Joel Broida
UCSD Fall 2009
Phys 130B
QM II
Homework Set 4 Solutions
1. (a) The exact solutions to the particle-in-a-box problem are
r
2
nπx
(0)
ψn (x) =
sin
l
l
together with
n2 π 2 ~2
.
2ml2
With H ′ = αδ(x − l/2), the first-order energy corrections are then given
by
Z
l
2α l 2 nπx
δ x−
sin
dx
En(1) = hψn(0) |H ′ ψn(0) i =
l 0
l
2
En(0) =
=
nπ
2α
sin2
.
l
2
So we see that
En(1) =
(
0
2α/l
for n even
.
for n odd
The reason there is no correction for n even is that in this case the wave
function always vanishes at x = l/2, and hence it isn’t affected by the
perturbation that only acts at that point.
(b) The first-order correction to the wave function is given by
ψn(1) =
X hm(0) |H ′ n(0) i
(0)
m6=n
(0)
En − Em
(0)
ψm
.
(1)
In this case, we want the first three nonzero terms in the expansion of
(1)
ψ1 . We have
Z
2α l
πx
l
mπx
(0)
′ (0)
hψm |H ψ1 i =
dx
sin
δ x−
sin
l 0
l
l
2
=
mπ
π
2α
sin
sin
l
2
2
=
2α
mπ
sin
l
2
1
so the first three nonzero terms occur for m = 3, 5, 7. (The sum does
not include m = 1 because we are working with the case n = 1.) The
(0)
(0)
corresponding differences En − Em are then given by
(0)
(0)
=
π 2 ~2
π 2 ~2
2
(1
−
3
)
=
−8
2ml2
2ml2
(0)
(0)
=
π 2 ~2
π 2 ~2
2
(1
−
5
)
=
−24
2ml2
2ml2
(0)
(0)
=
π 2 ~2
π 2 ~2
(1 − 72 ) = −48
2
2ml
2ml2
E1 − E3
E1 − E5
E1 − E7
so the wave function correction is
(1) (0)
(−1) (0)
2α 2ml2 (−1) (0)
(1)
ψ +
ψ +
ψ
ψ1 =
l π 2 ~2 (−8) 3
(−24) 5
(−48) 7
r 3πx 1
5πx 1
7πx
l
αm
sin
.
− sin
+ sin
= 2 2
π ~
2
l
3
l
6
l
(c) The second-order energy correction is given by
X hn(0) |H ′ m(0) i2
(2)
En =
(0)
(0)
En − Em
m6=n
(2)
where
hn(0) |H ′ m(0) i =
=
2α
l
Z
l
sin
0
mπx
l
nπx
dx
sin
δ x−
l
l
2
2α
nπ
mπ
sin
sin
.
l
2
2
This is zero unless m and n are both odd, in which case it is equal to
±2α/l. Therefore we have
2 X
2ml2 2α
1
En(2) = 2 2
for n odd
π ~
l
n 2 − m2
m6=n
m odd
and
En(2) = 0
for n even .
To sum the series, we first expand the summand as partial fractions:
1
A
B
1
=
=
+
n 2 − m2
(n − m)(n + m)
n−m n+m
so that
A(n + m) + B(n − m) = (A + B)n + (A − B)m = 1 .
2
Here the “variable” is m, so we equate powers of m on both sides to write
A − B = 0 and (A + B)n = 1. Therefore A = B = 1/2n and we have
X
1
1 X
1
1
=
+
n 2 − m2
2n
n−m n+m
m6=n
m odd
m6=n
m odd
=
1
1
1 X
.
−
2n
m+n m−n
m6=n
m odd
To see if we can see what this sums to, let us look at some particular
cases. For n = 1 we have
X
X 1
1
1
1
=
−
1 − m2
2 m=3,5,7,... m + 1 m − 1
m6=1
m odd
1 1 1 1
1 1 1 1
+ + + ···− − − − − ···
2 4 6 8
2 4 6 8
1
1
−
.
=
2
2
=
Now look at n = 3:
X
X 1
1
1
1
=
−
3 2 − m2
2 · 3 m=1,5,7,... m + 3 m − 3
m6=3
m odd
1
1
1 1 1 1
1
1 1 1
+ +
+ ···−
− − − − −
− ···
=
6 4 8 10
−2 2 4 6 8 10
1
1
−
.
=
6
6
And in general, we see that all terms will cancel except for the one term
−1/2n. Hence we can write
X
1
1
1
1
−
=− 2.
=
2
2
n −m
2n
2n
4n
m6=n
m odd
(2)
Putting this all together, we see that En = 0 for n even, and for n odd
we have
2mα2
En(2) = − 2 2 2 .
π ~ n
2. We first need the potential energy of the electron a distance r from the center
of the nucleus. From
F = −eE = −
3
e2
r̂ = −∇V
r2
and taking V (∞) = 0, we have, for r ≥ R
Z r
1
e2
−V (r) = −e2
dr =
.
2
r
∞ r
But for r ≤ R we have to be careful. Let the proton have radius R. Then at
a distance r ≤ R from the center, the amount of charge enclosed is
Q(r) =
e
4πr3
r3
e.
=
3 4πR3 /3
R3
Using this, the potential energy for r < R becomes
R
1
dr − e
2
∞ r
Z
e2 r
e2
− 3
r dr
=
R
R R
−V (r) = −e2
Z
=
e2
e2 r 2
e2
+
−
3
R
2R
2R
=
e2 r 2
3e2
.
−
2R
2R3
Z
r
R
Q(r)
dr
r2
The Hamiltonian is H = T + V , so for r ≤ R we have (adding and subtracting
the term e2 /r)
H=T+
=T−
e2 r 2
3e2
−
3
2R
2R
e2
+ H′ = H0 + H′
r
where H 0 is the usual hydrogen-atom Hamiltonian, and where
e2 r 2
3e2 e2
−
+
3
2R
2R
r
r2
2R
e2
−
3
+
=
2R R2
r
H′ =
for r ≤ R
and H ′ = 0 for r > R.
The ground-state wave function is
1 −3/2
ψ100 = √ a0 e−r/a0
π
4
and hence the first-order correction to the ground-state energy is given by
Z R 2
e2
2R −2r/a0 2
r
(1)
E0 = hψ100 |H ′ ψ100 i =
r drdΩ
e
−
3
+
2πRa30 0
R2
r
Z 2e2 R r4
2
=
− 3r + 2Rr e−2r/a0 dr .
Ra30 0
R2
Noting that 2R/a0 = 2 × 10−13 /(0.529 × 10−8 ) ∼ 4 × 10−5 , we see that we
can take the exponential factor in the integral to be essentially equal to 1 over
the nucleus. This leaves us with the integral
Z 2e2 R r4
2e2 R2
(1)
2
E0 =
−
3r
+
2Rr
dr =
.
3
2
Ra0 0
R
5a30
Using e = 4.8 × 10−10 statC, R = 10−13 cm and a0 = 0.529 × 10−8 cm we find
(1)
E0
= 6.22 × 10−21 erg = 3.89 × 10−9 eV
where we used 1 eV = 1.6 × 10−12 erg.
As a passing remark, the complete integral including the exponential in the
(1)
expression for E0 is straightforward to evaluate analytically, using our nowfamiliar trick of taking the derivative with respect to the constant in the
exponential. However, the algebra gets somewhat messy. If I use the Mathematica command Integrate to numerically evaluate the integral, I come up
(1)
with E0 = −3.10 × 10−10 eV. Since this is negative and quite small, my
guess is that the accuracy of the calculations as computed by Mathematica
isn’t good enough given the size of the numbers in the exponential, and hence
this answer is just wrong. However, if I use the command NIntegrate, then I
again obtain 3.89 × 10−9 eV.
3. From the zeroth-order wave function
ψ (0) =
Z 3 −Zr1 /a0 −Zr2 /a0
e
e
πa30
and the expansion
∞ X
l
l
X
4π r<
1
=
[Ylm (θ1 , φ1 )]∗ Ylm (θ2 , φ2 )
l+1
r12
2l + 1 r>
l=0 m=−l
we have the first-order energy correction
e2 E
D
ψ (0)
E (1) = ψ (0) r12
5
=
Z
Z
∞
l
Z 6 e2 X X 4π
2
r
dr
dΩ
r22 dr2 dΩ2
1
1 1
π 2 a60
2l + 1
l=0 m=−l
× e−2Zr1 /a0 e−2Zr2 /a0
l
r<
[Y m (θ1 , φ1 )]∗ Ylm (θ2 , φ2 ) .
l+1 l
r>
Now use the fact that Y00 (Y00 )∗ = 1/4π to write
E
(1)
Z ∞Z ∞
∞
l
l
1
16Z 6 e2 X X
−2Zr1 /a0 −2Zr2 /a0 r< 2 2
r r dr1 dr2
e
e
=
l+1 1 2
a60
2l + 1 0
r>
0
l=0 m=−l
Z
× dΩ1 [Ylm (θ1 , φ1 )]∗ Y00 (θ1 , φ1 )
Z
× dΩ2 [Y00 (θ2 , φ2 )]∗ Ylm (θ2 , φ2 ) .
But the Ylm are orthonormal, so the only term in the sum that doesn’t vanish
is the one with m = l = 0, and we are left with
Z Z
16Z 6e2 ∞ ∞ −2Zr1 /a0 −2Zr2 /a0 1 2 2
E (1) =
r r dr1 dr2 .
e
e
a60
r> 1 2
0
0
We first integrate over r1 , where for 0 ≤ r1 ≤ r2 we have r> = r2 , and where
for r2 ≤ r1 ≤ ∞ we have r> = r1 . Then
Z
16Z 6 e2 ∞ −2Zr2 /a0 2
E (1) =
r2
e
a60
0
Z r2
Z ∞
r2
r2
×
e−2Zr1 /a0 1 dr1 +
e−2Zr1 /a0 1 dr1 dr2
r2
r1
0
r2
=
16Z 6 e2
a60
Z
∞
e−2Zr2 /a0 r2
Z
r2
0
0
+
16Z 6 e2
a60
Z
e−2Zr1 /a0 r12 dr1 dr2
∞
e−2Zr2 /a0 r22
0
Z
∞
r2
e−2Zr1 /a0 r1 dr1 dr2 .
In order to evaluate these integrals, it is easiest to first tabulate some general
formulas. We have the indefinite integrals
Z
Z
x
1
∂ eαx
∂
= eαx
− 2
eαx dx =
xeαx dx =
∂α
∂α α
α α
Z
Z
2
2x
x2
∂2
αx
αx
+
−
e
dx
=
e
x2 eαx dx =
∂α2
α
α2
α3
And for n > −1 and α > 0 we also have
Z ∞
n Z ∞
n
n!
n −αx
n ∂
−αx
n ∂
x e
dx = (−1)
α−1 = n+1 .
e
dx
=
(−1)
n
n
∂α
∂α
α
0
0
6
Using these formulas, we have (letting α = +2Z/a0 )
2
Z r2
Z r2
r2
2
2
2r2
e−αr1 r12 dr1 = −e−αr2
e−2Zr1 /a0 r12 dr1 =
+ 2 + 3 + 3
α
α
α
α
0
0
and hence
Z ∞
0
e−αr2 r2
r2
Z
0
=
e−αr1 r12 dr1 dr2
∞
Z
0
=
Similarly,
Z
3
2r2
2r22
r2
−αr2 2r2
−2αr2
dr2
+ 2 + 3 +e
−e
α
α
α
α3
5
.
8α5
∞
e−2Zr1 /a0 r1 dr1 = e−αr2
r2
so that
Z
Z ∞
e−2Zr2 /a0 r22
0
∞
r2
1
r2
+ 2
α
α
Z
e−2Zr1 /a0 r1 dr1 dr2 =
∞
e−2αr2
0
=
r2
r23
+ 22
α
α
dr2
5
.
8α5
Putting together our results we have
5 Ze2
16Z 6 e2
5
5
(1)
=
E =
+
.
6
a0
8α5
8α5
8 a0
4. (a) We simply take the derivative of En = hn|H|ni and use the fact that
hn|ni = 1:
∂En
∂hn|
∂H
∂|ni
=
H|ni + hn|
|ni + hn|H
∂λ
∂λ
∂λ
∂λ
= En
∂H
∂|ni
∂hn|
|ni + hn|
|ni + En hn|
∂λ
∂λ
∂λ
= En
∂H
∂
hn|ni + hn|
|ni
∂λ
∂λ
= hn|
∂H
|ni .
∂λ
This result is called the Hellmann-Feynman theorem. Feynman proved
it as part of his senior thesis at MIT. (Hellmann proved it several years
earlier and published it in an obscure Russian journal.)
7
(b) From H = H 0 + λV we immediately have ∂H/∂λ = V . And from En =
(2)
(1)
(0)
En + λEn + λ2 En + · · · , we take the derivative with respect to λ to
write
∂En
= En(1) + 2λEn(2) + · · · .
∂λ
(1)
If we now let λ = 0, we obtain ∂En /∂λ = En . Using these two results
directly in the Hellmann-Feynman theorem, we obtain
En(1) = hn|V |ni
which is the same as the first-order perturbation results.
(c) The harmonic oscillator Hamiltonian is
H=−
1
~2 d2
+ kx2
2m dx2
2
so that letting λ = k we have
∂H
∂H
1
=
= x2 .
∂λ
∂k
2
The energy levels are given by
r 1
1
k
=~
n+
En = ~ω n +
2
m
2
so that
∂En
∂En
~
=
= √
∂λ
∂k
2 km
1
.
n+
2
Applying the Hellmann-Feynman theorem we have
1
~
1
1
√
n+
= hn| x2 |ni = hx2 i
2
2
2
2 km
so that
~
1
1
=
n+
.
n+
hx i = √
2
mω
2
km
2
~
5. (a) We want to know the effect of the perturbation
H′ =
eBz
Lz
2me c
on the n = 2 level of the hydrogen atom. The n = 2 level is four-fold
degenerate, and the states are
|n l mi = |2 0 0i, |2 1 0i, |2 1 1i, |2 1 −1i .
8
Since hn′ l′ m′ |n l mi = δnn′ δll′ δmm′ , we easily have the matrix elements
h2 l′ m′ |H ′ |2 l mi =
eBz
m~ δll′ δmm′ := βm δll′ δmm′
2me c
where we have defined
β=
e~Bz
2me c
for convenience. Choosing our basis {|2 l mi} in the order shown above,
(1)
′
the secular determinant det(Hij
− En δij ) = 0 becomes
−E (1)
0
0
0
2
(1)
−E2
0
0
0
= 0.
(1)
0
0
β − E2
0
(1) 0
0
0
−β − E2 (1)
This clearly has the roots E2 = 0, 0, β, −β, so we see that the four-fold
degeneracy is partially lifted, with the two m = 0 states still degenerate with their original energy eigenvalues. We thus have the following
situation:
|2 1 1i
|2 0 0i
|2 1 0i
|2 1 1i |2 1 −1i
|2 0 0i
|2 1 −1i
|2 1 0i
β=
e~Bz
2me c
β=
e~Bz
2me c
(b) Now the perturbation becomes
H′ =
eBx
Lx .
2me c
The problem is that the states |n l mi are eigenstates of L2 and Lz , not L2
and Lx . Because of this, we must write Lx = (1/2)(L+ + L− ) because we
know how the operators
L± act on the angular momentum states |l mi,
p
i.e., L± |l mi = ~ l(l + 1) − m(m ± 1) |l m ± 1i. So we have
√
√
L+ |0 0i = 0 L+ |1 0i = ~ 2 |1 1i L+ |1 1i = 0 L+ |1 −1i = ~ 2 |1 0i
and
√
√
L− |0 0i = 0 L− |1 0i = ~ 2 |1 −1i L− |1 1i = ~ 2 |1 0i L− |1 −1i = 0 .
From these, we see that if we choose our |l mi basis in the above order,
the matrix representations of L+ and L− with respect to this basis are
9
given by

0
0

0
0
L+ = 
√

0 ~ 2
0
0
0
0
0
0

0
√ 
~ 2


0 
0
0

0
L− = 
0

0

and
0
√
0 ~ 2
0

0
0
√

~ 2 0
.
0
0

0
0
(While in principle it doesn’t matter what order we choose for our basis
vectors, the matrix representation of L± , and hence also of Lx , will be
different. However, the eigenvalues will remain the same, although it may
be more difficult to factor the secular equation.) In any case, the matrix
representation of Lx = (1/2)(L+ + L− ) is


0
0
0
0

√
√ 
0
0
~/ 2 ~/ 2 


Lx = 
√

 0 ~/ 2

0
0


√
0 ~/ 2
0
0
so that the matrix representation of H ′

0 0

0 0
H′ = 
0 β

0
where here we define
β
becomes

0 0

β β

0 0

0
0
e~Bx
.
β= √
2 2 me c
Now the secular equation is
−E (1)
0
2
(1)
−E2
0
0
β
0
β
0
β = 0.
0 (1) −E2 0
β
(1)
−E2
0
Since this is in block diagonal form (a 1 × 1 determinant and a 3 × 3
determinant), we can easily evaluate it to obtain
(1)
(1)
(1)
(1)
(1)
−E2 ([−E2 ]3 + 2β 2 E2 ) = [E2 ]2 ([E2 ]2 − 2β 2 ) = 0 .
√
√
(1)
This has the roots E2 = 0, 0, ± 2 β. Since 2 β = e~Bx /2me c, we see
that the energy splitting is exactly the same as in part (a). This is to
10
be expected since in the absence of an external B field, the electron is in
a spherically symmetric potential. Hence it really doesn’t matter what
direction we choose our B field, the energy levels will be split the same
way.
11