Chapter 9 - THE HELIUM ATOM
Problems with solutions
9.1 Two identical, non-interacting spin-1/2 particles of mass m are in the one-dimensional
harmonic oscillator for which the Hamiltonian is
p x21 1
p x22 1
2 2
ˆ
H
 m x1 
 m 2 x22
2m 2
2m 2
a) Determine the ground state and first excited state kets and corresponding energies when the
two particles are in a total spin-0 state. What are the lowest energy states and corresponding kets
for the particles if they are in a total spin-1 state?
b) Suppose the two particles interact with a potential energy of interaction
V  x1  x2   V0 for x1  x2  a
0
elsewhere
Will the energies of part a) be raised or lowered as a result of V  x1  x2  ?
Solution:
a) The Hamiltonian is the sum of two parts, one containing only x1 and the other x2 where
Hˆ 1 , Hˆ 2  0 . Therefore, we can label the eigenstates by n1n2 SM s where SM s is the two-


particle total spin state. The energy is given by E  n1 , n2    n1  n2  1  .
The ground state energy corresponds to n1  0  n2 so that the ground state energy is
E0  E  00   . The ket corresponding to the ground state is n1n2 SM s  00 00 .
The first excited state energy is E1  E10  E01  2 with 4 possible kets:
1
1
 1

 1

10 
01  00 (singlet) and 
10 
01  S  1 M s  0  1 (triplet)

2
2
 2

 2

b) ) For the interaction potential given, the energy is lowest when the two are close together, i.e.
closer than a. Therefore, the states for which the particles come closest will have the lowest
energy. The particles come closest when they have anti-symmetric spin states, i.e. the ground
state. Therefore, the total spin-0 state will have the energy lowered more than that of the total
spin-1 state.
9.2 Make an order-of-magnitude estimate of the the singlet-triplet splitting of the energy levels of
helium due to a direct spin-spin interaction of the electrons by comparing with the magnitude of
the hyperfine interaction in hydrogen. By comparing this estimate with the observed splitting,
1eV , what conclusions can be drawn about the relative effects of exchange symmetry and
spin-spin interaction on the energy?
Solution:
Since the average separation between the two electrons in the helium atom is comparable with
that between the electron and proton in hydrogen, the size of the spin-spin interaction can be
inferred from the size of the hyperfine interaction in hydrogen. We must, however, replace the
magnetic moment of the proton with that of the electron. The magnetic moment of the electron
is greater than that of the proton by roughly the ratio of the masses m p / me . Therefore, the spinspin interaction should be roughly 1836 105 eV  102 eV which is much smaller than the ~1 eV
energy splittings that are observed. We conclude that the exchange symmetry of the spatial wave
functions is largely responsible for the splitting.
Solution to Chapter 9 problems
page 1
Chapter 9 - THE HELIUM ATOM
Problems with solutions
9.3 A helium atom is excited from the ground state to the autoionizing state 2s4p by absorption
of ultraviolet light. Assume that the 2s electron moves in the unscreened Coulomb field of the
nucleus and the 4p electron in the fully screened Coulomb potential
a) Obtain the energy of this autoionizing level and the corresponding wavelength of the
ultraviolet light required to effect this excitation. Make an energy level diagram showing this
level together with the ground states of neutral, singly ionized and doubly ionized helium atoms.
b) Find the velocity of the electron emitted in the autoionizing process in which the 2s4p state
decays into a free electron and a He+ ion in the ground state.
Solution:
Z2
22
1
a) 2s is unscreened. Therefore, (in atomic units) E 2 s   2  
  . Since the 4p is
2
2
2n
22
2
2
Z
1
1
fully screened the effective nuclear charge is 1 and E 4 p   2  
  . Therefore,
2
32
2n
24
1 1
17
E2s 4 p   
  au  14.5 eV . Graphically, we have
2 32
32
12,400
 192 Å
64.5
b) The electron kinetic energy is the difference between the autoionizing level and He+(1s):
1
1
v2
KEelectron  54.4  14.5  39.9 eV so mv 2  mc 2 2  39.9 eV . Therefore,
2
2
c
2
39.9 eV
v
 2
 1.6  10 4 so v  1.6 104 c  3.8 106 m/s
2
5
c
5.1  10 eV
Since the photon energy is 64.5 eV, the wavelength is  


9.4 Use the variational principle to calculate the ground state energy of a hydrogen atom (in
atomic units) using the normalized trial functions:

a)  r    3 /   exp  r  and b)  r   2 2 2 3 / 
parameters.
c) Which gives the most accurate answer? Why?
Solution:

1/ 4

 exp  r 2

where  and  are
Solution to Chapter 9 problems
page 2
Chapter 9 - THE HELIUM ATOM
Problems with solutions
1
1
1 1 d  2 d  1
a) Hˆ    2     2
r
  and
r
 2
 2  r dr  dr  r
 3  1  1 d  2 de  r  e  r 
 3    1  2   r



e . Therefore,
Hˆ   
r



 

  2  r 2 dr 
dr 
r 
  r
2 
 3     1  2  2 r 2
2


Hˆ  4

e
r
dr

  . Now, minimizing wrt , we have
 0  r
2 
2
d Hˆ
d
   1  0    1 which gives, as we should have expected, the exact grounds state
2
1
1
ˆ
    1   which is the exact
wave function. The energy is therefore, E0  H 
2
2
2
ground state energy.
b) The Hamiltonian is exactly the same as for part a). Differentiation and integration using the
3
2
given wavefunction yields Hˆ 
. Minimizing wrt  gives
2
2

d Hˆ
3
2 1
8
 
 1/ 2  0   
and
d
2
 
9
3
2
3 8
2 8
4
1  8 
 1
Hˆ 
2
 
2


      0.848  
2

2 9
 9
3
2  3 
 2
so that using this wavefunction gives a ~15% error.
c) Of course a) gives the most accurate answer because the minimization process produced the
exact ground state wavefunction.
9.5 A negative ion is formed when an electron attaches to an atom (or molecule), the result being
that the nucleus of charge Z binds  Z  1 electrons. The binding energy of the electron to the
neutral atom is referred to as the electron affinity of the atom. The electron affinity may also be
thought of as the ionization potential of the negative ion. Not surprisingly, the halogen atoms
form negative ions most readily. This means that the halogen atoms have the highest electron
affinities of all atoms (~3 eV). Hydrogen atoms also form negative ions. The electron affinity of
hydrogen is 0.75 eV. Use perturbation theory as it was applied to the helium atom to determine
the total energy of the hydrogen atom negative ion. Compare the answer with the actual value.
Compare the accuracy of the perturbation theory results for the hydrogen negative ion with the
result for the helium atom. Why is perturbation theory more accurate for helium? Note that
there are no "new" calculations necessary.
Solution:
Everything is the same as for helium except that Z  1 . Thus, the unperturbed energy is merely
0
twice the ground state energy of hydrogen, that is E1s1s 
E01   dr1dr2 1s  r1 
2


1 2
1
2


1 2
1
2
  1a.u. The integral
2
1
1s  r2  has already been calculated for an arbitrary value of Z.
r1  r2
Solution to Chapter 9 problems
page 3
Chapter 9 - THE HELIUM ATOM
Problems with solutions
5
5
1
It is E0    Z   . Therefore, the total energy of the hydrogen negative ion, as given by
8
8
5
3
perturbation theory is E0  E0 0  E01  1     10.2 eV . (This is unrelated to the
8
8
separation between the ground and first excited state of neutral hydrogen being 10.2 eV.)
The experimentally determined total energy is Eexp  13.6 eV  0.75eV  14.35eV so the error
is considerable, especially compared with that for helium. Evidently the reason is that the
perturbation term, the e-e repulsion, is a larger fraction of the unperturbed energy in hydrogen
than it is in helium.
That is, for hydrogen
E01
E01
3/ 8

 0.375 , while for helium  0
1
E0
 0
E0
H

H
5/8
 0.156 .
4
9.6 Assume the normalized trial wave function   1  A2   100  A 210  , where the kets
on the right hand side are spherical hydrogen atom eigenkets, represents a ground state hydrogen
atom in a constant electric field F. This wave function represents a state that has "ground state
character" and, assuming A  1 , a small amount of 210 character.
1
a) Using this trial wave function show that, if terms in Ĥ having powers greater than A2 can
be neglected, then second order perturbation theory, as applied in Chapter 9, yields the same
energy as the variational treatment. Note that if powers of A greater than two can be neglected in
Ĥ that powers of A greater than one may be neglected in  Hˆ / A .
b) Using this trial wave function estimate the dipole polarizability of ground state hydrogen and
compare the answer with the exact answer given in Chapter 9.
Solution:
a) We must first normalize  by finding K.
   1  K 2 1  A2   K 
1
1  A2
The Hamiltonian is Hˆ  Hˆ 0  Hˆ S  Hˆ 0  eFz as in Chapter 9.
1   0
Hˆ   Hˆ 0    Hˆ S  
E  A2 E2 0  2eFA 100 z 210 
2  1
1 A


Now minimize w.r.t A.
 Hˆ
1
2A

2 AE2 0  2eF 100 z 210 
2
A
1 A
1  A2






2
 E1 0  A2 E2 0  2eFA 100 z 210 


0
so
2 AE2 0  2eF 100 z 210 
2 A   0
E  A2 E2 0  2eFA 100 z 210   0
2  1
1 A


Retain only terms to the first power in A since A  1 . We have
Solution to Chapter 9 problems
page 4
Chapter 9 - THE HELIUM ATOM
Problems with solutions
2 AE2 0  2eF 100 z 210  2 AE1 0  0  A 
eF 100 z 210
E2 0  E1 0
Returning to Ĥ , we may re-write it in terms of our approximation that A  1 . Since we are
in Ĥ rather than its derivative we keep terms in A2 .
1
 E  0  A2 E2 0  2eFA 100 z 210 
2  1

1 A
Hˆ 


 1  A   E    A E    2eFA 100 z 210 
0
2

0
2
1
2

 E1 0  A2 E2 0  2eFA 100 z 210  A2 E1 0


 E1 0  A2 E2 0  E1 0  2eFA 100 z 210
Substituting for A we get
2
 eF 100 z 210 
 eF 100 z 210 
 0
 0
Hˆ  E1 0  
 E2  E1  2eF 
 100 z 210
 0
 0
 0
 0
 E2  E1

 E2  E1

 100 z 210 2 
 0
2 2 

 E1  e F 
 E2 0  E1 0 


Notice that this is precisely the same expression that would be obtained using second order
perturbation theory with the assumption that the only unperturbed state that contributes is n  2 .
d 2E
b) To find the polarizability we recall that  d   2 so we require the coefficient of F 2 . That
dF
is, the energy of the ground state hydrogen atom in the electric field will be
1
E1 2  E1 0   d F 2 where  d is the diple polarizability. (Recall that E11  0 ) Therefore, we
2
must compute the integral 100 z 210 .


 1

r
 1 3/ 2  r / a0 
100 z 210  all 
a0 e
a0 3/ 2 e  r / 2 a0 cos   dV
  r cos   
a0
space  

 4 2


3/
2
a
r


0
2


1

d  d sin  cos 2   drr 4e

4
0
0
0
4 a0 2
 cos3     5 
1

2





5
4 a0 4 2
 3   3 


 2a0 
0
5
a  2
28
 2
 0    4 3 2    5
a0
2 2  3
 3 3 2
Then, the energy resulting from application of the electric field is
Solution to Chapter 9 problems
page 5
Chapter 9 - THE HELIUM ATOM
Problems with solutions
  100 z 210 2 
1

  d F 2  e2 F 2 
2
 E2 0  E1 0 


so
2
 100 z 210 2 
8

1
218 2 2 1
2 
2 2
  2e 
 d  2e 
a


e a0  0
0
311
 E2 0  E1 0 
35 2  E  0  3 
E1



1 

 4
 e2  1
From Chapter 1, the energy may be written as E1 0   
so finally we have

4

2
a
0 
0

d  
218 2 2 1
218 2 2
e
a


e a0
0
311
311
E1 0
  4 0  2a0 
219 3
3


4


 
0  11 a0  2.96  4 0  a0
2
e
3


1
2
0
This may be compared with the exact answer  d  4.5  4 0  a03 . Since E1   E1    d F 2 we
2
see that the energy obtained using the variational method is indeed highet than the actual energy.
Solution to Chapter 9 problems
page 6