Degenerate perturbation theory

Let us now consider systems in which the eigenstates of the unperturbed Hamiltonian, $H_0$, 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) $L$. We can write

$$H_0 \vert n, l\rangle = E_n \vert n, l\rangle,$$ (651)

and

$$L \vert n,l\rangle = L_{nl} \vert n, l\rangle,$$ (652)

where $[H_0, L] = 0$. Here, the $E_n$ and the $L_{nl}$ are real numbers which depend on the quantum numbers $n$, and $n$ and $l$, respectively. It is always possible to find a sufficient number of operators which commute with the Hamiltonian in order to ensure that the $L_{nl}$ are all different. In other words, we can choose $L$ such that the quantum numbers $n$ and $l$ uniquely specify each eigenstate. Suppose that for each value of $n$ there are $N_n$ different values of $l$: i.e., the $n$th energy eigenstate is $N_n$-fold degenerate.

In general, $L$ does not commute with the perturbing Hamiltonian, $H_1$. This implies that the modified energy eigenstates are not eigenstates of $L$. In this situation, we expect the perturbation to split the degeneracy of the energy levels, so that each modified eigenstate $\vert n,l\rangle'$ acquires a unique energy eigenvalue $E_{nl}'$. Let us naively attempt to use the standard perturbation theory of Sect. 6.3 to evaluate the modified energy eigenstates and energy levels. A direct generalization of Eqs. (618) and (619) yields

$$E_{nl}' = E_n + e_{nlnl} + \sum_{n', l' \neq n,l} \frac{\vert e_{n'l'nl}\vert^2}{E_n - E_{n'}} + O(\epsilon^3),$$ (653)

and

$$\vert n, l\rangle' = \vert n,l\rangle + \sum_{n', l'\neq n, ... ...c{e_{n'l'nl}}{E_n-E_{n'}} \vert n',l'\rangle + O(\epsilon^2),$$ (654)

where

$$e_{n'l'nl} = \langle n',l'\vert H_1\vert n,l\rangle.$$ (655)

It is fairly obvious that the summations in Eqs. (653) and (654) are not well-behaved if the $n$th energy level is degenerate. The problem terms are those involving unperturbed eigenstates labeled by the same value of $n$, but different values of $l$: i.e., those states whose unperturbed energies are $E_n$. These terms give rise to singular factors $1/(E_n - E_n)$ in the summations. Note, however, that this problem would not exist if the matrix elements, $e_{nl'nl}$, of the perturbing Hamiltonian between distinct, degenerate, unperturbed energy eigenstates corresponding to the eigenvalue $E_n$ were zero. In other words, if

$$\langle n, l' \vert H_1\vert n, l\rangle = \lambda_{nl} \delta_{ll'},$$ (656)

then all of the singular terms in Eqs. (653) and (654) would vanish.

In general, Eq. (656) is not satisfied. Fortunately, we can always redefine the unperturbed energy eigenstates belonging to the eigenvalue $E_n$ in such a manner that Eq. (656) is satisfied. Let us define $N_n$ new states which are linear combinations of the $N_n$ original degenerate eigenstates corresponding to the eigenvalue $E_n$:

$$\vert n,l^{(1)}\rangle = \sum_{k=1}^{N_n} \langle n,k\vert n,l^{(1)}\rangle \vert n,k\rangle.$$ (657)

Note that these new states are also degenerate energy eigenstates of the unperturbed Hamiltonian corresponding to the eigenvalue $E_n$. The $\vert n,l^{(1)}\rangle$ are chosen in such a manner that they are eigenstates of the perturbing Hamiltonian, $H_1$. Thus,

$$H_1 \vert n, l^{(1)}\rangle = \lambda_{nl} \vert n, l^{(1)}\rangle.$$ (658)

The $\vert n,l^{(1)}\rangle$ are also chosen so that they are orthonormal, and have unit lengths. It follows that

$$\langle n, l'^{(1)} \vert H_1\vert n, l^{(1)}\rangle = \lambda_{nl} \delta_{ll'}.$$ (659)

Thus, if we use the new eigenstates, instead of the old ones, then we can employ Eqs. (653) and (654) 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

$$\sum_{l=1}^{N_n} \vert n,l\rangle \langle n,l\vert = 1,$$ (660)

where 1 denotes the identity operator in the sub-space of all unperturbed energy eigenkets corresponding to the eigenvalue $E_n$. Using this completeness relation, the operator eigenvalue equation (658) can be transformed into a straightforward matrix eigenvalue equation:

$$\sum_{l''=1}^{N_n}\langle n, l'\vert H_1\vert n, l''\rangle ... ...rangle = \lambda_{nl} \langle n, l'\vert n, l^{(1)}\rangle.$$ (661)

This can be written more transparently as

$${\bf U} {\bf x} = \lambda {\bf x},$$ (662)

where the elements of the $N_n\times N_n$ Hermitian matrix ${\bf U}$ are

$$U_{jk} = \langle n, j\vert H_1\vert n, k\rangle.$$ (663)

Provided that the determinant of ${\bf U}$ is non-zero, Eq. (662) can always be solved to give $N_n$ eigenvalues $\lambda_{nl}$ (for $l=1$ to $N_n$), with $N_n$ corresponding eigenvectors ${\bf x}_{nl}$. The eigenvectors specify the weights of the new eigenstates in terms of the original eigenstates: i.e.,

$$({\bf x}_{nl})_k = \langle n, k\vert n, l^{(1)}\rangle,$$ (664)

for $k=1$ to $N_n$. In our new scheme, Eqs. (653) and (654) yield

$$E_{nl}' = E_n + \lambda_{nl} + \sum_{n'\neq n, l'} \frac{\vert e_{n'l'nl}\vert^2}{E_n - E_{n'}} + O(\epsilon^3),$$ (665)

and

$$\vert n, l^{(1)}\rangle' = \vert n,l^{(1)}\rangle + \sum_{n'... ...c{e_{n'l'nl}}{E_n-E_{n'}} \vert n',l'\rangle + O(\epsilon^2).$$ (666)

There are no singular terms in these expressions, since the summations are over $n'\neq n$: i.e., they specifically exclude the problematic, degenerate, unperturbed energy eigenstates corresponding to the eigenvalue $E_n$. Note that the first-order energy shifts are equivalent to the eigenvalues of the matrix equation (662).