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Degenerate Perturbation Theory
Let us, rather naively, investigate the Stark effect in an excited (i.e.,
) state of the hydrogen atom using standard nondegenerate
perturbation theory. We can write

(940) 
since the energy eigenstates of the unperturbed Hamiltonian only depend
on the quantum number . Making use of the selection rules
(917) and (927), nondegenerate perturbation theory
yields the following expressions for the perturbed energy levels
and eigenstates [see Eqs. (909) and (910)]:

(941) 
and

(942) 
where

(943) 
Unfortunately, if then the summations in the above expressions
are not welldefined, because there exist nonzero matrix elements, , which couple degenerate eigenstates: i.e., there exist nonzero matrix elements which couple states with the same value of , but
different values of . These particular matrix elements give rise to
singular factors in the summations. This does not occur if
because, in this case, the selection rule , and the
fact that (since ), only allow to take the single value 1.
Of course, there is no state with .
Hence, there is only one coupled state corresponding to the
eigenvalue . Unfortunately, if then there are multiple
coupled states corresponding to the eigenvalue .
Note that our problem would disappear if the matrix elements of the
perturbed Hamiltonian corresponding to the same value of , but
different values of , were all zero: i.e., if

(944) 
In this case, all of the singular
terms in Eqs. (941) and (942) would reduce to zero.
Unfortunately, the above equation is not satisfied. Fortunately, we can
always redefine the unperturbed eigenstates corresponding to the
eigenvalue in such a manner that Eq. (944) is satisfied.
Suppose that there are coupled eigenstates belonging
to the eigenvalue . Let us define new states
which are linear combinations of our original degenerate eigenstates:

(945) 
Note that these new states are also degenerate energy eigenstates of the
unperturbed Hamiltonian, , corresponding to the eigenvalue . The
are chosen in such a manner that they are also
eigenstates of the perturbing Hamiltonian, :
i.e., they are simultaneous eigenstates of and . Thus,

(946) 
The
are also chosen so as to be orthonormal:
i.e.,

(947) 
It follows that

(948) 
Thus, if we use the new eigenstates, instead of the old ones, then we
can employ Eqs. (941) and (942) directly, since all
of the singular terms vanish. The only remaining difficulty
is to determine the new eigenstates in terms of the original ones.
Now [see Eq. (874)]

(949) 
where denotes the identity operator in the subspace of all coupled unperturbed
eigenstates corresponding to the eigenvalue . Using this completeness
relation, the eigenvalue equation (946) can be
transformed into a straightforward matrix equation:

(950) 
This can be written more transparently as

(951) 
where the elements of the Hermitian matrix
are

(952) 
Provided that the determinant of is nonzero, Eq. (951)
can always be solved to give eigenvalues (for to ), with corresponding eigenvectors . The
normalized eigenvectors specify the weights of the new eigenstates in terms of the
original eigenstates: i.e.,

(953) 
for to . In our new scheme, Eqs. (941) and (942) yield

(954) 
and

(955) 
There are no singular terms in these expressions, since the summations
are over : i.e., they specifically exclude the problematic,
degenerate, unperturbed energy eigenstates corresponding to the eigenvalue
. Note that the firstorder energy shifts are equivalent to the
eigenvalues of the matrix equation (951).
Next: Linear Stark Effect
Up: TimeIndependent Perturbation Theory
Previous: Quadratic Stark Effect
Richard Fitzpatrick
20100720