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The hydrogen atom Hamiltonian, with fine structure effects included, is $$H = \frac {p^2}{2m}-\frac{e^2}{r}-\frac{p^4}{8m^3c^2}+\frac{e^2}{2m^2c^2}\frac{\mathbf{S\cdot L}}{r^3}+\frac{\pi}{2}\frac{e^2\hbar ^2}{m^2c^2}\delta(r).$$ Where the first two terms are the unperturbed Hamiltonian, $H_0$, the third term is the relativistic correction, $H_{rel}$, the fourth term is the spin-orbit coupling, $H_{s.o}$, and the final term is the Darwin correction, $H_D$. Correct me if I'm wrong, but all of these terms commute with the square of the total angular momentum operator $\mathbf J^2= \mathbf S^2+\mathbf L^2 +2\mathbf {S\cdot L},$ and so the eigenfunctions $\psi_{njm}$ of $H_0$ (in the coupled basis) should diagonalize the entire Hamiltonian. Is this correct?

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  • $\begingroup$ not clear why one should think the term in $p^4$ and the term in $\delta(r)$ should commute with the unperturbed part $H_0$… $\endgroup$ Aug 16, 2022 at 4:20
  • $\begingroup$ Yes, it seems that you are right. Am I correct in saying that perturbation theory is not required for spin-orbit coupling? $\endgroup$ Aug 16, 2022 at 12:11

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You are not correct.

Your Hamiltonian commutes with $\vec J=\vec L+\vec S$ but $H_0$ does not commute with $p^4/8m^3c^2$ or with the $\delta(r)$ term.

With $(n\ell m)$ referring to unperturbed hydrogen states, the perturbation will mix $n$ values via the $p^4$ and $\delta(r)$ terms, and and the spin-orbit term will mix $\ell$ values because a given value of $j$ can occur as two states with $\Delta \ell=1$ (like $\ell_1=1$ and $\ell_2=2$) can combine with $s=1/2$ to form states with the same $j$ (like $j=3/2$).

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