11.6: Degenerate Perturbation Theory

Let us, rather naively, investigate the Stark effect in an excited (i.e., $n>1$) state of the hydrogen atom using standard non-degenerate perturbation theory. We can write $$H_0\,\psi_{nlm} = E_n\,\psi_{nlm},$$ because the energy eigenstates of the unperturbed Hamiltonian only depend on the quantum number $n$. Making use of the selection rules ([e12.63]) and ([e12.73]), non-degenerate perturbation theory yields the following expressions for the perturbed energy levels and eigenstates [see Equations ([e12.56]) and ([e12.57])]:

$$\label{e12.88} E_{nl}' = E_n + e_{nlnl} + \sum_{n',l'=l\pm 1}\frac{|e_{n'l'nl}|^{\,2}}{E_n-E_{n'}},$$ and

$$\label{e12.89} \psi'_{nlm} = \psi_{nlm} + \sum_{n',l'=l\pm 1}\frac{e_{n'l'nl}}{E_n-E_{n'}}\,\psi_{n'l'm},$$ where $$e_{n'l'nl} = \langle n',l',m|H_1|n,l,m\rangle.$$ Unfortunately, if $n>1$ then the summations in the previous expressions are not well defined, because there exist non-zero matrix elements, $e_{nl'nl}$, that couple degenerate eigenstates: that is, there exist non-zero matrix elements that couple states with the same value of $n$, but different values of $l$. These particular matrix elements give rise to singular factors $1/(E_n-E_n)$ in the summations. This does not occur if $n=1$ because, in this case, the selection rule $l'=l\pm 1$, and the fact that $l=0$ (because $0\leq l < n$), only allow $l'$ to take the single value 1. Of course, there is no $n=1$ state with $l'=1$. Hence, there is only one coupled state corresponding to the eigenvalue $E_1$. Unfortunately, if $n>1$ then there are multiple coupled states corresponding to the eigenvalue $E_n$.

Note that our problem would disappear if the matrix elements of the perturbed Hamiltonian corresponding to the same value of $n$, but different values of $l$, were all zero: that is, if

$$\label{e12.91} \langle n,l',m|H_1|n,l,m\rangle = \lambda_{nl}\,\delta_{ll'}.$$ In this case, all of the singular terms in Equations ([e12.88]) and ([e12.89]) would reduce to zero. Unfortunately, the previous equation is not satisfied in general. Fortunately, we can always redefine the unperturbed eigenstates corresponding to the eigenvalue $E_n$ in such a manner that Equation ([e12.91]) is satisfied. Suppose that there are $N_n$ coupled eigenstates belonging to the eigenvalue $E_n$. Let us define $N_n$ new states which are linear combinations of our $N_n$ original degenerate eigenstates: $$\psi_{nlm}^{(1)}= \sum_{k=1,N_n}\langle n,k,m|n,l^{(1)},m\rangle\,\psi_{nkm}.$$ Note that these new states are also degenerate energy eigenstates of the unperturbed Hamiltonian, $H_0$, corresponding to the eigenvalue $E_n$. The $\psi_{nlm}^{(1)}$ are chosen in such a manner that they are also eigenstates of the perturbing Hamiltonian, $H_1$: that is, they are simultaneous eigenstates of $H_0$ and $H_1$. Thus, $$\label{e12.93} H_1\,\psi_{nlm}^{(1)} = \lambda_{nl}\,\psi_{nlm}^{(1)}.$$ The $\psi_{nlm}^{(1)}$ are also chosen so as to be orthonormal: that is, $$\langle n,l'^{(1)},m|n,l^{(1)},m\rangle = \delta_{ll'}.$$ It follows that $$\langle n,l'^{(1)},m|H_1|n,l^{(1)},m\rangle =\lambda_{nl}\, \delta_{ll'}.$$ Thus, if we use the new eigenstates, instead of the old ones, then we can employ Equations ([e12.88]) and ([e12.89]) 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 [see Equation ([e12.20])] $$\sum_{l=1,N_n}|n,l,m\rangle\langle n,l,m|\equiv 1,$$ where $1$ denotes the identity operator in the sub-space of all coupled unperturbed eigenstates corresponding to the eigenvalue $E_n$. Using this completeness relation, the eigenvalue equation ([e12.93]) can be transformed into a straightforward matrix equation: $$\sum_{l''=1,N_n}\langle n,l',m|H_1|n,l'',m\rangle\,\langle n,l'',m|n,l^{(1)},m\rangle = \lambda_{nl}\,\langle n,l',m|n,l^{(1)},m\rangle.$$ This can be written more transparently as

$$\label{e12.100} {\bf U}\,{\bf x} = \lambda \,{\bf x},$$ where the elements of the $N_n\times N_n$ Hermitian matrix ${\bf U}$ are $$U_{jk} = \langle n,j,m|H_1|n,k,m\rangle.$$ Provided that the determinant of ${\bf U}$ is non-zero, Equation ([e12.100]) 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 normalized eigenvectors specify the weights of the new eigenstates in terms of the original eigenstates: that is, $$({\bf x}_{nl})_k = \langle n,k,m|n,l^{(1)},m\rangle,$$ for $k=1$ to $N_n$. In our new scheme, Equations ([e12.88]) and ([e12.89]) yield $$E_{nl}' = E_n +\lambda_{nl}+\sum_{n'\neq n,l'=l\pm 1}\frac{|e_{n'l'nl}|^{\,2}}{E_n-E_{n'}},$$ and $$\psi_{nlm}^{(1)'} = \psi_{nlm}^{(1)} + \sum_{n'\neq n,l'=l\pm 1} \frac{e_{n'l'nl}}{E_n-E_{n'}}\,\psi_{n'l'm}.$$ There are no singular terms in these expressions, because the summations are over $n'\neq n$: that is, 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 ([e12.100]).

Contributors and Attributions

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

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