Ph235-08
F. Merritt1
24-Oct-2008
Lecture 12 (October 24, 2008):
(version 1.0; 24-Oct-2008::00:00)
Fine Structure of Hydrogen
The Coulomb potential for the hydrogen atom leads to a Hamiltonian (CGS units)
H
 
2
2m
 2  Vc (r ) 
p 2 e2

2m r
(1.1)
But this needs to be corrected to include (a) the relativistically correct kinetic energy, and
(b) the potential energy of the electron, in its own rest frame, between the magnetic
dipole moment of the electron and the effective B-field.
(a) The Relativistic Correction:
The relativistic kinetic energy of the electron is
T  E  mc 2 
 mc    pc 
2 2
2
 mc 2
 1  p  2 1  p  4
 
 mc 1  


....
  1



 2  mc  8  mc 
 
p2 1 p4


 ....
2m 8 m3c 2
2
(1.2)
This leads to a perturbation hamiltonian
H  
1 p4
8 m3c 2
(1.3)
and an energy correction of
Er1  nlm)   
1
nlm p 4 nlm
3 2
8m c
(1.4)
But
nlm P4 nlm  2m  H 0  V  nlm  4m2  En2  2En V

Moreover, from the virial theorem we know
V n  2 T n
and
T n  V n  En
so
V n  12 En
2
nlm
 V2
Also, one can show [see Shankar 17.3.4(2) (HW5, problem 2)] that
nlm


(1.5)
(1.6)
(1.7)
(1.8)
Ph235-08
F. Merritt2
24-Oct-2008
V
2
n
e4
 2
r
n
e4 1
 3 2

n ao l  12
4  En0  n
2
(1.9)
l  12
Substituting these last two equations into (1.5) yields the total relativistic correction:
4m 2  En0  
4n 
E (nlm)  
 3 
 
3 2
8m c
l  12 

2
1
r
E 
0 2
n

4n 
3 

2mc  l  12 
2
(1.10)
(b) The Spin-Orbit Correction:
The magnetic dipole moment of the electron is given by the Dirac equation to be
e

S
(1.11)
mc
[It is interesting to note that classically, if we treat the electron as a current loop, its
e
e r
e
 r 2 / c 

L . The Dirac
magnetic moment would be   I  a / c 
2 r
2c 2mc
equation predicts a value that is larger by precisely a factor of 2.]
The magnetic field that the electron sees in its own instantaneous rest frame is

  e  e pr
e
B  E 
  2 rˆ  

L
(1.12)
3
c
c  r  mc r
mcr 3
The interaction of the electron dipole moment (due to electron spin) with the magnetic
field in the electron frame (due to its orbit) would appear to give a perturbation term
e2
H  2 2 3 S  L
(1.13)
mcr
However, the electron is not in an inertial frame since it is accelerating toward the
nucleus; this gives a relativistic correction which reduces H  by precisely a factor of 2
(also predicted by the Dirac equation). This gives a final Hamiltonian due to the spinorbit interaction of
e2

H so 
S L
(1.14)
2m 2 c 2 r 3
But to find the expectation value of this Hamiltonian in the nlm state, we need
to express it in terms of good quantum numbers for the hydrogen atom. When we
include H so , the Hamiltonian does not commute with either S or L , so these quantities
are not conserved; both L and S will precess around the total angular momentum J .
However, we don’t need to know the components of S and L in order to evaluate (1.14),
since
Ph235-08
F. Merritt3
24-Oct-2008
J  LS
 J 2  L2  2 L  S  S 2
3  (1.15)

 j ( j  1)  l  l  1  s ( s  1)    j ( j  1)  l  l  1  
2
2 
4
Substituting this into (1.14) and taking the expectation value gives
e2 2
1
1
(1.16)
Eso  nlm H so nlm 
j ( j  1)  l (l  1)  34  3

2 2 
4m c
r nlm
This last factor can be evaluated using the method described in Shankar’s Exercise 17.3.4
(HW5.2), giving
1
1
1
(1.17)
 3 3
3
r
n a0 l (l  1)(l  12 )
2
 LS 
2
When we substitute this into (1.16), and use the identities relating En for hydrogen to 
and to a0 , we obtain:
E
1
so
e 2 2 1  j ( j  1)  l (l  1)  34 



4m 2 c 2 n3 a03  l (l  1)(l  12 )

2
2 2
 e   c
1   j ( j  1)  l (l  1)  34 
 


2 
2 2 2
1
 2a0 n  2(mc ) a 0 n   l (l  1)(l  2 )

  2   j ( j  1)  l (l  1)  34 
 E   

1
 2n   l (l  1)(l  2 )

(1.18)
0
n
E 
0 2
n

mc 2
 j ( j  1)  l (l  1)  34 
n

1
 l (l  1)(l  2 )

Then the total fine-structure correction is the sum of equations (1.10) and (1.18):
Ef1s  Er1  Eso1



E 
0 2
n
 3 2n
 j ( j  1)  l (l  1)  34  



n



1
mc 2  2 l  12
 l (l  1)(l  2 )
 
E 

2n  2l (l  1)  j ( j  1)  l (l  1)  34  
3




2mc 2  l  12
l (l  1)

0 2
n
(1.19)
E 

2n   j ( j  1)  3l (l  1)  34  
3




2mc 2  l  12
l (l  1)

0 2
n
The second of the two terms within

can be simplified, since there are only two
possibilities: (a) l  j  12 , and (b) l  j  12 . Making the substitutions for each of these
Ph235-08
F. Merritt4
24-Oct-2008
cases, we find that we miraculously get the same result in both cases, provided we
express the result in terms of j. The final result is
2
E
1
fs

 En0  
4n 
3



2mc 2 
j  12 
(1.20)