Michael Fowler 2/16/06
If an atom (not necessarily in its ground state) is placed in an external electric field, the energy levels shift, and the wave functions are distorted. This is called the Stark effect. The new energy levels and wave functions could in principle be found by writing down a complete Hamiltonian, including the external field, and finding the eigenkets. This actually can be done in one case: the hydrogen atom, but even there, if the external field is small compared with the electric field inside the atom (which is billions of volts per meter) it is easier to compute the changes in the energy levels and wave functions with a scheme of successive corrections to the zero-field values. This method, termed perturbation theory, is the single most important method of solving problems in quantum mechanics, and is widely used in atomic physics, condensed matter and particle physics.
It should be noted that there are problems which cannot be solved using perturbation theory, even
when the perturbation is very weak, although such problems are the exception
rather than the rule. One such case is
the one-dimensional problem of free particles perturbed by a localized
potential of strength l. As we found earlier in the course, switching
on an arbitrarily weak attractive potential causes the
free particle wave
function to drop below the continuum of plane wave energies and become a
localized bound state with binding energy of order
. However, changing
the sign of l to give a repulsive
potential there is no bound state, the lowest energy plane wave state stays at
energy zero. Therefore the energy shift on
switching on the perturbation cannot be represented as a power series in l, the strength of the perturbation. This particular difficulty does not in
general occur in three dimensions, where arbitrarily weak potentials do not
give bound states—except for certain many-body problems (like the Cooper pair
problem) where the exclusion principle reduces the effective dimensionality of
the available states.
We begin with a
Hamiltonian
having known eigenkets
and eigenenergies:
![]()
The task is to find how these eigenkets and eigenenergies
change if a small term
(an external field,
for example) is added to the Hamiltonian, so:
![]()
That is to say, on switching on
,
![]()
The basic assumption in perturbation theory is that
is sufficiently small
that the leading corrections are the same order of magnitude as
itself, and the true energies can be better and better
approximated by a successive series of corrections, each of order
compared with the previous one.
The strategy, then, is to expand the true wave function and
corresponding eigenenergy as series in
. These series are
then fed into
, and terms of the same order of magnitude in
on the two sides are
set equal. The equations thus generated
are solved one by one to give progressively more accurate results.
To make it easier to identify terms of the same order in
on the two sides of
the equation, it is convenient to introduce a dimensionless parameter
which always goes with
, and then expand
as power series in
,
, etc. The ket
multiplied by
is therefore of order
This
is purely a bookkeeping device: we will set it equal to 1
when we are through! It’s just
there to keep track of the orders of magnitudes of the various terms.
Putting the series expansions for
in
![]()
we have
![]()
We’re now ready to match the two sides term by term in
powers of
.
The zeroth-order term, of course, just gives back
Matching the terms linear in l on both sides:
![]()
This equation is the key to finding the first-order change
in energy
. Taking the inner
product of both sides with
:
![]()
then using
, and
, we find
![]()
Taking now
, we have established that the first-order change in the
energy of a state resulting from adding a perturbing term
to the Hamiltonian is
just the expectation value of
in that state.
For example, we can estimate the ground state energy of the
helium atom by treating the electrostatic repulsion between the electrons as a
perturbation. The zeroth-order ground
state has the two (opposite spin) electrons in the ground state hydrogen-atom
wave function (scaled for the doubling of nuclear charge). The first-order
energy correction
is then given by
computing the expectation value
for this ground state
wave function.
The general expression for the first-order change in the wave function is found by taking the
inner product of the first-order equation with the bra
,
![]()
The last term is zero, since
, and in the first term
so

and therefore the wave function correct to first order is:

To find the second-order correction to the energy, it is necessary to match the second-order terms in
![]()
giving:
![]()
Taking the inner product with
yields:
![]()
The leading terms on the two sides cancel as before. What about the term
? Since
, and both
are normalized,
in leading order—that
is to say,
is pure
imaginary. That just means that if to
this order
has a component
parallel to
, that component has a small pure imaginary amplitude, and
can be written (to
this order) as
, with
small. But the phase factor can be eliminated by
redefining the phase of
, and with that redefinition
has no component in the
direction, we can
therefore drop the term ![]()
So the second-order correction to the energy is:

Perturbation theory involves evaluating matrix elements of operators. Very often, many of the matrix elements in a sum are zero—obvious tests are parity and the Wigner-Eckart theorem. These are examples of selection rules: tests to find if a matrix element may be nonzero.
When a hydrogen atom in its ground state is placed in an electric field, the electron cloud and the proton are pulled different ways, an electric dipole forms, and the overall energy is lowered.
The perturbing Hamiltonian from the electric field is
, where
is the electric field
strength, the field is in the z-direction,
the electron charge e is negative.
We shall denote the unperturbed eigenenergies of the
hydrogen atom by
, so in particular we denote
the ground state energy by E1.
The first-order correction to the ground state energy
, where

This first-order term is zero since there are equal contributions from positive and negative z.
The second-order term is

where we are now using
to denote the
unperturbed hydrogen atom wave functions, and here the
(in Rydbergs) are the unperturbed
energies.
Most of the terms in this infinite series are zero—the
selection rules help get rid of them as follows: since
is the
component of a
spherical vector and
is a zero angular
momentum state, it follows from the Wigner-Eckart theorem that
. This reduces the
second-order sum over states to:

This is still not easy to evaluate, but an upper bound
can be found be observing that
, so
![]()
where we have temporarily restored the full sum over n,
l, m, that is, we’ve put back all the zero terms. The reason for this seeming backward step is
that, having taken the energy-difference denominator outside the sum, we can
even include
in the
sum (it’s another zero
term) and in fact we can even include the plane-wave (ionized) states as well
as the bound states, since the plane waves all have energy greater than zero.
At this point, the
sum becomes a sum over all states, and therefore just becomes
the unit operator,
![]()
so
![]()
For the ground state hydrogen wave function,
,
, so
![]()
Furthermore, since all the terms in the series for
are negative, the first
term sets a lower bound on
:

This can be evaluated in straightforward fashion to
find
So, even though we have not actually evaluated the
second-order correction to the energy explicitly, we have it bracketed between
two values, the lower one being more than half the upper one. Other ingenious methods have been developed
(see Shankar or Sakurai) to find that the true answer is
, but in fact the whole problem can be solved exactly using
parabolic coordinates.
The above analysis works fine as long as the successive
terms in the perturbation theory form a convergent series. A necessary condition is that the matrix
elements of the perturbing Hamiltonian must be smaller than the corresponding
energy level differences of the original Hamiltonian. If
has different states
with the same energy, in other words degenerate energy levels, and the
perturbation has nonzero matrix elements between
these degenerate levels, then obviously the theory breaks down. To see just how it breaks down, and how to
fix it, we consider the two-dimensional simple harmonic oscillator:
![]()
Recall that for the one-dimensional simple harmonic oscillator the ground state wave function is
with
, and ![]()
The two-dimensional oscillator is simply a product of two one-dimensional
oscillators, so, writing
, the ground state is
, and the two (degenerate) next states, energy
above the ground state, are
![]()
Suppose now we add a small perturbation
![]()
with
a small parameter.
Notice that
, so according to naïve perturbation theory, there is no first-order correction to the
energies of these states.
However, on going to second-order in the energy correction,
the theory breaks down. The matrix
element
is nonzero, but the
two states
have the same
energy! This gives an infinite term in
the series for
.
Yet we know that a small term of this type will not wreck a
two-dimensional simple harmonic oscillator, so what is wrong with our
approach? It is helpful to plot the
original harmonic oscillator potential
together with the
perturbing potential
. The first of course
has circular symmetry, the second has axes in the directions
, climbing most steeply from the origin along x = y,
falling most rapidly in the directions x
= -y.
If we combine the two potentials into a single quadratic form, the
original circles of constant potential become ellipses, with their axes aligned
along
.
The problem arises even in the classical two-dimensional oscillator: picture a ball rolling backwards and forwards in a smooth saucer, a circular bowl. Now imagine the saucer is made slightly elliptical. The ball will still roll backwards and forwards through the center if it is released along one of the axes of the ellipse, although with different periods, as the axes differ in steepness. However, if it is released at a point off the axes, it will describe a complex path resolvable into components in the two axis directions having different periods.
For the quantum oscillator as for the classical one, as soon
as the perturbation is introduced, the eigenkets are in the direction of the
new elliptic axes. This is a large
change from the original x and y axes, and definitely not proportional to the small parameter
. But the original
unperturbed problem had circular symmetry, and there was no particular reason
to choose the x and y axes as we did. If we had instead chosen as our original axes
the lines
, the kets would not
have undergone large changes on switching on the perturbation.
The resolution of the problem is now clear: before switching on the perturbation, choose a set of basis kets in a degenerate subspace such that the perturbation is diagonal in that subspace.
In fact, for the simple harmonic oscillator example above, the problem can be solved exactly:

and it is clear that, despite the results of naïve first-order theory, there is indeed a first order shift in the energy levels,
![]()
The hydrogen atom, like the two-dimensional harmonic oscillator discussed above, has a nondegenerate ground state but degeneracy in its lowest excited states. Specifically, there are four n = 2 states, all having energy -1/4 Ryd :



Perturbing this system with an electric field in the z-direction,
, note first that
naïve perturbation theory predicts no
first-order shift in any of these energy levels. However, to second order, there is a nonzero
matrix element between two degenerate levels
. All the other matrix
elements between these basis kets in the four-dimensional degenerate subspace
are zero, so the only diagonalization necessary is within the two-dimensional degenerate subspace
spanned by
, where
![]()
with

Diagonalizing
within this subspace,
then, the new basis states are
with energy shifts
, linear in the perturbing electric field.
The states
are not changed by the
presence of the field to this approximation, so the complete energy map of the n = 2 states in the electric field has
two states at the original energy of -1/4Ryd, one state moved up from that
energy by
, and one down by
.
Notice that the new eigenstates
are not eigenstates of the parity operator—a
sketch of their wave functions reveals that in fact they have nonvanishing
electric dipole moment
, indeed this is the reason for the energy shift,
.