Physics 742 – Graduate Quantum Mechanics 2
Solutions to Chapter 18
5. [20] An electron is trapped in a 3D harmonic oscillator potential, H  P 2 2m  12 m02 R 2 .
It is in the ground state nx , ny , nz  0, 0, 0 . Photons are fired at the electron with
frequency   0 , going in the z-direction, and polarized in the x-direction, ε  xˆ .
Calculate the differential cross section d d  (summed over final polarizations, not
integrated), and the total cross section  .
The amplitude for the scattering, according to the notes, Eq. (18.18) is
 FI  
e2
 0V
mI
*
ε  rmI   ε I  rmI 
2
2  F
mI  

m
The initial photon is polarized in the x-direction, and therefore the relevant matrix element will
be


m  ax  ax†  000 
m 100
2m0
2m0
ε I  rmI  m X 000 
The only relevant intermediate state, then, is the 100 state. We therefore have
 FI
e2

 0V
100, I
*


e2
m  2   2  ε F  r100,I   ε I  r100,I     V 2m  2  0 2  Fx* .
100, I
0
0
0
By Fermi’s Golden rule, the rate for this transition is then
2
2
2  e 2 
2
2
2

 FI   EF  EI  
 Fx   F    .


2

  2 0Vm   2   2 
0
We wish to sum this over polarizations and final wave numbers of the photon. If we define  as
the angle between the final photon and the x-axis, then we would have


2 e 4 2
4 V m   
2
0
2
2
 2  2 2
2
0
Vcm   
2
2
0

2 2

 Fx  F    
2
kF ,


2 2
  4 c   2
2
2Vm   
2
2
2
 sin  d  F  F dF  F    
0
2
0
d 3k F

2 2
  2 
 2  2 4
Vcm   
2
2
0

3
2 2
sin 2  F   
 sin
2
 d F .
The rate is still inversely proportional to the volume. As usual, we convert this to a
cross-section by dividing by the number density (which means multiplying by V) and dividing by
the relative speed (c), so we have

 2  2 4
c 2 m 2 02  

2 2
 sin
2
 d F .
The unintegrated version of this is the differential cross-section, or we can go ahead and
integrate it for the total cross-section.
 2  2 4 sin 2 
8 2  2 4
d



,
and
.
2 2
2
2 2
d  c 2 m 2  2   2 2



3
c
m
 0 
0
As expected, there is a huge enhancement near resonance, when   0 .