I used to know that two operators can be simultaneously diagonalized, given they commute, they are hermitian and are non-degenerate (By simultaneous diagonalization, I mean they share a complete set of common eigenvectors). That's why the operators $L_z$ and $L^2$ commute but not all eigenvectors of $L^2$ (for instance the eigenvectors that it shares with $L_x$) are eigenvectors of $L_z$, this happens (that's what I knew) because of degeneracy, but today in a book that I follow it is argued on the contrary that the only necessity for Simultaneous diagonalization is that two operators must commute (they can be degenerate), can someone explain to me what's going on, because by that logic the eigenvectors of $L^2$ (including the ones it shares with $L_x$) will be eigenvectors of $L_z$ and that can't happen since $L_x$ and $L_z$ do not commute. Please Explain?
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2$\begingroup$ Related: These two operators commute…but their eigenvectors aren't all the same. Why? $\endgroup$– Qmechanic ♦Dec 11, 2021 at 21:15
1 Answer
The general statement is that if $A$ and $B$ commute, then they can be simultaneously diagonalized. If there are degeneracies in the spectrum say of $A$, there is freedom in choosing a basis for the degenerate subspace, since $\textit{any}$ vector in this subspace is an eigenvector of $A$ with the same eigenvalue, and one can be assured that there exists a choice such that both $A$ and $B$ become diagonal. To make it a little more concrete, let's look at the operator $L^2$. The eigenvectors are organized into degenerate subspaces labeled by $l=0,1,\cdots$, with degeneracy $2l+1$. Inside this $(2l+1)$-dimensional subspace, one can choose a basis labeled by the eigenvalues of $L_z$, then both $L_z$ and $L^2$ are diagonalized with this choice of eigenvectors. Or one can choose a basis to diagonalize $L_x$.
