Solutions - Chapter 4
Kevin S. Huang
Problem 4.1
i
Û (dt) = 1 − Ĥdt
h̄
Unitary:
Û † (t)Û (t) = 1
Neglect (dt)2 term:
i
i
i
i †
1 − Ĥdt = 1 + Ĥ † dt − Ĥdt = 1
1 + Ĥ dt
h̄
h̄
h̄
h̄
Ĥ = Ĥ †
Problem 4.2
i
i
i
Û (t) = lim 1 − Ĥ1 dt 1 − Ĥ2 dt ... 1 − Ĥn dt
n→∞
h̄
h̄
h̄
#
"
2
i
i
i
i
1 − Ĥ1 dt 1 − Ĥ2 dt = 1 − (Ĥ1 + Ĥ2 )dt +
Ĥ1 dtĤ2 dt
h̄
h̄
h̄
h̄
i
i
i
1 − Ĥ1 dt 1 − Ĥ2 dt 1 − Ĥ3 dt =
h̄
h̄
h̄
"
i
1 − (Ĥ1 + Ĥ2 + Ĥ3 )dt +
h̄
#
2 h
i i 3
i
Ĥ1 dt(Ĥ2 + Ĥ3 )dt + Ĥ2 dtĤ3 dt −
Ĥ3 dtĤ2 dtĤ1 dt
h̄
h̄
Z t
2 Z t
Z t1
i
i
Û (t) = 1 + −
Ĥ(t1 ) dt1 + −
Ĥ(t1 ) dt1
Ĥ(t2 ) dt2 + ...
h̄
h̄
0
0
0
n Z t
Z t1
Z tn−1
i
Uˆn = −
Ĥ(t1 ) dt1
Ĥ(t2 ) dt2 ...
Ĥ(tn ) dtn
h̄
0
0
0
[Ĥ(t1 ), Ĥ(t2 )] = 0:
1
Uˆn =
n!
i
−
h̄
n Z
t
0
0
Ĥ(t ) dt
0
1
n
i
Û (t) = exp −
h̄
Z
t
dt Ĥ(t )
0
0
0
Problem 4.3
Time-independent observable:
i
d
hAi = hψ(t)|[Ĥ, Â]|ψ(t)i
dt
h̄
Ĥ |Ei = E |Ei
hE|[Ĥ, Â]|Ei = hE|Ĥ Â − ÂĤ|Ei = hE|Ĥ Â|Ei − hE|ÂĤ|Ei
d
hAi = hE|E Â|Ei − hE|ÂE|Ei = E hE|Â|Ei − E hE|Â|Ei = 0
dt
Problem 4.4
µ̂ = −
gq
Ŝ
2mc
~ = B0 ı̂
B
~ = ge Sˆx B0 = ω0 Sˆx
Ĥ = −µ̂ · B
2mc
ˆ
ˆ
Û (t) = e−iĤt/h̄ = e−iω0 Sx t/h̄ = e−iSx φ/h̄ = R̂(φı̂)
| h−z|R̂(φî)| + zi |2 =
1
4
φ
φ
1
| h−z| cos
|+zi − i sin
|−zi i | =
2
2
2
1
φ
sin
=
2
2
φ=
π
3
t=
l0
v0
ω0
l0
π
=
v0
3
l0 =
2
πv0
3ω0
Problem 4.5
~ = B0 sin θî + B0 cos θẑ
B
ω0
~ = ω0 (Sˆx sin θ + Sˆz cos θ) = ω0 Sˆn (φ = 0)
Ŝ · B
B0
Below we use φ as defined by ω0 t:
2
2
iφ
iφ ˆ
1
−iφSˆn /h̄
−
Û (t) = e
=1+ −
Sn +
Sˆn + ...
h̄
2!
h̄
"
#
2
3
φ
1 φ
φ
φ
1 φ
+ ... + (−iσn )
−
+ ... = cos
− iσn sin
=1−
2! 2
2
3! 2
2
2
cos θ
sin θ
σn →
sin θ − cos θ
Sz
cos(φ/2) − i sin(φ/2) cos θ
−i sin(φ/2) sin θ
Û (t) →
−i sin(φ/2) sin θ
cos(φ/2) + i sin(φ/2) cos θ
Sz
Ĥ =
1
i
h+y| = √ h+z| − √ h−z|
2
2
φ
φ
φ
Û (t) |+zi = cos
− i sin
cos θ |+zi − i sin
sin θ |−zi
2
2
2
φ
φ
φ
1
1
− i sin
cos θ − √ sin
sin θ
h+y|Û (t)| + zi = √ cos
2
2
2
2
2
h̄
1 − sin(ω0 T ) sin θ
P Sy =
= | h+y|Û (T )| + zi |2 =
2
2
θ = 0:
P =
1
2
θ = π/2:
P =
1 − sin(ω0 T = π2 )
=0
2
Problem 4.6
|ψ(t)i =
e−iω0 t/2
eiω0 t/2
√
|+zi + √ |−zi
2
2
hSz i = 0
d
i
∂ Sˆz
hSz i = hψ(t)|[Ĥ, Sˆz ]|ψ(t)i + hψ(t)|
|ψ(t)i
dt
h̄
∂t
3
∂ Sˆz
=0
∂t
Ĥ = ω0 Sˆz :
[Ĥ, Sˆz ] = 0
hSx i =
h̄
cos ω0 t
2
d
i
∂ Sˆx
hSx i = hψ(t)|[Ĥ, Sˆx ]|ψ(t)i + hψ(t)|
|ψ(t)i
dt
h̄
∂t
∂ Sˆx
=0
∂t
Ĥ = ω0 Sˆz :
ω0 h̄2
ω0 h̄2
[Ĥ, Sˆx ] =
[σz , σx ] =
(2iσy ) = ω0 h̄iSˆy
4
4
d
hSx i = −ω0 hψ(t)|Sˆy |ψ(t)i
dt
−iω0 t/2
iω0 t/2
ih̄
e
e
√
Sˆy |ψ(t)i =
|−zi − √ |+zi
2
2
2
−ω0 ih̄
−ω0 hψ(t)|Sˆy |ψ(t)i =
2
−ω0 ih̄
=
2
−eiω0 t + e−iω0 t
2
iω0 t/2
eiω0 t/2
e−iω0 t/2
e
e−iω0 t/2
√ h+z| + √
h−z|
− √ |+zi + √
|−zi
2
2
2
2
−ω0 ih̄
=
2
d ˆ
d
hSx i =
dt
dt
− cos ω0 t − i sin ω0 t + cos ω0 t − i sin ω0 t
2
h̄
ω0 h̄
cos ω0 t = −
sin ω0 t
2
2
hSy i =
h̄
sin ω0 t
2
d
i
∂ Sˆy
hSy i = hψ(t)|[Ĥ, Sˆy ]|ψ(t)i + hψ(t)|
|ψ(t)i
dt
h̄
∂t
∂ Sˆy
=0
∂t
Ĥ = ω0 Sˆz :
4
=−
ω0 h̄
sin ω0 t
2
[Ĥ, Sˆx ] =
ω0 h̄2
ω0 h̄2
[σz , σy ] =
(−2iσx ) = −ω0 h̄iSˆx
4
4
d
hSy i = ω0 hψ(t)|Sˆx |ψ(t)i
dt
h̄ e−iω0 t/2
eiω0 t/2
ˆ
√
Sx |ψ(t)i =
|−zi + √ |+zi
2
2
2
ω0 h̄
ω0 hψ(t)|Sˆx |ψ(t)i =
2
ω0 h̄
=
2
eiω0 t + e−iω0 t
2
iω0 t/2
eiω0 t/2
e−iω0 t/2
e
e−iω0 t/2
√ h+z| + √
√ |+zi + √
h−z|
|−zi
2
2
2
2
ω0 h̄
=
2
cos ω0 t + i sin ω0 t + cos ω0 t − i sin ω0 t
2
d ˆ
d
hSy i =
dt
dt
=
ω0 h̄
cos ω0 t
2
ω0 h̄
h̄
sin ω0 t =
cos ω0 t
2
2
Problem 4.7 - Skipped
Problem 4.8
|ψ(0)i = cos
|ψ(t)i = e−iω0 t/2 cos
θ
θ
|+zi + sin |−zi
2
2
θ
θ
|+zi + eiω0 t/2 sin |−zi
2
2
hSx i = hψ(t)|Sˆx |ψ(t)i
θ
θ
θ
θ
iω0 t/2
−iω0 t/2
−iω0 t/2
iω0 t/2
ˆ
cos h+z| + e
= e
sin h−z| Sx e
cos |+zi + e
sin |−zi
2
2
2
2
h̄
=
2
θ
θ
θ
θ
iω0 t/2
−iω0 t/2
−iω0 t/2
iω0 t/2
e
cos h+z| + e
sin h−z|
e
cos |−zi + e
sin |+zi
2
2
2
2
h̄
=
2
θ
θ
θ
θ
h̄
θ
θ
iω0 t
−iω0 t
e sin cos + e
sin cos
= sin cos (2 cos ω0 t)
2
2
2
2
2
2
2
hSx i =
h̄
sin θ cos ω0 t
2
hSy i = hψ(t)|Sˆy |ψ(t)i
5
θ
θ
θ
θ
iω0 t/2
−iω0 t/2
−iω0 t/2
−iω0 t/2
ˆ
= e
cos h+z| + e
sin h−z| Sy e
sin |−zi
cos |+zi + e
2
2
2
2
ih̄
=
2
θ
θ
θ
θ
iω0 t/2
−iω0 t/2
−iω0 t/2
iω0 t/2
e
cos h+z| + e
sin h−z|
e
cos |−zi − e
sin |+zi
2
2
2
2
ih̄
=
2
−iω0 t
e
θ
θ
θ
θ
sin cos − eiω0 t sin cos
2
2
2
2
hSy i =
=
ih̄
θ
θ
sin cos (−2i sin ω0 t)
2
2
2
h̄
sin θ sin ω0 t
2
hSz i = hψ(t)|Sˆz |ψ(t)i
θ
θ
θ
θ
−iω0 t/2
−iω
t/2
−iω
t/2
iω0 t/2
sin h−z| Sˆz e 0 cos |+zi + e 0 sin |−zi
= e
cos h+z| + e
2
2
2
2
h̄
=
2
θ
θ
θ
θ
−iω0 t/2
−iω0 t/2
iω0 t/2
iω0 t/2
sin h−z|
e
cos |+zi − e
sin |−zi
e
cos h+z| + e
2
2
2
2
h̄
=
2
h̄
2 θ
2 θ
cos − sin
= cos θ
2
2
2
hSz i =
h̄
cos θ
2
Problem 4.9
Ĥ = ω0 Sˆz + ω1 (cos ωt)Sˆx
|ψ(0)i = |+zi
d |ψ(t)i
dt
h̄
ȧ(t)
ω0
ω1 cos ωt
a(t)
= ih̄
ω1 cos ωt
−ω0
b(t)
ḃ(t)
2
Ĥ |ψ(t)i = ih̄
Approximation: B1 B0 , ω1 ω0
a(t)
c(t)e−iω0 t/2
=
b(t)
d(t)eiω0 t/2
6
Approximation: ω ≈ ω0
ω1 i(ω0 −ω)t
e
d(t)
4
˙ = ω1 ei(ω−ω0 )t c(t)
id(t)
4
iċ(t) =
ω1
˙
[i(ω0 − ω)ei(ω0 −ω)t d(t) + ei(ω0 −ω)t d(t)]
4
¨ = ω1 [i(ω − ω0 )ei(ω−ω0 )t c(t) + ei(ω−ω0 )t ċ(t)]
id(t)
4
ω1
i(ω0 −ω)t iω1 i(ω−ω0 )t
i(ω0 −ω)t 4i i(ω−ω0 )t
ic̈(t) =
i(ω0 − ω)e
e
ċ(t) − e
e
c(t)
4
ω1
4
4i
ω1
ω1
(ω0 − ω) ċ(t) − c(t)
c̈(t) =
4
ω1
4
ω 2
1
c̈(t) = i(ω0 − ω)ċ(t) −
c(t)
4
ω 2
1
c̈ − i(ω0 − ω)ċ +
c=0
4
Characteristic equation:
ω 2
1
=0
r2 + [i(ω − ω0 )]r +
4
p
i(ω0 − ω) ± i (ω0 − ω)2 + (ω1 2 /4)
r=
2
!
p
p
2 + (ω 2 /4)
2 + (ω 2 /4)
(ω
−
ω)
(ω
−
ω)
0
1
0
1
t + B cos
t
c(t) = ei(ω0 −ω)t/2 A sin
2
2
ic̈(t) =
c(0) = 1, d(0) = 0:
i(ω − ω0 )
A= p
(ω0 − ω)2 + (ω1 2 /4)
B=1
c(t) = ei(ω0 −ω)t/2
ċ(t) = e
p
i(ω − ω0 )
p
sin
(ω0 − ω)2 + (ω1 2 /4)
i(ω0 −ω)t/2
(ω0 − ω)2 + (ω1 2 /4)
t + cos
2
p
ω1 2 /4
p
sin
2 (ω0 − ω)2 + (ω1 2 /4)
7
!
p
(ω0 − ω)2 + (ω1 2 /4)
t
2
!
(ω0 − ω)2 + (ω1 2 /4)
t
2
d(t) = ie−i(ω0 −ω)t/2
p
ω1 /2
p
sin
(ω0 − ω)2 + (ω1 2 /4)
!
(ω0 − ω)2 + (ω1 2 /4)
t
2
ω1 2 /4
sin2
| h−z|ψ(t)i | = b (t)b(t) = d (t)d(t) =
2
2
(ω0 − ω) + (ω1 /4)
∗
2
∗
p
(ω0 − ω)2 + (ω1 2 /4)
t
2
Problem 4.10
1
1
|Ii = √ |1i + √ |2i
2
2
1
1
|IIi = √ |1i − √ |2i
2
2
1
1
h1|Ii = √ , h2|Ii = √
2
2
1
1
h1|IIi = √ , h2|IIi = − √
2
2
hI|Ĥ|Ii =
X
hI|ai ha|Ĥ|a0 i ha0 |Ii =
1,2
h1|Ĥ|1i + h1|Ĥ|2i + h2|Ĥ|1i + h2|Ĥ|2i
= E0 − A
2
hI|Ĥ|IIi =
h1|Ĥ|1i − h1|Ĥ|2i + h2|Ĥ|1i − h2|Ĥ|2i
~
= µe |E|
2
hII|Ĥ|Ii =
h1|Ĥ|1i + h1|Ĥ|2i − h2|Ĥ|1i − h2|Ĥ|2i
~
= µe |E|
2
h1|Ĥ|1i − h1|Ĥ|2i − h2|Ĥ|1i + h2|Ĥ|2i
= E0 + A
2
~ 0 | cos ωt
E0 − A
µe |E
Ĥ →
~ 0 | cos ωt
I,II
µe |E
E0 + A
hII|Ĥ|IIi =
Compare with:
h̄
Ĥ →
+z,−z 2
ω0
ω1 cos ωt
ω1 cos ωt
−ω0
Analogous Hamiltonian:
E+ = E0 + A
E− = E0 − A
8
~ 0 | cos ωt
E+
µe |E
Ĥ →
~ 0 | cos ωt
II,I
µe |E
E−
~ 0|
µe |E
ċ(t)
d(t)ei(E+ −E− )t/h̄
=
i
cos ωt
˙
c(t)e−i(E+ −E− )t/h̄
d(t)
h̄
+z → II, −z → I
E+ =
h̄ω0
→ E0 + A
2
E− = −
h̄ω0
→ E0 − A
2
E+ − E− = h̄ω0 → 2A
h̄ω1
~ 0|
→ µe |E
2
Analogue of Rabi’s formula:
ψ(0) = |IIi
q
~ 0 |2 /h̄2
µe |E
2
2
| hI|ψ(t)i | =
2
~ 0 |2 /h̄2
(2A/h̄ − ω)2 + µe 2 |E
sin
~ 0 |2 /h̄2
(2A/h̄ − ω)2 + µe 2 |E
2
Problem 4.11
gq ~
µ
~=
S
2mc
~ = B0 k̂
B
~ = − gq B0 Sˆz = ω0 Sˆz
Ĥ = −~µ · B
2mc
Ĥ |1, 1i = ω0 Sˆz |1, 1i = h̄ω0 |1, 1i = E1 |1, 1i
Ĥ |1, 0i = ω0 Sˆz |1, 0i = 0ω0 |1, 0i = E0 |1, 0i
Ĥ |1, −1i = ω0 Sˆz |1, −1i = −h̄ω0 |1, −1i = E−1 |1, −1i
√
i 2
1
1
|ψ(0)i = |1, 1iy = |1, 1i +
|1, 0i − |1, −1i
2
2
2
!
√
1
i
2
1
|ψ(t)i = e−iĤt/h̄
|1, 1i +
|1, 0i − |1, −1i
2
2
2
9
t
√
e−iE1 t/h̄
i 2e−iE0 t/h̄
e−iE−1 t/h̄
=
|1, 1i +
|1, 0i −
|1, −1i
2
2
2
√
e−iω0 t
i 2
eiω0 t
|ψ(t)i =
|1, 1i +
|1, 0i −
|1, −1i
2
2
2
√
1
2
1
|1, 1ix = |1, 1i +
|1, 0i + |1, −1i
2
2
2
√
√
2
2
|1, 0ix =
|1, 1i −
|1, −1i
2
2
√
1
2
1
|1, −1ix = |1, 1i −
|1, 0i + |1, −1i
2
2
2
√
1
i 2
1
|1, 1iy = |1, 1i +
|1, 0i − |1, −1i
2
2
2
√
√
2
2
|1, 0iy =
|1, 1i +
|1, −1i
2
2
√
1
i 2
1
|1, −1iy = |1, 1i −
|1, 0i − |1, −1i
2
2
2
2
i
(1 − sin ω0 t)2
e−iω0 t − eiω0 t
+
=
| h1, 1|x |ψ(t)i | =
4
2
4
√ 2
cos2 ω0 t
(e−iω0 t + eiω0 t ) 2
2
| h1, 0|x |ψ(t)i | =
=
4
2
2
i
e−iω0 t − eiω0 t
| h1, −1|x |ψ(t)i | =
−
4
2
2
2
hSx i =
=
(1 − sin ω0 t)2
(1 + sin ω0 t)2
(h̄) + 0 +
(−h̄) = −h̄ sin ω0 t
4
4
| h1, 1|y |ψ(t)i |2 =
e−iω0 t + eiω0 t 1
+
4
2
2
=
√
−iω0 t
iω0 t
(e
−
e
)
2
| h1, 0|y |ψ(t)i |2 =
4
| h1, −1|y |ψ(t)i |2 =
hSy i =
(1 + sin ω0 t)2
4
e−iω0 t + eiω0 t 1
−
4
2
(1 + cos ω0 t)2
4
2
=
2
=
sin2 ω0 t
2
(1 − cos ω0 t)2
4
(1 + cos ω0 t)2
(1 − cos ω0 t)2
(h̄) + 0 +
(−h̄) = h̄ cos ω0 t
4
4
10
1
1
hSz i = (h̄) + 0 + (−h̄) = 0
4
4
Problem 4.12
Ĥ = ω0 Sˆx
√
2
1
1
|1, 0ix + |1, −1ix
|ψ(0)i = |1, 1i = |1, 1ix +
2
2
2
√ −iE t/h̄
2e 0
e−iE1 t/h̄
e−iE−1 t/h̄
|ψ(t)i =
|1, 1ix +
|1, 0ix +
|1, −1ix
2
2
2
√
e−iω0 t
2
eiω0 t
=
|1, 1ix +
|1, 0ix +
|1, −1ix
2
2
2
√
1
2
1
h1, −1| = h1, 1|x −
h1, 0|x + h1, −1|x
2
2
2
| h1, −1|ψ(t)i |2 =
e−iω0 t + eiω0 t 1
−
4
2
2
=
(1 − cos ω0 t)2
4
Problem 4.13

E0

0
Ĥ →
1,2,3
A
0
E1
0

A
0 
E0
Find eigenstates:
E0 − λ
0
A
0
E1 − λ
0
A
0
E0 − λ
= (E0 − λ)(E1 − λ)(E0 − λ) − A2 (E1 − λ) = 0
(λ − E0 )2 − A2 = 0
λ = E1 , E0 ± A



E0 − λ
0
A
a

 b  = 0
0
E1 − λ
0
A
0
E0 − λ
c
λ = E1 :
(E0 − E1 )a + Ac = 0
Aa + (E0 − E1 )c = 0
a = c = 0, b = 1
|E1 i = |2i
11
λ = E0 + A:
−Aa + Ac = 0
(E1 − E0 − A)b = 0
Aa − Ac = 0
1
a = c = √ ,b = 0
2
1
1
|E0 + Ai = √ |1i + √ |3i
2
2
λ = E0 − A:
Aa + Ac = 0
(E1 − E0 + A)b = 0
Aa + Ac = 0
1
a = −c = √ , b = 0
2
1
1
|E0 − Ai = √ |1i − √ |3i
2
2
a)
|ψ(0)i = |2i
|ψ(t)i = e−iĤt/h̄ |ψ(0)i = e−iE1 t/h̄ |E1 i
b)
1
1
|ψ(0)i = |3i = √ |E0 + Ai − √ |E0 − Ai
2
2
|ψ(t)i = e−iĤt/h̄ |ψ(0)i =
e−i(E0 +A)t/h̄
e−i(E0 −A)t/h̄
√
√
|E0 + Ai −
|E0 − Ai
2
2
Problem 4.14
0 −iE0
Ĥ →
iE0
0
x,y
h̄ω0
0 −i
Ĥ →
i 0
x,y 2
1
i
|+h̄ω0 /2i = |+yi = √ |+zi + √ |−zi
2
2
1
i
|−h̄ω0 /2i = |−yi = √ |+zi − √ |−zi
2
2
a) By analogy, the eigenstates and eigenvalues are:
12
1
|+E0 i = √ |xi +
2
1
|−E0 i = √ |xi −
2
i
√ |yi
2
i
√ |yi
2
b)
1
1
|ψ(0)i = |xi = √ |+E0 i + √ |−E0 i
2
2
|ψ(t)i = e−iĤt/h̄ |ψ(0)i =
e−iE0 t/h̄
eiE0 t/h̄
√
|+E0 i + √ |−E0 i
2
2
e−iE0 t/h̄ + eiE0 t/h̄
| hx|ψ(t)i | =
2
2
2
= cos2
E0 t
h̄
i
i
|yi = − √ |+E0 i + √ |−E0 i
2
2
−ie−iE0 t/h̄ + ieiE0 t/h̄
| hy|ψ(t)i | =
2
2
2
= sin2
E0 t
h̄
The photon polarization is oscillating between the x and y states.
Problem 4.15
[Â, B̂] = iĈ → ∆A∆B ≥
|hCi|
2
h[Â, B̂]i = ihĈi
|hĈi| = |h[Â, B̂]i|
dhAi
i
i
= hψ(t)|[Ĥ, Â]ψ(t)i = h[Ĥ, Â]i
dt
h̄
h̄
|h[Ĥ, Â]i| = h̄|dhAi/dt|
|h[Ĥ, Â]i|
2
∆A
h̄
∆E
≥
|dhAi/dt|
2
∆E∆A ≥
∆t is time needed for expected value of observable to change significantly.
Problem 4.16 - Skipped
13