Physics 742 – Graduate Quantum Mechanics 2
Solutions to Chapter 18
2. [15] A hydrogen atom is initially in a 3d state, specifically, n, l , m  3, 2, 2 .
(a) [9] Find all non-zero matrix elements of the form n, l , m R 3, 2, 2 , where n  n .
Which state(s) will it decay into?
The position operator is a rank one tensor, so by the Wigner-Eckart theorem, the final
value of l’ must be in the range l   l  1, l , l  1 , or l’ = 3, 2, or 1. But since n’ < 3, and l’ < n’,
the only possibility is l’ = 1, which in turn implies n’ = 2. Furthemore, if we look at the matrix
1
elements of the various components of R, which in spherical tensor notation would be Rq  , then
2,1, m Rq1 3, 2, 2 is only non-vanishing for m  q  2 which has the unique solution
m  q  1 . So there is only one matrix element, 2,1, 1 R11 3, 2, 2 , and it must decay only to
the state 2,1, 1 .
We now need the non-zero matrix elements 2,1,1 R 3, 2, 2 , which we simply work out
directly:
2,1,1 X 3, 2, 2   d 3rR21  r  Y21  ,   R32  r  Y22  ,  r sin  cos 
*
2



3 5
21134
3
5
i
cos


sin




R
r
R
r
r
dr
e
d
d
a0 ,




21
32
0
0
16 0
57
2,1,1 Y 3, 2, 2   d 3rR21  r  Y21  ,   R32  r  Y22  ,  r sin  sin 
*
2


3 5
21134
2
5
i

sin


sin




R
r
R
r
rr
dr
e
d
d
ia0 ,




21
32
0
0
16 0
57
2,1,1 Z 3, 2, 2   d 3rR21  r  Y21  ,   R32  r  Y22  ,  r cos 
*
2


3 5

R21  r  R32  r  rr 2 dr  e  i e 2i d  sin 4  cos  d  0.

16 0
0
0
I used my hydrogen Maple routine to do the r and  integral; the  integrals are , i, and 0
respectively.
> simplify(int(radial(3,2)*radial(2,1)*r^3,r=0..infinity)
*int(sin(theta)^5,theta=0..Pi)*3*sqrt(5)/16);
(b) [6] Calculate the decay rate in s-1.
We start with our standard expression, namely
4IF3 rFI
I  F  
3c 2
2



4  222  38 IF3 a02
225  37 IF3 a02
2
2
1
.


i

3  514 c 2
514 c 2
The frequency is given by the difference in energy between the two states, namely
IF 
I  F


1  mc 2 2 mc 2 2  5mc 2 2
.



2  4  23  32 
  29
The Bohr radius is a0    mc . Substituting each of these expressions into the rate equation, we
have
3
225  37   5mc 2 2     3  216  5 mc 2
  I  F   14 2  3 2  
 
5 c  2  3     mc 
511 

2
3  216  511, 000 eV
5 137.036   6.582 10
11
5
16
eV  s 
 6.469  107 s 1 .