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In pag. 270 of Griffith's "Introduction to Quantum Mechanics" a perturbative method for finding relativist correction to the energy levels of the Hydrogen athom is exposed.

It is asserted, if I understand well, that the operator consisting in $\hat{L}^2\hat{L}_z$ commutes with the Hamiltonian (both the perturbed one and the unperturbed) and has different eigenvalues for each of the $n^2$ eigenstates that have the same $E_n$. This justifies the use of the nondegenerate theory.

However, it seems to me that the degeneracy is really $2n^2$, because of the spin. This problem can indeed be solved if we consider the operator $\hat{L}^2\hat{L}_z\hat{S}_z$, but Griffith makes no mention of this. Am I missing something?

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Note that the underlying point of the paragraph you mention is to justify the use of nondegenerate perturbation theory. Earlier in the chapter he shows that the nondegenerate perturbation theory is justified to use even for a problem $H^0$ that has degenerate states \begin{align} H^0|a^0\rangle&=E_a|a^0\rangle,\\H^0|b^0\rangle&=E_b|b^0\rangle,\\E_a&=E_b \end{align}

iff the perturbation matrix elements in the degenerate subspace disappear, i.e. $\langle a^0|H^1|b^0\rangle=0$.

On the other hand this is precisely what happens if you happen to find a hermitian operator $A$ which commutes with both the original ($H^0$) and perturbation ($H^1$) Hamiltonians, and has distinct eigenvalues in the degenerate subspace ($A|a^0\rangle=a|a^0\rangle,A|b^0\rangle=b|b^0\rangle$). This is because

$$ \langle a^0|[A,H^1]|b^0\rangle=(a-b)\langle a^0|H^1|b^0\rangle $$

and since $[A,H^1]=0$, and $a\neq b$ it follows that $\langle a^0|H^1|b^0\rangle=0$. In other words it is beneficial to find a symmetry that you haven't exploited in a degenerate problem.

Now, since the lowest-order relativistic correction $H^1=-p^4/8m^3c^2$ is spherically symmetric, it will commute with $L^2$ and $L_z$. And since $L^2$ and $L_z$ already commute with the unperturbed Hamiltonian according to the previous discussion it is justified to use nondegenerate perturbation theory (with the small caveat addressed by footnote 9 in Griffith's book). There is never any mention of an operator $L^2L_z$.

Finally, as you correctly point out, the degeneracy is $2n^2$ if we include spin. Again having in mind the previous discussion (and the fact that the correction is kinetic in nature), it would be justified to use nondegenerate perturbation theory by using the basis that diagonalizes $H^0$,$L^2$,$L_z$ and $S_z$. But the real point to make here is that since this relativistic correction is just a kinetic addition, it has nothing to do with spin, and one need not include it in the calculation - it will just be a complication.

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