The full Hamiltonian of the hydrogen atom (including fine structure) is $$H = H_0 + H_{rel} + H_{s.o},$$ where $H_0 = \frac{p^2}{2m} -\frac {e^2}r$, $H_{rel}$ (resp. $H_{s.o}$) is the relativistic (resp. spin-orbit) correction ( I did not include the Darwin correction, since degenerate perturbation theory needn't be applied in that case). The corrections are defined by ($V(r) = \frac {e^2}r$) $$H_{s.o} = k \frac 1 r \frac{\partial V}{\partial r}L\cdot S$$ where $k$ is a constant, and $$H_{rel}= -\frac {p^4}{8m^3c^2}.$$The eigenstates $|nlm_lm_s\rangle$ of $H_0$ are $n^2$-fold degenerate, so we must use degenerate perturbation theory. However, it doesn't seem like the degeneracy will ever be lifted completely, as none of the fine-structure corrections involve the magnetic quantum number (that is, states with different $m_l$ but equal $l$ or different $m_s$ (equivalently different $m$ and identical $j$, using total angular momentum) will have equal corrections.) How is it that we can get away with using 1st order degenerate perturbation theory on the hydrogen atom, when the degeneracies aren't lifted to 1st order (or possibly any order)? Is it just by using the $|nljm\rangle$ basis which diagonalizes each perturbation that everything works out?
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$\begingroup$ I suggest writing the expressions for the corrections and/or their matrix elements explicitly, since otherwise the question lacks the essential information (which likelyreadily answer it). $\endgroup$– Roger V.Commented Jun 5, 2022 at 19:35
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1$\begingroup$ @RogerVadim I added the explicit expressions. $\endgroup$– Daniel WatersCommented Jun 5, 2022 at 20:58
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You answered your own question already: "Is it just by using the $|nljm⟩$ basis which diagonalizes each perturbation that everything works out?" Yes, indeed. The whole purpose of perturbation theory, degenerate or not, is to try to diagonalize a Hamiltonian to a given order. If you work with a basis in which $H$ is already diagonal, you essentially bypass perturbation theory.
