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# 11.5: 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

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

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