Physics 742 – Graduate Quantum Mechanics 1
Solutions to Chapter 13
2. [15] Complete the computation of the spin-orbit splitting for hydrogen for the 2p, 3p,
and 3d states of hydrogen. Write your answers as multiples of  2 En , where  is the
fine structure constant and En is the unperturbed binding energy of this state.
Starting from equation (13.8) in the class notes, we have
E 
 2  2l  1  2  1 dVc  r   2
 2  2l  1 ke e 2
r
dr
R
r




 nl
4m 2 c 2 0
4m 2 c 2
 r dr 


0
dr 2
Rnl  r  .
r
Recalling that the energy of the n’th state is En  ke e 2 2n 2 a0 , this can be rewritten as
 2  2l  1 a0 n 2  dr 2
 2  2l  1 n 2  dr 3 2
E  En
0 r Rnl  r   En 2m2c 2 a02 0 r a0 Rnl  r  .
2m 2 c 2
Substituting a0   2 ke2 m , and then using   ke e 2 c , this can be rewritten as

ke2 e 4  2l  1 n 2  dr 3 2
dr
2
2
1
E  En
a0 Rnl  r   2  En n  2l  1  a03 Rnl2  r  .
2 2 2

r
r
2m c 
0
0
Only the final integral remains. We find



dr 3 2
1 dr r 2  r a0 1
1
x
0 r a0 R21  r   a02 0 r 24 e  24 0 xe dx  24 ,
2


dr 3 2
1 dr 25 r 2 
r   2 r 3a0 25  3 


 7 
a
R
r
1



 e
0 r 0 31
a02 0 r 37  6a0 
3 2


2 
2
 x  x
0 x 1  4  e dx
8  22 6  1
 1 
  ,
243 
4 16  81

23  3 
1 dr 23 r 4  2 r 3a0
dr 3 2
 9  
0 r a0 R32  r   a04 0 r 39  5 e
3 5  2 
4 
x e
3 x
dx 
0
6
1
.

3 10 405
5
We now just substitute these numbers into our previous equations, to find
E2 p  12  2 E2 22  3 241  14  2 E2 ,
E3 p  12  2 E3 32  3 811  16  2 E3 ,
1
E3d  12  2 E3 32  5  405
 181  2 E3 .
3. [5] Prove, as asserted in section C, that
  3rˆ rˆ    d   0 .
j k
jk
This is actually nine
formulas, but only six of them are independent.
We simply write rˆ  r r   sin  cos  ,sin  sin  , cos   and work out all the components.
We’ll do it as a matrix for fun.
 3sin 2  cos 2   1 3sin 2  cos  sin  3sin  cos  cos  


2
2
2
 sin   d 
  3rˆj rˆk   jk  d     3sin  cos  sin  3sin  sin   1 3sin  cos
 3sin  cos  cos  3sin  cos  sin 
3cos 2   1 

We start with the integral over . It is easy to see that the integral of cos  or sin  is just zero,
and since cos  sin   12 sin  2  , this can be seen to integrate to zero. The integral of 1 is trivial,
and we can use Carlson’s rule on the cos 2  and sin 2  (they are each half of the size of the
interval), so we have
 32 sin 2   1

  3rˆj rˆk   jk  d   2 0  0

0


 0

0 
0 0


0

3
0
2 sin   1
 sin  d
2
0
3cos   1
0
2
0 1

0   1  3cos 2   d  cos    0.
2  1