Verification of proof of complete set of commuting operators

Question

Hi I am interested in the validity of the following proof. I am interested in the validity of this particular proof as I am aware of how to prove this result in a different way.

Theorem: If two observables are compatible, their corresponding self-adjoint operators possess a set of common (or simultaneous) eigenstates (this theorem holds for both degenerate and nondegenerate eigenstates).

Proof: If $|\psi_n \rangle$ is a nondegenerate eigenstate of $\hat{A}$, $\hat{A}|\psi_n \rangle = a_{n}| \psi_n \rangle$, we have

$$\langle \psi_m | [\hat{A},\hat{B}]| \psi_n \rangle = (a_m - a_n)\langle \psi_m | \hat{B} | \psi_n \rangle = 0$$ since $\hat{A}$ and $\hat{B}$ commute. So $\langle \psi_m | \hat{B} | \psi_n \rangle$ must vanish unless $a_n = a_m$. That is, $$\langle \psi_m | \hat{B} | \psi_n \rangle = \langle \psi_n | \hat{B} | \psi_n \rangle ~\propto~ \delta_{nm}.$$

Hence the $|\psi_n \rangle$ are joint or simultaneous eigenstates of $\hat{A}$ and $\hat{B}$.

I don't see why it can't be that $\langle \psi_m | \hat{B} | \psi_n \rangle = 0$ even if $a_n = a_m$.

Also why does $\langle \psi_m | \hat{B} | \psi_n \rangle \propto \delta_{nm}$ imply that $|\psi_n\rangle$ is an eigenstate of $\hat{B}$? The following implication only goes in one direction $\hat{B}|\psi_n \rangle = b_n |\psi_n \rangle \implies \langle \psi_n| \hat{B} | \psi_n \rangle = b_n$, hence we don't necessarily have that $| \psi_n \rangle$ is an eigenstate of $\hat{B}$.

Answer 1

I think the proof can actually work, but it needs to be formulated a bit better and it needs a bit more of explaining. Take $|\psi_n \rangle$ as a non-degenerate Eigenstate of $\hat{A}$. Then for all other Eigenvectors $|\psi_m \rangle$: $$\langle \psi_m | [\hat{A},\hat{B}]| \psi_n \rangle = (a_m - a_n)\langle \psi_m | \hat{B} | \psi_n \rangle = 0$$ So: $\hat{B} | \psi_n \rangle$ is not in the linear hull of all the other eigenvectors $| \psi_m \rangle$. If they do span our whole Hilbert space, it follows that $\hat{B} | \psi_n \rangle$ needs to be in the linear hull of $| \psi_n \rangle$, thus this is an eigenvector of $\hat{B}$.

If we have a set of degenerate Eigenvectors $|\psi_{n_1} \rangle, ..., |\psi_{n_k} \rangle$, then $\hat{B}|\psi_{n_1} \rangle, ..., \hat{B}|\psi_{n_k} \rangle$ are by the same argument in the linear hull of $|\psi_{n_1} \rangle, ..., |\psi_{n_k} \rangle$. The choice of the basis in this linear subspace is of course free and can be taken to be $|\psi_{n_1} \rangle, ..., |\psi_{n_k} \rangle$.

Summarising, the eigenvectors of $\hat{A}$ are eigenvectors of $\hat{B}$ if they span the whole Hilbert space

I am not quite sure what happens if $\hat{A}$ and $\hat{B}$ are only defined on subspaces of the whole Hilbert space and I hope I did not make any stupid mistake ...