Physics 742 – Graduate Quantum Mechanics 2
Solutions to Chapter 18
3. [20] An electron is trapped in a 3D harmonic oscillator potential, H  P 2 2m  12 m02 R 2 .
It is in the quantum state nx , n y , nz  0, 0, 2 . It is going to decay directly into the
ground state 0, 0, 0
(a) [1] Convince yourself that it cannot go there via the electric dipole transition. It
can, however, go there via the electric quadrupole transition.
The position operators R only raise or lower one of the three eigenvalues, and hence
0, 0, 0 R 0, 0, 2 vanishes.
(b) [3] Calculate every non-vanishing matrix element of the form 0, 0, 0 Ri R j 0, 0, 2 .
There is only one non-vanishing component, namely
0, 0, 0 ZZ 0, 0, 2 


0, 0, 0 az2 0, 0, 2 
2m0
2m0
(c) [12] Calculate the polarized differential decay rate d  pol  002  000  d  for this
decay. This will require, among many other things, converting a sum to an integral
in the infinite volume limit.
We start with the equation after (18.5) from the notes, but keep only the electric
quadrupole term.
e
000 eik R ε*  P  i  k  ε*   S  002   12 eFI 000  ε*  R   k  R  002
m
e 20 *
e *
ε z kz  
ε z kz .

m 2
2 2m0
We now substitute this into the equation in the middle of p. 316
WFI   
1
e *

ε z kz
2 0V 
m 2
where we have substituted n = 1, since there is no photon in the initial state. Substituting this
into (18.1), we have
2
2
2 e *

2
  EF  EI 
I  F  
 FI   EF  EI  

ε z kz
 0V 

 2m

 e22 2 2
k z ε z     20 
2 0Vm 2
As usual, the rate vanishes in the large volume limit, but this is easily dealt with by summing
over the final state wave numbers and then taking the infinite volume limit.
I  F  
 e22 2 2
k z ε z     20 
2 0Vm 2

k F , F
 e2 2
d 3k F
2

k 2 ε z   ck  20 
3 z
2

2 0Vm   2  
F

e
2
16
2
k   ck  2  dk  cos
 m
4
0
2
0
F
2
 d F  ε z
0
2
F
e   20 c 
e 2 03
2
2
2

ε

cos
cos 2  d  F  ε z .
d


F
z
2
2
2
5 2 

16  0 cm  20 
2  0 c m
F
F
2


2
2
4
As we can see from the delta function, k  20 c . This, then, is the unpolarized differential
decay rate,
d  2  203
2

cos 2  ε z ,
2 4
d  m c
where we have, in typical fashion, introduced the fine structure constant,   e 2 4 0c .
(d) [4] Sum it over polarizations and integrate it over angles to determine the total
decay rate   002  000  .
If we sum it over polarizations, the only polarization that will contribute is the one
partially along the z-axis, for which we find ε z  sin  , so we have
d  2  203

cos 2  sin 2  .
d   m2c 4
Integrating this over  just gives a factor of 2, and then the remaining integral yields

4  203
16  203
2
2


cos

sin

sin

d

.
m 2 c 4 0
15m 2 c 4