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To decide whether I have to use non-degenerate or degenerate perturbation theory, I have to look only on whether the energy level I am calculating corrections to is degenerate, the degree of degeneracy of the other levels is always immaterial for this, correct?

For instance, if I calculate corrections to ground state $n=1$ of the hydrogen atom, I use non-degenerate perturbation theory, but for all other levels $n\geq 2$ I use degenerate perturbation theory and might get splitting of energy levels, depending on the perturbation (e.g. the Stark effect of hydrogen atom).

The motivation behind this question is that although it seems plausible, in textbook chapters on non-degenerate perturbation theory it is often assumed that no eigenvalue of the unperturbed Hamiltonian is degenerate. Maybe there is something more behind this I overlook?

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You can consider degenerate perturbation theory as the more general case. So that you can think of the non-degenerate case, as a degenerate Hilbert subspace of dimension one.

In this case there is no fundamental distinction, and the methods do not depend on whether other subspaces are degenerate or not. Possibly what the statement in books mean is to simplify expressions when summing over other energy eigenstates in the correction to the wave function, so that for every energy there is one state in the sum, and we do not have to choose a basis for all other states and expand, before they have explained the procedure of degenerate perturbation theory.

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  • $\begingroup$ This is a nice answer. Maybe something like this should be somehow briefly mentioned in the books, though yes, it is fairly obvious. $\endgroup$ – wondering Mar 1 '15 at 9:25

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