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The quadratic Stark effect
Suppose that a oneelectron atom [i.e., either a hydrogen atom, or an alkali metal
atom (which possesses one valance electron orbiting outside a closed, spherically
symmetric shell)] is subjected to a uniform electric field in the positive
direction. The Hamiltonian of the system can be split into two
parts. The unperturbed Hamiltonian,

(621) 
and the perturbing Hamiltonian,

(622) 
It is assumed that the unperturbed energy eigenvalues and eigenstates are completely
known. The electron spin is irrelevant in this problem (since the spin operators
all commute with ), so we can ignore the spin degrees of freedom of the system.
This implies that the system possesses no degenerate energy eigenvalues. This is
not true for the energy levels of the hydrogen atom, due to the special
properties of a pure Coulomb potential.
It is necessary to deal with this case separately, because
the perturbation theory presented in Sect. 6.3 breaks down for degenerate
unperturbed energy levels.
An energy eigenket of the unperturbed Hamiltonian is characterized by three quantum numbersthe radial quantum number , and the two angular quantum numbers and
(see Sect. 5.6). Let us denote such a ket , and let its
energy level be . According to Eq. (618), the change in this
energy level induced by a small electric field is given by
Now, since

(624) 
it follows that

(625) 
Thus,

(626) 
giving

(627) 
since is, by definition, an eigenstate of with eigenvalue
. It is clear, from the above relation, that
the matrix element
is zero unless .
This is termed the selection rule for the quantum number .
Let us now determine the selection rule for . We have
where use has been made of Eqs. (297)(302).
Similarly,
Thus,
This reduces to
However, it is clear from Eqs. (297)(299) that

(633) 
Hence, we obtain

(634) 
Finally, the above expression expands to give

(635) 
Equation (635) implies that

(636) 
This expression yields
which reduces to

(638) 
According to the above formula, the matrix element
vanishes unless or . This matrix element can be written

(639) 
where
. Recall, however,
that the wavefunction of an state is spherically symmetric (see Sect. 5.3):
i.e.,
. It follows from Eq. (639)
that the matrix element
vanishes by symmetry when . In conclusion, the matrix element
is zero unless . This is
the selection rule for the quantum number .
Application of the selection rules to Eq. (623) yields

(640) 
Note that all of the terms in Eq. (623) which vary linearly with
the electric fieldstrength
vanish by symmetry, according to the selection rules.
Only those terms which vary quadratically with the
fieldstrength survive. The polarizability of an atom is defined in terms
of the energyshift of the atomic state as follows:

(641) 
Consider the ground state of a hydrogen atom. (Recall, that we cannot address
the excited states because they are degenerate, and our theory cannot
handle this at present). The polarizability of this state is given by

(642) 
Here, we have made use of the fact that
for a hydrogen atom.
The sum in the above expression can be evaluated approximately by noting that
[see Eq. (416)]

(643) 
for a hydrogen atom,
where

(644) 
is the Bohr radius. We can write

(645) 
Thus,

(646) 
However,
where we have made use of the fact that the wavefunctions of a hydrogen atom
form a complete set. It is easily demonstrated from the
actual form of the ground state wavefunction
that

(648) 
Thus, we conclude that

(649) 
The true result is

(650) 
It is actually possible to obtain this answer, without recourse to perturbation
theory, by solving Schrödinger's equation exactly in parabolic coordinates.
Next: Degenerate perturbation theory
Up: Approximation methods
Previous: Nondegenerate perturbation theory
Richard Fitzpatrick
20060216