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)

  \( \newcommand {\ltapp} {\stackrel {_{\normalsize<}}{_{\normalsize \sim}}}\) \(\newcommand {\gtapp} {\stackrel {_{\normalsize>}}{_{\normalsize \sim}}}\) \(\newcommand {\btau}{\mbox{\boldmath$\tau$}}\) \(\newcommand {\bmu}{\mbox{\boldmath$\mu$}}\) \(\newcommand {\bsigma}{\mbox{\boldmath$\sigma$}}\) \(\newcommand {\bOmega}{\mbox{\boldmath$\Omega$}}\) \(\newcommand {\bomega}{\mbox{\boldmath$\omega$}}\) \(\newcommand {\bepsilon}{\mbox{\boldmath$\epsilon$}}\)