Question: For $\vec L$ the orbital angular momentum of an electron, $\bar S$ its spin, and $\vec J:=\vec L+\vec S$ the sum, do $\vec J^2$ and $\vec L^2$ commute?

I assume it does: $[\vec J^2,\vec L^2]=\sum_i[(L_i+S_i)^2,\vec L^2]=2\sum_i[L_iS_i,\vec L^2]$ (as both $L_i^2$ and $S_i^2$ commute with $\vec L^2$) and $[L_iS_i,\vec L^2]=L_i[S_i,\vec L^2]+[L_i,\vec L^2]S_i=0$.

But in a lecture note, it was said (I paraphrase and translate):

Even in the finestructure case, where $H$ depends not only on $n$ but on $j$ as well, one can cum grano salis speak of an orbital momentum of the stationary solution. E.g., for $n=2$, one can still speak of the states $2{s_{1/2}}$, $2p_{1/2}$ and $2p_{3/2}$, even if the states really are a superposition of states with $l=j-1/2$ and $l=j+1/2$. The reason is that Dirac theory implies that in this superposition one of the parts is dominant.

If $\vec J^2$ and $\vec L^2$ commute anyway, I cannot make much sense out of that quote. What is going on here?

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    $\begingroup$ Of course they do commute. The problem is that $\vec{L}^2$ and $\vec{S}^2$ may not commute with the Hamiltonian $H$. If the fine structure splitting of the hydrogen atom is treated exactly by the methods of relativistic quantum theory, the solutions are superpositions of states of orbital angular momenta ($\ell = 0, 2, 4, ...$ or $\ell = 1, 3, 5, ...$), because the relativistic Hamiltonian does not commute with $\vec{L}^2$. $\endgroup$ – QuantumDot Mar 13 '15 at 14:47
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    $\begingroup$ @QuantumDot that could be an answer. $\endgroup$ – Robin Ekman Mar 13 '15 at 14:49
  • $\begingroup$ @QuantumDot: Well, this is what I do not understand: The Hamiltonian(at least in the approximation that is used in the lecture note) only depends on $n$ (principal quantum number) and $\vec J^2$. So when I have a state that corresponds to, say, $n=2$ and $j=1/2$, then this determines the energy. Additionally $\bar J^2$ collumtes with $\bar L^2$, and also $n$ ``commutes'' with $\bar L^2$ (I assume by that I mean a projection to the $n$ principal number states), so why should there be a problem to fix a $2s_{1/2}$-state? $\endgroup$ – Jakob Mar 13 '15 at 19:26

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