PHYS 40202/6402 Advanced QM

Examples 3
(Starred questions are MSc coursework)

These exercises cover the section of the course on time-dependent perturbation

theory.
1. * A one-dimensional harmonic oscillator of mass m and angular frequency ω
in its ground state is subject to a small constant force F acting for a time
interval τ . What value of τ gives the greatest chance that the oscillator will be
found in its first excited state thereafter? The ground and first excited state
wavefunctions of the oscillator are
s
mωx2
1/4 !
mω 2mω

φ0 (x) = exp − , φ1 (x) = xφ0 (x)
πh̄ 2h̄ h̄
respectively.
2. * (a) The time-dependent Hamiltonian Ĥ = Ĥ0 + V̂ (t) with
eB0 eB1  
Ĥ0 = Ŝz , V̂ (t) = Ŝx cos ωt + Ŝy sin ωt
m m
describes a spin S = 1/2 system which is subject to the static magnetic field
(0, 0, B0 ) and the rotating magnetic field (B1 cos ωt, B1 sin ωt, 0). B0 and B1
are uniform throughout space. Initially, at time t = 0, the spin of the system
is pointing in the direction of the negative z-axis. Assuming B1 is weak and
taking V̂ (t) as perturbation, calculate the probability that at time t > 0 the
spin points in the positive z-axis.
(b) Discuss at what value of ω (i.e. resonance), the perturbation theory breaks
down for sufficiently large t, however weak B1 . Suggest a criterion in t for the
perturbation theory to hold.
(c) Solve exactly the time-dependent Schrödinger equation at the resonance for
the given initial condition. Discuss the relation between the exact results and
the first-order perturbation results obtained earlier.
3. A hydrogen atom is placed in a uniform magnetic field B in z direction. The
corresponding weak-field Zeeman energy shift is given by
∆E = gµB BMJ
where the Landé g factor is defined by
J(J + 1) + S(S + 1) − L(L + 1)
g =1+ ,
2J(J + 1)

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 with usual notations for angular momentum quantum numbers MJ , J, L, and
S. List the weak-field Zeeman shifts for all the n = 1 and n = 2 states of
hydrogen and draw a level diagram. Mark all the transitions consistent with
electric dipole selection rules.

4. * Which of the following transitions in carbon are electric dipole transitions?

(a) (2s)2 (2p)(3d) 3 D → (2s)2 (2p)2 3 P

(b) (2s)2 (2p)(3s) 3 P → (2s)2 (2p)2 1 S
(c) (2s)2 (2p)(3d) 1 D → (2s)2 (2p)(3s) 1 P
(d) (2s)(2p)3 3 D → (2s)2 (2p)2 3 P
(e) (2s)2 (2p)(3p) 3 P → (2s)2 (2p)2 3 P
(f) (2s)2 (2p)(3d) 1 D → (2s)2 (2p)2 1 S

5. The K1 and K2 mesons have slightly different masses m1 and m2 and lifetimes
τ1 = 0.9 × 10−10 s and τ2 = 0.5 × 10−7 s respectively. The K 0 and K̄ 0 mesons
ar the superpositions
1 1
|K 0 i = √ (|K1 i + |K2 i), |K̄ 0 i = √ (|K1 i − |K2 i).
2 2

A K 0 meson is produced in the process π − + p → Λ0 + K 0 at time t = 0. Show

that the probability that at a time t the stationary (relativistic) K 0 meson has
turned into a K̄ 0 meson is
(m1 − m2 )c2 t
!
1 −t/τ1 1 1 t
   
P = e + e−t/τ2 − 2 exp − + cos .
4 τ1 τ2 2 h̄

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 PHYS 40202 Advanced QM
Examples 3 Solutions

1. We write the perturbation potential as

(
−F x, 0<t<τ
V̂ (t) =
0, otherwise.

The probability amplitude for the transition from the ground state

mωx2
1/4 !
mω

|0i → φ0 = exp −
πh̄ 2h̄

of the oscillator to its first excited state

s
2mω
|1i → φ1 = xφ0
h̄
is
−F Z τ
c (τ ) = h1| x |0i dt eiωt
ih̄ s 0

−F h̄ eiωτ − 1
= · ·
ih̄ 2mω iω
where we have done the integral
s s
2mω Z ∞ h̄
h1| x |0i = dx x2 φ20 = .
h̄ −∞ 2mω
Therefore
2F 2 ωτ
 
|c (τ )|2 = 3
sin2
mh̄ω 2
2F 2
and this probability attains its maximum value of mh̄ω 3
at ωτ = (2n − 1) π,
with n = 1, 2, 3, · · · .

2. (a) Let |↑i and |↓i be the spin-up and spin-down eigenstates of Ŝz with eigen-
values
h̄ h̄
Ŝz |↑i = |↑i , Ŝz |↓i = − |↓i .
2 2
Hence the eigenvalues for the Hamiltonian Ĥ0 = eB 0
m z
Ŝ are E↑ = eB0 h̄/2m and
E↓ = −eB0 h̄/2m. At t = 0, the system is in the state |↓i. According to the

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 (1)
first-order perturbation theory, the probability amplitude c↑ (t) that at time
t > 0 the system is in the spin-up state |↑i is given by

(1) 1 Zt E↑ − E↓ eB0
c↑ (t) = V↑↓ (t0 ) exp (iω↑↓ t0 ) dt0 , ω↑↓ = =
ih̄ 0 h̄ m
where
eB1  
V↑↓ (t0 ) = h↑| V (t0 ) |↓i = h↑| Ŝx cos ωt + Ŝy sin ωt |↓i .
m
Using
1 +  1  + 
Ŝx = Ŝ + Ŝ − , Ŝy = Ŝ − Ŝ −
2 2i
and

Ŝ + |↑i = Ŝ − |↓i = 0
Ŝ + |↓i = h̄ |↑i , Ŝ − |↑i = h̄ |↓i

we have
h̄eB1
V↑↓ (t0 ) = exp (−iωt0 ) .
2m
We can also express spin operators Ŝi in terms of Pauli matrices Ŝi = h̄2 σi , and
the eigenstates ! !
1 0
| ↑i = , | ↓i = ,
0 1
and then use matrix calculations for evaluation of h↑| V (t0 ) |↓i to obtain the
same results. Hence
−ieB1 /m −i(ω−eB0 /m)/2 1
 
(1)
c↑ (t) = e sin (ω − eB0 /m) t
ω − eB0 /m 2
and the probability is

(eB1 /m)2 2 1
 2  
 (1)
c↑ (t) = 2 sin (ω − eB0 /m) t

(ω − eB0 /m) 2

(b) At the resonant frequency, namely ω = ω↑↓ = eB0 /m, we have

(1) 1 Zt eB1
c↑ (t) = V↑↓ (t0 ) exp (iω↑↓ t0 ) dt0 = t
ih̄ 0 2im
and

 (1)
2 e2 B12 2
c↑ (t) = t .

4m2
4
  2
 (1)
For t > 2 |m/eB1 |, the probability c↑ (t) is greater than one, which is obvi-


ously nonsense. For perturbation theory to hold, we must have

m 
 

t  2  .
eB1 
(c) Substitute
e−iEσ t/h̄ cσ (t)|σi
X
Ψ=
σ=↑,↓

into the time-dependent Schrödinger equation

∂
ih̄ |Ψi = (Ĥ0 + V̂ )|Ψi
∂t
and using the eigen equations Ĥ0 |σi = Eσ |σi, we have
!
∂cσ −iEσ t/h̄
V̂ cσ e−iEσ t/h̄ |σi.
X X
ih̄ e |σi =
σ ∂t σ

Take inner product from LHS with h↑ | and with h↓ |, we obtain, respectively
∂c↑ h̄eB1 i(eB0 /m−ω)t
ih̄ = h↑ |V̂ | ↑ic↑ + h↑ |V̂ | ↓iei(E↑ −E↓ )t/h̄ c↓ = e c↓ ;
∂t 2m
∂c↓ h̄eB1 −i(eB0 /m−ω)t
ih̄ = h↓ |V̂ | ↑iei(E↓ −E↑ )t/h̄ c↑ + h↓ |V̂ | ↓ic↓ = e c↑ ,
∂t 2m
where we have used
h↑ |V̂ | ↑i = h↓ |V̂ | ↓i = 0.
At the resonance, ω = eB0 /m, substitute the first into second equation, we have
2
d 2 c↑ eB1

= − c↑ .
dt2 2m
With the initial condition c↑ (0) = 0, we have the physical solution
eB1 t
c↑ (t) = sin ,
2m
with the first order result for the probability (eB1 t/2m)2 , in agreement with the
first-order perturbation result obtained earlier in part (b).

3. The energy shift for a general state in the weak field limit is glsj µB Bmj . Thus,
both (1s) and (2s) states, or the term symbol 2 S1/2 , are shifted by ±µB B, the
2
P1/2 states shifted by ±µB B/3 and the 2 P3/2 states shifted by ±2µB B/3 or

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 ±2µB B (depending on mj ). Taking E1/2 and E3/2 to be the energies in the
absence of the external magnetic field, the transitions are, in the order shown
in the diagram,

E3/2 + µB B, E3/2 − µB B/3, E3/2 + 5µB B/3, E3/2 − 5µB B/3, E3/2 + µB B/3,
E3/2 − µB B, E1/2 − 2µB B/3, E1/2 + 4µB B/3, E1/2 − 4µB B/3, E1/2 + 2µB B/3

mj ∆Ε/(µΒB)
2 3/2 2
P3/2 1/2 2/3
−1/2 −2/3
−3/2 −2
n=2 2
P1/2 1/2 1
1/2 1/3
2 −1/2 −1/3
S1/2 −1/2 −1

2
S1/2 1/2 1
n=1
−1/2 −1

Note that there are no transitions from the n = 2 2 S1/2 state, since it cannot
decay via an electric dipole transitions.

4. According the dipole selection rules, the allowed transitions are (a) and (d).
Note that in (b), ∆S 6= 0, in (c) and (e), the parities of the initial and final
states are the same; and ∆L = 2 for (f).

5. The general state of the K 0 meson can be written as

|K 0 i = a1 (t)|K1 i + a2 |K2 i

The initial condition is

1
a1 (0) = a2 (0) = √ .
2
For t > 0, we write the decaying coefficients

aj (t) = aj (0) exp(−imj c2 t/h̄ − t/2τj ), j = 1, 2

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The probability amplitude hK̄ 0 |K 0 i is then determined as
 
1 1 X
hK̄ 0 |K 0 i = √ (hK1 | − hK2 )  √ exp(−imj c2 t/h̄ − t/2τj )|Kj i
2 2 j=1,2
1 1
= exp(−im1 c2 t/h̄ − t/2τ1 ) − exp(−im2 c2 t/h̄ − t/2τ2 )
2 2
where we have used the fact hKi |Kj i = δij . Hence the probability is

(m1 − m2 )c2 t
!
1 −t/τ1 1 1 t
   
P = e + e−t/τ2 − 2 exp − + cos .
4 τ1 τ2 2 h̄