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# Degenerate Perturbation Theory

Let us now consider systems in which the eigenstates of the unperturbed Hamiltonian, , possess degenerate energy levels. It is always possible to represent degenerate energy eigenstates as the simultaneous eigenstates of the Hamiltonian and some other Hermitian operator (or group of operators). Let us denote this operator (or group of operators) . We can write

 (7.66)

and

 (7.67)

where . Here, the and the are real numbers that depend on the quantum numbers , and and , respectively. It is always possible to find a sufficient number of operators that commute with the Hamiltonian in order to ensure that the are all different. In other words, we can choose such that the quantum numbers and uniquely specify each eigenstate. Suppose that for each value of there are different values of . In other words, the th energy eigenstate is -fold degenerate.

In general, does not commute with the perturbing Hamiltonian, . In this situation, we expect the perturbation to couple degenerate states with the same value of , but different values of . Let us naively attempt to use the standard perturbation theory of Section 7.3 to evaluate the modified energy eigenstates and energy levels. A direct generalization of Equations (7.32) and (7.33) yields

 (7.68)

and

 (7.69)

where

 (7.70)

It is fairly obvious that the summations in Equations (7.69) and (7.70) are not well behaved if the th energy level is degenerate. The problem terms are those that involve coupling to unperturbed eigenstates labeled by the same value of , but different values of : that is, those states whose unperturbed energies are . These terms give rise to singular factors in the summations. Note, however, that this problem would not exist if the matrix elements, , of the perturbing Hamiltonian between distinct, degenerate, unperturbed energy eigenstates corresponding to the eigenvalue were zero. In other words, if

 (7.71)

then all of the singular terms in Equations (7.69) and (7.70) would vanish.

In general, Equation (7.72) is not satisfied. Fortunately, we can always redefine the unperturbed energy eigenstates belonging to the eigenvalue in such a manner that Equation (7.72) is satisfied. Let us define new states that are linear combinations of the original degenerate eigenstates corresponding to the eigenvalue :

 (7.72)

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 eigenstates of the perturbing Hamiltonian, . Thus,

 (7.73)

The are also chosen so that they are orthonormal. It follows that

 (7.74)

Thus, if we use the new eigenstates, instead of the old ones, then we can employ Equations (7.69) and (7.70) directly, because all of the singular terms vanish. The only remaining difficulty is to determine the new eigenstates in terms of the original ones.

Now,

 (7.75)

where 1 denotes the identity operator in the sub-space of all unperturbed energy eigenkets corresponding to the eigenvalue . Using this completeness relation, the operator eigenvalue equation (7.74) can be transformed into a straightforward matrix eigenvalue equation:

 (7.76)

This can be written more transparently as

 (7.77)

where the elements of the Hermitian matrix are

 (7.78)

Provided that the determinant of is non-zero, Equation (7.78) can always be solved to give eigenvalues (for to ), with corresponding eigenvectors [92]. The eigenvectors specify the weights of the new eigenstates in terms of the original eigenstates: that is,

 (7.79)

for to . In our new scheme, Equations (7.69) and (7.70) yield

 (7.80)

and

 (7.81)

There are no singular terms in these expressions, because the summations are over . In other words, they specifically exclude the problematic, degenerate, unperturbed energy eigenstates corresponding to the eigenvalue . Note that the first-order energy-shifts are equivalent to the eigenvalues of the matrix equation (7.78).

Incidentally, it is clear, from the previous analysis, that if the perturbing Hamiltonian, , commutes with and the operator (or group of operators) then there is no danger of singular terms appearing in the perturbation expansion to second order, because . Another way of saying this is that there are no singular terms if the simultaneous eigenstates of and are also eigenstates of .

Next: Linear Stark Effect Up: Time-Independent Perturbation Theory Previous: Quadratic Stark Effect
Richard Fitzpatrick 2016-01-22