Chapter 7 - FINE STRUCTURE OF THE HYDROGEN ATOM
Problems with solutions
7.1 Show that no combination of quantum numbers can conspire to make the fine structure
correction to the Bohr energy vanish.
Solution:
2 
n
3
1
EFS
  for j   1/ 2
 n, j   En 0 2 
n   j  1/ 2  4 
To vanish, the quantity in the square bracket must vanish. Therefore, we must have
3
4n  3 j 

n
3
2 0
 0 

4  j  1/ 2 
  j  1/ 2  4 
Since the denominator is manifestly positive we must have 4n  3 j 
3
.
2
The possible values that j can take on are given by
1
j  n   2k  1   where k  0,1, 2...  n  1 . For example, if k  0 then j  jmax  n  1/ 2 . If
 2
k   n  1 then j  jmin  1/ 2 .
3
 0 becomes
2

 1  3
4n  3 n   2k  1      0
 2  2

Then 4n  3 j 
3 3
 0

n  3k
2 2
Since n must be positive this last equation is impossible.
n  3k 
7.2 Suppose the proton is approximated as a spherical shell of radius R  104 nm . Calculate the
first order correction to the energy of the ground state, E 01 , due to the finite size of the nucleus.
Is it positive or negative? Use the fact that R / a 0  10 5 to make the approximation e 2 R / a0  1 .
Estimate the value of E 01 in terms of the unperturbed ground state energy E0 0 and compare with
the fine structure correction to the ground state.
Solution:
First we must find the potential energy function to which the electron is subjected. Using Gauss'
law we find:
1 e2


V r 

rR
40 R
e2


rR
40 r
The perturbation is simply the difference between the above potential energy and that for a point
nucleus. Thus, the perturbing Hamiltonian is:
1
Solution to Chapter 7 problems
page 1
Chapter 7 - FINE STRUCTURE OF THE HYDROGEN ATOM
Problems with solutions
 e2 e2 
 rR
 
40  r
R 
0 rR
Hˆ ' r  
1
The first order correction to the energy of the ground state is therefore  100 Hˆ '  100 , or
E01 
1
e2

a0 3 40

R
0
 e 2 e 2  2 r / a0
 
  e
4r 2 dr But, e 2 R / a0  1 so
r
R


R1
 1  1
1
 
E01  
 e 2     4r 2 dr
3
0
r R
 40  a0
R
 1  4
 r2 r3 
  3  e 2  
 

 2 3R  0
 40  a0
2
 1  2 e2  R 
     
 
4

0  3 a0  a0 

which is clearly positive. Since the (unperturbed) energy of the hydrogen atom may be written
2
 1  e2
4 0 R
1
0
 
We may write E 01 in terms of E 00  : E0   E0    1010 E0 
En  
3
 a0 
 40  2a0
1
Compared with the magnitude of the fine structure corrections, EFS
 n  1  2 E00 105 E00 ,
this quite small.
0 
7.3 A particle of rest mass m0 is confined to an infinite one-dimensional potential well, i.e.
V  x    x  0 and x  L
0
otherwise
1
Find the first order correction to E n , the energy of the nth level of this particle-in-a-box, due to
the relativistic kinetic energy of the electron. Put your answer in terms of the unperturbed
energies and the rest energy of the particle. Under what circumstances will the validity of E n1 be
questionable? Consider two cases, an electron in a 0.1 nm box (an atom) and a proton in a 10-4
nm box (a nucleus).
Solution:
The relativistic energy is:
1/ 2

p2 
p2 1 p4
T  E  m0 c  m0 c  p c  m0 c  m0 c 1  2 2   m0 c 2 
 
2 m0 8 m0 3 c 2
 m0 c 
The first term is, of course, the non-relativistic kinetic energy so the second term may be
1 p4
ˆ
considered the perturbation due to the relativistic motion of the particle, i.e. H '    3 2 .
8 m0 c
2
2
4
2
2
2
Now, the unperturbed energies are: E n0  
2
 2 2n2
2m0 L2
. In the well the unperturbed Hamiltonian is
Solution to Chapter 7 problems
page 2
Chapter 7 - FINE STRUCTURE OF THE HYDROGEN ATOM
Problems with solutions
p2
given by Hˆ 0 
so we may write Ĥ ' in terms of Ĥ 0 , i.e.
2m0


2
1 p4
1
Hˆ '    3 2  
2m0 Hˆ 0 . The first order correction to the energy is thus:
3 2
8 m0 c
8m 0 c
E n1   n0  Hˆ '  n0   
 
0 
1
0  ˆ
ˆ  0    E n

H
|
H
n
0
0
n
2 m0 c 2
2 m0 c 2
2
Note that E n1 can also be evaluated by performing the integral, i.e. so that
1
0 
En   n
Hˆ '  n0   
1
3
8m0 c
0 
2
0 
n p n
4

1
3
8m0 c
2

L
0
2
 nx   d 
sin 
  
L  L  i dx 
4
2
 nx 
sin 
dx
L  L 
L
 nx 
4  n 



sin 2 


dx
3 2

0
4m0 c L
 L 
 L 
n
L
x ; dx 
dy ; x  0  y  0 ; x  L  y  n . Therefore,
Let y 
L
n
4
4
1
1
 L  n
 L   n 
1
4  n 
2
4  n 
En  
 
 
 
  0 sin ydy  
 


3 2
3 2
4m0 c L
4m 0 c L
 L   n 
 L   n   2 
4
1

 2  2 n 2 

E n0  


2 
2m 0 c 2
 2m0 L 
If m0 is the mass of an electron and L = 0.1 nm, then E n0   10 eV , i.e. typical of atomic energy
1
1
 
2 m0 c 2
2
2
levels. Since m0 c 2  0.51 MeV E n1 will be small unless n is very large. On the other hand, if a
proton for which m0 c 2  1000 MeV is confined to a box with L  10 4 nm , the result for E n1
will be questionable.
7.4 A particle of rest mass m0 is confined to a one-dimensional harmonic oscillator potential
V  x   kx 2 . Find the first order correction to E01 the energy of the ground state due to the
2
1
relativistic kinetic energy of the electron. Put your answer in terms of the unperturbed energies
and the rest energy of the particle. Suppose the vibrations of a diatomic molecule are
approximated by the harmonic oscillator. Show that, because the separations between molecular
vibrational levels are typically the order of tenths of eV, the correction due to the relativistic
motion of the electron is small. There are (at least) three ways to work this problem.
Solution:
Method #1:
The perturbing Hamiltonian is the same as in the previous problem
1
Hˆ '  A  p 4 where A  
. Simply compute the integral 0 H ' 0 using the known
8m03c 2
ground state wave function in coordinate space and making the substitution p  i  d / dx  .
Thus, the Gaussian ground state wave function will be differentiated four times and the integral
evaluated. Uninteresting!
Method #2:
Solution to Chapter 7 problems
page 3
Chapter 7 - FINE STRUCTURE OF THE HYDROGEN ATOM
Problems with solutions


m0 
m0 
i
i
p  and a 
p  so that, as
x
x
2 
m0 
2 
m0 
and a n  n n 1 where the n are the harmonic oscillator
Use the ladder operators: a  
usual, a n 
 n  1
n 1
eigenfunctions corresponding to the quantum number n. Subtracting the defining equations of
m 
the ladder operators we get p   0   a  a   . The first order correction that we seek is
 2 
4
4
m 
E0  A  0  0 a  a  0 . The expansion of  a  a   is simplified because a 0  0 .
 2 
Also, since the kets are orthogonal there must be the same number of raising as lowering
operators to keep the integral from vanishing. We have
2
1

0 a  a

4



0 a
0  0 a 2  aa   a  a  a  2

4

2
0

 a 3a   a 2 a  a  a 2 a  2  aa  a 2  aa  aa   a  aa  a  aa  3 0


0 a  a 3  a  a 2 a   a  aa  a  a  aaa  2  a  2 a 2  a  2aa   a  3a  a  4 0
Only two of these terms survive the above criteria. We have

0 a  a

4
0  0 a 2 a 2 0  0 aa  aa  0
 2 0 a 2 2  0 aa  2
 2 2 1
3
Finally,
3  
m 
E0  A  0   3   
32 m0 c 2
 2 
Method #3:
In momentum space
2
2
1

E01  0 H ' 0  A 0*  p  p 40  p  dp

where 0  p  is the ground state wave function in
momentum space. We can look this up or calculate it easily as follows.


i
a 0 0  x
p  0  p   0 Making the substitution x  i  d / dp  we have
m0 

d0  p 

1
p2 

p  0  p   C exp  
 . To normalize the wave function we
0  p 
m 
 2m0  

p
p2 
2 
 dp  m0  dy so
compute the integral 1  C  exp  
dp Let


m

m

0


0
1 C
2



m0   exp  y 2 dy . Using the integral



0
x m e  ax
2
 m 1


2 

dx 
we have
 m 1 

2a 

2 
Solution to Chapter 7 problems
page 4
Chapter 7 - FINE STRUCTURE OF THE HYDROGEN ATOM
Problems with solutions
1/ 4
 1 
C 

  m0  
E01  A
p
m0 
. Finally, we have


1
p2  4
exp


 p dp .
 m0  
 m0  
Making the same substitution as above we have
 dp  m0  dy which leads to


1 
1
2
E01   
exp  y 2  m0   y 4
3 2 


 8m0 c   m0 
 5 / 2

1  1
2
 
 m0   2 
3 2 
2
 8m0 c  

1  1
2 3
 
 m0   
3 2 
4
 8m0 c  



m0  dy

3  
 
32 m0 c 2
2
  101 eV the correction for a diatomic molecule will be
Since m0c 2  5 105 eV and
108 eV .
7.5 Starting with the Hamiltonian for the fine-structure correction to the energy of the hydrogen
atom
 1 

e2
Hˆ SO  
S  L    r    S  L

2 2 3 
 4 0  2me c r 
find the energy difference between to the j   
1
levels.
2
Solution:
We found that
Hˆ SO    r    S  L  so that
1
ESO
 Hˆ SO    r    S  L 
  r  
2
 j  j  1 
2 
2

   r    j  j  1 
2 

 1  s  s  1 

3
 1  
4
Then
Solution to Chapter 7 problems
page 5
Chapter 7 - FINE STRUCTURE OF THE HYDROGEN ATOM
Problems with solutions
1
1
1 
1 
ESO  ESO
    ESO   
2
2


2
 
1 
3
   r         
2  
2 
2

 
1 
1
 1        
2 
2
 


 1  

1 
3
1

      
2
2 
2
2
2
1 
3
1

   r         
2
2 
2
2
  r  
2
  r  
2
2
2
 1
7.6 An electron of mass m is bound by an isotropic harmonic oscillator potential
1
1
1
V r   k x 2  y 2  z 2   kr 2  m 2 r 2
2
2
2
Find the corrections to the ground and first two excited state energies due to the spin-orbit
2
1 dV r  ˆ ˆ
V

 
L S .
correction SO
2 2
2m c r dr
Solution:
Because the Hamiltonion is separable, i.e. Hˆ  Hˆ x  Hˆ y  Hˆ z the eigenvalues are the sums of


the eigenvalues for each oscillator. So En  n  3 / 2 where n  n x  n y  n z . The first
three energies are 3 / 2 , 5 / 2 and 7 / 2 .
2

 
1 ˆ 2 ˆ2 ˆ 2
ˆ
ˆ
Inserting V(r) into VSO we get VSO   L  S   J  L  S where  
.
2mc 2
2
1
The first order correction to the energy is VSO    j  j  1    1  s s  1 .
2
Now examine the values of  for each of the ground and first excited states by comparing the
degeneracies of the levels recalling that for a given  the degeneracy is (2+1).
3

Writing the energy eigenvalues as En   nx  n y  nz     n  we have:
2

n  0 : n x , n y , n z   0,0,0 degeneracy = 1




n  1 : n x , n y , n z   1,0,0, 0,1,0, 0,0,1 degeneracy =3
n  2 : n x , n y , n z   2,0,0, 0,2,0, 0,0,2 , 1,1,0, 1,0,1, 0,1,1 degeneracy = 6
1
In general, the degeneracy is  2  n  1 n  2   . This may be looked up, but it is relatively easy


to show as follows:
n  nx  n y  nz Since nx  0,1, 2...n there are  n  1 possible values for nx and the sum
n y  nz  n  nx . Once nx is chosen, there are  n  nx  1 ways to choose n y . After both nx and
n y are chosen of course nz is set so the degeneracy is the number of possible ways that
Solution to Chapter 7 problems
page 6
Chapter 7 - FINE STRUCTURE OF THE HYDROGEN ATOM
Problems with solutions
 n  nx  1
can be constructed. That is
degeneracy 
n
n
n
n
n
nx  0
nx  0
nx  0
nx  0
nx  0
  n  nx  1  n  1   nx   1  n  n  1   nx   n  1
1
1  1
2

  n  1  n  n  1   n  1  n  1  n    n  1 n  2 
2
2  2

n
Note that
n
nx  0
x
 n  n  1 was evaluated using Gauss' trick.
1
2
The key to determining the allowed values of  for each value of n is to recall that for any central
potential (such as this isotropic oscillator) the degeneracy for each value of  is  2  1 . Thus,
we must determine how to "assemble" the above degeneracies from this fact.
Ground state: n = 0,  = 0 only because this is the only way to get a degeneracy of 1.
Since  = 0 and j = s for the ground state, there is no correction to the ground state.
First excited state:
For n = 1,  = 1 only because this is the only way to get a degeneracy of 3.
 = 1, and j = 1 ± 1/2 = 1/2, 3/2. Therefore, for
1
5
1
j  3 / 2; VSO    so E j 3 / 2    
2
2
2
5
j  1 / 2; VSO   so E j 1 / 2    
2
Second excited state:
For n = 2,  = 0 and 2. The degeneracy of 6 is the result of adding 1   2  2  1 .
As for the ground state there will be no correction for  = 0.
For  =2, and j = 2 ± 1/2 = 3/2, 5/2.
7
j  5 / 2; VSO   so E j 5 / 2     .
2
3
7
3
j  3 / 2; VSO    so E j 3 / 2    
2
2
2
Note: Generalizing, we see that the eigenstates are also eigenstates of the parity operator because
even n's have only even 's.
Solution to Chapter 7 problems
page 7