Quantum Mechanics PHYS 212B
Problem Set 4
Due Tuesday, February 9, 2016
Exercise 4.1
Consider a potential
V cos ωt =
V −iωt
e
+ eiωt
2
turned on slowly with a factor eα5 . Use the perturbation expansion developed in class and take α → 0 to
calculate the first-order transition probability between initial state |0i and |n|i, where as before will take
|n|i to be a continuum of states.
Solution 4.1
The first-order perturbation
Z
0
0
0
1 t
1 −iωt0
hn|ψt |i =
dt0 ei(n −0 )t /~
e
+ eiωt eαt hn|V |0i
i~ −∞
2
Z t
i
h
0
it0
it
hn|V |0i
=
dt0 e ~ (n −0 −~ω)+αt + e ~ (n −0 +~ω)+αt
2i~
−∞
"
#
it
it
αt
e hn|V |0i
e ~ (n −0 +~ω)
e ~ (n −0 −~ω)
=
+
2
0 − n + ~ω + iα~ 0 − n − ~ω + iα~
Let α = 0 and the probability is
|hn|V |0i|2
1
1
2 cos(2ωt)
P (|ni = |hn|ψt |i| =
+
+
4
(0 − n + ~ω)2
(0 − n − ~ω)2
(0 − n )2 − ~2 ω 2
2
Exercise 4.2
We can regard the vector potential as an operator at some spacetime point (rt):
−ik·r+iωt
X
eik·r−iωt
† ∗e
+ Akˆ ˆ √
A(rt) =
Akˆ ˆ √
Vol
Vol
kˆ
where k is the wavenumber and ˆ is the
vector
and we
h polarization
i
h
i can regard the coefficient Akˆ , etc. as
†
†
†
0 ˆ
operators. Show that [Akˆ , Ak0 ˆ0 ] = 0, Akˆ , Ak0 ˆ0 = 0, Akˆ , Ak0 ˆ0 = 2π~c
ˆ0∗ .
|k| δkk Solution 4.2
The canonical momentum of free EM field is (Derivation can be found here)
1
1
1 ∂A(rt)
p(rt) = −
E(rt) = −
−
4πc
4πc
c ∂t
ik·r−iωt
−ik·r+iωt
X
iω
e
† ∗e
√
√
=−
A
ˆ
−
A
ˆ
kˆ
kˆ
4πc2
Vol
Vol
kˆ
Based on the fundamental commutation relation
[A(rt), p(r0 t)] = i~δ(r − r0 ),
1
(1)
we have
"
"
##
ik0 ·r0 −iωt
−ik0 ·r0 +iωt
−ik·r+iωt
X
iω X
eik·r−iωt
† ∗e
†
0e
‘∗ e
√
Ak0 ˆ0 ˆ √
[A(rt), p(r t)] =
+ Akˆ ˆ √
,−
− Ak0 ˆ0 ˆ
Akˆ ˆ √
4πc2 0 0
Vol
Vol
Vol
Vol
k ˆ
kˆ
n
h
i
X
X
0 0
0 0
iω
[Akˆ , Ak0 ˆ0 ] ˆˆ0 eik·r+ik ·r −2iωt − A†kˆ , A†k0 ˆ0 ˆ∗ ˆ‘∗ e−ik·r−ik ·r +2iωt
=−
4πc2 Vol
kˆ
k0 ˆ0
i
h
i
o
h
0 0
0 0
− Akˆ , A†k0 ˆ0 ˆˆ‘∗ eik·r−ik ·r + A†kˆ , Ak0 ˆ0 ˆ∗ ˆ0 e−ik·r+ik ·r
i
h
XXn
0 0
iω
†
†
0 ik·r+ik0 ·r0 −2iωt
0
0] =−
ˆ∗ ˆ‘∗ e−ik·r−ik ·r +2iωt
[A
,
A
ˆ
ˆ
e
−
A
,
A
0
0 kˆ
k
ˆ
kˆ
k
ˆ
4πc2 Vol
kˆ
k0 ˆ0
h
i
o
0 0
−2 Akˆ , A†k0 ˆ0 ˆˆ‘∗ eik·r−ik ·r
0
The RHS of Eq.(1) doesn’t contains t, so the coefficients of e−2iωt should be zero,
[Akˆ , Ak0 ˆ0 ] = 0,
h
i
A†kˆ , A†k0 ˆ0 = 0.
We know
δ(r − r0 ) =
1 X ik(r−r0 )
e
,
Vol
k
substitute it into Eq.(1), we have
i 2π~c2
δkk0 ˆˆ0∗
Akˆ , A†k0 ˆ0 =
ω
Replace ω = c|k|, we have the comutation relation we need,
h
i 2π~c
h
Akˆ , A†k0 ˆ0 =
δkk0 ˆˆ0∗
|k|
Exercise 4.3 A hydrogen atom in its ground state is placed between the plates of a capacitor and at time
t = 0 a time-dependent, but spatially uniform, electric filed E = E0 e−t/τ is applied. Take E0 to be in the
positive z-direction. What is the probability for the atom to be found in each of the three 2p states when
t τ?
Solution 4.3
The perturbation Hamiltonian is
H 0 = er · E = ezE0 e−t/τ
We know [z, Lz ] = 0, so
0 = hnf , lf , mf |[z, Lz ]|ni , li , mi i
= (mi − mf )~hnf , lf , mf |z|ni , li , mi i
Therefore, the selection rule for z operator is ∆m = 0,
h2, 1, ±1|H 0 |100i = 0
2
and
h210|H 0 |100i =
Z
d3 rψ210 H 0 ψ100
1
= eE0 e−t/τ √
4 2πa40
=
Z
+∞
Z
π
Z
2π
dφe−3r/2a0 r4 cos2 θ sin θ
dθ
dr
0
0
0
eE0 e−t/τ 4π 4!
√
5
4 2πa40 3
3
2a0
=
where
Z
0
π
2
15/2
a0 e
E0 e−t/τ
35
2
dθ cos θ sin θ = ,
3
2
+∞
Z
0
dre−3r/2a0 r4 = 4!
3
2a0
5 .
Then
1
i~
Z
+∞
it
h210|H 0 |100ie ~ (E2 −E1 ) dt
0
Z
t
i(E2 − E − 1)
215/2 a0 eE0 +∞
exp − +
t dt
=
35 i~
τ
~
0
h210|ψt i =
=
215/2 a0 eE0
35 i~
1
1
τ
1
− i E2 −E
~
The probability to find the 2p states
P (|2, 1, ±1i) = 0
P (|210i) = |h210|ψt i|2 =
where (E2 − E1 )/~ = 3e2 /8a0 ~.
3
215 a20 e2 E02
310 ~2
1
1
τ2
+
E2 −E1 2
~