Physics 742 – Graduate Quantum Mechanics 2
Solutions to Chapter 15
5. [20] A hydrogen atom is in the 1s ground state while being bathed in light of sufficient
frequency to excite it to the n = 2 states. The light is traveling in the +z direction and
has circular polarization, ε  12  xˆ  iyˆ  .
(a) [10] Calculate all relevant dipole moments rFI for final states 2lm .
The matrix elements we need are
2lm R  ε 100 
1
2
2lm  X  iY  100 
1
2
 d r  r   r  r sin   cos   i sin  
3
*
2 lm
100


1
2

 r drR  r  R  r  d Y  ,   Y  ,   sin  e .
3
m
2l
10
*
l
0
0
i
0
Of course, Y00  1
4 , and comparison of the final factors with spherical harmonics makes it
clear that it is simply sin  ei   8 3Y11  ,   , so when we put this all together, we have
2lm R  ε 100  
1
2

8 3
*
r drR2l  r  R10  r   d Yl m  ,   Y10  ,  

3 0
1
4





1
2
 l1 m1  r 3drR21  r  R10  r   
  r 3 dr re r 2 a0 e r a0
5 3 l1 m1 
3
3  24a0 a0
0
0
5

1
128 2
2 
a0 l1 m1 .
  4! a0   
4 l 1 m1
243
3 2a0
3 
Doing it by hand seemed easier than using Maple. We used orthogonality of the Y’s to do the
angular integral.
(b) [4] Find a formula for the rate at which the atom makes this transition.
We now simply use equation (15.30) from the notes, to obtain
2
 128 2 
217  2 a02
a0  
 FI  .
  I  F   4    FI  
310 
 243

1
2
(c) [6] What is the wavelength required for this transition? Assume at this wavelength
the power is      100 W/m 2 /nm . Find the rate at which the atom converts. (Note
the footnote on p. 280)
.
The energy involved is E 
wavelength from
3
4
13.60 eV   10.20 eV .
We can then calculate the relevant

c hc 1240 nm  eV


 121.5 nm .
f
E
10.20 eV
We then calculate the intensity per unit angular frequency as
121.5 nm  100 W/m 2 /nm 
2
   
   
 7.84  1013 W  s/m 2 .
8
9
2 c
2  2.998  10 m/s 10 nm/m 
2
Substituting this into our expression, we find
217  2  5.29 1011 m   7.84  1013 J/m 2 
2
  100  211  
10
3
137  1.054 10
34
J  s
 3.33 s 1 .
So even with this rather weak source, an atom will undergo such a transition several times per
second. The reverse rate would be the same, except it can be shown that spontaneous emission, a
process we have not yet accounted for, is a much faster process.