19.4 Excited States of the Helium Atom. Degenerate Perturbation Theory 803
the electric field. How important do you think that this
electric field is for this particle? Is the first-order
perturbation method applicable in this case?b.Repeat the calculation forEx 1. 00 × 107 Vm−^1 .Is
the first-order perturbation method applicable in this
case?19.20Using first-order perturbation, calculate the energy of
the ground state of an anharmonic oscillator with
potential energyVkx^2 / 2 +bx, wherekandb
are constants. Compare your answer with that of
Problem 19.5.
19.21Using first-order perturbation, calculate the energy of the
ground state of an anharmonic oscillator with potential
energy functionVkz^2 / 2 +cz^4 , wherek576Jm−^2 ,
c 1. 00 × 1020 Jm−^4 , andm 8. 363 × 10 −^28 kg.
19.22Using first-order perturbation, find a formula for the
energy of a hydrogen atom in the 1sstate that is exposed
to an electric field in thezdirection, denoted byEz.
19.23Consider a system such that one of the zero-order wave
functions happens to be an eigenfunction of the
perturbation term in the Hamiltonian. Show that
first-order perturbation theory gives an exact solution to
the Schrödinger equation.19.4 Excited States of the Helium Atom.
Degenerate Perturbation Theory
Excited states are more difficult to treat than ground states. The variation method is
usually not used because the variation theorem applies only to ground states. There is
anextended variation theorem, which states that the calculated variation energy will be
no lower than the correct energy of the first excited state if the variation trial function
is orthogonal to the correct ground-state energy eigenfunction. It will be no lower than
the energy of the second excited state if the variation trial function is orthogonal to
both the ground state and the first excited state, and so on.^12 Unfortunately, the correct
ground-state energy eigenfunction is not generally known, so that choosing a family of
functions exactly orthogonal to it might be impossible. Some calculations have been
made in which a family of functions is chosen that is orthogonal to an approximate
ground-state variation function. This family of functions might be nearly orthogonal to
the correct ground-state function and the minimum variation energy from this family
might be a good approximation to the energy of the first excited state. In other cases,
even if the correct ground-state wave function is not known, some known property,
such as being spherically symmetric, might permit construction of a trial function that
is orthogonal to it.
The perturbation method as described in the previous section does not apply if sev-
eral wave functions correspond to the same zero-order energy (the degenerate case).
For example, the zero-order orbital energies of the 2sand 2phydrogen-like orbitals
are all equal, so that all of the states of the (1s)(2s) and (1s)(2p) helium configurations
have the same energy in zero order. A version of the perturbation method has been
developed to handle this case. We will describe this method only briefly and present
some results for some excited states of the helium atom.^13 There is additional infor-
mation in Appendix G.
In the degenerate case there is no guarantee that the wave functions that we obtain
with the zero-order Schrödinger equation are in one-to-one correspondence with the
correct wave functions. If not, the smooth dependence on the parameterλdepicted
in Figure 19.2 will not occur. The first task of the degenerate perturbation method is
to find the zero-order wave functions that are in one-to-one correspondence with the
exact wave functions. We call them thecorrect zero-order wave functions.(^12) I. N. Levine,op. cit., p. 212ff (note 2).
(^13) I. N. Levine,op. cit., p. 259ff (note 2).