 0,
π
2
, PlotRange → .001, 105
104
100
��������
1
0.01
0.0
0.5
1.0
1.5
372
CHAPTER 11. QUANTUM DYNAMICS
Chapter 11
Quantum Dynamics
Problem 11.1
(a) This equation can be solved by separation of variables:
df
= kf
dt
df
= k dt
f
Z
Z
df
= k dt
f
ln f = k t + C
f (t) = f0 ek t ,
where we’ve defined the new constant f0 = eC in the last line.
(b) If k is a function of time, the same procedure holds. Skipping to the third line in the derivation of part
(a):
Z f
df 0
=
0
f0 f
Z t
k(t0 ) dt0
0
Z t
k(t0 ) dt0
0
Z t
0
0
f (t) = f0 exp
k(t ) dt ,
ln f − ln f0 =
0
where this time I’ve done the integrals as definite integrals rather than specify the integration constant at
the end.
(d) These two are not the same because Ĥ is an operator and for operators  and B̂
eÂ+B̂ = eÂ+B̂
only if  and B̂ commute.
CHAPTER 11. QUANTUM DYNAMICS
373
Problem 11.2
ψnlm = Rnl Ylm . From Tables 4.3 and 4.7:
1 r −r/2a
1
e−r/a ; ψ200 = √
1−
e
;
ψ100 = √
2a
πa3
8πa3
ψ210 = √
1
r −r/2a
e
cos θ;
ψ21±1 = ∓ √
32πa3 a
1
r −r/2a
e
sin θ e±iφ .
64πa3 a
But r cos θ = z and r sin θe±iφ = r sin θ(cos φ ± i sin φ) = r sin θ cos φ ± ir sin θ sin φ = x ± iy. So |ψ|2 is an
R
even function of z in all cases, and hence z|ψ|2 dx dy dz = 0, so Hii0 = 0. Moreover, ψ100 is even in z, and
0
so are ψ200 , ψ211 , and ψ21−1 , so Hij
= 0 for all except
1
1
1
0
√
H100,210
= eE √
3
3
πa
32πa a
eE
= √
4 2πa4
or
Z ∞
4 −3r/2a
r e
Z
Z π
dr
0
e
eE
z d r= √
4 2πa4
−r/a −r/2a 2
e
2
Z 2π
cos θ sin θ dθ
0
3
0
Z
e−3r/2a r2 cos2 θ r2 sin θ dr dθ dφ
eE
4!
dφ = √
4 2πa4
2a
3
5
2
2π =
3
28
√
35 2
eEa,
0.7449 eEa.
Problem 11.3
i 0 −iω0 t
ċa = − Hab
e
cb ;
~
i 0 iω0 t
ċb = − Hba
e
ca . Differentiating with respect to t :
~
i 0 i 0 −iω0 t
i 0 iω0 t
i 0 iωo t
iω0 t
iω0 t
c̈b = − Hba iω0 e
ca + e
ċa = iω0 − Hba e
ca − Hba e
− Hab e
cb , or
~
~
~
~
1
1
0 2
0 2
|Hab
| cb . Let α2 ≡ 2 |Hab
| . Then c̈b − iω0 ċb + α2 cb = 0.
2
~
~
This is a linear differential equation with constant coefficients, so it can be solved by a function of the form
cb = eλt :
q
q
1
i
iω0 ± −ω02 − 4α2 = (ω0 ± ω) , where ω ≡ ω02 + 4α2 .
λ2 − iω0 λ + α2 = 0 =⇒ λ =
2
2
c̈b = iω0 ċb −
The general solution is therefore
cb (t) = Aei(ω0 +ω)t/2 + Bei(ω0 −ω)t/2 = eiω0 t/2 Aeiωt/2 + Be−iωt/2 , or
cb (t) = eiω0 t/2 [C cos (ωt/2) + D sin (ωt/2)] . But cb (0) = 0,
so C = 0,
and hence
cb (t) = Deiω0 t/2 sin (ωt/2) . Then
i
h
iω0 iω0 t/2
ω iω0 t/2
ω
ω0
i 0 iω0 t
ċb = D
e
sin (ωt/2) + e
cos (ωt/2) = Deiω0 t/2 cos (ωt/2) + i sin (ωt/2) = − Hba
e
ca .
2
2
2
ω
~
i
i~ ω −iω0 t/2 h
ω0
i~ ω
ca = 0
e
D cos (ωt/2) + i sin (ωt/2) . But ca (0) = 1, so
0 2 D = 1. Conclusion:
Hba 2
ω
Hba
374
CHAPTER 11. QUANTUM DYNAMICS
h
i
ω0
ca (t) = e−iω0 t/2 cos (ωt/2) + i sin (ωt/2) ,
ω
0
2Hba
iω0 t/2
cb (t) =
e
sin (ωt/2) ,
i~ω
where ω ≡
q
ω02 + 4
0 |2
|Hab
~2 .
ω02
4|H 0 |2
sin2 (ωt/2) + 2 ab2 sin2 (ωt/2)
2
ω ~ ω
0 2
|H
|
1
sin2 (ωt/2) = cos2 (ωt/2) + sin2 (ωt/2) = 1. X
= cos2 (ωt/2) + 2 ω02 + 4 ab
ω
~2
|ca |2 + |cb |2 = cos2 (ωt/2) +
[In light of the Comment you might question the initial conditions. If the perturbation includes a factor θ(t),
are we sure this doesn’t alter ca (0) and cb (0)? That is, are we sure ca (t) and cb (t) are continuous at a step
function potential? The answer is “yes”, for if we integrate Eq. 11.17 from − to ,
Z i 0
ca () − ca (−) = − Hab
e−iω0 t cb (t) dt.
~
0
But |cb (t)| ≤ 1, so the integral goes to zero as → 0, and hence ca (−) = ca (). The same goes for cb , of course.]
Problem 11.4
This is a tricky problem, and I thank Prof. Onuttom Narayan for showing me the correct solution. The safest
approach is to represent the delta function as a sequence of rectangles:
(1/2), − < t < ,
δ (t) =
0,
otherwise.
Then Eq. 11.17 ⇒

t < − :








t > :
ca (t) = 1, cb (t) = 0,
ca (t) = a, cb (t) = b,


iα −iω0 t

e
cb ,
 ċa = − 2~




−
<
t
<
:


∗


iω0 t
ċb = − iα
ca .
2~ e
In the interval − < t < ,
d2 cb
iα∗
−iα −iω0 t
iα∗
i2~ dcb
iα
|α|2
dcb
iω0 t
iω0 t
=
−
iω
e
c
+
e
e
c
=
−
iω
−
c
=
iω
−
cb .
0
a
b
0
b
0
dt2
2~
2~
2~
α∗ dt
2~
dt
(2~)2
Thus cb satisfies a homogeneous linear differential equation with constant coefficients:
|α|2
d2 cb
dcb
− iω0
+
cb = 0.
2
dt
dt
(2~)2
Try a solution of the form cb (t) = eλt :
iω0 ±
|α|2
λ − iω0 λ +
=0⇒λ=
2
(2~)
2
or
λ=
iω0
iω
± , where ω ≡
2
2
q
p
−ω02 − |α|2 /(~)2
,
2
ω02 + |α|2 /(~)2 .
CHAPTER 11. QUANTUM DYNAMICS
375
The general solution is
cb (t) = eiω0 t/2 Aeiωt/2 + Be−iωt/2 .
But
cb (−) = 0 ⇒ Ae−iω/2 + Beiω/2 = 0 ⇒ B = −Ae−iω ,
so
cb (t) = Aeiω0 t/2 eiωt/2 − e−iω(+t/2) .
Meanwhile
iω 2i~ −iω0 t
2i~ −iω0 t/2
iω0 iωt/2
−iω(+t/2)
iωt/2
−iω(+t/2)
e
ċ
=
e
A
+
e
−
e
e
+
e
b
α∗
α∗
2
2
i
h
~ −iω0 t/2
= − ∗e
A (ω + ω0 )eiωt/2 + (ω − ω0 )e−iω(+t/2) .
α
ca (t) =
But
~ i(ω0 −ω)/2
2~ω
α∗ i(ω−ω0 )/2
e
A [(ω + ω0 ) + (ω − ω0 )] = − ∗ ei(ω0 −ω)/2 A, so A = −
e
.
∗
α
α
2~ω
i
1 −iω0 (t+)/2 h
ca (t) =
e
(ω + ω0 )eiω(t+)/2 + (ω − ω0 )e−iω(t+)/2
2ω
ω0
ω(t + )
ω(t + )
−iω0 (t+)/2
+ i sin
;
= e
cos
2
ω
2
i
α∗ iω0 (t−)/2 h iω(t+)/2
iα∗ iω0 (t−)/2
ω(t + )
−iω(t+)/2
cb (t) = −
e
e
−e
=−
e
sin
.
2~ω
~ω
2
ca (−) = 1 = −
Thus
i
h
ω0
a = ca () = e−iω0 cos(ω) + i sin(ω) ,
ω
b = cb () = −
iα∗
sin(ω).
~ω
This is for the rectangular pulse; it remains to take the limit → 0: ω → |α|/~,
a → cos
|α|
~
ω0 ~
+i
sin
|α|
|α|
~
→ cos
|α|
~
,
iα∗
b→−
sin
|α|
and we conclude that for the delta function
ca (t) =
cb (t) =
1,
t < 0,
cos(|α|/~), t > 0;

 0, r
 −i
t < 0,
α∗
α
sin(|α|/~), t > 0.
Obviously, |ca (t)|2 + |cb (t)|2 = 1 in both time periods. Finally,
Pa→b = |b|2 = sin2 (|α|/~).
so
|α|
~
,
376
CHAPTER 11. QUANTUM DYNAMICS
Problem 11.5
(a)

i
0
0 −iω0 t 

Eq. 11.14 =⇒ ċa = − ca Haa + cb Hab e
~
 (these are exact, and replace Eq. 11.15).
i
0
0 iω0 t

Eq. 11.17 =⇒ ċb = − cb Hbb
+ ca Hba
e
~
ca (0) = 1,
Initial conditions:
cb (0) = 0.
Zeroth order: ca (t) = 1, cb (t) = 0.

Z

i t 0 0 0

i

0
Haa (t ) dt
 ċa = − Haa
=⇒ ca (t) = 1 − ~
~
0
Z
First order:
t
i 0 iω0 t
0

0

=⇒ cb (t) = − i

Hba
(t0 )eiω0 t dt0
 ċb = − ~ Hba e
~ 0
Z t
2
Z
Z
i t 0 0 0
i t 0 0 0
1
0
|ca |2 = 1 −
Haa (t ) dt
1+
Haa (t ) dt = 1 +
Haa
(t0 ) dt0 = 1 (to first order in H 0 ).
~ 0
~ 0
~ 0
Z t
Z t
0
0
i
i
|cb |2 = −
H 0 (t0 )eiω0 t dt0
H 0 (t0 )e−iω0 t dt0 = 0 (to first order in H 0 ).
~ 0 ba
~ 0 ab
So |ca |2 + |cb |2 = 1 (to first order).
(b)
t
0
0
0
i
d˙a = e ~ 0 Haa (t ) dt
R
Rt 0
0
0
i
i 0
i
0
0 −iω0 t
Haa ca + e ~ 0 Haa (t ) dt ċa . But ċa = − ca Haa
+ cb Hab
e
~
~
Two terms cancel, leaving
t
0
0
0
i
i i t 0 0 0
0 −iω0 t
d˙a = − e ~ 0 Haa (t ) dt cb Hab
e
. But cb = e− ~ 0 Hbb (t ) dt db .
~
0
0
0
i
i i Rt 0 0
0 −iω0 t
0 −iω0 t
e
db .
= − e ~ 0 [Haa (t )−Hbb (t )]dt Hab
e
db , or d˙a = − eiφ Hab
~
~
R
R
Similarly,
t
0
0
0
i
d˙b = e ~ 0 Hbb (t ) dt
R
Rt 0 0
0
i
i 0
i
0
0 iω0 t
Hbb cb + e ~ 0 Hbb (t ) dt ċb . But ċb = − cb Hbb
+ ca Hba
e
.
~
~
Rt 0
0
0
i
i i Rt 0 0 0
0 iω0 t
= − e ~ 0 Hbb (t ) dt ca Hba
e
. But ca = e− ~ 0 Haa (t ) dt da .
~
0
0
0
i i Rt 0 0
i
0 iω0 t
0 iω0 t
= − e ~ 0 [Hbb (t )−Haa (t )]dt Hba
e
da = − e−iφ Hba
e
da . QED
~
~
(c)
Initial conditions:
Zeroth order:
First order:
ca (0) = 1 =⇒ da (0) = 1;
da (t) = 1,
cb (0) = 0 =⇒ db (0) = 0.
db (t) = 0.
t
0
0
0
i
d˙a = 0 =⇒ da (t) = 1 =⇒ ca (t) = e− ~ 0 Haa (t ) dt .
R
CHAPTER 11. QUANTUM DYNAMICS
377
i
i
0 iω0 t
d˙b = − e−iφ Hba
e
=⇒ db = −
~
~
i i Rt 0 0 0
cb (t) = − e− ~ 0 Hbb (t )dt
~
Z t
Z t
0
0
0
e−iφ(t ) Hba
(t0 )eiω0 t dt0 =⇒
0
0
0
0
e−iφ(t ) Hba
(t0 )eiω0 t dt0 .
0
These don’t look much like the results in (a), but remember, we’re only working to first order in H 0 ,
Rt 0 0 0
0
(t ) dt (to this order), while for cb , the factor Hba
in the integral means it is
so ca (t) ≈ 1 − ~i 0 Haa
−iφ
already firstR order and hence both the exponential factor in front and e
should be replaced by 1. Then
0
t
0
cb (t) ≈ − ~i 0 Hba
(t0 )eiω0 t dt0 , and we recover the results in (a).
Problem 11.6
(0)
c(0)
cb (t) = b.
a (t) = a,

Z
i 0 −iω0 t
ib t 0 0 −iω0 t0 0

(1)

ċ
=
−
H
e
b
=⇒
c
(t)
=
a
−
H (t )e
dt .
 a
a
~ ab
~ Z0 ab
First order:
t
0

 ċb = − i H 0 eiω0 t a =⇒ c(1) (t) = b − ia

H 0 (t0 )eiω0 t dt0 .
ba
b
~
~ 0 ba
Z
i 0 −iω0 t
ia t 0 0 iω0 t0 0
Second order: ċa = − Hab
e
Hba (t )e
dt =⇒
b−
~
~ 0
Zeroth order:
ib
c(2)
a (t) = a −
~
Z t
0
0
Hab
(t0 )e−iω0 t dt0 −
0
Z t
a
~2
"Z 0
t
0
0
Hab
(t0 )e−iω0 t
0
#
00
0
Hba
(t00 )eiω0 t dt00
dt0 .
0
To get cb , just switch a ↔ b (which entails also changing the sign of ω0 ):
ia
(2)
cb (t) = b −
~
Z t
0
0
Hba
(t0 )eiω0 t dt0 −
0
b
~2
Z t
0
0
0
Hba
(t0 )eiω0 t
"Z
t0
#
00
0
Hab
(t00 )e−iω0 t dt00
dt0 .
0
Problem 11.7
0
For H independent of t, Eq. 11.21
i 0
(2)
(1)
=⇒ cb (t) = cb (t) = − Hba
~
0
t
Z t
0
eiω0 t dt0 =⇒
0
0
Hba
eiω0 t − 1 . Meanwhile Eq. 11.22 =⇒
~ω0
0
"Z 0
#
Z t
Z t
t
0
1
1
(2)
0 2
−iω0 t0
iω0 t00 00
0 2 1
ca (t) = 1 − 2 |Hab |
e
e
dt dt0 = 1 − 2 |Hab
|
1 − e−iω0 t dt0
~
~
iω0 0
0
0
!
t
0
i
i
e−iω0 t
1
0 2
0
0 2
−iω0 t
= 1+
|H |
t +
|H | t +
e
−1 .
= 1+
ω0 ~2 ab
iω0
ω0 ~2 ab
iω0
i 0 eiω0 t
(2)
cb (t) = − Hba
~
iω0
= −
0
378
CHAPTER 11. QUANTUM DYNAMICS
For comparison with the exact answers (Problem 11.3), note first that cb (t) is already first order (because of
0
the Hba
in front), whereas ω differs from ω0 only in second order, so it suffices to replace ω → ω0 in the exact
formula to get the second-order result:
0
0
H0
2Hba
2Hba
1 iω0 t/2
cb (t) ≈
e
− e−iω0 t/2 = − ba eiω0 t − 1 ,
eiω0 t/2 sin (ω0 t/2) =
eiω0 t/2
i~ω0
i~ω0
2i
~ω0
in agreement with the result above. Checking ca is more difficult. Note that
s
0 2
0 |2
4|Hab
|Hab
|
|H 0 |2
ω0
|H 0 |2
≈
ω
1
+
2
= ω0 + 2 ab2 ;
≈ 1 − 2 2ab 2 .
ω = ω0 1 +
0
2
2
2
2
ω0 ~
ω0 ~
ω0 ~
ω
ω0 ~
Taylor expansion:

0 2
0 2
|Hab
| t
ω0 t |Hab
| t


≈
cos
(ω
t/2)
−
sin (ω0 t/2)
+
 cos(x + ) = cos x − sin x =⇒ cos (ωt/2) = cos
0
2
ω0 ~ ω0 ~2
2
0 2
ω0 t |Hab
|H 0 |2 t
| t


≈ sin (ω0 t/2) + ab 2 cos (ω0 t/2)
+
 sin(x + ) = sin x + cos x =⇒ sin (ωt/2) = sin
2
2
ω0 ~
ω0 ~
0 2
0 2
0 2
ω0 t
ω0 t
ω0 t
|Hab
| t
|Hab
|
|Hab
| t
ω0 t
−
sin
+i 1−2 2 2
+
cos
sin
ca (t) ≈ e
cos
2
ω 0 ~2
2
ω0 ~
2
ω 0 ~2
2
0 2
ω0 t
ω0 t
ω0 t
ω0 t
ω0 t
|Hab |
2i
−iω0 t/2
= e
cos
sin
+ i sin
−
t sin
− i cos
+
2
2
ω0 ~2
2
2
ω0
2
0 2
|H |
2i 1
= e−iω0 t/2 eiω0 t/2 − ab2 −iteiω0 t/2 +
eiω0 t/2 − e−iω0 t/2
ω0 ~
ω0 2i
0 2
|
1
i
1
|Hab
−iω0 t
0 2
−iω0 t
−it +
1−e
=1+
|H | t +
e
− 1 , as above. X
= 1−
ω 0 ~2
ω0
ω0 ~2 ab
iω0
−iω0 t/2
Problem 11.8
(a) Using Equation (11.21),
(1)
cb (t) = −
i
~
Z t
0
0
Hba
(t0 )eiω0 t dt0 = −
−∞
iα
√
~τ π
Z t
0
2
0
e−(t /τ ) eiω0 t dt0 .
−∞
0 2
0
2 t
t
ω0 τ
ω0 τ 2
0
. Let t → ∞ and t0 /τ ≡ u:
Complete the square: −
+ iω0 t = −
−i
−
τ
τ
2
2
Z ∞
2
2√
2
2
iα
iα
iα
(1)
e−u du = − √ e−(ω0 τ /2) π = − e−(ω0 τ ) /4 .
cb (∞) = − √ e−(ω0 τ /2) τ
~
~τ π
~
π
−∞
The probability of a transition, then, is
(1)
Pa→b ≈ |cb (∞)|2 =
α 2
~
1
2
exp − (ω0 τ ) .
2
(b) In the limit τ → 0,
Pa→b ≈
α 2
.
~
The exact result (from Problem 11.4) is sin2 (α/~), and expanding this as a power series we get Pa→b =
(α/~)2 + · · · . First-order perturbation theory is precisely the lowest-order term.
CHAPTER 11. QUANTUM DYNAMICS
379
(c) In this limit Pa→b = 0 (no transition occurs). [Interestingly, it is not zero for all times—only as t → ∞
(where the perturbation turns off). In fact, if you compute the probability of a transition at a finite time
you get:
α 2
2
1
t
ω0 τ 2
e−(ω0 τ ) /2 1 − erfc
−i
.
Pa→b (t) =
~
2
τ
2
For large ω0 τ , the rise and fall of the transition probability can also be computed from the overlap of the
instantaneous energy eigenstate with the unperturbed state ψb .]
Problem 11.9
(a)
ċa = −
i
Vab eiωt e−iω0 t cb ;
2~
ċb = −
i
Vba e−iωt eiω0 t ca .
2~
Differentiate the latter, and substitute in the former:
i
Vba h
i(ω0 − ω)ei(ω0 −ω)t ca + ei(ω0 −ω)t ċa
2~
Vba i(ω0 −ω)t
Vab −i(ω0 −ω)t
Vba i(ω0 −ω)t
|Vab |2
e
ca − i
e
−i
e
cb = i(ω0 − ω)ċb −
cb .
= i(ω0 − ω) −i
2~
2~
2~
(2~)2
c̈b = −i
d2 cb
dcb
|Vab |2
|Vab |2
λt
2
+
i(ω
−
ω
)
+
c
=
0.
Solution
is
of
the
form
c
=
e
:
λ
+
i(ω
−
ω
)λ
+
= 0.
0
b
b
0
dt2
dt
4~2
4~2
"
#
r
1
|Vab |2
(ω − ω0 )
2
λ=
−i(ω − ω0 ) ± −(ω − ω0 ) −
=i −
± ωr , with ωr defined in Eq. 11.37.
2
~2
2
i
h
(ω−ω0 )
+ωr t
i −
2
General solution: cb (t) = Ae
+ Be
h
i
(ω−ω0 )
i −
−ωr t
2
= e−i(ω−ω0 )t/2 Aeiωr t + Be−iωr t ,
or, more conveniently: cb (t) = e−i(ω−ω0 )t/2 [C cos(ωr t) + D sin(ωr t)] . But cb (0) = 0, so C = 0 :
ω0 − ω
i(ω0 −ω)t/2
i(ω0 −ω)t/2
cb (t) = De
sin(ωr t). ċb = D i
e
sin(ωr t) + ωr e
cos(ωr t) ;
2
2~ i(ω−ω0 )t
2~ i(ω−ω0 )t/2
ω0 − ω
ca (t) = i
e
ċb = i
e
D i
sin(ωr t) + ωr cos(ωr t) . But ca (0) = 1 :
Vba
Vba
2
2~
−iVba
1=i
Dωr , or D =
.
Vba
2~ωr
i(ω0 −ω)t/2
cb (t) = −
i
Vba ei(ω0 −ω)t/2 sin(ωr t),
2~ωr
ω0 − ω
ca (t) = ei(ω−ω0 )t/2 cos(ωr t) + i
sin(ωr t) .
2ωr
(b)
2
Pa→b (t) = |cb (t)| =
|Vab |
2~ωr
2
sin2 (ωr t).
The largest this gets (when sin2 = 1) is
|Vab |2 /~2
,
4ωr2
380
CHAPTER 11. QUANTUM DYNAMICS
and the denominator,
4ωr2 = (ω−ω0 )2 +|Vab |2 /~2 , exceeds the numerator, so P ≤ 1 (and 1 only if ω = ω0 ).
2
2
|Vab |
ω0 − ω
2
sin (ωr t) +
sin2 (ωr t)
|ca | + |cb | = cos (ωr t) +
2ωr
2~ωr
(ω − ω0 )2 + (|Vab |/~)2
= cos2 (ωr t) +
sin2 (ωr t) = cos2 (ωr t) + sin2 (ωr t) = 1. X
4ωr2
2
(c) If
2
2
2
2
2
|Vab | ~ (ω − ω0 ) ,
0
|Vab |2 sin2 ω−ω
2 t
,
and Pa→b ≈
~2
(ω − ω0 )2
1
then ωr ≈ |ω − ω0 |,
2
Eq. 11.35.
confirming
(d) ωr t = π =⇒ t = π/ωr .
Problem 11.10
Spontaneous emission rate (Eq. 11.63): A =
R=
π
|℘|2 ρ(ω),
30 ~2
with
ρ(ω) =
ω 3 |℘|2
. Thermally stimulated emission rate (Eq. 11.54):
3π0 ~c3
~
ω3
π 2 c3 (e~ω/kB T − 1)
(Eq. 11.59).
So the ratio is
ω 3 |℘|2 30 ~2 π 2 c3 e~ω/kB T − 1
A
=
·
·
= e~ω/kB T − 1.
R
3π0 ~c3 π|℘|2
~ω 3
The ratio is a monotonically increasing function of ω, and is 1 when
e~ω/kb T = 2,
or
~ω
= ln 2,
kB T
ν=
ω=
kB T
ln 2,
~
or
ν=
ω
kB T
=
ln 2. For T = 300 K,
2π
h
(1.38 × 10−23 J/K)(300 K)
ln 2 = 4.35 × 1012 Hz.
(6.63 × 10−34 J · s)
For higher frequencies, (including light, at 1014 Hz), spontaneous emission dominates.
Problem 11.11
(a) Look for solutions of the form f (x, y, z, t) = X(x)Y (y)Z(z)T (t). Plug this into the wave equation,
d2 T
d2 X
d2 Y
d2 Z
1
XY Z 2 − Y Z 2 + XZ 2 + XY 2 T = 0.
c2
dt
dx
dy
dz
Divide by XY ZT :
1 d2 X
1 d2 Y
1 d2 Z
1 1 d2 T
−
−
−
= 0.
c2 T dt2
X dx2
Y dy 2
Z dz 2
Since they are functions of independent variables, each term must be a constant:
d2 T
= −ω 2 T,
dt2
d2 X
= −kx2 X,
dx2
d2 Y
= −ky2 Y,
dy 2
d2 Z
= −kz2 Z,
dz 2
CHAPTER 11. QUANTUM DYNAMICS
381
with
ω2
− 2 + kx2 + ky2 + kz2 = 0.
c
The general solution for T (t) is A sin(ωt) + B cos(ωt), but we may as well start the clock so that T (t) =
B cos(ωt). The general solution for X(x) is Ax sin(kx x) + Bx cos(kx x), but the boundary condition at
x = 0 means Bx = 0, and the boundary condition at x = l then says sin(kx l) = 0, so kx = nx π/l, for some
integer nx . Negative nx doesn’t give us a new solution (it just switches the sign, and we can absorb that
into Ax ), and nx = 0 gives us no solution at all, so in fact nx = 1, 2, 3, . . .. The same applies to Y (y) and
Z(z), so (combining the overall constants):
n π n π n π x
y
z
f (x, y, z, t) = A cos(ωt) sin
x sin
y sin
z
l
l
l
with
πc q 2
ω=
nx + n2y + n2z , where nx , ny , nz are positive integers.
l
q
(b) As in Figure 5.3, one octant of a spherical shell of radius n = n2x + n2y + n2z and thickness dn contains
2(1/8)4πn2 dn = πn2 dn modes, and hence energy
3
l
dE = (~ω)πn2 dn = π~ω
ω 2 dω
πc
So
ρ0 (ω) =
⇒
dE
=
l3
~ω 3
pi2 c3
dω.
~ω 3
.
π 2 c3
(c) Plug this into Eq. 11.54:
Rb→a =
3
π
ω 3 |℘|2
2 ~ω
|℘|
=
,
30 ~2
π 2 c3
3π0 ~c3
reproducing Eq. 11.63.
Problem 11.12
N (t) = e−t/τ N (0) (Eqs. 11.65 and 11.66).
et/τ , so t/τ = ln 2, or t1/2 = τ ln 2.
After one half-life,
N (t) = 12 N (0), so
1
2
= e−t/τ ,
or
2=
Problem 11.13
28
√ a. As
35 2
for x and y, we noted that |1 0 0i, |2 0 0i, and |2 1 0i are even (in x, y), whereas |2 1 ± 1i is odd. So the only
non-zero matrix elements are h1 0 0|x|2 1 ± 1i and h1 0 0|y|2 1 ± 1i. Using the wave functions in Problem 11.2:
Z
1
∓1
1
√
h1 0 0|x|2 1 ± 1i = √
e−r/a re−r/2a sin θ e±iφ (r sin θ cos φ)r2 sin θ dr dθ dφ
3
3
a
πa
8 πa
Z ∞
Z π
Z 2π
1
3
4 −3r/2a
= ∓
r
e
dr
sin
θ
dθ
(cos φ ± i sin φ) cos φ dφ
8πa4 0
0
0
" # 5
∓1
2a
4
27
=
4!
(π)
=
∓
a.
8πa4
3
3
35
In Problem 11.2 we calculated the matrix elements of z; all of them are zero except h1 0 0|z|2 1 0i =
382
CHAPTER 11. QUANTUM DYNAMICS
" # Z
5
2π
∓1
2a
4
h1 0 0|y|2 1 ± 1i =
(cos φ ± i sin φ) sin φ dφ
4!
4
8πa
3
3
0
" # 5
2a
∓1
27
4
4!
a.
=
(±iπ)
=
−i
8πa4
3
3
35
√
27 2
27 h1 0 0|r|2 0 0i = 0; h1 0 0|r|2 1 0i =
a
k̂;
h1
0
0|r|2
1
±
1i
=
a
∓
î
−
i
ĵ
, and hence
35
35
215
℘2 = 0 (for |2 0 0i → |1 0 0i), and |℘|2 = (qa)2 10 (for |2 1 0i → |1 0 0i and |2 1 ± 1i → |1 0 0i).
3
E2 − E1
1 E1
3E1
Meanwhile, ω =
=
− E1 = −
, so for the three l = 1 states:
~
~ 4
4~
2
1
29 E13 e2 a2
210 E1
c
33 E13 (ea)2 215
=− 8
= 8
A =− 6 3
10
3
4
3
2
2 ~
3
3π0 ~c
3 π 0 ~ c
3
mc
a
2
10
8
2
(3.00 × 10 m/s)
13.6
1
= 8
= 6.27 × 108 /s; τ =
= 1.60 × 10−9 s
3
0.511 × 106
(0.529 × 10−10 m)
A
for the three l = 1 states (all have the same lifetime); τ = ∞
for the l = 0 state.
Problem 11.14
hn0 `0 m0 |(Lz x − xLz )|n`mi = ~(m0 − m)hn0 `0 m0 |x|n`mi = i~hn0 `0 m0 |y|n`mi,
hn0 `0 m0 |(Lz y − yLz )|n`mi = ~(m0 − m)hn0 `0 m0 |y|n`mi = −i~hn0 `0 m0 |x|n`mi,
hn0 `0 m0 |(Lz z − zLz )|n`mi = ~(m0 − m)hn0 `0 m0 |z|n`mi = 0.
It follows that

0

then hn0 `0 m0 |x|n`mi = hn0 `0 m0 |y|n`mi = 0,
if m = m,
0
if m = m ± 1, then hn0 `0 m0 |x|n`mi = ±ihn0 `0 m0 |y|n`mi = 0, hn0 `0 m0 |z|n`mi = 0,


otherwise,
hn0 `0 m0 |x|n`mi = hn0 `0 m0 |y|n`mi = hn0 `0 m0 |z|n`mi = 0.
To establish the last line, note that
(m0 − m)2 hn0 `0 m0 |x|n`mi = (m0 − m)ihn0 `0 m0 |y|n`mi = i(−i)hn0 `0 m0 |x|n`mi = hn0 `0 m0 |x|n`mi,
so unless (m0 − m)2 = 1, hn0 `0 m0 |x|n`mi = 0 (and the same goes for y).
Problem 11.15
(a)
[L2 , z] = [L2x , z] + [L2y , z] + [L2z , z] = Lx [Lx , z] + [Lx , z]Lx + Ly [Ly , z] + [Ly , z]Ly + Lz [Lz , z] + [Lz , z]Lz

 [Lx , z] = [ypz − zpy , z] = [ypz , z] − [zpy , z] = y[pz , z] = −i~y,
[Ly , z] = [zpx − xpz , z] = [zpx , z] − [xpz , z] = −x[pz , z] = i~x,
But

[Lz , z] = [xpy − ypx , z] = [xpy , z] − [ypx , z] = 0.
CHAPTER 11. QUANTUM DYNAMICS
383
[L2 , z] = Lx (−i~y) + (−i~y)Lx + Ly (i~x) + (i~x)Ly = i~(−Lx y − yLx + Ly x + xLy ).
Lx y = Lx y − yLx + yLx = [Lx , y] + yLx = i~z + yLx ,
But
Ly x = Ly x − xLy + xLy = [Ly , x] + xLy = −i~z + xLy .
So:
So: [L2 , z] = i~(2xLy − i~z − 2yLx − i~z) =⇒ [L2 , z] = 2i~(xLy − yLx − i~z). X
2 2 L , [L , z] = 2i~ [L2 , xLy ] − [L2 , yLx ] − i~[L2 , z]
= 2i~ [L2 , x]Ly + x[L2 , Ly ] − [L2 , y]Lx − y[L2 , Lx ] − i~(L2 z − zL2 ) .
But
[L2 , Ly ] = [L2 , Lx ] = 0 (Eq. 4.102), so
2 2 L , [L , z] = 2i~ 2i~ (yLz − zLy − i~x) Ly − 2i~ (zLx − xLz − i~y) Lx − i~ L2 z − zL2 , or
2 2 2
L , [L , z] = −2~ 2yLz Ly
−2zL2y − 2zL2x
{z
|
2
2
−2i~xLy + 2xLz Lx + 2i~yLx − L z + zL
}
−2z(L2x +L2y +L2z )+2zL2z
2
2yLz Ly − 2i~xLy + 2xLz Lx + 2i~yLx + 2zL2z − 2zL2 − L2 z + zL2
2
2
2
2
= 2~ zL + L z − 4~ (yLz − i~x) Ly + (xLz + i~y) Lx + zLz Lz
{z
}
|
{z
}
|
= −2~
Lz y
= 2~
2
2
2
Lz x
zL + L z − 4~ (Lz yLy + Lz xLx + Lz zLz ) = 2~2 (zL2 + L2 z). QED
|
{z
}
2
Lz (r·L)=0
(b) Making the obvious generalization of part (a):
2 2 L , [L , r] = 2~2 (rL2 + L2 r).
As before, we sandwich this commutator between hn0 `0 m0 | and |n`mi to derive the selection rule:
hn0 `0 m0 | L2 , [L2 , r] |n`mi = 2~2 hn0 `0 m0 |(rL2 + L2 r)|n`mi
= 2~4 [`(` + 1) + `0 (`0 + 1)]hn0 `0 m0 |r|n`mi = hn0 `0 m0 |(L2 [L2 , r] − [L2 , r]L2 )|n`mi
= ~2 [`0 (`0 + 1) − `(` + 1)]hn0 `0 m0 |[L2 , r]|n`mi = ~2 [`0 (`0 + 1) − `(` + 1)]hn0 `0 m0 |(L2 r − rL2 )|n`mi
= ~4 [`0 (`0 + 1) − `(` + 1)]2 hn0 `0 m0 |r|n`mi.
Conclusion: Either
2[`(` + 1) + `0 (`0 + 1)] = [`0 (`0 + 1) − `(` + 1)]2 ,
or else
hn0 `0 m0 |r|n`mi = 0.
But
[`0 (`0 + 1) − `(` + 1)] = (`0 + ` + 1)(`0 − `)
and
2[`(` + 1) + `0 (`0 + 1)] = (`0 + ` + 1)2 + (`0 − `)2 − 1,
so the first possibility can be written in the form
[(`0 + ` + 1)2 − 1][(`0 − `)2 − 1] = 0.
The first factor cannot be zero (unless `0 = ` = 0, but hn0 00|r|n00i is trivially zero, since the integrand is
odd), so the condition simplifies to `0 = ` ± 1. Thus we obtain the selection rule for `:
No transitions occur unless ∆` = ±1.
384
CHAPTER 11. QUANTUM DYNAMICS
Problem 11.16
(a)


 |2 1 0i 
|2 1 1i
|3 0 0i →
→ |1 0 0i.


|2 1−1i
(|3 0 0i → |2 0 0i and |3 0 0i → |1 0 0i violate ∆l = ±1 rule.)
(b)
From Eq. 11.76:
h2 1 0|r|3 0 0i = h2 1 0|z|3 0 0i k̂.
h2 1 ± 1|r|3 0 0i = h2 1 ± 1|x|3 0 0i î + h2 1 ± 1|y|3 0 0i ĵ.
±h2 1 ± 1|x|3 0 0i = ih2 1 ± 1|y|3 0 0i.
Thus
|h2 1 0|r|3 0 0i|2 = |h2 1 0|z|3 0 0i|2
and |h2 1 ± 1|r|3 0 0i|2 = 2|h2 1 ± 1|x|3 0 0i|2 ,
so there are really just two matrix elements to calculate.
ψ21m = R21 Y1m ,
ψ300 = R30 Y00 . From Table 4.3:
r r
√ √ π
Z π
Z 2π
Z
3
1
3
cos3 θ
3 2
1
2
0 0
(2π) =
cos θ sin θ dθ
dφ =
−
=√ .
Y1 Y0 cos θ sin θ dθ dφ =
4π 4π 0
4π
3
2
3
3
0
0
r r
Z
Z π
Z 2π
∗
3
1
Y1±1 Y00 sin2 θ cos φ dθ dφ = ∓
sin3 θ dθ
cos φe∓iφ dφ
8π 4π 0
0
r Z 2π
Z 2π
1
1
1
3 4
= ∓
cos2 φ dφ ∓ i
cos φ sin φ dφ = ∓ √ (π ∓ 0) = ∓ √ .
4π 2 3
π
6
6
0
0
From Table 4.7:
Z ∞
Z ∞
1
2
r −r/2a
2r
2 r 2 −r/3a 3
√
K ≡
R21 R30 r3 dr = √
e
1−
+
e
r dr
3 a 27 a
24a3/2 27a3/2 0 a
0
" 6
7 #
Z ∞
5
2 2
a
2
6
2
6
1
2
6
4
4 −5u/6
du = √ 4!
− 5!
+ 6!
= √
a
1− u+ u u e
3
3
27
5
3
5
27
5
9 2a
9 2
0
5
5
7 4√
a 4! 6
2
2
a 4! 6
2 3
= √
5 − 6 · 5 + 63 = √
= 6 2 a.
6
6
3
27
5
9 2 5
9 2 5
So:
1
R21 (Y1±1 )∗ (r sin θ cos φ)R30 Y00 r2 sin θ dr dθ dφ = K ∓ √
.
6
Z
1
h2 1 0|z|3 0 0i =
R21 Y10 (r cos θ)R30 Y00 r2 sin θ dr dθ dφ = K √
.
3
Z
h2 1 ± 1|x|3 0 0i =
|h2 1 0|r|3 0 0i|2 = |h2 1 0|z|3 0 0i|2 = K 2 /3;
|h2 1 ± 1|r|3 0 0 i|2 = 2|h2 1 ± 1|x|3 0 0 i|2 = K 2 /3.
Evidently the three transition rates are equal, and hence 1/3 go by each route.
CHAPTER 11. QUANTUM DYNAMICS
(c) For each mode,
ω 3 e2 |hri|2
;
A=
3π0 ~c3
385
E3 − E2
1
ω =
=
~
~
here
E1
E1
−
9
4
=−
5 E1
,
36 ~
so the total
decay rate is
9 2 3
2
c
1 27 34 √
2
5 E1
e2
E1
R=3 −
2a = 6
36 ~
3π0 ~c3 3
56
5
mc2
a
9 2 2
3 × 108
1
13.6
=6
/s = 6.32 × 106 /s. τ =
= 1.58 × 10−7 s.
6
−10
5
0.511 × 10
0.529 × 10
R
Problem 11.17
(a) R =
2π Vif
~ 2
2
ρ(Ef ). From Eqs. 11.31 and 11.39, Vif = hψi |V |ψf i, where V = eE0 z, so
1
1
√
Vif = eE0 hψ0 |z|ψf i = eE0 √
3
πa
l3
Z
e
−r/a
ik·r
ze
eE0
d r= √
πa3 l3
3
Z
∂
−i
e−r/a eik·r d3 r.
∂kz
To evaluate the integral, set the z axis along k, so
Z π
Z
Z
Z ∞
−r/a ik·r 4
−r/a ikr cos θ 2
2 −r/a
ikr cos θ
e
e
d r= e
e
r sin θ dr dθ dφ = 2π
r e
e
sin θ dθ dr
0
0
)
(
π
Z
Z ∞
2πi ∞ −r/a −ikr
eikr cos θ
dr =
re
e
− eikr dr
= 2π
r2 e−r/a −
ikr
k 0
0
Z ∞
Z0 ∞
2πi
=
re−r(1/a+ik) dr −
re−r(1/a−ik) dr =
k
0
0
1
1
2πa2 i (1 − ika)2 − (1 + ika)2
2πi
=
−
=
k (1/a + ik)2
(1/a − ik)2
k
[1 + (ka)2 ]2
(−4ika
8πa3
2πa2 i
=
.
=
2
2
k
[1 + (ka) ]
[1 + (ka)2 ]2
ieE0 ∂
Vif = − √
πa3 l3 ∂kz
!
8πa3
2
[1 + k 2 a2 ]
Finally, quoting Eq. 11.85,
3 p 3
2m Ef
l
ρ(Ef ) =
dΩ,
2π
~3
r
= −8ieE0
where
πa3 −4a2 kz
32ieE0 a2 k cos θ
=
3
3
l3 [1 + k 2 a2 ]
[1 + k 2 a2 ]
~2 k 2
Ef =
,
2m
so
ρ(Ef ) =
l
2π
3
mk
dΩ.
~2
Putting it all together,
π
R=
2~
=
32eE0 a2 k cos θ
[1 + k 2 a2 ]
3
r
πa3
l3
!2 256
k 3 a6
2
0 E02
6 cos θ dΩ.
~
[1 + k 2 a2 ]
l
2π
3
mk
dΩ,
~2
or, using
me2 =
r
4π0 ~2
,
a
πa3
.
l3
386
CHAPTER 11. QUANTUM DYNAMICS
(b) Integrating over all angles, using dΩ = sin θ dθ dφ,
Z
4π
256
256
k 3 a6
k 3 a6
2
2
Ri→all =
cos
θ
sin
θ
dθ
dφ
=
0 E02
E
,
0 0
6
6
2
2
2
2
~
~
3
[1 + k a ]
[1 + k a ]
and noting that
~ω =
~2
~2
~2 k 2
=
(1 + k 2 a2 ),
+
2
2m
2ma
2ma2
the cross-section is
1
256
k 3 a6
2
σ(k) = 1
E
0
0
6
2
~
[1 + k 2 a2 ]
2 0 E0 c
4π
3
~2
1024 π~
k 3 a4
2 2
(1
+
k
a
)
=
.
2ma2
3mc [1 + k 2 a2 ]5
(c)
λ=
2πc
ω
But ma2 =
⇒
~2
,
2|E1 |
(1 + k 2 a2 ) =
~ω = ~
⇒
(1 + k 2 a2 ) =
2ma2 ~2πc
4πma2 c
=
.
~2
λ
~λ
so
4πc ~2
2π~c
2π(1.055 × 10−34 )(3 × 108 )
=
=
= 4.15.
~λ 2|E1 |
λ|E1 |
(220 × 10−10 )(13.6)(1.6 × 10−19 )
(ka)2 = 3.15
σ=
2πc
~2
=
(1 + k 2 a2 )
λ
2ma2
⇒
ka = 1.77.
1024 π(1.055 × 10−34 )(1.77)3 (5.29 × 10−11 )
= 0.986 × 10−22 m2 = 1 Mb.
3(9.11 × 10−31 )(3 × 108 )(4.15)5
Problem 11.18
r
r
π nπ n 2 π 2 ~2
2
2
(a) New allowed energies: En =
; Ψ(x, 0) =
sin
x , ψn (x) =
sin
x .
2
2m(2a)
a
a
2a
2a
√ Z a
√ Z an
π nπ h n
πx i
h n
πx io
2
2
cn =
sin
x sin
x dx =
cos
−1
− cos
+1
dx
a 0
a
2a
2a 0
2
a
2
a
(
) a
sin n2 − 1 πx
sin n2 + 1 πx
1
a
π
πa
=√
−
(for n 6= 2)
n
n
−
1
+
1
2a
2
a
2
a
0
(
#
)
"
sin n2 − 1 π
sin n2 + 1 π
sin n2 + 1 π
1
1
1
− n
√
=√
−
=
n
n
n
2π
2π
2 −1
2 +1
2 −1
2 +1
)
(
√
0, √
if n is even
4 2 sin n2 + 1 π
=
=
.
4 2
± π(n
π
(n2 − 4)
2 −4) , if n is odd
√ Z a
√ Z a
2
2
1
1
2 π
c2 =
sin
x dx =
dx = √ . So the probability of getting En is
a 0
a
a 0 2
2
CHAPTER 11. QUANTUM DYNAMICS
387
1
 2,

if n = 2 
Pn = |cn |2 = π2 (n32
.
2 −4)2 , if n is odd


0,
otherwise
Most probable: E2 =
π 2 ~2
(same as before).
2ma2
Probability: P2 = 1/2.
π 2 ~2
32
, with probability P1 =
= 0.36025.
2
8ma
9π 2
~2 d2 Ra
R
π
(c) hHi = Ψ∗ HΨ dx = a2 0 sin πa x − 2m
dx2 sin a x dx, but this is exactly the same as before the wall
(b) Next most probable: E1 =
moved – for which we know the answer:
π 2 ~2
.
2ma2
Problem 11.19
The new allowed energies are En0 = (n + 12 )~ω 0 = 2(n + 21 )~ω = ~ω, 3~ω, 5~ω, . . .. So the probability of
1
is zero. The probability of getting ~ω (the new ground state energy) is P0 = |c0 |2 , where c0 =
Rgetting 2 ~ω
Ψ(x, 0)ψ00 dx, with
Ψ(x, 0) = ψ0 (x) =
mω 1/4
π~
mω
2
e− 2~ x ,
ψ0 (x)0 =
m2ω
π~
1/4
So
1/4
c0 = 2
r
mω
π~
Z ∞
e
2
− 3mω
2~ x
1/4
dx = 2
−∞
Therefore
P0 =
r
mω √
2 π
π~
1
2
r
2~
3mω
2
m2ω
e− 2~ x .
!
r
1/4
=2
2
.
3
2√
2 = 0.9428.
3
Problem 11.20
To show:
i~
∂χ
= Hχ,
∂t
where χ is given by Eq. 11.103 and H is given by Eq. 11.97.
∂χ
=
∂t
 h
i
−iωt/2 iω h
(ω1 −ω)
λ
λt
λt
α
−
sin
−
i
cos
cos
− 2 cos
2
λ
2
2 e
2


h
i
iωt/2 iω h
(ω1 +ω)
λ
λt
λt
α
−
sin
−
i
cos
sin
+ 2 cos
2
2
λ
2
2 e
−iωt/2 
e


i

i(ω
+ω)
λt
λt
α
1
iωt/2
−
sin
sin
e
2
λ
2
2
λt
2
− i(ω1λ−ω) sin
λt
2
i
cos
α
2
Hχ =

h
i
h
i

(ω1 −ω)
i(ω1 +ω)
λt
α −iωt/2
λt
λt
α iωt/2
−iωt
)
−
i
sin(
)
cos
e
+
e
sin
α
cos(
)
−
sin(
)
sin
e
cos α cos( λt
2
λ
2
2
2
λ
2
2

~ω1 


h
i
h
i

2  iωt
(ω
−ω)
i(ω
+ω)
α −iωt/2
α iωt/2
1
1
e sin α cos( λt
sin( λt
− cos α cos( λt
sin( λt
2 )−i
λ
2 ) cos 2 e
2 )−
λ
2 ) sin 2 e
388
CHAPTER 11. QUANTUM DYNAMICS
(1) Upper elements:
λt
λt
λ
(ω1 − ω)
λt
α iω
i(ω1 − ω)
λt
α
− sin
cos
i~
−i
cos
cos −
−
sin
cos
2
λ
2
2
λ
2
2
2
2
2
α
λt
(ω1 − ω)
λt
α
λt
(ω1 + ω)
λt
? ~
ω1
=
cos
−i
sin
cos α cos + cos
−i
sin
sin
| {zα} sin 2 ,
2
λ
2
2
λ
2
2
2
F
α
α
where F = 2 sin cos
2 2
The sine terms:
λt
iω(ω1 − ω) iω1 (ω1 − ω)
iω1 (ω1 + ω)
α ?
sin
−iλ −
+
cos α +
2 sin2
= 0.
2
λ
λ
λ
2
h
i
i
λt
2
2 − ω 2 + 2ωω cos α − ωω + sin
−ω
ω
+ (ω12 − ωω1 ) cos α + (ω12 + ωω1 )(1 − cos α)
1
1
1
λ
2
h
i
λt
i
2
2
2
2
−ω
+
2ωω
cos
α
−
ωω
+
ω
cos
α
−
ωω
cos
α
+
ω
+
ωω
−
ω
cos
α
−
ωω
cos
α
= 0. X
= − sin
1
1
1
1
1
1
1
1
1
λ
2
The cosine terms:
h
i
λt
λt
2 α
(ω1 − ω) + ω − ω1 cos α − ω1 2 sin
= −ω1 cos
[−1 + cos α + (1 − cos α)] = 0.
cos
2
2
2
X
(2) Lower elements:
α
α
iω λ
λt
λt
λt
λt
(ω1 + ω)
i(ω1 + ω)
sin
sin
i~
−
sin
−
i
cos
+
cos
−
sin
2
λ
2
2
λ
2
2
2
2
2
λt
λt
λt
λt
(ω1 − ω)
α
i(ω1 + ω)
α
? ~
ω1
2 α
sin
− cos
sin
cos
−i
2 sin cos
−
cos α sin .
=
2
λ
2
2
2
λ
2
2
2
2
The sine terms:
iω (ω + ω)
λt
iω(ω1 + ω) iω1 (ω1 − ω)
?
1
1
2 α
sin
−iλ +
+
2 cos
−
cos α = 0.
2
λ
λ
2
λ
h
i
i
λt
2
2
sin
−
ω
− ω12 + 2ωω1 cos α + ωω1 + ω
+ (ω12 − ωω1 )(1 + cos α) − (ω12 + ωω1 ) cos α
λ
2
h
i
i
λt
2
=
sin
−ω12 + 2ωω1 cos α + ωω
ωω
ω12
cos
α − ωω1 cos α − ω12
cos
α − ωω1 cos α = 0. X
1 + ω1 − 1 +
λ
2
The cosine terms:
h
i
α
λt
λt
(ω1 + ω) − ω − ω1 2 cos2 + ω1 cos α = cos
[ω1 − ω1 (1 + cos α) + ω1 cos α] = 0. X
cos
2
2
2
As for Eq. 11.105:
λt
(ω1 − ω cos α)
λt
ω
λt
cos α
sin α2
cos
−i
sin
e−iωt/2 iωt 2 α + i
sin α sin
e−iωt/2
e sin 2
−eiωt cos α2
2
λ
2
λ
2
α
, with
=
β
CHAPTER 11. QUANTUM DYNAMICS
389
λt
λt
iω1
λt
α iω
α
α
α=
cos
sin
−
sin
cos +
cos α cos + sin α sin
e−iωt/2
2
λ
2
2
λ |
2 {z
2}
2
α
cos(α− α
2 )=cos 2
i(ω1 − ω)
λt
α
λt
−
sin
cos e−iωt/2
= cos
2
λ
2
2
(confirming the top entry).
λt
λt
iω1
λt
α iω
α
α
sin
β=
cos
−
sin
sin +
cos α sin − sin α cos
eiωt/2
2
λ
2
2
λ |
2 {z
2}
2
α
sin( α
2 −α)=− sin 2
λt
i(ω1 + ω)
λt
α
= cos
−
sin
sin eiωt/2
2
λ
2
2
(confirming the bottom entry).
λt
λt
(ω1 − ω cos α)2
ω2
λt
2
2
2
sin
+
+ 2 sin α sin
|c+ | + |c− | = cos
2
λ2
2
λ
2
λt
1
λt
2
2
2
2
2
2
2
= cos
+ 2 ω1 − 2ωω1 cos α + ω cos α + ω sin α sin
{z
}
2
λ |
2
2
2
2
ω 2 +ω12 −2ωω1 cos α=λ2
= cos2
λt
2
+ sin2
λt
2
= 1. X
Problem 11.21
In the adiabatic regime, ω ω1 , Eq. 11.104 becomes
s
2
ω
ω
ω
λ = ω1 1 − 2 cos α +
cos α = ω1 − ω cos α,
≈ ω1 1 −
ω1
ω1
ω1
and Eq. 11.105 says
λt
λt
ω
λt
χ(t) ≈ cos
− i sin
e−iωt/2 χ+ (t) + i
sin α sin
eiωt/2 χ− (t)
2
2
ω1 − ω cos α
2
ω1 t
ω
sin α sin
e+iωt/2 χ− (t).
= e−i(ω1 −ω cos α)t/2 e−iωt/2 χ+ (t) + i
ω1
2
As ω/ω1 → 0 the second term drops out completely; the electron remains in the state χ+ (t), acquiring a total
phase
[ω cos α − (ω + ω1 )]
t
.
2
The dynamic phase is
Z
1 t
ω1 t
θ+ (t) = −
E+ (t0 ) dt0 = −
,
~ 0
2
(where E+ = ~ω1 /2, from Eq. 11.101), so the geometric phase is
γ+ (t) = [ω cos α − (ω + ω1 )]
t
ω1 t
ωt
+
= (cos α − 1) .
2
2
2
For a complete cycle T = 2π/ω, and therefore Berry’s phase is γ+ (T ) = π(cos α − 1).
390
CHAPTER 11. QUANTUM DYNAMICS
Problem 11.22
According to the adiabatic theorem (Equation 11.93),
Ψ(x, t) = ei θ(t) ei γ(t) ψ(x, t)
[F]
where ψ(x, t) is the instantaneous eigenstate, in this case (Equation 2.132)
p
m α(t)
m α(t) |x|
,
ψ(x, t) =
exp −
~
~2
[FF]
and where θ is the dynamical phase:
1
θ(t) = −
~
Z t
1
θ̇ = − E(t)
~
E(t0 ) dt0
0
[FFF]
with, in this case (Equation 2.132),
2
E(t) = −
m α(t)
.
2 ~2
[FFFF]
We may plug [F] in the time-dependent Schrödinger equation to obtain
Ĥ(t) Ψ(x, t) = i ~
∂Ψ
h∂t
i
ei θ(t) ei γ(t) Ĥ(t) ψ(x, t) = i ~ i θ̇ ψ(x, t) + i γ̇ ψ(x, t) + ψ̇(x, t) ei θ(t) ei γ(t)
h
i
Ĥ(t) ψ(x, t) = i ~ i θ̇ ψ(x, t) + i γ̇ ψ(x, t) + ψ̇(x, t)
E(t) ψ(x, t) = E(t) ψ(x, t) − ~ γ̇ ψ(x, t) + i ~ ψ̇(x, t)
∂ψ
γ̇ ψ(x, t) = i
[FFFFF]
∂t
Taking the derivative of Equation [FF] gives
∂ψ
∂ψ dα
=
∂t
dt
∂α √
√
1 m −m α |x|/~2
m α m |x| −m α |x|/~2
√ e
=
−
e
α̇
2 ~ α
~
~2
1
m |x|
=
−
α̇ ψ
2α
~2
and from Equation [FFFFF] we have
1 α̇ m α̇ |x|
γ̇ = i
−
.
2 α
~2
But in the adiabatic limit α̇ → 0, so γ̇ → 0, so γ is a constant, and at t = 0 γ = 0, so γ = 0 . [Note, by
contrast, that θ̇ = −E(t) /~ is not small, regardless of α̇.]
CHAPTER 11. QUANTUM DYNAMICS
391
Meanwhile, from [FFF] and [FFFF]:
Z
Z
1 α2
dα
1 t
0
0
E(t ) dt = −
E(α)
θ(t) = −
~ 0
~ α1
dα/dt0
Z α2
m
dα
=
α2
2 ~3 α1
c
m
α2 3 α1 3
=
−
6c
~
~
Problem 11.23
(a) Using Equation 11.108 we have
i~
∂Ψ
∂
= i ~ eĜ(t) Ψ(0)
∂t
∂t ∂
1
1
= i~
1 + Ĝ(t) + Ĝ(t) Ĝ(t) + Ĝ(t) Ĝ(t) Ĝ(t) + · · · Ψ(0)
∂t
2
3!
and noting that i ~ Ġ = H, this is
∂Ψ
1 ˙
1
1 ˙
˙
˙
i~
= i ~ 0 + Ĝ(t) + Ĝ(t) Ĝ(t) + Ĝ(t) Ĝ(t) + Ĝ(t) Ĝ(t) Ĝ(t)
∂t
2
2
3!
1
1
˙
˙
= + Ĝ(t) Ĝ(t) Ĝ(t) + Ĝ(t) Ĝ(t) Ĝ(t) + · · · Ψ(0)
3!
3!
1
1
1
= Ĥ(t) + Ĥ(t) Ĝ(t) + Ĝ(t) Ĥ(t) + Ĥ(t) Ĝ(t) Ĝ(t)
2
2
3!
1
1
= + Ĝ(t) Ĥ(t) Ĝ(t) + Ĝ(t) Ĝ(t) Ĥ(t) + · · · Ψ(0) .
3!
3!
h
i
Now if Ĝ, Ĥ = 0 this becomes
∂Ψ
1
i~
= Ĥ(t) 1 + Ĝ(t) + Ĝ(t) Ĝ(t) + · · · Ψ(0)
∂t
2
= Ĥ(t) eĜ(t) Ψ(0)
= Ĥ(t) Ψ(t) .
so Eq. 11.108 is a solution to the Schrödinger equation if G and H commute which is not typically the
case for time-dependent Hamiltonians.
(b) Now we try Equation 11.109. Taking the derivative and recalling that
Z x
d
f (u) du = f (x)
dx a
we have
i~
Z t
i
0 + Ĥ(t) + −
Ĥ(t)
Ĥ(t2 ) dt2
~
0
)
2
Z t2
Z t
i
=+ −
Ĥ(t)
Ĥ(t2 )
Ĥ(t3 ) dt3 dt2 Ψ(0) .
~
0
0
∂Ψ
=
∂t
392
CHAPTER 11. QUANTUM DYNAMICS
In this case all the Ĥ(t)’s are on the left so we don’t have to worry about operators commuting. t2 and
t3 are dummy variables so we can rename them to t1 and t2 and we have
Z t
∂Ψ
i
i~
= Ĥ(t) 1 + −
Ĥ(t1 ) dt1
∂t
~
0
)
Z t1
2 Z t
i
Ĥ(t2 ) dt1 dt1 Ψ(0) .
Ĥ(t1 )
=+ −
~
0
0
which looking back to Equation 11.109 is
i~
∂Ψ
= Ĥ(t) Ψ(t) ,
∂t
as required.
(c) We consider
"
#
2 Z t
Z t
i
00
00
0
0
Ĥ(t ) dt
Ĥ(t ) dt
T Ĝ(t) Ĝ(t) = T −
~
0
0
2 Z t Z t h
i
i
= −
T Ĥ(t0 ) Ĥ(t00 ) dt0 dt00
~
0
0
#
2 "Z t Z t00
Z t Z t0
i
0
00
00
0
00
0
0
00
Ĥ(t ) Ĥ(t ) dt dt
Ĥ(t ) Ĥ(t ) dt dt +
= −
~
0
0
0
0
i
h
Replacing the variables with t0 → t2 and t00 → t1 in the first integral and t0 → t1 and t00 → t2 in the
second integral we have
2 Z t
Z t1
i
i
Ĥ(t2 ) dt2 dt1 ,
T Ĝ(t) Ĝ(t) = 2 −
Ĥ(t1 )
~
0
0
h
the desired result.
The generalization for Gn is
n Z t
Z tn−1
Z t1
h i
i
T Ĝn = n! −
Ĥ(t1 )
Ĥ(t2 ) · · ·
Ĥ(tn ) dtn dtn−1 · · · dt2 dt1 .
~
0
0
0
Note that it is n! (not say, n) because there are n! ways to order the n times t1 , t2 , . . . , tn .
Problem 11.24
(a)
Ψ(t) =
X
X
−iEn t/~
cn e
cn (t)e−iEn t/~ ψn .
En ψn +
X
cn e
HΨ = i~
∂Ψ
;
∂t
−iEn t/~
0
H = H0 + H 0 (t);
H ψn = i~
X
ċn e
−iEn t/~
H0 ψn = En ψn . So
i X
ψn + i~ −
cn En e−iEn t/~ ψn .
~
CHAPTER 11. QUANTUM DYNAMICS
393
The first and last terms cancel, so
X
X
cn e−iEn t/~ H 0 ψn = i~
ċn e−iEn t/~ ψn . Take the inner product with ψm :
X
cn e−iEn t/~ hψm |H 0 |ψn i = i~
X
ċn e−iEn t/~ hψm |ψn i.
hψm |ψn i = δmn , and define
Assume orthonormality of the unperturbed states,
X
0
cn e−iEn t/~ Hmn
= i~ ċm e−iEm t/~ ,
(b) Zeroth order:
i 0
ċN = − HN
N,
~
ċm = −
or
0
Hmn
≡ hψm |H 0 |ψn i.
iX
0
cn Hmn
ei(Em −En )t/~ .
~ n
cm (t) = 0 for m 6= N . Then in first order:
Z
i t 0
cN (t) = 1 −
H (t0 ) dt0 , whereas for m 6= N :
~ 0 NN
cN (t) = 1,
or
i 0
ċm = − HmN
ei(Em −EN )t/~ ,
~
or
i
cm (t) = −
~
Z t
0
0
HmN
(t0 )ei(Em −EN )t /~ dt0 .
0
(c)
"
#t
i(EM −EN )t/~
i(EM −EN )t0 /~
e
−1
e
i
0
0
i(EM −EN )t /~
0
= −HM N
e
dt = − HM N
~
i(EM − EN )/~
EM − EN
0
0
0
EM − EN
HM
N
ei(EM −EN )t/2~ 2i sin
t .
= −
(EM − EN )
2~
i 0
cM (t) = − HM
N
~
Z t
0
2
0
4|HM
N|
PN →M = |cM | =
sin2
(EM − EN )2
2
EM − EN
t .
2~
(d)
Z
0
0
i
1 t iωt0
cM (t) = − VM N
e
+ e−iωt ei(EM −EN )t /~ dt0
~
2 0
"
#t
0
0
ei(−~ω+EM −EN )t /~
iVM N
ei(~ω+EM −EN )t /~
.
+
=−
2~
i(~ω + EM − EN )/~ i(−~ω + EM − EN )/~
0
If EM > EN , the second term dominates, and transitions occur only for ω ≈ (EM − EN )/~:
iVM N
1
EM − EN − ~ω
i(EM −EN −~ω)t/2~
cM (t) ≈ −
e
2i sin
t , so
2~ (i/~)(EM − EN − ~ω)
2~
|VM N |2
EM − EN − ~ω
2
PN →M = |cM |2 =
sin
t
.
(EM − EN − ~ω)2
2~
If EM < EN the first term dominates, and transitions occur only for ω ≈ (EN − EM )/~:
iVM N
1
EM − EN + ~ω
i(EM −EN +~ω)t/2~
e
2i sin
t , and hence
cM (t) ≈ −
2~ (i/~)(EM − EN + ~ω)
2~
394
CHAPTER 11. QUANTUM DYNAMICS
|VM N |2
sin2
PN →M =
(EM − EN + ~ω)2
EM − EN + ~ω
t .
2~
Combining the two results, we conclude that transitions occur to states with energy EM ≈ EN ± ~ω, and
|VM N |2
EM − EN ± ~ω
2
PN →M =
sin
t
.
(EM − EN ± ~ω)2
2~
Vba = −℘E0 (Eq. 11.41).
(e) For light,
RN →M =
The rest is as before (Section 11.2.3), leading to Eq. 11.54:
π
|℘|2 ρ(ω), with ω = ±(EM − EN )/~
30 ~2
(+ sign ⇒ absorption, − sign ⇒ stimulated emission).
Problem 11.25
For example (c):
i 0
cN (t) = 1 − HN
N t;
~
0
HmN
cm (t) = −2i
ei(Em −EN )t/2~ sin
(Em − EN )
Em − EN
t (m 6= N ).
2~
0
1
|2
|HmN
Em − EN
2
0
2 2
2
|cN | = 1 + 2 |HN N | t , |cm | = 4
sin
t , so
~
(Em − EN )2
2~
0
X
X
|2
t2 0 2
|HmN
Em − EN
2
|cm |2 = 1 + 2 |HN
|
+
4
sin
t
.
N
~
(Em − EN )2
2~
m
2
m6=N
This is plainly greater than 1! But remember: The c’s are accurate only to first order in H 0 ; to this order the
|H 0 |2 terms do not belong. Only if terms of first order appeared in the sum would there be a genuine problem
with normalization.
For example (d):
Z t
t
i
i
i
sin(ωt0 )
=⇒ cN (t) = 1 −
cN = 1 − VN N
VN N sin(ωt).
cos(ωt0 ) dt0 = 1 − VN N
~
~
ω
~ω
0
0
cm (t) = −
|cN |2 = 1 +
VmN
2
i(Em −EN +~ω)t/~
e
− 1 ei(Em −EN −~ω)t/~ − 1
+
(m 6= N ).
(Em − EN + ~ω)
(Em − EN − ~ω)
|VN N |2
sin2 (ωt);
(~ω)2
So
and in the rotating wave approximation
|VmN |2
Em − EN ± ~ω
2
sin
t
(m 6= N ).
(Em − EN ± ~ω)2
2~
P
Again, ostensibly
|cm |2 > 1, but the “extra” terms are of second order in H 0 , and hence do not belong (to
first order).
P
You would do better to use 1 − m6=N |cm |2 . Schematically: cm = a1 H + a2 H 2 + · · · , so |cm |2 =
|cm |2 =
a21 H 2 + 2a1 a2 H 3 + · · · , whereas cN = 1 + b1 H + b2 H 2 + · · · , so |cN |2 = 1 + 2b1 H + (2b2 + b21 )H 2 + · · · .
Thus knowing cm to first order (i.e., knowing a1 ) gets you |cm |2 to second order, but knowing cN to first order
(i.e., b1 ) does not get you |cN |2 to second order (you’d also need
P b22). It is precisely this b2 term that would
cancel the “extra” (second-order) terms in the calculations of
|cm | above.
CHAPTER 11. QUANTUM DYNAMICS
395
Problem 11.26
(a)
Equation 11.116 ⇒ ċm = −
i
ċm = − cm V0 (t);
~
cm (t) = cm (0)e
2
− ~i
iX
0
0
cn Hmn
ei(Em −En )t/~ . Here Hmn
= hψm |V0 (t)|ψn i = δmn V0 (t).
~ n
dcm
i
i
= − V0 (t) dt ⇒ ln cm = −
cm
~
~
Rt
0
V0 (t0 ) dt0
1
. Let Φ(t) ≡ −
~
Z t
Z
V0 (t0 ) dt0 + constant.
V0 (t0 ) dt0 ;
cm (t) = eiΦ cm (0). Hence
0
1
Φ(T ) = −
~
2
|cm (t)| = |cm (0)| , and there are no transitions.
Z T
V0 (t) dt.
0
(b)

Z
i t

0

Eq. 11.118 ⇒ cN (t) ≈ 1 −
V0 (t ) dt = 1 + iΦ.

cN (t) = 1 + iΦ(t),
~
0
Z t
0
i
 cm (t) = 0 (m 6= N ).

Eq. 11.119 ⇒ cm (t) = −
δmN V0 (t0 )ei(Em −EN )t /~ dt0 = 0 (m 6= N ). 
~ 0
The exact answer is cN (t) = eiΦ(t) , cm (t) = 0, and they are consistent, since eiΦ ≈ 1 + iΦ, to first order.
Problem 11.27
Use result of Problem 11.24(c).
Here
En =
n 2 π 2 ~2
,
2ma2
so
E2 − E1 =
3π 2 ~2
.
2ma2
Z a/2
π 2π
x V0 sin
x dx
a
a
0
"
# a/2
sin 3π
x
2V0 sin πa x
V0
π
1
3π
4V0
a
=
−
=
sin
− sin
=
.
a
2(π/a)
2(3π/a)
π
2
3
2
3π
0
H12
=
2
a
sin
0
Eq. 11.120 =⇒ P1→2 = 4
4V0
3π
2 2ma2
3π 2 ~2
2
2
sin
3π 2 ~
t
4ma2
=
16ma2 V0
sin
9π 3 ~2
3π 2 ~T
4ma2
2
.
Problem 11.28
Spontaneous absorption would involve taking energy (a photon) from the ground state of the electromagnetic
field. But you can’t do that, because the gound state already has the lowest allowed energy.
396
CHAPTER 11. QUANTUM DYNAMICS
Problem 11.29
(a)
H = −γB · S = −γ (Bx Sx + By Sy + Bz Sz ) ;
~
γ~
0 1
0 −i
1 0
H = −γ (Bx σx + By σy + Bz σz ) = −
Bx
+ By
+ Bz
1 0
i 0
0 −1
2
2
γ~
γ~
Bz
Bx − iBy
B0
Brf (cos ωt + i sin ωt)
=−
= −
−Bz
−B0
2 Bx + iBy
2 Brf (cos ωt − i sin ωt)
γ~
B0
Brf eiωt
= −
.
−iωt
−B0
2 Brf e
(b) i~χ̇ = Hχ ⇒
γ~
γ~ B0 a + Brf eiωt b
ȧ
B0
Brf eiωt
a
i~
=−
=−
b
ḃ
2 Brf e−iωt −B0
2 Brf e−iωt a − B0 b


 ȧ = i γ B0 a + Brf eiωt b = i Ωeiωt b + ω0 a ,
2
2
i
γ

 ḃ = −i B0 b − Brf e−iωt a =
Ωe−iωt a − ω0 b .
2
2
⇒
(c) You can decouple the equations by differentiating with respect to t, but it is simpler just to check the quoted
results. First of all, they clearly satisfy the initial conditions: a(0) = a0 and b(0) = b0 . Differentiating a:
iω
i
ω0
ω0
a + −a0 sin(ω 0 t/2) + 0 [a0 (ω0 − ω) + b0 Ω]
cos(ω 0 t/2) eiωt/2
2
2
ω
2
i iωt/2
ω
ωa0 cos(ω 0 t/2) + i 0 [a0 (ω0 − ω) + b0 Ω] sin(ω 0 t/2)
= e
2
ω
+ iω 0 a0 sin(ω 0 t/2) + [a0 (ω0 − ω) + b0 Ω] cos(ω 0 t/2)
ȧ =
Equation 11.124 says this should be equal to
i
i
Ω
Ωeiωt b + ω0 a = eiωt/2 Ωb0 cos(ω 0 t/2) + i 0 [b0 (ω − ω0 ) + a0 Ω] sin(ω 0 t/2)
2
2
ω
ω0
+ ω0 a0 cos(ω 0 t/2) + i 0 [a0 (ω0 − ω) + b0 Ω] sin(ω 0 t/2) .
ω
By inspection the cos(ω 0 t/2) terms in the two expressions are equal; it remains to check that
i
Ω
ω0
ω
[a0 (ω0 − ω) + b0 Ω] + iω 0 a0 = i 0 [b0 (ω − ω0 ) + a0 Ω] + i 0 [a0 (ω0 − ω) + b0 Ω] ,
ω0
ω
ω
which is to say
a0 ω(ω0 − ω) + b0 ωΩ + a0 (ω 0 )2 = b0 Ω(ω − ω0 ) + a0 Ω2 + a0 ω0 (ω0 − ω) + b0 ω0 Ω,
or
a0 ωω0 − ω 2 + (ω 0 )2 − Ω2 − ω02 + ω0 ω = b0 [Ωω − ω0 Ω + ω0 Ω − ωΩ] = 0.
CHAPTER 11. QUANTUM DYNAMICS
397
Substituting Eq. 11.125 for ω 0 , the coefficient of a0 on the left becomes
2ωω0 − ω 2 + (ω − ω0 )2 + Ω2 − Ω2 − ω02 = 0. X
The check of b(t) is identical, with a ↔ b, ω0 → −ω0 , and ω → −ω.
(d)
b(t) = i
Ω
sin(ω 0 t/2)e−iωt/2 ;
ω0
P (t) = |b(t)|2 =
Ω
ω0
2
sin2 (ω 0 t/2).
(e)
P(t)
1
1/2
6t
t0
t
The maximum (Pmax = 1) occurs (obviously) at ω = ω0 .
P = 21 ⇒ (ω − ω0 )2 = Ω2 ⇒ ω = ω0 ± Ω,
so
∆ω = ω+ − ω− = 2Ω.
(f ) B0 = 10, 000 gauss = 1 T; Brf = 0.01 gauss = 1 × 10−6 T.
gp e
7.89, γ =
, where gp = 5.59. So
2mp
ω0 = γB0 . Comparing Eqs. 4.156 and
ω0
gp e
(5.59)(1.6 × 10−19 )
=
(1) = 4.26 × 107 Hz.
B0 =
2π
4πmp
4π(1.67 × 10−27 )
∆ω
Ω
γ
2Brf
∆ν =
=
=
2Brf = νres
= (4.26 × 107 )(2 × 10−6 ) = 85.2 Hz.
2π
π
2π
B0
νres =
Problem 11.30
(a)
∂A
E0
=−
cos(k · r − ωt)(−ω) = E0 cos(k · r − ωt).
∂t
ω
1
1
B = ∇ × A = sin(k · r − ωt)∇ × E0 − E0 × ∇ (sin(k · r − ωt))
ω
ω
1
k × E0
1
= 0 − E0 × [cos(k · r − ωt)k] =
cos(k · r − ωt) = k × E
ω
ω
ω
E = −∇ϕ −
(note that E0 is a constant, so its derivatives are zero).
398
CHAPTER 11. QUANTUM DYNAMICS
(b) From Eq. 4.188
1
p2
q
q2 2
(p − qA) · (p − qA) + qϕ + V =
+V −
(p · A + A · p) +
A
2m
2m
2m
2m
q2 2
q
(p · A + A · p) +
A .
= H0 −
2m
2m
H=
Now, acting on a test function f (r),
(p · A + A · p)f = −i~ [∇(Af ) + A · (∇f )] = −i~ [(∇ · A)f + A · (∇f ) + A · (∇f )] .
But
1
1
∇ · [E0 sin(k · r − ωt)] = [(∇ · E)0) sin(k · r − ωt) + E0 · (∇ sin(k · r − ωt))]
ω
ω
1
= [0 + E0 · cos(k · r − ωt)k] = 0
ω
∇·A=
(since E0 · k = 0.) So (dropping the test function)
(p · A + A · p) = 2A · p.
Ignoring the A2 term, then,
H0 =
e
e
e
e
A·p=
sin(k · r − ωt)E0 · p =
eik·r E0 · pe−iωt −
e−ik·r E0 · peiωt . X
m
mω
2imω
2imω
(c) Referring to Eq. 11.36, we have
Vba =
e
E0 hb|pz |ai
imω
But (using Eqs. 2.52 and 3.65)
[H 0 , z] =
1 2
1 2
1
−i~
[p , z] =
[p , z] =
(pz [pz , z] + [pz , z]pz ) =
pz ,
2m
2m z
2m
m
so (referring to Eqs. 11.18 and 11.40)
e
im
eE0
eE0
E0
hb|[H 0 , z]|ai =
hb|(H 0 z − zH 0 )|ai =
(Eb − Ea )hb|z|ai
imω
~
~ω
~ω
eE0
ω0
=
~ω0 hb|z|ai = − E0 ℘. X
~ω
ω
Vba =
This differs from Eq. 11.41, which did not include the ratio of the ω’s, but it does not affect our conclusions
in Sections 11.2.3 or 11.3, because we only considered transitions at resonance, ω = ω0 (the probability
of a transition at other driving frequencies being negligible).
Problem 11.31
(a)
H 0 = −qE · r = −q(E0 · r)(k · r) sin(ωt).
Write E0 = E0 n̂, k =
ω
k̂.
c
Then
CHAPTER 11. QUANTUM DYNAMICS
399
E0 ω
qE0 ω
0
(n̂ · r)(k̂ · r) sin(ωt). Hba
=−
hb|(n̂ · r)(k̂ · r)|ai sin(ωt).
c
c
0
This is the analog to Eq. 11.40: Hba
= −qE0 hb|n̂ · r|ai cos ωt. The rest of the analysis is identical to
the dipole case (except that it is sin(ωt) instead of cos(ωt), but this amounts to resetting the clock, and
clearly has no effect on the transition rate). We can skip therefore to Eq. 11.63, except for the factor of
1/3, which came from the averaging in Eq. 11.53:
H 0 = −q
A=
q2 ω5
ω3 q2 ω2
2
|hb|(n̂
·
r)(
k̂
·
r)|ai|
|hb|(n̂ · r)(k̂ · r)|ai|2 .
=
π0 ~c3 c2
π0 ~c5
(b) Let the oscillator lie along the x direction, so (n̂ · r) = n̂x x and k̂ · r = k̂x x. For a transition from n to n0 ,
we have
2
~
q2 ω5 k̂x n̂x |hn0 |x2 |ni|2 . From Example 2.5, hn0 |x2 |ni =
hn0 |(a2+ +a+ a− +a− a+ +a2− )|ni,
A=
5
π0 ~c
2mω̄
where ω̄ is the frequency of the oscillator, not to be confused with ω, the frequency of the electromagnetic
wave. Now, for spontaneous emission the final state must be lower in energy, so n0 < n, and hence the
only surviving term is a2− . Using Eq. 2.67:
p
~ p
~
hn0 | n(n − 1)|n − 2i =
n(n − 1) δn0 ,n−2 .
2mω̄
2mω̄
Evidently transitions only go from |ni to |n − 2i, and hence
hn0 |x2 |ni =
1
En − En−2
=
(n + 12 )~ω̄ − (n − 2 + 21 )~ω̄ = 2ω̄.
~
~
2
p
~
q2 ω5
2 ~
(
k̂
n̂
)
n(n − 1).
hn0 |x2 |ni =
n(n − 1) δn0 ,n−2 ; Rn→n−2 =
x
x
mω
π0 ~c5
m2 ω 2
ω=
It remains to calculate the average of (k̂x n̂x )2 . It’s easiest to reorient the oscillator along a direction r̂,
making angle θ with the z axis, and let the radiation be incident
from the z direction (so k̂x → k̂r = cosθ).
Averaging over the two polarizations (î and ĵ): hn̂2r i = 21 î2r + ĵr2 = 12 sin2 θ cos2 φ + sin2 θ sin2 φ =
2
1
2 sin θ. Now average overall directions:
Z
Z π
1
1
1
2
2
sin θ cos θ sin θ dθ dφ =
2π
(1 − cos2 θ) cos2 θ sin θ dθ
4π
2
8π
π
0
cos3 θ cos5 θ
1 2 2
1
1
=
−
+
=
−
=
.
4
3
5
4
3
5
15
0
hk̂r2 n̂2r i =
R=
1 q 2 ~ω 3
n(n − 1).
15 π0 m2 c5
Comparing Eq. 11.70:
R(forbidden)
2
~ω
= (n − 1) 2 .
R(allowed)
5
mc
For a nonrelativistic system, ~ω mc2 ; hence the term “forbidden”.
(c) If √
both the initial state and the final state have l = 0, the wave function is independent of angle (Y00 =
1/ 4π), and the angular part of the integral is:
Z
4π
ha|(n̂ · r)(k̂ · r)|bi = · · · (n̂ · r)(k̂ · r) sin θ dθ dφ = · · ·
(n̂ · k̂) (Eq. 6.95).
3
But n̂ · k̂ = 0, since electromagnetic waves are transverse.
for forbidden transitions.
So R = 0 in this case, both for allowed and
400
CHAPTER 11. QUANTUM DYNAMICS
Problem 11.32
`0 = ` + 1
From Eq. 11.76, hn0 `0 m0 |z|n`mi = 0 unless m0 = m, so the only nonzero z term is
Z
0 0
∗
m 2
hn ` m|z|n`mi =
Rn0 `0 (Y`m
r dr sin θ dθ dφ
0 ) r cos θ Rn` Y`
s
s
Z π
(2`0 + 1) (`0 − |m|)! (2` + 1) (` − |m|)!
m
2π
P`m
= I
0 P` cos θ sin θ dθ.
4π
(`0 + |m|)!
4π (` + |m|)!
0
[F1]
This is independent of the sign of m, so we might as well assume m ≥ 0. The integral (changing variables to
x ≡ cos θ) is (using Eq. 11.134)
Z 1
Intθ =
m
x P`+1
(x)P`m (x) dx =
−1
Z 1
Z 1
1
m
m
m
m
(` + m)
P`+1 P`−1 dx + (` − m + 1)
P`+1 P`+1 dx .
(2` + 1)
−1
−1
Now, it follows from Eq. 4.33 that the associated Legendre functions satisfy the orthogonality relation
Z 1
m
P`m
0 (x)P` (x) dx =
−1
(` + |m|)!
2
δ``0 ,
(2` + 1) (` − |m|)!
[F2]
so
Intθ =
(` − m + 1)
2
(` + 1 + m)!
2
(` + 1 + m)!
=
.
(2` + 1) (2` + 3) (` + 1 − m)!
(2` + 1)(2` + 3) (` − m)!
Putting this into Eq. [F1]:
s
s
I
(` + 1 − m)!
(` − m)!
2
(` + 1 + m)!
hn (` + 1)m|z|n`mi =
(2` + 3)
(2` + 1)
2
(` + 1 + m)!
(` + m)! (2` + 1)(2` + 3) (` − m)!
s
(` + 1)2 − m2
= I
.
[F3]
(2` + 1)(2` + 3)
0
From Eq. 11.76, hn0 `0 m0 |x|n`mi = 0 unless m0 = m ± 1; let’s start with m + 1:
Z
hn0 `0 (m + 1)|x|n`mi = Rn0 `0 (Y`m+1
)∗ r sin θ cos φ Rn` Y`m r2 dr sin θ dθ dφ
0
s
s
Z
Z 2π
(2`0 + 1) (`0 − m − 1)! (2` + 1) (` − m)! π m+1 m 2
= I
P
P
sin
θ
dθ
cos φe−i(m+1)φ eimφ dφ.[F4]
0
`
4π
(`0 + m + 1)!
4π (` + m)! 0 `
0
Intφ =
1
2
Z 2π
0
(eiφ + e−iφ )e−iφ dφ =
1
2
Z 2π
(1 + e−2iφ ) dφ = π.
[F5]
0
Changing variables (x ≡ cos θ), and using Eqs. 11.135 and [F2]:
Z 1
Z 1p
Z 1
1
m+1
m+1 m+1
m+1 m+1
m
2
Intθ =
1 − x P`+1 (x)P` (x) dx =
P
P
dx −
P`+1 P`−1 dx
(2` + 1) −1 `+1 `+1
−1
−1
2
(` + m + 2)!
=
.
(2` + 1)(2` + 3) (` − m)!
CHAPTER 11. QUANTUM DYNAMICS
401
Thus [F4] becomes
I
hn0 (` + 1)(m + 1)|x|n`mi =
2
s
I
=
2
s
(` − m)!
(2` + 3)
(` + m + 2)!
s
(2` + 1)
(` − m)!
1
(` + m + 2)!
(` + m)! (2` + 1)(2` + 3) (` − m)!
(` + m + 2)(` + m + 1)
.
(2` + 1)(2` + 3)
[F6]
Now we do the same for m0 = m − 1:
Z
hn0 `0 (m − 1)|x|n`mi = Rn0 `0 (Y`m−1
)∗ r sin θ cos φ Rn` Y`m r2 dr sin θ dθ dφ
0
s
s
Z
Z 2π
(2`0 + 1) (`0 − m + 1)! (2` + 1) (` − m)! π m−1 m 2
P
cos φe−i(m−1)φ eimφ dφ.[F7]
P
sin
θ
dθ
=I
0
`
4π
(`0 + m − 1)!
4π (` + m)! 0 `
0
Intφ =
1
2
Z 2π
(eiφ + e−iφ )eiφ dφ =
0
1
2
Z 2π
(e2iφ + 1) dφ = π.
[F8]
0
Changing variables (x ≡ cos θ), and using Eqs. 11.135 and [F2]:
Z 1
Z 1p
Z 1
1
m−1
m
m
m
m m
2
Intθ =
1 − x P`+1 (x)P` (x) dx =
P P dx −
P` P` dx
(2` + 3) −1 `+2 `
−1
−1
(` + m)!
2
= −
,
(2` + 1)(2` + 3) (` − m)!
and [F7] becomes
I
hn (` + 1)(m − 1)|x|n`mi = −
4
s
I
= −
2
s
0
(` − m + 2)!
(2` + 3)
(` + m)!
s
(2` + 1)
(` − m)!
2
(` + m)!
(` + m)! (2` + 1)(2` + 3) (` − m)!
(` − m + 2)(` − m + 1)
.
(2` + 1)(2` + 3)
[F9]
Meanwhile, Eq. 11.76 says |hn0 `0 m0 |y|n`mi|2 = |hn0 `0 m0 |y|n`mi|2 , so
|hn0 (` + 1)(m + 1)|r|n`mi|2 + |hn0 (` + 1)m|r|n`mi|2 + |hn0 (` + 1)(m − 1)|r|n`mi|2
s
#2 " s
#2
"
#2
" s
I (` − m + 2)(` − m + 1)
(` + 1)2 − m2
I (` + m + 2)(` + m + 1)
+ I
+2 −
=2
2
(2` + 1)(2` + 3)
(2` + 1)(2` + 3)
2
(2` + 1)(2` + 3)
2
2
2
I
(` + m + 2)(` + m + 1) + 2[(` + 1) − m ] + (` − m + 2)(` − m + 1)
=
2
(2` + 1)(2` + 3)
2
(2` + 5` + 3)
(` + 1)
= I2
= I2
.
[F10]
(2` + 1)(2` + 3)
(2` + 1)
℘|2 (summed over the three allowed transitions) is e2 I 2 (`+1)/(2`+1), and the spontaneous emission
Therefore, |℘
rate (Eq. 11.63) is
A`→`+1 =
`0 = ` − 1
e2 ω 3 I 2 (` + 1)
.
3π0 ~c3 (2` + 1)
[F11]
402
CHAPTER 11. QUANTUM DYNAMICS
Return to Eq. [F1]. This time the integral is
Z 1
Z 1
Z 1
1
m
m
m
m
m
P`−1
P`+1
dx
x P`−1
(x)P`m (x) dx =
P`−1
P`−1
dx + (` − m + 1)
Intθ =
(` + m)
(2` + 1)
−1
−1
−1
(` + m)
2
(` − 1 + m)!
2
(` + m)!
=
=
.
(2` + 1) (2` − 1) (` − 1 − m)!
(2` − 1)(2` + 1) (` − m − 1)!
Therefore
s
s
I
2
(` + m)!
(` − 1 − m)!
(` − m)!
hn (` − 1)m|z|n`mi =
(2` − 1)
(2` + 1)
2
(` − 1 + m)!
(` + m)! (2` − 1)(2` + 1) (` − m − 1)!
s
`2 − m2
= I
.
[F12]
(2` − 1)(2` + 1)
0
From hn0 `0 m0 |x|n`mi with m0 = m + 1, Eqs. [F4] and [F5] are unchanged; this time
Z 1
Z 1p
Z 1
1
m+1 m+1
m+1
m+1 m+1
m
2
P
1 − x P`−1 (x)P` (x) dx =
P
Intθ =
dx −
P`−1 P`−1 dx
(2` + 1) −1 `−1 `+1
−1
−1
2
(` + m)!
,
= −
(2` − 1)(2` + 1) (` − m − 2)!
and [F4] becomes
I
hn (` − 1)(m + 1)|x|n`mi = −
2
s
I
= −
2
s
0
(` − m − 2)!
(2` − 1)
(` + m)!
s
(2` + 1)
(` − m)(` − m − 1)
.
(2` − 1)(2` + 1)
(` − m)!
1
(` + m)!
(` + m)! (2` − 1)(2` + 1) (` − m − 2)!
[F13]
Now we do the same for m0 = m − 1. Eqs. [F7] and [F8] are unchanged, the θ integral is
Z 1
Z 1p
Z 1
1
m−1
m
m m
m
m
2
Intθ =
1 − x P`−1 (x)P` (x) dx =
P P dx −
P`−2 P` dx
(2` − 1) −1 ` `
−1
−1
(` + m)!
2
,
=
(2` − 1)(2` + 1) (` − m)!
and [F7] becomes
I
hn (` − 1)(m − 1)|x|n`mi =
2
s
I
=
2
s
0
(` − m)!
(2` − 1)
(` + m − 2)!
(` + m)(` + m − 1)
.
(2` − 1)(2` + 1)
s
(2` + 1)
(` − m)!
1
(` + m)!
(` + m)! (2` − 1)(2` + 1) (` − m)!
[F14]
Thus
|hn0 (` − 1)(m + 1)|r|n`mi|2 + |hn0 (` − 1)m|r|n`mi|2 + |hn0 (` − 1)(m − 1)|r|n`mi|2
s
"
#2 " s
#2
" s
#2
I (` − m)(` − m − 1)
`2 − m2
I (` + m)(` + m − 1)
=2 −
+ I
+2
.
2
(2` − 1)(2` + 1)
(2` − 1)(2` + 1)
2
(2` − 1)(2` + 1)
I 2 (` − m)(` − m − 1) + 2(`2 − m2 ) + (` + m)(` + m − 1)
=
2
(2` − 1)(2` + 1)
2
(2` − `)
`
= I2
= I2
,
[F15]
(2` − 1)(2` + 1)
(2` + 1)
CHAPTER 11. QUANTUM DYNAMICS
403
and the emission rate is
A`→`−1 =
`
e2 ω 3 I 2
.
3π0 ~c3 (2` + 1)
[F16]
(Of course, I is different for the two cases ` → ` ± 1.)
Problem 11.33
Choosing the triplet configuration |1i = |↑e ↑p i we have
1
h1 | Se | 0i = √ [h↑e ↑p | Se | ↑e ↓p i − h↑e ↑p | S | ↓e ↑p i] .
2
Since Se doesn’t effect the spin of the proton the first term vanishes and we have
1
h1 | Se | 0i = − √ h↑e | Se | ↓e i
2

 1
h↑e | Se+ + Se− | ↓e i
2
1
= − √  21i h↑e | Se+ − Se− | ↓e i 
2
h↑e | Sez | ↓e i
 
i
i~
= √ 1 .
2 2 0
We can then compute
2
|h1 | Se | 0i| =
~2
~2
(1 + 1) =
.
8
4
Using the expression given in the problem we then have
A=
~2
ω03 e2
3 π 0 ~ c5 m2e 4
The frequency ω0 is that of the 21cm line transition (see Equation 7.97),
ω0 =
∆E
4 gp ~3
=
,
~
3 mp m2e c2 a4
giving
τ=
1
81 1 1
=
A
64 α13 gp3
mp
me
3
~
= 1.1 × 107 yr .
me c2
Problem 11.34
(a) Schrödinger equation:
~2 d2 ψ
−
= Eψ,
2m dx2
or
d2 ψ
= −k 2 ψ
dx2
(k ≡
√
2mE/~)
(
0 < x < 12 a + ,
1
2 a + < x < a.
404
CHAPTER 11. QUANTUM DYNAMICS
Boundary conditions:
ψ(0) = ψ( 21 a + ) = ψ(a) = 0.
Solution:
(1) 0 < x < 12 a + :
ψ(x) = A sin kx + B cos kx. But ψ(0) = 0 ⇒ B = 0, and
(
k( 12 a + ) = nπ (n = 1, 2, 3, . . .) ⇒ En = n2 π 2 ~2 /2m(a/2 + )2 ,
1
ψ( 2 a + ) = 0 ⇒
or else A = 0.
(2) 21 a + < x < a :
ψ(x) = F sin k(a − x) + G cos k(a − x). But ψ(a) = 0 ⇒ G = 0, and
(
k( 12 a − ) = n0 π (n0 = 1, 2, 3, . . .) ⇒ En0 = (n0 )2 π 2 ~2 /2m(a/2 − )2 ,
1
ψ( 2 a + ) = 0 ⇒
or else F = 0.
The ground state energy is



either


or else
π 2 ~2
(n = 1), with F = 0,
2m( 12 a + )2
2 2
π ~
E10 =
(n0 = 1), with A = 0.
2m( 12 a − )2
E1 =
Both are allowed energies, but E1 is (slightly) lower (assuming is positive), so the ground state is
ψ(x) =
q

2
1
2 a+
sin
πx
1
2 a+
, 0 ≤ x ≤ 21 a + ;
1
2 a + ≤ x ≤ a.
0,
s(x)
ax
a_ a_ + ¡
2 2
(b)
~2 d2 ψ
−
+f (t)δ(x− 12 a−)ψ = Eψ
2m dx2
⇒
ψ(x) =
A sin kx,
0 ≤ x < 21 a + ,
F sin k(a − x), 12 a + < x ≤ a,
√
where
k≡
2mE
.
~
A sin k
1
2a + Continuity in ψ at x = 12 a + :
A sin k
1
2a + = F sin k a − 12 a − = F sin k
1
2a − ⇒F =A
sin k
sin k
1
2 a + .
1
2a − Discontinuity in ψ 0 at x = 21 a + (Eq. 2.128):
2mf
−F k cos k(a−x)−Ak cos kx = 2 A sin kx ⇒ F cos k
~
1
2 a − +A cos k
1
2a + =−
2mf
~2 k
.
CHAPTER 11. QUANTUM DYNAMICS
1
2 a + cos k
1
2a − 405
2T
A sin k 21 a + .
+ A cos k
=−
z
2T
sin k 12 a + sin k 12 a − .
sin k 12 a + cos k 21 a − + cos k 12 a + sin k 12 a − = −
z
2T 1 cos k 12 a + − 21 a + − cos k 12 a + + 12 a − .
sin k 12 a + + 12 a − = −
z
2
sin k
A
sin k
sin ka = −
1
2a − 1
2a + T
(cos 2k − cos ka)
z
⇒
z sin z = T [cos z − cos(zδ)].
(c)
sin z =
T
(cos z − 1)
z
z
cos z − 1
=
= − tan(z/2)
T
sin z
⇒
⇒
z
tan(z/2) = − .
T
Plot tan(z/2) and −z/T on the same graph, and look for intersections:
tan(z/2)
/
3/
2/
z
-z/T
As t : 0 → ∞, T : 0 → ∞, and the straight line rotates counterclockwise from 6 o’clock to 3 o’clock,
~2 π 2
so the smallest z goes from π to 2π, and the ground state energy goes from ka = π ⇒ E(0) =
2ma2
~2 π 2
(appropriate for a well of width a/2.
(appropriate to a well of width a) to ka = 2π ⇒ E(∞) =
2m(a/2)2
(d) Mathematica yields the following table:
(e) Pr =
Ir
1
=
,
Ir + Il
1 + (Il /Ir )
T
z
0
3.14159
1
3.67303
5
4.76031
20
5.72036
100
6.13523
where
a/2+
1
1
x−
sin(2kx)
2
4k
0
0
h
i a 1 a
1
a
2
1
2
2
2
= A
+ −
sin 2k
+
= A 1+
−
sin ka + ka
2 2
4k
2
4
a
ka
a
a
1
= A2 1 + δ − sin(z + zδ) .
4
z
Z a/2+
Il =
Z a
A2 sin2 kx dx = A2
F 2 sin2 k(a − x) dx.
Ir =
Let
u ≡ a − x, du = −dx.
a/2+
= −F 2
Z 0
a/2−
sin2 ku du = F 2
Z a/2−
0
sin2 ku du =
a 2
1
F 1 − δ − sin(z − zδ) .
4
z
1000
6.21452
406
CHAPTER 11. QUANTUM DYNAMICS
sin2 [z(1 − δ)/2]
sin2 k(a/2 − )
Il
A2 [1 + δ − (1/z) sin(z + zδ)]
A2
=
.
=
= 2
. But (from (b))
Ir
F [1 − δ − (1/z) sin(z − zδ)]
F2
sin2 k(a/2 + )
sin2 [z(1 + δ)/2]
I+
1
1
Pr =
, where I± ≡ 1 ± δ − sin z(1 ± δ) sin2 [z(1 ∓ δ)/2].
.
=
I−
z
1 + (I+ /I− )
Using δ = 0.01 and the z’s from (d), Mathematica gives
T
Pr
0
0.490001
1
0.486822
5
0.471116
20
0.401313
100
0.146529
1000
0.00248443
As t : 0 → ∞ (so T : 0 → ∞), the probability of being in the right half drops from almost 1/2 to zero—the
particle gets sucked out of the slightly smaller side, as it heads for the ground state in (a).
(f )
T=0
T=1
T=5
T=20
T=100
T=1000
Problem 11.35
(a)
r
Let
2
(mvx
− 2Eni at)/2~w = φ(x, t).
Φn =
nπ 2
sin
x eiφ ,
w
w
so
r h
r
nπ ∂φ nπ nπ i
√
1 1
nπx
∂Φn
2
2
iφ
iφ
eiφ
= 2 −
x e +
x e +
sin
x
i
v sin
− 2 v cos
∂t
2 w3/2
w
w
w
w
w
w
∂t
nπ v
nπxv
∂φ
1
v
∂φ
2Eni a
Ei a v
2
i
= −
−
cot
x
+
i
=
−
Φ
.
−
mvx
−
2E
at
= − n − φ.
n
n
2
2
2w
w
w
∂t
∂t
2~
w
w
~w
w
nπ v
∂Φn
nπxv
Ei a
v
i~
= −i~
+
cot
x + i n + i φ Φn .
2
∂t
2w
w
w
~w
w
r h
r
nπ i
nπ ~ 2 ∂ 2 Φn
∂Φn
2 nπ
2
∂φ
iφ
iφ
HΦn = −
.
=
cos
x e +
sin
x e
i
.
2m ∂x2
∂x
w w
w
w
w
∂x
h nπ
nπ ∂φ
mvx
∂Φn
mvx i
=
.
=
cot
x +i
Φn .
∂x
~w
∂x
w
w
~w
imv h nπ
nπ ∂ 2 Φn
nπ 2
mvx i2
2 nπ
=
−
csc
x
+
Φ
+
cot
x
+
i
Φn .
n
∂x2
w
w
~w
w
w
~w
So the Schrödinger equation (i~∂Φn /∂t = HΦn ) is satisfied ⇔
nπ v
nπxv
Eni a
v
+
cot
x +i
+i φ
−i~
2w
w2
w
~w
w
CHAPTER 11. QUANTUM DYNAMICS
407
~2
=−
2m
nπ nπ 2 2 nπ imv h nπ
mvx i2
csc
−
x +
+
cot
x +i
w
w
~w
w
w
~w
nπxv ~2 nπ mvx nπvx
?
Cotangent terms: − i~
=−
2 i
= −i~ 2 . X
2
w
2m
w ~w
w
Remaining trig terms on right:
nπ 2
nπ nπ 2
nπ nπ 2
nπ 2 1 − cos2 (nπx/w) −
csc2
cot2
=
−
.
x +
x =−
w
w
w
w
w
w
sin2 (nπx/w)
This leaves:
v
Ei a
v mvx2 − 2Eni at
nπ 2 imv m2 v 2 x2
? ~
+
i
+i n +i
=
−
− 2 2
2w
~w
w
2~w
2m
w
~w
~ w
2
2
iv Eni a mv 2
x
vEni at ? ~n2 π 2
x
iv mv 2
−
− +
=−
+ − ~
~w
2mw
2
2 2~w
2~w
i
i 2
2 2
E a
E a ? ~n π
n2 π 2 ~2 a2
~n2 π 2
− n (w − vt) = − n = −
⇔−
=
−
= r.h.s. X
~w
~w
2mw
2ma2 ~w
2mw
So Φn does satisfy the Schrödinger equation, and since Φn (x, t) = (· · · ) sin (nπx/w), it fits the boundary
conditions: Φn (0, t) = Φn (w, t) = 0.
(b)
Equation 11.137 =⇒ Ψ(x, 0) =
X
cn Φn (x, 0) =
X
r
cn
nπ 2
2
sin
x eimvx /2~a .
a
a
r
0 2
nπ
2
sin
x e−imvx /2~a and integrate:
a
a
r Z a
0 0 nπ X 2Z a
2
nπ
nπ
−imvx2 /2~a
x e
dx =
cn
x sin
x dx = cn0 .
Ψ(x, 0) sin
sin
a 0
a
a 0
a
a
{z
}
|
Multiply by
δnn0
r Z a
nπ 2
2
e−imvx /2~a sin
x Ψ(x, 0)dx. In this particular case,
So, in general: cn =
a 0
a
Z
nπx π 2 a −imvx2 /2~a
π
a
mvx2
mvz 2 a2
mva 2
cn =
e
sin
sin
x dx. Let
x ≡ z; dx = dz;
=
=
z .
a 0
a
a
a
π
2~a
2~a π 2
2π 2 ~
Z
2 π −iαz2
cn =
e
sin(nz) sin(z) dz. QED
π 0
(c)
w(Te ) = 2a ⇒ a + vTe = 2a ⇒ vTe = a ⇒ Te = a/v;
e−iE1 t/~ ⇒ ω =
E1
2π
~
⇒ Ti =
= 2π , or
~
ω
E1
2π~
4 ma2
4ma2
a
4ma2
4 mav
2
2ma
=
.
T
=
.
Adiabatic
⇒
T
T
⇒
⇒
1, or
i
e
i
2
2
π ~
π ~
π~
v
π~
π ~
Z
mav 2 π
8π
= 8πα 1, so α 1. Then cn =
sin(nz) sin(z)dz = δn1 . Therefore
2π 2 ~
π 0
Ti =
408
CHAPTER 11. QUANTUM DYNAMICS
r
Ψ(x, t) =
πx 2
i
2
ei(mvx −2E1 at)/2~w ,
sin
w
w
which (apart from a phase factor) is the ground state of the instantaneous well, of width w, as required
2
mva
by the adiabatic theorem. (Actually, the first term in the exponent, which is at most mv(2a)
2~(2a) = ~ 1
and could be dropped, in the adiabatic regime.)
(d)
t
Z t
1
π2 ~
π 2 ~2
1
1
0
dt = −
−
0 2
2m
2m
v a + vt0
0 (a + vt )
0
2
2
2
π ~ 1
1
π ~ vt
π ~t
=−
−
=−
=−
.
2mv a w
2mv aw
2maw
r
πx i
2
mvx2
e−iE1 at/~w
term, as explained in (c)) Ψ(x, t) =
sin
So (dropping the
2~w
w
w
r
πx E1i at
π 2 ~2 at
π 2 ~t
2
(since −
=−
=−
= θ):
sin
Ψ(x, t) =
eiθ .
~w
2ma2 ~w
2maw
w
w
1
θ(t) = −
~
can be written
This is exactly what one would naively expect: For a fixed well (of width a) we’d have Ψ(x, t) =
Ψ1 (x)e−iE1 t/~ ; for the (adiabatically) expanding well, simply replace a by the (time-dependent) width
w, and integrate to get the accumulated phase factor, noting that E1 is now a function of t.
Problem 11.36
(a) Check the answer given:
Rt
xc = ω 0 f (t0 ) sin [ω(t − t0 )] dt0 =⇒ xc (0) = 0. X
ẋc = ωf (t) sin [ω(t − t)] + ω 2
Z t
f (t0 ) cos [ω(t − t0 )] dt0 = ω 2
0
ẍc = ω 2 f (t) cos [ω(t − t)] − ω 3
Z t
f (t0 ) cos [ω(t − t0 )] dt0 ⇒ ẋc (0) = 0. X
0
Z t
f (t0 ) sin [ω(t − t0 )] dt0 = ω 2 f (t) − ω 2 xc .
0
Now the classical equation of motion is m(d2 x/dt2 ) = −mω 2 x + mω 2 f . For the proposed solution,
m(d2 xc /dt2 ) = mω 2 f − mω 2 xc , so it does satisfy the equation of motion, with the appropriate boundary
conditions.
(b) Let z ≡ x − xc
(so ψn (x − xc ) = ψn (z), and z depends on t as well as x).
∂Ψ
dψn
i
1
xc
m
mω 2
=
(−ẋc )ei{} + ψn ei{}
− (n + )~ω + mẍc (x − ) − ẋ2c +
f xc
∂t
dz
~
2
2
2
2
1
mω 2
ẋ2
[ ] = −(n + )~ω +
2x(f − xc ) + x2c − c2 .
2
2
ω
dψn i{}
1
mω 2
ẋ2c
∂Ψ
2
= −ẋc
e + iΨ −(n + )ω +
2x(f − xc ) + xc − 2
.
∂t
dz
2
2~
ω
2
∂Ψ
dψn i{}
∂2Ψ
d2 ψn i{}
dψn i{} i
mẋc
i{} i
=
e + ψn e
(mẋc );
=
e +2
e
(mẋc ) −
ψn ei{} .
∂x
dz
~
∂x2
dz 2
dz
~
~
CHAPTER 11. QUANTUM DYNAMICS
409
~2 ∂ 2 Ψ 1
+ mω 2 x2 Ψ − mω 2 f xΨ
2m ∂x2
2
2
2 2
~2 dψn i{} imẋc
1
~2 mẋc
~ d ψn i{}
e
−
Ψ + mω 2 x2 Ψ − mω 2 f xΨ.
2
e
+
=−
2m dz 2
2m dz
~
2m
~
2
HΨ = −
1
~2 d 2 ψ n
1
+ mω 2 z 2 ψn = (n + )~ωψn , so
2
2m dz
2
2
1 1
dψ
m
1
n i{}
HΨ = (n
+ )~ωΨ − mω 2 z 2 Ψ − i~ẋ
e + ẋ2c Ψ + mω 2 x2 Ψ − mω 2 f xΨ
c
dz
2
2
2
2
dψ
1 2
1 mω 2
∂Ψ
?
n i{}
2
= −i~
ẋc
e − ~Ψ −(n
(2xf − 2xxc + xc − 2 ẋc )
+ )ω +
= i~
∂t
2~
ω
dz
2
!
2
1
1
m
? mω
− 2xxc + x2c − 1 ẋ2c
2
− mω 2 z 2 + ẋ2c + mω 2 x2 − 2xf
mω
fx = −
2
2
2
2
ω2
But −
?
z 2 − x2 = −2xxc + x2c ;
?
z 2 = (x2 − 2xxc + x2c ) = (x − xc )2 . X
(c)
1
~2 ∂ 2
1
Eq. 11.140 ⇒ H = −
+ mω 2 x2 − 2xf + f 2 − mω 2 f 2 . Shift origin: u ≡ x − f.
2
2m ∂x
2
2
~2 ∂ 2
1
1
H= −
+ mω 2 u2 − mω 2 f 2 .
2m ∂u2
2
2
The first term is a simple harmonic oscillator in the variable u; the second is a constant (with respect
to position). So the eigenfunctions are ψn (u), and the eigenvalues are harmonic oscillator ones, (n +
1
less the constant: En = (n + 12 )~ω − 12 mω 2 f 2 .
2 )~ω,
Z t
1 d
d
0
0
(d) Note that sin [ω(t − t )] =
cos [ω(t − t )], so xc (t) =
f (t0 ) 0 cos [ω(t − t0 )] dt0 , or
ω dt0
dt
0
Z t Z t t
df
df
0
0
xc (t) = f (t0 ) cos [ω(t − t0 )] −
cos
[ω(t
−
t
)]
dt
=
f
(t)
−
cos [ω(t − t0 )] dt0
0
0
dt
dt
0
0
0
(since f (0) = 0). Now, for an adiabatic process we want df /dt very small; specifically:
df
ωf (t)
dt0
(0 < t0 ≤ t). Then the integral is negligible compared to f (t), and we have xc (t) ≈ f (t). (Physically,
this says that if you pull on the spring very gently, no fancy oscillations will occur; the mass just moves
along as though attached to a string of fixed length.)
(e) Put xc ≈ f
into Eq. 11.142, using Eq. 11.143:
i
Ψ(x, t) = ψn (x, t)e ~
i
h
2 Rt 2 0
−(n+ 21 )~ωt+mf˙(x−f /2)+ mω
f (t )dt0
2
0
.
The dynamic phase (Eq. 11.92) is
Z
Z
1 t
1
mω 2 t 2 0 0
0
0
θn (t) = −
En (t ) dt = −(n + )ωt +
f (t ) dt ,
~ 0
2
2~ 0
so
Ψ(x, t) = ψn (x, t)eiθn (t) eiγn (t) ,
˙
confirming Eq. 11.144, with the geometric phase given (ostensibly) by γn (t) = m
~ f (x − f /2). But the
eigenfunctions here are real, and this means the geometric phase should be zero. The point is that (in the
˙
adiabatic approximation) f˙ is extremely small (see above), and hence in this limit m
~ f (x − f /2) ≈ 0 (at
least, in the only region of x where ψn (x, t) is nonzero).
410
CHAPTER 11. QUANTUM DYNAMICS
Problem 11.37
(a) The probability that the system is still in the upper state at time t is
Pb (t) = 1 − α t2 ,
and after two measurements
2
2
Pb (t) = 1 − α t2 ≈ 1 − 2 α t2 .
However, if we make a single measurement at time 2 t, the probability that the system is still in the upper
state is instead
2
Pb (2 t) = 1 − α (2 t) = 1 − 4 α t2 .
So the system is more likely to be in the upper state if we repeatedly measure it.
(b) Now imagine that we make measurements at n intervals of time T /n between 0 and T . The probability
that the system is still in the upper state is
"
2 # n
T
Pb (T ) = 1 − α
n
and by the binomial expansion this is
"
2 # n
2
4
T
T
n (n − 1) 2 T
1−α
= 1 − nα
+
α
+ ...
n
n
2
n
=1−
1 (n − 1) 2 4
1
α T + ...
αT2 +
n
2 n3
In the limit that n → ∞, all terms except the first vanish and the probability is 1.
Problem 11.38
(a) Let θ ≡ ∆tH/~. Equation 11.151 (the exact expression) is
U = e−iθ = 1 − iθ −
θ2
θ3
θ4
+i +
+ ··· ,
2
6
24
whereas Eq. 11.153 (the approximation) says
U≈
1 − iθ/2
=
1 + iθ/2
1−i
θ
2
θ θ2
θ3
θ4
θ2
θ3
θ4
1−i −
+i +
+ · · · = 1 − iθ −
+i +
+ ··· ,
2
4
8
16
2
4
8
which matches the exact expression up through the θ2 term.
As for the unitarity of Eq. 11.153, note that θ is hermitian, so
1 − iθ/2
1 + iθ/2
†
UU =
= 1. X
1 + iθ/2
1 − iθ/2
X
CHAPTER 11. QUANTUM DYNAMICS
411
(b) Here’s the Mathematica code:
Problem 11.38 Numerics
Physical constants
�������
m = 1; ω = 1; ℏ = 1;
Numerical constants
�������
Nx = 100; xmin = - 10.; xmax = 10.; dx =
Nt = 100; dt =
2π / Ω
Nt
xmax - xmin
Nx + 1
;
;
Construct H
The Identity Matrix
�������
OpenOne = SparseArray[{i_, i_} → 1., {Nx, Nx}];
The second derivative operator
�������
D2 =
1.0
dx2
SparseArray[{{i_, i_} → - 2., {i_, j_} /; Abs[i - j] ⩵ 1 → 1.}, {Nx, Nx}];
The position operator
�������
X = SparseArray[{i_, i_} → xmin + dx i, {Nx, Nx}];
The Hamiltonian
1
H =-
�������
Take[Reverse[Eigenvalues[H]], 2]
�������
{0.498772, 1.49385}
2
D2 +
1
�������
2
X.X;
These differ from the exact values of 1/2 and 3/2 by less than 1%.
The ground and first excited states (dividing by Sqrt[dx] normalizes them):
��������
EVE = Eigenvectors[H]
Sqrt[dx]
412
2
CHAPTER 11. QUANTUM DYNAMICS
11P38.nb
��������
ListLinePlot[EVE[[100]], PlotRange → {0, 1}]
1.0
0.8
0.6
��������
0.4
0.2
2
11P38.nb
0
��������
20
40
60
80
100
ListLinePlot[EVE[[99]], PlotRange → {- 1, 1}]
1.0
��������
ListLinePlot[EVE[[100]], PlotRange → {0, 1}]
1.0
0.5
0.8
��������
0.6
20
40
60
80
100
��������
0.4
-0.5
0.2
-1.0
0
Construct U
��������
20
40
60
80
100
ListLinePlot[EVE[[99]], PlotRange → {- 1, 1}]
The denominator
in Cayley’s form
1.0
��������
1
Uplus = OpenOne +
2
I H dt;
0.5
The numerator in Cayley’s form
��������
Uminus
= OpenOne ��������
20
(c) Continuing:
1
2
I H dt;
40
60
80
100
Solve the
t-dep Schrödinger equation
-0.5
The ground and first excited states and our wave function at t=0
��������
-1.0
ψ0 = Eigenvectors[H][[Nx]]
Sqrt[dx];
ψ1 = Eigenvectors[H][[Nx - 1]]
Psi0
= ψ0 + ψ1U
Construct
Sqrt[dx];
Sqrt[2];
The denominator in Cayley’s form
��������
1
Uplus = OpenOne +
2
I H dt;
The numerator in Cayley’s form
��������
Uminus = OpenOne -
1
2
I H dt;
Solve the t-dep Schrödinger equation
The ground and first excited states and our wave function at t=0
��������
ψ0 = Eigenvectors[H][[Nx]]
Sqrt[dx];
ψ1 = Eigenvectors[H][[Nx - 1]]
Psi0 =
ψ0 + ψ1
Sqrt[2];
Sqrt[dx];
CHAPTER 11. QUANTUM DYNAMICS
11P38.nb
Analytical form of eigenstates
��������
psi[n_, x_] :=
mω
1
1/4
2n n !
πℏ
��������
HermiteH n,
mω
x Exp -
ℏ
1
mω
2
ℏ
2
x
Psi = Psi0;
Data =
Table
(* Step the wave function forward in time *)
Psi = LinearSolve[Uplus, Uminus.Psi];
If
(* Only Plot 100 frames regardless of number of time steps *)
Mod i, Nt
100 ⩵ 0,
(* The exact solution. Minus sign is due to
different sign convention in numeric
g[x_] = psi[0, x] Exp - I ω i dt
2
analytic states *)
- psi[1, x] Exp - 3 I ω i dt
2
Sqrt[2];
(* Plotting commands *)
Show[
{Plot[{Re[g[x]], Im[g[x]], Abs[g[x]]}, {x, - 3, 3}, PlotRange → {- 1, 1},
AxesLabel → {"x", "|Ψ|"}, PlotStyle → {Blue, Red, Black}],
ListPlot[{Re[Psi], Im[Psi], Abs[Psi]}, PlotRange → {- 1, 1},
DataRange → {xmin + dx, xmax - dx},
PlotStyle → PointSize[Medium]]
}
],
Nothing
,
4
11P38.nb
{i, 1, Nt}
;
ListAnimate[Data]
|Ψ|
1.0
0.5
-3
-2
1
-1
-0.5
-1.0
2
3
x
3
413
414
CHAPTER 11. QUANTUM DYNAMICS
Problem 11.39
(a) Ω = ω/5:
The exact solution for the driven oscillator which we’ll compare to
The exact solution for the driven oscillator which we’ll compare to
Clear[m, ω, ℏ, Ω, A];
Clear[m, ω, ℏ, Ω, A];
������� f[t_] = A Sin[٠t];
������� f[t_] = A Sin[٠t];
������� xc[t_] = Simplify[ω Integrate[f[tp] Sin[ω (t - tp)], {tp, 0, t}]];
������� xc[t_] = Simplify[ω Integrate[f[tp] Sin[ω (t - tp)], {tp, 0, t}]];
2
m ω 1/4
1
mω
1
mω
2
������� psi[n_, x_] :=
m ω 1/4
1 HermiteH n,
m ω x Exp - 1
m ωx
ℏ
ℏ
HermiteH n,
x Exp -2
x
������� psi[n_, x_] := π ℏ
2n n !
n
πℏ
ℏ
2
ℏ
2 n!
������� Ψ[n_, x_, t_] =
������� Ψ[n_, x_, t_] =
I
1
xc[t]
Simplify psi[n, x - xc[t]] Exp I - n + 1 ℏ ω t + m xc '[t] x - xc[t] +
Simplify psi[n, x - xc[t]] Exp ℏ
- n +2
ℏ ω t + m xc '[t] x - 2
+
ℏ
2
2
m ω2
2
Integrate[f[tp]
xc[tp],
{tp,
0,
t}]
;
mω
2
Integrate[f[tp] xc[tp], {tp, 0, t}]
;
2
�������
�������
Numerics
NumericsΩΩ==ωω//55
Physical constants
Physical constants
�������
m = 1; ω = 1; ℏ = 1; Ω = .2; A = 1;
m = 1; ω = 1; ℏ = 1; Ω = .2; A = 1;
�������
Numerical constants
Numerical constants
�������
xmax - xmin
Nx = 100; xmin = - 10.; xmax = 10.; dx = xmax - xmin
;
Nx + 1
Nx = 100; xmin = - 10.; xmax = 10.; dx =
;
Nx
+
1
2π / Ω
Nt = 1000; dt = 2 π / Ω
;
Nt = 1000; dt = Nt
;
Nt
�������
Construct H
Construct H
The Identity Matrix
The Identity Matrix
�������
OpenOne = SparseArray[{i_, i_} → 1., {Nx, Nx}];
OpenOne = SparseArray[{i_, i_} → 1., {Nx, Nx}];
�������
The second derivative operator
The second derivative operator
�������
1.0
D2 = 1.0SparseArray[{{i_, i_} → - 2., {i_, j_} /; Abs[i - j] ⩵ 1 → 1.}, {Nx, Nx}];
D2 = dx2 SparseArray[{{i_, i_} → - 2., {i_, j_} /; Abs[i - j] ⩵ 1 → 1.}, {Nx, Nx}];
dx2
�������
The position operator
The position operator
��������
X = SparseArray[{i_, i_} → xmin + dx i, {Nx, Nx}];
X = SparseArray[{i_, i_} → xmin + dx i, {Nx, Nx}];
��������
The Hamiltonian
The Hamiltonian
��������
2
ℏ2
1
H[t_] = - ℏ2D2 + 1m ω2 X.X - m ω2 X f[t] ;
H[t_] = -2 m D2 +2 m ω2 X.X - m ω2 X f[t] ;
2m
2
11P39.nb
��������
Construct the time-evolution operator
The numerator in Cayley’s form
��������
Uplus[t_] = OpenOne + I H t +
dt
dt
2
2
dt
dt
2
2
;
The denominator in Cayley’s form
��������
Uminus[t_] = OpenOne - I H t +
;
Solve the t-dep Schrödinger equation
��������
Psi0 = Eigenvectors[H[0]][[Nx]]
��������
Psi = Psi0;
Sqrt[dx];
Data =
Table
(* Step the wave function forward in time *)
Psi = LinearSolve Uplus
i - 1 dt , Uminus
i - 1 dt .Psi ;
If
(* Only Plot 100 frames regardless of number of time steps *)
Construct the time-evolution operator
The numerator in Cayley’s form
��������
Uplus[t_] = OpenOne + I H t +
dt
dt
2
2
dt
dt
2
2
;
The denominator in Cayley’s form
CHAPTER 11. QUANTUM DYNAMICS
��������
Uminus[t_] = OpenOne - I H t +
415
;
Solve the t-dep Schrödinger equation
��������
Psi0 = Eigenvectors[H[0]][[Nx]]
��������
Psi = Psi0;
Sqrt[dx];
Data =
Table
(* Step the wave function forward in time *)
Psi = LinearSolve Uplus
i - 1 dt , Uminus
i - 1 dt .Psi ;
If
(* Only Plot 100 frames regardless of number of time steps *)
Mod i, Nt
100 ⩵ 0,
(* The exact solution *)
g[x_] = Ψ[0, x, i dt];
(* The instantaneous ground state *)
h[x_] = psi[0, x - f[i dt]];
(* Plotting commands *)
Show[
{Plot[{Re[g[x]], Im[g[x]], Abs[g[x]], Abs[h[x]]},
{x, - 3, 3}, PlotRange → {- 1, 1}, AxesLabel → {"x", "|Ψ|"},
PlotStyle → {Blue, Red, Black, {Black, Dashed}}],
ListPlot[{Re[Psi], Im[Psi], Abs[Psi]}, PlotRange → {- 1, 1},
DataRange → {xmin + dx, xmax - dx},
PlotStyle → PointSize[Medium]]
}
],
Nothing
,
{i, 1, Nt}
11P39.nb
;
��������
ListAnimate[Data]
|Ψ|
1.0
0.5
��������
-3
-2
1
-1
2
3
x
-0.5
-1.0
Numerics Ω = 5 ω
Physical constants
��������
m = 1; ω = 1; ℏ = 1; Ω = 5; A = 1;
Numerical constants
��������
Nx = 100; xmin = - 10.; xmax = 10.; dx =
Nt = 1000; dt =
2π / Ω
Nt
xmax - xmin
Nx + 1
;
;
Construct H
The Identity Matrix
��������
OpenOne = SparseArray[{i_, i_} → 1., {Nx, Nx}];
The second derivative operator
��������
D2 =
1.0
dx2
SparseArray[{{i_, i_} → - 2., {i_, j_} /; Abs[i - j] ⩵ 1 → 1.}, {Nx, Nx}];
The position operator
3
11P39.nb
��������
ListAnimate[Data]
416
CHAPTER 11. QUANTUM DYNAMICS
|Ψ|
1.0
(b) Ω = 5ω:
0.5
��������
-3
-2
1
-1
2
3
x
-0.5
-1.0
Numerics Ω = 5 ω
Physical constants
��������
m = 1; ω = 1; ℏ = 1; Ω = 5; A = 1;
Numerical constants
��������
Nx = 100; xmin = - 10.; xmax = 10.; dx =
Nt = 1000; dt =
2π / Ω
Nt
xmax - xmin
Nx + 1
;
;
Construct H
The Identity Matrix
��������
OpenOne = SparseArray[{i_, i_} → 1., {Nx, Nx}];
The second derivative operator
��������
1.0
D2 =
dx2
SparseArray[{{i_, i_} → - 2., {i_, j_} /; Abs[i - j] ⩵ 1 → 1.}, {Nx, Nx}];
The position operator
��������
X = SparseArray[{i_, i_} → xmin + dx i, {Nx, Nx}];
The Hamiltonian
��������
4
H[t_] = -
ℏ2
2m
D2 +
1
2
m ω2 X.X - m ω2 X f[t] ;
11P39.nb
Construct the time-evolution operator
The numerator in Cayley’s form
��������
Uplus[t_] = OpenOne + I H t +
dt
dt
2
2
dt
dt
2
2
;
The denominator in Cayley’s form
��������
3
Uminus[t_] = OpenOne - I H t +
;
Solve the t-dep Schrödinger equation
��������
Psi0 = Eigenvectors[H[0]][[Nx]]
��������
Psi = Psi0;
Sqrt[dx];
Data =
Table
(* Step the wave function forward in time *)
Psi = LinearSolve Uplus
i - 1 dt , Uminus
i - 1 dt .Psi ;
If
(* Only Plot 100 frames regardless of number of time steps *)
Construct the time-evolution operator
The numerator in Cayley’s form
��������
Uplus[t_] = OpenOne + I H t +
dt
dt
2
2
dt
dt
2
2
;
The denominator in Cayley’s form
CHAPTER 11. QUANTUM DYNAMICS
��������
Uminus[t_] = OpenOne - I H t +
417
;
Solve the t-dep Schrödinger equation
��������
Psi0 = Eigenvectors[H[0]][[Nx]]
��������
Psi = Psi0;
Sqrt[dx];
Data =
Table
(* Step the wave function forward in time *)
Psi = LinearSolve Uplus
i - 1 dt , Uminus
i - 1 dt .Psi ;
If
(* Only Plot 100 frames regardless of number of time steps *)
Mod i, Nt
100 ⩵ 0,
(* The exact solution *)
g[x_] = Ψ[0, x, i dt];
(* The instantaneous ground state *)
h[x_] = psi[0, x - f[i dt]];
(* Plotting commands *)
Show[
{Plot[{Re[g[x]], Im[g[x]], Abs[g[x]], Abs[h[x]]},
{x, - 3, 3}, PlotRange → {- 1, 1}, AxesLabel → {"x", "|Ψ|"},
PlotStyle → {Blue, Red, Black, {Black, Dashed}}],
ListPlot[{Re[Psi], Im[Psi], Abs[Psi]}, PlotRange → {- 1, 1},
DataRange → {xmin + dx, xmax - dx},
PlotStyle → PointSize[Medium]]
}
],
Nothing
,
11P39.nb
{i, 1, Nt}
;
��������
ListAnimate[Data]
|Ψ|
1.0
0.5
��������
-3
-2
1
-1
2
3
x
-0.5
-1.0
Numerics Ω = 6/5 ω
Physical constants
��������
m = 1; ω = 1; ℏ = 1; Ω = 6
5; A = 1;
Numerical constants
��������
Nx = 100; xmin = - 10.; xmax = 10.; dx =
Nt = 1000; dt =
2π / Ω
Nt
xmax - xmin
Nx + 1
;
;
Construct H
The Identity Matrix
��������
OpenOne = SparseArray[{i_, i_} → 1., {Nx, Nx}];
The second derivative operator
��������
D2 =
1.0
dx2
SparseArray[{{i_, i_} → - 2., {i_, j_} /; Abs[i - j] ⩵ 1 → 1.}, {Nx, Nx}];
The position operator
5
11P39.nb
��������
ListAnimate[Data]
418
CHAPTER 11. QUANTUM DYNAMICS
|Ψ|
1.0
(c) Ω = 6ω/5:
0.5
��������
-3
-2
1
-1
2
3
x
-0.5
-1.0
Numerics Ω = 6/5 ω
Physical constants
��������
m = 1; ω = 1; ℏ = 1; Ω = 6
5; A = 1;
Numerical constants
��������
Nx = 100; xmin = - 10.; xmax = 10.; dx =
Nt = 1000; dt =
2π / Ω
Nt
xmax - xmin
Nx + 1
;
;
Construct H
The Identity Matrix
��������
OpenOne = SparseArray[{i_, i_} → 1., {Nx, Nx}];
The second derivative operator
��������
D2 =
1.0
dx2
SparseArray[{{i_, i_} → - 2., {i_, j_} /; Abs[i - j] ⩵ 1 → 1.}, {Nx, Nx}];
The position operator
��������
X = SparseArray[{i_, i_} → xmin + dx i, {Nx, Nx}];
The Hamiltonian
��������
6
H[t_] = -
ℏ2
2m
D2 +
1
2
m ω2 X.X - m ω2 X f[t] ;
11P39.nb
Construct the time-evolution operator
The numerator in Cayley’s form
��������
Uplus[t_] = OpenOne + I H t +
dt
dt
2
2
dt
dt
2
2
;
The denominator in Cayley’s form
��������
5
Uminus[t_] = OpenOne - I H t +
;
Solve the t-dep Schrödinger equation
��������
Psi0 = Eigenvectors[H[0]][[Nx]]
��������
Psi = Psi0;
Sqrt[dx];
Data =
Table
(* Step the wave function forward in time *)
Psi = LinearSolve Uplus
i - 1 dt , Uminus
i - 1 dt .Psi ;
If
(* Only Plot 100 frames regardless of number of time steps *)
Construct the time-evolution operator
The numerator in Cayley’s form
��������
Uplus[t_] = OpenOne + I H t +
dt
dt
2
2
;
The denominator in Cayley’s form
CHAPTER 11. QUANTUM
DYNAMICS
dt
Uminus[t_] = OpenOne - I H t +
��������
2
dt
2
419
;
Solve the t-dep Schrödinger equation
��������
Psi0 = Eigenvectors[H[0]][[Nx]]
��������
Psi = Psi0;
Sqrt[dx];
Data =
Table
(* Step the wave function forward in time *)
Psi = LinearSolve Uplus
i - 1 dt , Uminus
i - 1 dt .Psi ;
If
(* Only Plot 100 frames regardless of number of time steps *)
Mod i, Nt
100 ⩵ 0,
(* The exact solution *)
g[x_] = Ψ[0, x, i dt];
(* The instantaneous ground state *)
h[x_] = psi[0, x - f[i dt]];
(* Plotting commands *)
Show[
{Plot[{Re[g[x]], Im[g[x]], Abs[g[x]], Abs[h[x]]},
{x, - 3, 3}, PlotRange → {- 1, 1}, AxesLabel → {"x", "|Ψ|"},
PlotStyle → {Blue, Red, Black, {Black, Dashed}}],
ListPlot[{Re[Psi], Im[Psi], Abs[Psi]}, PlotRange → {- 1, 1},
DataRange → {xmin + dx, xmax - dx},
PlotStyle → PointSize[Medium]]
}
],
Nothing
,
11P39.nb
{i, 1, Nt}
;
��������
ListAnimate[Data]
|Ψ|
1.0
0.5
��������
-3
-2
1
-1
-0.5
-1.0
2
3
x
7
420
CHAPTER 12. AFTERWORD
Chapter 12
Afterword
Problem 12.1
Suppose, on the contrary, that
α|φa (1)i|φb (2)i + β|φb (1)i|φa (2)i = |ψr (1)i|ψs (2)i,
for some one-particle states |ψr i and |ψs i. Because |φa i and |φb i constitute a complete set of one-particle states
(this is a two-level system), any other one-particle state can be expressed as a linear combination of them. In
particular,
|ψr i = A|φa i + B|φb i, and |ψs i = C|φa i + D|φb i,
for some complex numbers A, B, C, and D. Thus
α|φa (1)i|φb (2)i + β|φb (1)i|φa (2)i = A|φa (1)i + B|φb (1)i C|φa (2)i + D|φb (2)i
= AC|φa (1)i|φa (2)i + AD|φa (1)i|φb (2)i + BC|φb (1)i|φa (2)i + BD|φb (1)i|φb (2)i.
(i)
(ii)
(iii)
(iv)
Take the inner product with hφa (1)|hφb (2)|:
Take the inner product with hφa (1)|hφa (2)|:
Take the inner product with hφb (1)|hφa (2)|:
Take the inner product with hφb (1)|hφb (2)|:
α = AD.
0 = AC.
β = BC.
0 = BD.
(ii) ⇒ either A = 0 or C = 0. But if A = 0, then (i) ⇒ α = 0, which is excluded by assumption, whereas if
C = 0, then (iii) ⇒ β = 0, which is likewise excluded. Conclusion: It is impossible to express this state as a
product of one-particle states. QED
Problem 12.2
(a) “The particle must have been in box B1 all along, since there is no way that opening B1 could have
instantaneously affected the outcome of opening faraway box B2 .”
(b) The measurement at B1 collapses the wave function to a narrow spike at the particle’s measured position;
the wave function is now zero everywhere in box B2 , and the box will be empty when it is opened.
CHAPTER 12. AFTERWORD
421
Problem 12.3
(a) Reading from the figure, a = î, b = cos η î + sin η ĵ, Sa = Sa (sin θ cos φ î + sin θ sin φ ĵ + cos θ k̂) = −Sb , so
a · Sa = Sa sin θ cos φ,
b · Sb = −Sa sin θ(cos φ cos η + sin φ sin η) = −Sa sin θ cos(φ − η).
But sin θ ≥ 0 (since θ is between 0 and π), so
A = sign(a·Sa ) = sign(cos φ), B = −sign((cos(φ − η)) , and hence A(a, λ)B(b, λ) = −sign [cos φ cos(φ − η)] . X
Z
Z 2π
1
2
(b) P (a, b) = hA(a, λ)B(b, λ)i =
sign[cos φ cos(φ − η)] dφ.
[A(a, λ)B(b, λ)] sin θ dθ dφ = −
4π
4π 0
π
3π
Now, cos φ is positive unless π2 < φ < 3π
2 , and cos(φ − η) is positive unless 2 + η < φ < 2 + η.
Assume first that 0 < η < π2 , and chop the integral into eight segments, writing (++) to indicate that cos φ
is positive and cos(φ − η) is positive, etc.:
(Z
Z π/2
Z π/2+η
Z π
η
1
P (a, b) = −
(++) dφ +
(++) dφ +
(−+) dφ +
(−−) dφ
2π
0
η
π/2
π/2+η
)
Z π+η
Z 3π/2
Z 3π/2+η
Z 2π
+
(−−) dφ +
(−−) dφ +
(+−) dφ +
(++) dφ
π
π+η
3π/2
3π/2+η
(
π
π
π π
3π
1
(η − 0) +
−η −
+η−
+ π − − η + (π + η − π) +
−π−η
=−
2π
2
2
2
2
2
)
3π
3π
3π
1
2η
−
+η−
+ 2π −
−η
= − (2π − 4η) =
− 1.
2
2
2
2π
π
That’s for 0 < η < π2 ; for π2 < η < π we have:
(Z
Z π/2
Z η
Z π
η−π/2
1
(+−) dφ +
(++) dφ +
(−+) dφ +
(−+) dφ
P (a, b) = −
2π
0
η−π/2
π/2
η
)
Z η+π/2
Z 3π/2
Z η+π
Z 2π
+
(−+) dφ +
(−−) dφ +
(+−) dφ +
(+−) dφ
π
1
=−
2π
(
η+π/2
3π/2
η+π
π
π
3π
π
− (η − − 0) +
−η+
− η−
− (π − η) − (η + − π) +
−η−
2
2
2
2
2
2
2
)
3π
1
2η
− η+π−
− (2π − η − π) = − (2π − 4η) =
− 1 (same as before).
2
2π
π
π
π
π
The average is an even function of η, so the general result (for − π2 < η < π2 ) is
(c) The quantum prediction (Equation 12.4) is P (a, b) = − cos η; plotting the two:
2|η|
− 1.
π
422
CHAPTER 12. AFTERWORD
�������
2 Abs[x]
Plot
- 1, - Cos[x] , {x, - π, π}, PlotRange → {- 1, 1}
π
1.0
0.5
�������
-3
-2
1
-1
2
3
-0.5
-1.0
Plot[- Cos[x],
{x, - π, π}, PlotRange → {- 1, 1}]
They agree at�������η =
0, ± π2 , and ±π.
1.0
(d) In this example Bell’s inequality says
2
2
2
|ηab | − 1 − |ηac | + 1 ≤ 1 + |ηbc | − 1, or
π
π
π
0.5
|ηab | − |ηac�������
| ≤ |ηbc |,
-3
-2
1
-1
2
3
where all angles are in the range [−π, π]. Now, first assume that |ηab | > |ηac |. In that case
-0.5
|ηab | ≤ |ηac | + |ηcb | .
-1.0
On the other hand, if |ηac | > |ηab |
�������
Show {%3, %4}
1.0
|ηac | ≤ |ηab | + |ηbc | .
�������
1.0
0.5
Show
-3 -2 -1
-0.5
1 2 3
, Show
0.5
-3 -2 -1
-0.5
1 2 3
In either case, the equation says
that one angle is-1.0smaller than the sum of the other two. This is clearly correct:
-1.0
the vectors a, b, and c define points a, b, and c on a unit sphere, and the angles between the vectors are
proportional to the lengths of the great arcs connecting those points. So the inequality just expresses the fact
that walking “straight” from a to b is shorter than walking from a to c and c to b (unless c happens to lie on
your path in which case it is the same distance).
Problem 12.4
(a) Equation 12.17: ρ̂2 = |Ψi hΨ | Ψi hΨ|, and since Ψ is normalized, ρ̂2 = |Ψi 1 hΨ| = ρ̂.
X
Equation 12.18: To find the Hermitian conjugate of an operator in Dirac notation, we reverse the order
of the terms, switch bras and kets, place a † on every operator and a ∗ on every number:
†
ρ̂† = (|Ψi hΨ|) = |Ψi hΨ| = ρ . X
Equation 12.19:
Tr(ρ) =
P
i ρii =
P
i hei | Ψi hΨ | ei i = hΨ| (
P
i |ei i hei |) |Ψi = hΨ | Ψi = 1.
X
CHAPTER 12. AFTERWORD
423
Equation 12.20:
Tr(ρ A) =
X
(ρ A)ii =
i
=
i
XX
i
XX
ρij Aji =
XX
j
D
hΨ | ej i ej  ei
i
E
E
D
hei | Ψi hΨ | ej i ej  ei
j


X
hei | Ψi = hΨ| 
|ej i hej | Â
j
j
!
X
|ei i hei |
E
D
|Ψi = Ψ Â Ψ
i
= hAi . X
E
D d
ρ̂ = i ~ Ψ̇ hΨ| + |Ψi Ψ̇ . From the Schrödinger equation: i ~ |Ψ̇i = Ĥ |Ψi, and, taking the
dt
adjoint: −i ~ hΨ̇| = hΨ| Ĥ. Thus
(b) i ~
i~
h
i
d
ρ̂ = Ĥ |Ψi hΨ| − |Ψi hΨ| Ĥ = Ĥ ρ̂ − ρ̂ Ĥ = Ĥ, ρ̂ . X
dt
Problem 12.5
For an electron with spin-down along the y direction (Problem 4.32(a)):
1
|Ψi = √
2
1
−i
;
1
hΨ| = √ 1 i .
2
So
1
1
1
1
1
10
1i
= (1)(1) = ,
ρ11 =
−i
0
2
2
2
1
0
1
i
1
10
1i
ρ12 =
= (1)(i) =
−i
1
2
2
2
1
1
1
1
i
01
1i
ρ21 =
= (−i)(1) = −
−i
0
2
2
2
1
0
1
1
1
01
1i
ρ22 =
= (−i)(i) =
−i
1
2
2
2
and
ρ=
1
2
1 i
−i 1
.
Or, more efficiently,
ρ = |Ψi hΨ| =
1
2
1 1 i
1
1 i =
.
−i
2 −i 1
424
CHAPTER 12. AFTERWORD
Problem 12.6
!†
X
†
(a) Equation 12.31: As in Problem 12.4(a), ρ̂ =
pk |Ψk i hΨk |
=
X
k
|Ψk i hΨk | p∗k . Since pk is real,
k
ρ̂† = ρ̂, X
Equation 12.32:
+
*
Tr [ρ] =
=
X
ρii =
XX
pk hei | Ψk i hΨk | ei i =
i
i
X
pk hΨk | Ψk i =
k
X
pk
Ψk
X
X
k
|ei i hei | Ψk
i
k
pk = 1 . X
k
In the last two steps I used the fact that each wave function is normalized, and Equation (12.30).
Equation 12.33:
Tr(ρ A) =
D
E
X
XX
XXX
(ρ A)ii =
ρij Aji =
pk hei | Ψk i hΨk | ej i ej  ei
i
i
j
i

=
X
pk hΨk | 
X
k
!
X
|ej i hej | Â
j
k
j

|ei i hei |
|Ψk i =
X
i
pk
D
E
Ψk  Ψk = hAi . X
k
Equation 12.34: As in Problem 12.4(b):
i~
X E
D
X d
pk Ĥ |Ψk i hΨk | − |Ψk i hΨk | Ĥ
ρ̂ = i ~
pk Ψ̇k hΨk | + |Ψk i Ψ̇k =
dt
k
k
h
i
= Ĥ ρ̂ − ρ̂ Ĥ = Ĥ, ρ̂ . X
(b)
XXX
X 2
Tr ρ2 =
ρ ii =
pk pj hei | Ψk i hΨk | Ψj i hΨj | ei i
i
i
k
j
!
=
XX
k
pk pj hΨj |
j
X
|ei i hei |
i
|Ψk i hΨk | Ψj i =
XX
k
pk pj |hΨk | Ψj i|2
j
The wave functions are normalized, so |hΨk |Ψj i| ≤ 1, with equality if and only if k = j. Therefore, unless
this is a pure state
X X
XX
Tr ρ2 <
pk pj =
pk
pj = 1 . QED
k
j
k
j
(c) We already know that ρ2 = ρ for a pure state. In part (b) we proved that Tr ρ2 < 1 for a non-pure
state, whereas from (a) Tr(ρ) = 1 for any density matrix. Therefore, for a non-pure state ρ2 6= ρ (they
have different traces). QED
CHAPTER 12. AFTERWORD
425
Problem 12.7
(a) From Example 12.1, the density matrix for an electron in the state spin up along x is
1/2 1/2
ρx+ =
;
1/2 1/2
from Problem 12.5 the density matrix for an electron in the state spin down along y is
1/2 i/2
ρy− =
.
−i/2 1/2
Therefore the density matrix for the state in question is
1
2
1/2
1/6 + i/3
ρ = ρx+ + ρy− =
.
1/6 − i/3
1/2
3
3
(b) From Equation 4.147,
~
Sy =
2
0 −i
,
i 0
so
~
~
1/2
1/6 + i/3
0 −i
i/6 − 1/3
−i/2
= Tr
hSy i = Tr(ρSy ) = Tr
1/6 − i/3
1/2
i 0
i/2
−i/6 − 1/3
2
2
~ i
~
1
i
1
~
2
=
− − −
=
−
= − .
2 6 3 6 3
2
3
3
Problem 12.8
(a) The density matrix for a spin-1/2 state is a two by two matrix which can be written
c c
ρ= 1 2
c3 c4
for complex numbers c1 , c2 , c3 , c4 . Because the matrix is Hermitian we know that c1 and c4 must be real
and c2 and c3 must be complex conjugates. Thus
b1
b2 − i b3
ρ=
b2 + i b3
b4
for real numbers b1 , b2 , b3 , b4 . In addition, since the trace is 1 we know that b4 = 1−b1 . Defining a1 = 2 b2 ,
a2 = 2 b3 , and a3 = 2 b1 − 1 then gives
1
1
1 + a3 a1 − i a2
1 0
0 1
0 −i
1 0
ρ=
=
+ a1
+ a2
+ a3
0 1
1 0
i 0
0 −1
2 a1 + i a2 1 − a3
2
1
1
= (1 + a1 σx + a2 σy + a3 σz ) = (1 + a · σ) . X
2
2
(b)
2
Tr ρ
1
1 + a3 a1 − i a2 1
1 + a3 a1 − i a2
= Tr
2 a1 + i a2 1 − a3
2 a1 + i a2 1 − a3
2
2
1
1
a1 + a22 +(1 + a3 )
2 (a1 − ia2 )
=
1 + |a|2 .
= Tr
2
2
2
4
2
2 (a1 + i a2 )
a1 + a2 +(1 − a3 )
426
CHAPTER 12. AFTERWORD
By virtue of Problem 12.6c we see that |a| ≤ 1 in order for this to be a density matrix, and it describes a
pure state if and only if |a| = 1.
(c) We can calculate the expectation value of Sz and, since there are only two possibilities,
~
~
~
~
1
hSz i = P+ + P− −
= P+ +(1 − P+ ) −
= ~ P+ −
.
2
2
2
2
2
This tells us the probability that a measurement will return spin up along z. Now
~
~
1
1 0
1 + a3 a1 − i a2
hSz i = Tr
σz ρ = Tr
0 −1 2 a1 + i a2 1 − a3
2
2
~
~
1 + a3 a1 − i a2
= Tr
= a3 .
−a
−
i
a
−1
+
a
4
2
1
2
3
Combining these two results we have
P+ =
1 + a3
.
2
For the given directions we have: (i) P+ = 1, (ii) P+ = 1/2, and (iii) P+ = 0.
(d) For a point on the equator at azimuthal angle φ, we have a1 = cos φ, a2 = sin φ, a3 = 0, so
1
1 e−i φ
ρ=
.
2 ei φ 1
a
For a spinor χ =
the density matrix is
b
2
a
|a| a b∗
∗ ∗
a b =
ρ = χχ =
,
b
b a∗ |b|2
†
so, up to an arbitrary overall phase,
1
χ= √
2
1
ei φ
.
APPENDIX. LINEAR ALGEBRA
427
Appendix A
Linear Algebra
Problem A.1
(a) Yes; two-dimensional.
(b) No; the sum of two such vectors has az = 2, and is not in the subset. Also, the null vector (0,0,0) is not
in the subset.
(c) Yes; one-dimensional.
Problem A.2
(a) Yes;
1, x, x2 , . . . , xN −1 is a convenient basis. Dimension: N.
(b) Yes;
1, x2 , x4 , . . ..
(c) No.
The sum of two such “vectors” is not in the space.
(d) Yes;
(x − 1), (x − 1)2 , (x − 1)3 , . . . , (x − 1)N −1 .
(e) No.
The sum of two such “vectors” would have value 2 at x = 0.
Dimension N/2 (if N is even) or (N + 1)/2 (if N is odd).
Dimension: N − 1.
Problem A.3
Suppose |αi = a1 |e1 i + a2 |e2 i + · · · an |en i and |αi = b1 |e1 i + b2 |e2 i + · · · + bn |en i. Subtract: 0 = (a1 − b1 )|e1 i +
(a2 − b2 )|e2 i + · · · + (an − bn )|en i. Suppose aj 6= bj for some j; then we can divide by (aj − bj ) to get:
|ej i = −
(a1 − b1 )
(a2 − b2 )
(an − bn )
|e1 i −
|e2 i − · · · − 0|ej i − · · · −
|en i,
(aj − bj )
(aj − bj )
(aj − bj )
so |ej i is linearly dependent on the others, and hence {|ej i} is not a basis. If {|ej i} is a basis, therefore, the
components must all be equal (a1 = b1 , a2 = b2 , . . . , an = bn ). QED
428
APPENDIX. LINEAR ALGEBRA
Problem A.4
(i)
he1 |e1 i = |1 + i|2 + 1 + |i|2 = (1 + i)(1 − i) + 1 + (i)(−i) = 1 + 1 + 1 + 1 = 4. ke1 k = 2.
|e01 i =
i
1
1
(1 + i) î + ĵ + k̂.
2
2
2
(ii)
he01 |e2 i =
1
1
(1 − i)(i) + (3) +
2
2
−i
2
1=
1
(i + 1 + 3 − i) = 2.
2
|e002 i ≡ |e2 i − he01 |e2 i|e01 i = (i − 1 − i)î + (3 − 1)ĵ + (1 − i)k̂ = (−1)î + (2)ĵ + (1 − i)k̂.
he002 |e002 i = 1 + 4 + 2 = 7.
1
|e02 i = √ [−î + 2ĵ + (1 − i)k̂].
7
(iii)
he01 |e3 i =
1
28 = 14;
2
√
2
he02 |e3 i = √ 28 = 8 7.
7
|e003 i = |e3 i − he01 |e3 i|e01 i − he02 |e3 i|e02 i = |e3 i − 7|e1 i − 8|e002 i
= (0 − 7 − 7i + 8)î + (28 − 7 − 16)ĵ + (0 − 7i − 8 + 8i)k̂ = (1 − 7i)î + 5ĵ + (−8 + i)k̂.
ke003 k2 = 1 + 49 + 25 + 64 + 1 = 140.
1
|e03 i = √ [(1 − 7i)î + 5ĵ + (−8 + i)k̂].
2 35
Problem A.5
hα|βi
hα|βi
From Eq. A.21: hγ|γi = hγ| |βi −
|αi = hγ|βi −
hγ|αi. From Eq. A.19:
hα|αi
hα|αi
hα|βi
|hα|βi|2
hα|βi
|αi = hβ|βi −
hβ|αi = hβ|βi −
, which is real.
hγ|βi∗ = hβ|γi = hβ| |βi −
hα|αi
hα|αi
hα|αi
hα|βi
hα|βi
∗
hγ|αi = hα|γi = hα| |βi −
|αi = hα|βi −
hα|αi = 0. hγ|αi = 0. So (Eq. A.20) :
hα|αi
hα|αi
hγ|γi = hβ|βi −
|hα|βi|2
≥ 0, and hence |hα|βi|2 ≤ hα|αihβ|βi. QED
hα|αi
Problem A.6
hα|βi = (1 − i)(4 − i) + (1)(0) + (−i)(2 − 2i) = 4 − 5i − 1 − 2i − 2 = 1 − 7i; hβ|αi = 1 + 7i;
r
1 + 49
1
hα|αi = 1 + 1 + 1 + 1 = 4; hβ|βi = 16 + 1 + 4 + 4 = 25; cos θ =
=√ ;
θ = 45◦ .
4 · 25
2
APPENDIX. LINEAR ALGEBRA
429
Problem A.7
Let
|γi ≡ |αi + |βi;
hγ|γi = hγ|αi + hγ|βi.
hγ|αi∗ = hα|γi = hα|αi + hα|βi =⇒ hγ|αi = hα|αi + hβ|αi.
hγ|βi∗ = hβ|γi = hβ|αi + hβ|βi =⇒ hγ|βi = hα|βi + hβ|βi.
k(|αi + |βi)k2 = hγ|γi = hα|αi + hβ|βi + hα|βi + hβ|αi.
But
p
hα|βi + hβ|αi = 2Re(hα|βi) ≤ 2|hα|βi| ≤ 2 hα|αihβ|βi (by Schwarz inequality), so
k(|αi + |βi)k2 ≤ kαk2 + kβk2 + 2kαkkβk = (kαk + kβk)2 , and hence k(|αi + |βi)k ≤ kαk + kβk. QED
Problem A.8

1
1
0
2
1
3 .
3i (3 − 2i) 4

(a)



(−2 + 0 − 1) (0 + 1 + 3i) (i + 0 + 2i)
−3 (1 + 3i)
3i
 (4 + 0 + 3i) (0 + 0 + 9) (−2i + 0 + 6) = (4 + 3i)
9
(6 − 2i) .
6i
(6 − 2i)
6
(4i + 0 + 2i) (0 − 2i + 6) (2 + 0 + 4)

(b)
 

0
0 0
(−2 + 0 + 2) (2 + 0 − 2) (2i + 0 − 2i)
0 3 .
BA =  (0 + 2 + 0) (0 + 0 + 0) (0 + 3 + 0)  =  2
(6 + 3i) −3i 12
(−i + 6 + 4i) (i + 0 − 4i) (−1 + 9 + 4)


−3
(1 + 3i)
3i
9
(3 − 2i) .
[A, B] = AB − BA =  (2 + 3i)
(−6 + 3i) (6 + i)
−6

(c)
(d)


−1 2 2i
 1 0 −2i .
i 3 2

(e)

−1 1 −i
 2 0 3 .
−2i 2i 2
(f )


−1 2 −2i
 1 0 2i  .
−i 3 2
(g) 4 + 0 + 0 − 1 − 0 − 0 = 3.
430
APPENDIX. LINEAR ALGEBRA
(h)

1
B−1 = C̃;
3
| 13 02 |
C = − 03 −i
2
0 −i
1 0
− | 0i 20 |
2 −i
i 2
− 20 −i
0
 

| 0i 31 |
2 0 −i
− | 2i 30 | = −3i 3 −6 .
i 0 2
| 20 01 |


2 −3i i
1
B−1 =  0 3 0 .
3
−i −6 2



 

(4 + 0 − 1) (−6i + 0 + 6i) (2i + 0 − 2i)
3 0 0
1 0 0
1
1
(0 + 0 + 0)  = 0 3 0 = 0 1 0 . X
BB−1 =  (0 + 0 + 0) (0 + 3 + 0)
3
3
(2i + 0 − 2i) (3 + 9 − 12) (−1 + 0 + 4)
0 0 3
0 0 1
det A = 0 + 6i + 4 − 0 − 6i − 4 = 0.
No;
A does not have an inverse.
Problem A.9
(a)




−i + 2i + 2i
3i
 2i + 0 + 6  = 6 + 2i .
−2 + 4 + 4
6
(b)


2
−i −2i 2 1 − i = −2i − 2i(1 − i) + 0 = −2 − 4i.
0
(c)




2
4
2 0 −i
i 2i 2 0 1 0  1 − i = i 2i 2 1 − i = 4i + 2i(1 − i) + 2(3 − i) = 8 + 4i.
0
3−i
i 3 2

(d)


 
i
2i (−1 + i) 0
2i 2 (1 + i) 0 = 4i (−2 + 2i) 0 .
4 (2 + 2i) 0
2
Problem A.10
(a) S =
1
(T + T̃);
2
A=
1
(T − T̃).
2
(b) R =
1
(T + T∗ );
2
M=
1
(T − T∗ ).
2
(c) H =
1
(T + T† );
2
K=
1
(T − T† ).
2
APPENDIX. LINEAR ALGEBRA
431
Problem A.11
f ki = (ST)ik =
(ST)
n
X
Sij Tjk =
j=1
†
∗
n
X
f = T̃S̃. QED
T̃kj S̃ji = (T̃S̃)ki ⇒ ST
j=1
∗
∗ ∗
f = (T̃S̃) = T̃ S̃ = T† S† . QED
(ST) = (ST)
(T−1 S−1 )(ST) = T−1 (S−1 S)T = T−1 T = I
⇒ (ST)−1 = T−1 S−1 . QED
U† = U−1 , W† = W−1 ⇒ (WU)† = U† W† = U−1 W−1 = (WU)−1 ⇒ WU is unitary.
H = H† , J = J† ⇒ (HJ)† = J† H† = JH;
the product is hermitian ⇔ this is HJ, i.e. ⇔ [H, J] = 0 (they commute).
?
(U + W)† = U† + W† = U−1 + W−1 = (U + W)−1 .
No; the sum of two unitary matrices is not unitary.
(H + J)† = H† + J† = H + J. Yes; the sum of two hermitian matrices is hermitian.
Problem A.12
U† U = I
=⇒
(U† U)ik = δik =⇒
n
X
†
Uij
Ujk =
j=1
n
X
∗
Uji
Ujk = δik .
j=1
Construct the set of n vectors a(j) i ≡ Uij (a(j) is the j-th column of U; its i-th component is Uij ). Then
a(i)† a(k) =
n
X
∗
a(i) j a(k) j =
j=1
n
X
∗
Uji
Ujk = δik ,
j=1
so these vectors are orthonormal. Similarly,
UU† = I
=⇒
(UU† )ik = δik
=⇒
n
X
†
Uij Ujk
=
j=1
n
X
∗
Ukj
Uij = δki .
j=1
This time let the vectors b(j) be the rows of U: b(j) i ≡ Uji . Then
b(k)† b(i) =
n
X
j=1
∗
b(k) j b(i) j =
n
X
∗
Ukj
Uij = δki ,
j=1
so the rows are also orthonormal.
Problem A.13
H† = H (hermitian) ⇒ det H = det(H† ) = det(H̃∗ ) = (det H̃)∗ = (det H)∗ ⇒ det H is real.
X
U† = U−1 (unitary) ⇒ det(UU† ) = (det U)(det U† ) = (det U)(det Ũ)∗ = | det U|2 = det I = 1, so | det U| = 1. X
S̃ = S−1 (orthogonal) ⇒ det(SS̃) = (det S)(det S̃) = (det S)2 = 1, so det S = ±1. X
432
APPENDIX. LINEAR ALGEBRA
Problem A.14
(a)
î0 = cos θ î + sin θ ĵ;
ĵ 0 = − sin θ î + cos θ ĵ;
k̂ 0 = k̂.


cos θ − sin θ 0
Ta =  sin θ cos θ 0 .
0
0
1
y
y'
e
x'
e
x
z, z'
(b)
î0 = ĵ;
ĵ 0 = k̂;
k̂ 0 = î.


0 0 1
Tb = 1 0 0 .
0 1 0
y, x'
x, z'
z, y'
(c)
î0 = î;
ĵ 0 = ĵ;
k̂ 0 = −k̂.


1 0 0
Tc = 0 1 0  .
0 0 −1
(d)


 

cos θ sin θ 0
cos θ − sin θ 0
1 0 0
T̃a Ta = − sin θ cos θ 0  sin θ cos θ 0 = 0 1 0 . X
0
0 1
0
0
1
0 0 1
APPENDIX. LINEAR ALGEBRA
433


 

010
001
1 0 0
T̃b Tb = 0 0 1 1 0 0 = 0 1 0 . X
100
010
0 0 1


 

1 0 0
1 0 0
1 0 0
T̃c Tc = 0 1 0  0 1 0  = 0 1 0 . X
0 0 −1
0 0 −1
0 0 1
det Ta = cos2 θ + sin2 θ = 1.
det Tb = 1.
det Tc = -1.
Problem A.15
y
y'
e
x, x'
e
z
î0 = î;
z'
ĵ 0 = cos θ ĵ + sin θ k̂;


1 0
0
Tx (θ) = 0 cos θ − sin θ .
0 sin θ cos θ
k̂ 0 = cos θ k̂ − sin θ ĵ.
y, y'
x'
e
x
z
e
z'

î0 = cos θ î − sin θ k̂;
ĵ 0 = ĵ;
k̂ 0 = cos θ k̂ + sin θ î.

cos θ 0 sin θ
1 0 .
Ty (θ) =  0
− sin θ 0 cos θ
434
APPENDIX. LINEAR ALGEBRA

î = −ĵ 0 ;
ĵ = î0 ;
k̂ = k̂ 0 ,
so (Eq. A.61)

0 1 0
S = −1 0 0 ;
0 0 1


0 −1 0
S−1 = 1 0 0 .
0 0 1




0 10
1 0
0
0 −1 0
STx S−1 = −1 0 0 0 cos θ − sin θ 1 0 0
0 01
0 sin θ cos θ
0 0 1


 

0 10
0 −1
0
cos θ 0 − sin θ
0  = Ty (−θ).
= −1 0 0 cos θ 0 − sin θ =  0 1
0 01
sin θ 0 cos θ
sin θ 0 cos θ




0 10
cos θ 0 sin θ
0 −1 0
1 0  1 0 0
STy S−1 = −1 0 0  0
0 01
− sin θ 0 cos θ
0 0 1


 

0 10
0 − cos θ sin θ
1 0
0
0
0  = 0 cos θ − sin θ = Tx (θ).
= −1 0 0 1
0 01
0 sin θ cos θ
0 sin θ cos θ
Is this what we would expect? Yes, for rotation about the x axis now means rotation about the −y 0 axis, and
rotation about the y axis has become rotation about the x0 axis.
Problem A.16
From Eq. A.64 we have
Af Bf = SAe S−1 SBe S−1 = S(Ae Be )S−1 = SCe S−1 = Cf . X
Suppose S† = S−1 and He = He† (S unitary, He hermitian). Then
Hf † = (SHe S−1 )† = (S−1 )† He† S† = SHe S−1 = Hf , so Hf is hermitian. X
In an orthonormal basis, hα|βi = a† b (Eq. A.50). So if {|fi i} is orthonormal, hα|βi = af † bf . But bf = Sbe
(Eq. A.63), and also af † = ae† S† . So hα|βi = ae† S† Sbe . This is equal to ae† be (and hence {|ei i} is also
orthonormal), for all vectors |αi and |βi ⇔ S† S = I, i.e. S is unitary.
Problem A.17
Tr(T1 T2 ) =
n X
n
n
n X
n
n
X
X
X
X
(T2 )ji (T1 )ij =
(T2 T1 )jj = Tr(T2 T1 ).
(T1 )ij (T2 )ji =
(T1 T2 )ii =
i=1
Is
Tr(T1 T2 T3 ) = Tr(T2 T1 T3 )?
i=1 j=1
No. Counterexample:
01
0 0
T1 =
, T2 =
,
00
1 0
01
00
10
0 1
0 0
1 0
=
=
00
10
00
0 0
1 0
0 0
00
01
10
0 0
0 0
0 0
T2 T1 T3 =
=
=
10
00
00
1 0
0 0
0 0
T1 T2 T3 =
j=1
j=1 i=1
T3 =
1 0
.
0 0
=⇒ Tr(T1 T2 T3 ) = 1.
=⇒ Tr(T2 T1 T3 ) = 0.
APPENDIX. LINEAR ALGEBRA
435
Problem A.18
Eigenvalues:
(cos θ − λ) − sin θ
= (cos θ − λ)2 + sin2 θ = cos2 θ − 2λ cos θ + λ2 + sin2 θ = 0, or λ2 − 2λ cos θ + 1 = 0.
sin θ
(cos θ − λ)
2 cos θ ±
λ=
√
p
4 cos2 θ − 4
= cos θ ± − sin2 θ = cos θ ± i sin θ = e±iθ .
2
So there are two eigenvalues, both of them complex. Only if sin θ = 0 does this matrix possess real eigenvalues,
i.e., only if θ = 0 or π.
Eigenvectors:
cos θ − sin θ
α
±iθ α
=e
=⇒ cos θ α − sin θ β = (cos θ ± i sin θ)α ⇒ β = ∓iα. Normalizing:
sin θ cos θ
β
β
1
a(1) = √
2
1
;
−i
1
a(2) = √
2
1
.
i
Diagonalization:
1
(1)
(S−1 )11 = a1 = √ ;
2
1
=√
2
−1
S
−1
STS
1 1
;
−i i
1
(2)
(S−1 )12 = a1 = √ ;
2
1
inverting: S = √
2
−i
(1)
(S−1 )21 = a2 = √ ;
2
i
(2)
(S−1 )22 = a2 = √ .
2
1 i
.
1 −i
1 1 i
1 1 i
cos θ − sin θ
(cos θ + i sin θ) (cos θ − i sin θ)
1 1
=
=
sin θ cos θ
(sin θ − i cos θ) (sin θ + i cos θ)
−i i
2 1 −i
2 1 −i
iθ
iθ
−iθ
iθ
1 2e
1 1 i
e
e
0
e
0
=
=
=
. X
−ieiθ ie−iθ
0 2e−iθ
0 e−iθ
2 1 −i
2
Problem A.19
(1 − λ)
1
= (1 − λ)2 = 0 =⇒ λ = 1
0
(1 − λ)
11
α
α
=
=⇒ α + β = α =⇒ β = 0;
01
β
β
(only one eigenvalue).
a=
1
0
(only one eigenvector—up to an arbitrary constant factor). Since the eigenvectors do not span the
space,
this
1 0
matrix cannot be diagonalized. [If it could be diagonalized, the diagonal form would have to be
, since
0 1
the only eigenvalue is 1. But in that case I = SMS−1 . Multiplying from the left by S−1 and on the right by
S : S−1 IS = S−1 SMS−1 S = M. But S−1 IS = S−1 S = I. So M = I, which is false.]
436
APPENDIX. LINEAR ALGEBRA
Problem A.20
Expand the determinant (Eq. A.72) by minors, using the first column:
(T22 − λ) . . .
...
n
X
..
..
.
.
+
Tj1 cofactor(Tj1 ).
det(T − λ1) = (T11 − λ)
..
j=2
.
(Tnn − λ)
But the cofactor of Tj1 (for j > 1) is missing two of the original diagonal elements: (T11 − λ) (from the first
column), and (Tjj − λ) (from the j-th row). So its highest power of λ will be (n − 2). Thus terms in λn and
λn−1 come exclusively from the first term above. Indeed, the same argument applied now to the cofactor of
(T11 − λ) – and repeated as we expand that determinant – shows that only the product of the diagonal elements
contributes to λn and λn−1 :
(T11 − λ)(T22 − λ) · · · (Tnn − λ) = (−λ)n + (−λ)n−1 (T11 + T22 + · · · + Tnn ) + · · ·
Evidently then, Cn = (−1)n , and Cn−1 = (−1)n−1 Tr(T). To get C0 – the term with no factors of λ – we simply
set λ = 0. Thus C0 = det(T). For a 3 × 3 matrix:
(T11 − λ)
T12
T13
T21
(T22 − λ)
T23
T31
T32
(T33 − λ)
= (T11 − λ)(T22 − λ)(T33 − λ) + T12 T23 T31 + T13 T21 T32
− T31 T13 (T22 − λ) − T32 T23 (T11 − λ) − T12 T21 (T33 − λ)
= −λ3 + λ2 (T11 + T22 + T33 ) − λ(T11 T22 + T11 T33 + T22 T33 ) + λ(T13 T31 + T23 T32 + T12 T21 )
+ T11 T22 T33 + T12 T23 T31 + T13 T21 T32 − T31 T13 T22 − T32 T23 T11 − T12 T21 T33
= −λ3 + λ2 Tr(T) + λC1 + det(T),
with
C1 = (T13 T31 + T23 T32 + T12 T21 ) − (T11 T22 + T11 T33 + T22 T33 ).
Problem A.21
The characteristic equation is an n-th order polynomial, which can be factored in terms of its n (complex) roots:
(λ1 − λ)(λ2 − λ) · · · (λn − λ) = (−λ)n + (−λ)n−1 (λ1 + λ2 + · · · + λn ) + · · · + (λ1 λ2 · · · λn ) = 0.
Comparing Eq. A.92, it follows that Tr(T) = λ1 + λ2 + · · · λn and det(T) = λ1 λ2 · · · λn .
QED
Problem A.22
(a)
1 1
M =
;
1 −i
†
†
MM =
2
(1 − i)
,
(1 + i)
2
†
M M=
2
(1 + i)
;
(1 − i)
2
†
[M, M ] =
(b) Find the eigenvalues:
(1 − λ)
1
= (1 − λ)(i − λ) − 1 = i − λ(1 + i) + λ2 − 1 = 0;
1
(i − λ)
0 −2i
2i 0
6= 0.
No.
APPENDIX. LINEAR ALGEBRA
437
√
(1 + i)2 − 4(i − 1)
(1 + i) ± 4 − 2i
λ=
=
.
2
2
Since there are two distinct eigenvalues, there must be two linearly independent eigenvectors, and that’s
enough to span the space. So this matrix is diagonalizable, even though it is not normal.
(1 + i) ±
p
Problem A.23
[Tf1 , Tf2 ] = Tf1 Tf2 − Tf2 Tf1 = STe1 S−1 STe2 S−1 − STe2 S−1 STe1 S−1 = STe1 Te2 S−1 − STe2 Te1 S−1 = S[Te1 , Te2 ]S−1 = 0.
X
Problem A.24
V ã(1) = d11 V a(1) + d21 V a(2)
= d11 c11 a(1) + c21 a(2) + d21 c12 a(1) + c22 a(2)
= (c11 d11 + c12 d21 ) a(1) +(c21 d11 + c22 d21 ) a(2)
= ν1 d11 a(1) + ν1 d21 a(2) = ν1 ã(1)
(and the same goes for ã(2) ).
Problem A.25
(a) Since A† = A and B† = B, the matrices are clearly normal (Equation A.82), and hence diagonalizable.


 


1 4 1
1 −2 −1
1 −2 −1
1 4 1
[A, B] = 4 −2 4 −2 2 −2 − −2 2 −2 4 −2 4
1 4 1
−1 −2 1
−1 −2 1
1 4 1

 

−8 4 −8
−8 4 −8
=  4 −20 4  −  4 −20 4  = 0. X
−8 4 −8
−8 4 −8
(b) To get the eigenvalues, use the characteristic equation:
det(A − λ1) =
1−λ
4
1
4 −2 − λ 4
= −(1 − λ)2 (2 + λ) + 16 + 16 + (2 + λ) − 16(1 − λ) − 16(1 − λ)
1
4
1−λ
= −(1 − 2λ + λ2 )(2 + λ) + 2 + 33λ = −λ3 + 36λ = 0,
so the eigenvalues are 0, 6, and -6 (nondegenerate). As for the eigenvectors:

 
 
1 4 1
v1
v1
4 −2 4 v2  = λ v2 
1 4 1
v3
v3
⇒


v1 + 4v2 + v3 = λv1 ,
4v1 − 2v2 + 4v3 = λv2 ,


v1 + 4v2 + v3 = λv3 ,
438
APPENDIX. LINEAR ALGEBRA
(The third equation is redundant, for each of the eigenvalues.)
λ = 0 v1 + 4v2 + v3 = 0, 4v1 − 2v2 + 4v3 = 0
⇒
λ = 6 v1 + 4v2 + v3 = 6v1 , 4v1 − 2v2 + 4v3 = 6v2
λ = −6
v1 = −v3 , v2 = 0.
⇒
v1 + 4v2 + v3 = −6v1 , 4v1 − 2v2 + 4v3 = −6v2
So the (normalized) eigenvectors of A are
 
 
1
1
1  
1  
0
1
λ 1 = 0 : v1 = √
λ2 = 6 : v2 = √
2 −1
3 1
v1 = v2 = v3 .
⇒
v1 = v3 , v2 = −2v1 .
 
1
1  
−2
λ3 = −6 : v3 = √
6
1
(c)

 
 
1 −2 −1
2
1
1
1
Bv1 = √ −2 2 −2  0  = √  0  = 2v1 ,
2 −1 −2 1
2 −2
−1

 
 
1 −2 −1
1
−2
1
1
Bv2 = √ −2 2 −2 1 = √ −2 = −2v2 ,
3 −1 −2 1
3 −2
1
 

 
1
1 −2 −1
4
1
1
Bv3 = √ −2 2 −2 −2 = √ −8 = 4v3 .
6 −1 −2 1
6
1
4
(Evidently the eigenvalues of B are 2, -2, and 4.)
Problem A.26
(a) Since A† = A and B† = B, the matrices are normal (Equation A.82), and hence diagonalizable.


 


2 2 −1
2 −1 2
2 −1 2
2 2 −1
[A, B] =  2 −1 2  −1 5 −1 − −1 5 −1  2 −1 2 
−1 2 2
2 −1 2
2 −1 2
−1 2 2

 

0 9 0
0 9 0
= 9 −9 9 − 9 −9 9 = 0. X
0 9 0
0 9 0
(b) To get the eigenvalues, use the characteristic equation:
det(A − λ1) =
2−λ
2
−1
2 −1 − λ 2
= −(2 − λ)2 (1 + λ) − 4 − 4 + (1 + λ) − 4(2 − λ) − 4(2 − λ)
−1
2
2−λ
= −(4 − 4λ + λ2 )(1 + λ) − 23 + 9λ = −(λ + 3)(λ − 3)2 = 0,
so the eigenvalues are 3, 3, and -3 (degenerate). As for the eigenvectors:


 
 

2 2 −1
v1
v1
2v1 + 2v2 − v3 = λv1 ,
 2 −1 2  v2  = λ v2  ⇒
2v1 − v2 + 2v3 = λv2 ,


−1 2 2
v3
v3
−v1 + 2v2 + 2v3 = λv3 ,
APPENDIX. LINEAR ALGEBRA
439
(The third equation is redundant, for each of the eigenvalues.)
λ = −3
2v1 + 2v2 − v3 = −3v1 , 2v1 − v2 + 2v3 = −3v2
λ = 3 2v1 + 2v2 − v3 = 3v1 , 2v1 − v2 + 2v3 = 3v2
⇒
⇒
v1 = v3 , v2 = −2v1 .
2v2 = v1 + v3 ,
but in the latter case there is no condition on v1 or v3 —any vector of the form


2a
v = a + b
2b
is an eigenvector of A, with eigenvalue 3. We might choose the (orthonormal) eigenvectors as follows:
So the (normalized) eigenvectors of A are
 
1
1  
−2
λ1 = −3 : v1 = √
6
1
 
1
1  
0
λ 2 = 3 : v2 = √
2 −1
 
1
1  
1
λ3 = 3 : v3 = √
3 1
(c)
 



1
2 −1 2
6
1
1 
−1 5 −1 −2 = √ −12 = 6 v1 ,
Bv1 = √
6
6
1
2 −1 2
6
 

 
1
2 −1 2
0
1
1
Bv2 = √ −1 5 −1  0  = √ 0 = 0 v2 ,
2
2 0
2 −1 2
−1
 

 
1
2 −1 2
3
1
1 
−1 5 −1 1 = √ 3 = 3 v3 .
Bv3 = √
3
3 3
1
2 −1 2
I got lucky! It just happens that the v2 and v3 I picked are also eigenvectors of B (whose eigenvalues are
evidently 0, 3, and 6). But the most general eigenvector of A with eigenvalue 3 is not an eigenvector of B:


 



2a
2 −1 2
2a
3(a + b)
−1 5 −1 a + b = 3(a + b) 6= λ a + b ;
3(a + b)
2b
2b
2 −1 2
the middle line works only if λ = 3 (in which case the top line requires a = b—this is my v3 ), or b = −a
(this is my v2 ).
Problem A.27
Let |γi = |αi + c|βi, for some complex number c. Then
hγ|T̂ γi = hα|T̂ αi + chα|T̂ βi + c∗ hβ|T̂ αi + |c|2 hβ|T̂ βi, and
hT̂ γ|γi = hT̂ α|αi + c∗ hT̂ β|αi + chT̂ α|βi + |c|2 hT̂ β|βi.
440
APPENDIX. LINEAR ALGEBRA
Suppose hT̂ γ|γi = hγ|T̂ γi for all vectors. For instance, hT̂ α|αi = hα|T̂ αi and hT̂ β|βi = hβ|T̂ βi), so
chα|T̂ βi + c∗ hβ|T̂ αi = chT̂ α|βi + c∗ hT̂ β|αi, and this holds for any complex number c.
In particular, for c = 1: hα|T̂ βi+hβ|T̂ αi = hT̂ α|βi+hT̂ β|αi, while for c = i: hα|T̂ βi−hβ|T̂ αi = hT̂ α|βi−hT̂ β|αi.
(I canceled the i’s). Adding: hα|T̂ βi = hT̂ α|βi. QED
Problem A.28
(a)
T† = T̃∗ =
(b)
1 1−i
= T. X
1+i 0
√
1± 1+8
1±3
(1 − λ) (1 − i)
2
=
.
= −(1 − λ)λ − 1 − 1 = 0; λ − λ − 2 = 0; λ =
(1 + i) (0 − λ)
2
2
λ1 = 2, λ2 = −1.
(c)
α
α
=2
=⇒ α + (1 − i)β = 2α =⇒ α = (1 − i)β.
β
β
1
1 1−i
|α|2 + |β|2 = 1 =⇒ 2|β|2 + |β|2 = 1 =⇒ β = √ . a(1) = √
.
1
3
3
1
1
(1 − i)
α
α
=−
=⇒ α + (1 − i)β = −α; α = − (1 − i)β.
(1 + i)
0
β
β
2
r
3
1
2
1 i−1
2|β|2 + |β|2 = 1 =⇒ |β|2 = 1; β =
. a(2) = √
.
2
4
2
3
6
(i − 1)
1
1
(1)† (2)
a a = √ (1 + i) 1
= √ (i − 1 − 1 − i + 2) = 0. X
2
3 2
3 2
1
(1 − i)
(1 + i)
0
(d)
1
1
(1)
(2)
Eq. A.81 =⇒ (S−1 )11 = a1 = √ (1 − i); (S−1 )12 = a1 = √ (i − 1);
3
6
1
2
(1)
(2)
(S−1 )21 = a2 = √ ; (S−1 )22 = a2 = √ .
3
6
√ 1 (1 − i) (i − 1)/ 2
1
(1 + i)√ √1
√
S−1 = √
; S = (S−1 )† = √
.
2
1
3
3 (−i − 1)/ 2 2
√ 1
(1 + i)√ √1
1
(1 − i)
(1 − i) (i −√
1)/ 2
−1
STS =
(1 + i)
0
1
2
3 −(1 + i)/ 2 2
√ 1
1 6 0
(1 + i)√ √1
2(1 − i) (1 − √
i)/ 2
2 0
=
=
=
. X
0 −1
2
− 2
3 −(1 + i)/ 2 2
3 0 −3
(e)
Tr(T) = 1;
det(T) = 0 − (1 + i)(1 − i) = −2.
Tr(STS−1 ) = 2 − 1 = 1. X
det(STS−1 ) = −2. X
APPENDIX. LINEAR ALGEBRA
441
Problem A.29
(a)
det(T) = 8 − 1 − 1 − 2 − 2 − 2 = 0.
Tr(T) = 2 + 2 + 2 = 6.
(b)
(2 − λ)
i
1
−i (2 − λ)
i
= (2 − λ)3 − 1 − 1 − (2 − λ) − (2 − λ) − (2 − λ) = 8 − 12λ + 6λ2 − λ3 − 8 + 3λ = 0.
1
−i (2 − λ)
−λ3 + 6λ2 − 9λ = −λ(λ2 − 6λ + 9) = −λ(λ − 3)2 = 0.
λ1 + λ2 + λ3 = 6 = Tr(T). X
λ1 λ2 λ3 = 0 = det(T). X
λ1 = 0, λ2 = λ3 = 3.


0 0 0
Diagonal form: 0 3 0 .
0 0 3
(c)
 
α
2 i 1
−i 2 i  β  = 0 =⇒ 2α + iβ + γ = 0
.
−iα + 2β + iγ = 0 =⇒ α + 2iβ − γ = 0
γ
1 −i 2

Add the two equations: 3α + 3iβ = 0 =⇒ β = iα; 2α − α + γ = 0 =⇒ γ = −α.
 
 
α
1
1
1
a(1) =  iα  . Normalizing: |α|2 + |α|2 + |α|2 = 1 =⇒ α = √ . a(1) = √  i  .
3
3 −1
−α

 
 
α
α
=⇒ −α + iβ + γ = 0,
2 i 1
 2α + iβ + γ = 3α
−i 2 i  β  = 3 β  =⇒ −iα + 2β + iγ = 3β =⇒ α − iβ − γ = 0,

γ
γ
α − iβ + 2γ = 3γ
=⇒ α − iβ − γ = 0.
1 −i 2

The three equations are redundant – there is only one condition here: α − iβ − γ = 0. We could pick
γ = 0, β = −iα, or β = 0, γ = α. Then


 
α
α
(2)
(3)
a0 = −iα ; a0 =  0  .
0
α
(2)
But these are not orthogonal, so we use the Gram-Schmidt procedure (Problem A.4); first normalize a0 :


1
1
a(2) = √ −i .
2 0
 
 
 


1
1
1
1/2
† (3)
† (3)
α
α
α
(3)
a(2) a0 = √ 1 i 0 0 = √ . So a0 − (a(2) a0 ) a(2) = α 0 − −i = α  i/2  .
2
2
2
1
1
0
1
442
APPENDIX. LINEAR ALGEBRA
|α|2
Normalize:
1 1
+ +1
4 4
3
= |α|2 = 1
2
r
=⇒
α=
2
.
3
 
1
1
a(3) = √  i  .
6 2
Check orthogonality:
(1) † (2)
a
a
(1) † (3)
a
a
 
1
1
1
= √ 1 −i −1 −i = √ (1 − 1 + 0) = 0. X
6
6
0
 
1
1
1
= √ 1 −i −1  i  = √ (1 + 1 − 2) = 0. X
3 2
3 2
2
(d) S−1 is the matrix whose columns are the eigenvectors of T (Eq. A.81):
√

√
2
3 1
√
√
1
S−1 = √  √
2 i − 3 i i ;
6 − 2 0 2
√
√ 
√
2 −√ 2 i − 2
√
1
S = (S−1 )† = √  3
3i
0 .
6
1
−i
2
√
√
√ 
 √

 

√

2
3
1
2
−
2
i
−
2
0
0
0
2
i
1
0
0
0
√
√
1 √ √
1
STS−1 =  3
2 i − 3 i i  = 0 18 0  = 0 3 0 . X
3i
0  −i 2 i   √
6
6
0 0 3
1 −i 2
0 0 18
1
−i
2
− 2 0 2
|
{z
}


√
3
0 3 √3


0 −3 3 i 3i


0
0
6
Problem A.30
(a) hÛ α|Û βi = hÛ † Û α|βi = hα|βi. X
(b) Û |αi = λ|αi =⇒ hÛ α|Û αi = |λ|2 hα|αi. But from (a) this is also hα|αi.
(c) Û |αi = λ|αi, Û |βi = µ|βi =⇒ |βi = µÛ −1 |βi, so Û † |βi =
So |λ| = 1.
X
1
|βi = µ∗ |βi (from (b)).
µ
hβ|Û αi = λhβ|αi = hÛ † β|αi = µhβ|αi, or (λ − µ)hβ|αi = 0. So if λ 6= µ, then hβ|αi = 0. QED
Problem A.31
(a) (i)


004
M2 = 0 0 0 ;
000


0 0 0
M3 = 0 0 0 ,
0 0 0
so



 



0 0 4
1 1 5
100
013
1
eM = 0 1 0 + 0 0 4 + 0 0 0 = 0 1 4 .
2
001
000
0 0 0
0 0 1
APPENDIX. LINEAR ALGEBRA
443
(ii)
2
M =
−θ2 0
0 −θ2
= −θ2 I;
M3 = −θ2 M;
M4 = θ4 I;
etc.
1 2
θ3 0 1
θ4
0 1
e =I+θ
− θ I−
+ I + ···
−1 0
2
3! −1 0
4!
2
4
3
5
θ
θ
θ
θ
0 1
= 1−
+
− ··· I + θ −
+
− ···
−1 0
2
4!
3!
5!
cos θ sin θ
10
0 1
.
= cos θ
+ sin θ
=
− sin θ cos θ
01
−1 0
M
(b)

d1
..

SMS−1 = D = 
0
M −1
Se S
0


 for some S.
.
dn
1 2
1 3
= S I + M + M + M + · · · S−1 . Insert SS−1 = I :
2
3!
1
1
SeM S−1 = I + SMS−1 + SMS−1 SMS−1 + SMS−1 SMS−1 SMS−1 + · · ·
2
3!
1
1
= I + D + D2 + D3 + · · · = eD . Evidently
2
3!
det(eD ) = det(SeM S−1 ) = det(S) det(eM ) det(S−1 ) = det(eM ). But


 k

 3
 2
d1
0
d1
0
d1
0






3
k
D2 =  . . .
,
 , D =  ...
 , D =  ...
3
2
k
0
dn
0
dn
0
dn
so


 2

 3

 d

d1
0
d1
0
d1
0
e 1
0
 1  ..



 1  ..

..
eD = I +  . . .
.
+ 
 + ··· = 
+ 
.
.
.
2
3!
2
3
dn
0
dn
0
dn
0
dn
0
e
det(eD ) = ed1 ed2 · · · edn = e(d1 +d2 +···dn ) = eTr D = eTr M (Eq. A.68), so det(eM ) = eTr M . QED
(c) Matrices that commute obey the same algebraic rules as ordinary numbers, so the standard proofs of
ex+y = ex ey will do the job. Here are two:
(i) Combinatorial Method: Use the binomial theorem (valid if multiplication is commutative):
e
M+N
=
∞
X
1
n=0
n!
n
(M + N) =
∞
n X
1 X n
n=0
n! m=0 m
Mm Nn−m =
∞ X
n
X
1
Mm Nn−m .
m!(n
−
m)!
n=0 m=0
444
APPENDIX. LINEAR ALGEBRA
m
n
Instead of summing vertically first, for fixed n (m : 0 → n), sum horizontally first, for fixed m (n :
m → ∞, or k ≡ n − m : 0 → ∞)—see diagram (each dot represents a term in the double sum).
eM+N =
∞
∞
X
1 mX 1 k
M
N = eM eN . QED
m!
k!
m=0
k=0
(ii) Analytic Method: Let
S(λ) ≡ eλM eλN ;
dS
= MeλM eλN + eλM NeλN = (M + N)eλM eλN = (M + N)S.
dλ
(The second equality, in which we pull N through eλM , would not hold if M and N did not commute.)
Solving the differential equation: S(λ) = Ae(M+N)λ , for some constant A. But S(0) = I, so A = 1,
and hence eλM eλN = eλ(M+N) , and (setting λ = 1) we conclude that eM eN = e(M+N) . [This method
generalizes most easily when M and N do not commute—leading to the famous Baker-CampbellHausdorf lemma.]
0 1
0 0
As a counterexample when [M, N] 6= 0, let M =
, N=
. Then M2 = N2 = 0, so
0 0
−1 0
eM = I + M =
11
1 0
, eN = I + N =
;
01
−1 1
But (M + N) =
eM eN =
1 1
1 0
0 1
=
.
0 1
−1 1
−1 1
0 1
cos(1) sin(1)
, so (from a(ii)): eM+N =
.
−1 0
− sin(1) cos(1)
The two are clearly not equal.
(d)
eiH =
∞
X
1
n!
n=0
in Hn =⇒ (eiH )† =
∞
X
1
n!
n=0
(−i)n (H† )n =
(eiH )† (eiH ) = e−iH eiH = ei(H−H) = I, using (c).
∞
X
1
n!
n=0
(−i)n Hn = e−iH (for H hermitian).
So eiH is unitary. X
APPENDIX. LINEAR ALGEBRA
445
PROBLEM CONCORDANCE
Second and Third Editions
Introduction to Quantum Mechanics
Chapter 1
3rd Edition
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.10
1.11 (new)
1.12 (new)
1.13 (new) C
1.14
1.15
1.16
1.17
1.18
2nd Edition
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.10
1.14
1.16
1.17
1.15
1.18 (modified)
Chapter 2
3rd Edition
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
2.10
2.11
2.12
2.13
2.14
2.15
2.16
2.17
2.18
2nd Edition
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8 (modified)
2.9
2.10
2.11
2.12
2.13
2.15
2.16
2.17
2.18
2.19
2.19
2.20
2.21
2.22
2.23
2.24
2.25 (new)
2.26
2.27
2.28
2.29
2.30
2.31
2.32
2.33
2.34
2.35
2.36
2.37
2.38
2.39
2.40
2.41
2.42
2.43
2.44
2.45 (new)
2.46
2.47
2.48
2.49
2.50
2.51 (new)
2.52
2.53
2.54
2.55
2.56
2.57
2.58 (new)
2.59 (new)
2.60 (new)
2.61 (new)
2.62 (new)
2.63 (new)
2.64 (new)
2.20
2.21
2.22
2.23
2.24
2.25
2.26
2.27 (modified)
2.28
2.29
2.30
2.31
2.32
2.33
2.34
2.35
2.36
2.37
2.39
2.48
2.41 (modified)
2.42
2.43 (modified)
2.44
2.45
2.46
2.47
2.40
2.49
2.50
2.51
2.52
2.53
2.54
2.55
2.56
446
APPENDIX. LINEAR ALGEBRA
Chapter 3
3rd Edition
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
3.10
3.11
3.12 (new)
3.13
3.14
3.15
3.16
3.17
3.18
3.19 (new)
3.20
3.21
3.22
3.23
3.24 (new)
3.25
3.26
3.27
3.28 (new)
3.29 (new)
3.30 (new)
3.31
3.32
3.33
3.34
3.35
3.36
3.37
3.38
3.39
3.40 (new)
3.41
3.42
3.43
3.44
3.45 (new)
3.46
3.47 (new)
3.48 (new)
3.49
2nd Edition
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
3.10
3.11
3.12
3.13 (modified)
3.14
3.15
3.16
3.17
3.18
3.19
3.20
3.21
3.23
3.22
2.24 (modified)
3.25
3.26 (modified)
3.27
3.28 (modified)
3.29
3.30
3.31
3.32
3.33
3.34
3.35 (modified)
3.36
3.37
3.38
3.40
APPENDIX. LINEAR ALGEBRA
Chapter 4
3rd Edition
4.1
4.2
4.3 (new)
4.4
4.5
4.6 (new)
4.7
4.8
4.9
4.10
4.11
4.12
4.13
4.14
4.15
4.16
4.17 (new)
4.18
4.19
4.20
4.21
4.22
4.23
4.24
4.25
4.26
4.27
4.28
4.29
4.30
4.31
4.32
4.33
4.34
4.35
4.36
4.37
4.38
4.39 (new)
4.40
4.41
4.42
4.43
4.44
4.45
2nd Edition
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
4.10
4.11
4.12
4.13
4.14
4.15
4.16
4.17
4.18 (modified)
4.19
4.20
4.21
4.22
4.23
4.24 (modified)
4.25
4.26
4.27
4.28
4.29
4.30
4.31
4.32
4.33
4.34
4.35
4.36
4.37
4.59
4.60
4.61 (modified)
10.7
447
4.46
4.47
4.48
4.49
4.50
4.51 (new)
4.52
4.53
4.54
4.55
4.56
4.57
4.58
4.59
4.60
4.61
4.62
4.63
4.64
4.65 (new)
4.66
4.67 (new)
4.68 (new)
4.69 (new)
4.70 (new)
4.71 (new)
4.72 (new)
4.73 (new)
4.74 (new)
4.75 (new)
4.76 (new)
4.38
4.39 (modified)
4.40
4.41
4.42
4.43
4.44
4.45
4.46
4.47
4.48
4.49
4.50
4.51
4.52
4.53
4.54
4.55
4.58
448
APPENDIX. LINEAR ALGEBRA
Chapter 5
3rd Edition
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9(new)
5.10(new)
5.11 (new)
5.12
5.13
5.14
5.15
5.16 (new)
5.17
5.18
5.19
5.20
5.21
5.22 (new)
5.23
5.24
5.25
5.26
5.27
5.28 (new)
5.29
5.30
5.31 (new)
5.32 (new)
5.33 (new)
5.34 (new)
5.35
5.36
5.37 (new)
5.38 (new)
2nd Edition
5.1
5.2
5.3
5.4
5.5
5.6
5.32 (modified)
5.7
5.8 (modified)
5.9
5.10
5.11
5.12
5.13
5.14
5.15
5.16
5.17
5.18
5.19
5.20
5.21
5.33
5.34
5.35
5.36
Chapter 6
All new.
Chapter 7
3rd Edition
7.1
7.2
7.3
7.4 (new)
7.5
7.6
7.7 (new)
7.8
7.9
7.10 (new)
7.11
7.12
7.13
7.14
7.15
7.16
7.17
7.18
7.19
7.20
7.21
7.22
7.23
7.24
7.25 (new)
7.26
7.27
7.28
7.29
7.30
7.31
7.32
7.33
7.34 (new)
7.35 (new)
7.36
7.37
7.38
7.39 (new)
7.40 (new)
7.41 (new)
7.42
7.43
7.44
7.45
2nd Edition
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
6.10
6.11
6.12
6.13
6.14
6.15
6.16
6.17
6.18
6.19
6.20
6.21
6.22
6.23
6.24
6.25
6.26
6.27
6.28
6.29
6.30
6.31
6.32
6.33
6.34
6.35
6.36
APPENDIX. LINEAR ALGEBRA
7.46
7.47
7.48
7.49 (new)
7.50 (new)
7.51
7.52 (new)
7.53 (new)
7.54 (new)
7.55 (new)
7.56 (new)
7.57 (new)
6.37
6.38
6.39
6.40
Chapter 8
3rd Edition
8.1
8.2
8.3
8.4
8.5 (new)
8.6
8.7
8.8
8.9
8.10
8.11
8.12 (new)
8.13 (new)
8.14 (new)
8.15 (new)
8.16 (new)
8.17
8.18
8.19
8.20 (new)
8.21
8.22
8.23
8.24
8.25
8.26
8.27
8.28 (new)
8.29 (new)
8.30(new)
2nd Edition
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
7.9
7.10
7.11
7.12
7.13
7.14
7.15
7.16
7.17
7.18
7.19
7.20
449
Chapter 9
3rd Edition
9.1
9.2
9.3
9.4
9.5 (new)
9.6
9.7
9.8
9.9
9.10
9.11
9.12 (new)
9.13
9.14
9.15
9.16
9.17
9.18
9.19
9.20 (new)
2nd Edition
8.1
8.2
8.3
8.4
8.5
8.6
8.7
8.8
8.9
8.10
8.11
8.12
8.13
8.14 (modified)
8.15
8.16
8.17
450
APPENDIX. LINEAR ALGEBRA
Chapter 10
3rd Edition
10.1
10.2
10.3
10.4
10.5
10.6
10.7
10.8
10.9
10.10
10.11
10.12
10.13
10.14
10.15
10.16
10.17
10.18
10.19
10.20
10.21 (new)
10.22 (new)
10.23 (new)
2nd Edition
11.1
11.2
11.3
11.4
11.5
11.6
11.7
11.8
11.9
11.10
11.11
11.12
11.13
11.14
11.15
11.16
11.17
11.18
11.19
11.20
Chapter 11
3rd Edition
11.1 (new)
11.2
11.3
11.4
11.5
11.6
11.7
11.8 (new)
11.9
11.10
11.11
11.12
11.13
11.14 (new)
11.15 (new)
11.16
11.17 (new)
11.18
11.19
11.20
11.21
11.22
11.23 (new)
11.24
11.25
11.26
11.27
11.28
11.29
11.30 (new)
11.31
11.32
11.33 (new)
11.34
11.35
11.36
11.37 (new)
11.38 (new)
11.39 (new)
2nd Edition
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
9.9 (modified)
9.10
9.11
(from text)
(from text)
9.14
2.38
2.14
10.2
Example 11.4
10.4
9.15
9.16
9.17
9.18
9.19
9.20
9.21
9.22
10.8
10.1
10.9
APPENDIX. LINEAR ALGEBRA
451
Chapter 12
3rd Edition
12.1
12.2 (new)
12.3 (new)
12.4 (new)
12.5 (new)
12.6 (new)
12.7 (new)
12.8 (new)
2nd Edition
12.1
Appendix
3rd Edition
A.1
A.2
A.3
A.4
A.5
A.6
A.7
A.8
A.9
A.10
A.11
A.12
A.13
A.14
A.15
A.16
A.17
A.18
A.19
A.20
A.21
A.22
A.23
A.24 (new)
A.25 (new)
A.26 (new)
A.27
A.28
A.29
A.30
A.31
2nd Edition
A.1
A.2
A.3
A.4
A.5
A.6
A.7
A.8
A.9
A.10
A.11
A.12
A.13
A.14
A.15
A.16
A.17
A.18
A.19
A.20
A.21
A.23
A.22
A.24
A.25
A.26
A.27
A.28