# Are the unperturbed hydrogen wavefunctions solutions of the fine structure Schrödinger equation?

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?

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

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 $$j=3/2$$).