0
$\begingroup$

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 $\hat{B}$.

$\endgroup$
  • 1
    $\begingroup$ Some statements are surely wrong. For example, before $\propto \delta_{mn}$, the expression doesn't even depend on $m$. $\endgroup$ – Luboš Motl Jun 2 '16 at 15:56
  • 1
    $\begingroup$ The key word here is "nondegenerate": If $a_m = a_n$, $\psi_n$ is definitely not a nondegenerate eigenstate of ${\hat A}$. As for why $\psi_n$-s are eigenstates of $\hat B$: they diagonalize it. $\endgroup$ – udrv Jun 2 '16 at 15:58
  • 1
    $\begingroup$ Actually the reason comes from $\langle \psi_m | {\hat B} | \psi_n \rangle = \delta_{mn} \langle \psi_n | {\hat B} | \psi_n\rangle$ : since the $|\psi_n\rangle$-s are a basis (complete set of eigenstates of $\hat A$), you have ${\hat B} |\psi_n\rangle = \sum_m{\langle \psi_m | {\hat B} | \psi_n \rangle | \psi_m \rangle} = \sum_m{\delta_{mn} \langle \psi_n | {\hat B} | \psi_n\rangle | \psi_m \rangle }= \langle \psi_n | {\hat B} | \psi_n\rangle | \psi_n\rangle$. $\endgroup$ – udrv Jun 2 '16 at 16:17
  • 1
    $\begingroup$ Alex: I think that $\propto$ should instead be a simple product. Otherwise, no, I don't think that it's an OK proof even with this fix. For example, it assumes that there is some non-degenerate eigenstate of A. But the theorem claims to claim that it works even for degenerate spectrum and that case isn't discussed by the proof at all. $\endgroup$ – Luboš Motl Jun 2 '16 at 16:18
  • 2
    $\begingroup$ The actual natural proof is trivial. Just take all eigenvalues of A, separate the Hilbert space to eigenspaces according to the eigenvalues, and for every eigenspace, B acts within this eigenspace because A,B commute, so divide the eigenspace to smaller eigenspaces according to the eigenvalues of B, and then pick any basis in any simultaneous eigenspace of A,B, that's it. I don't understand the purpose of debating some not quite correctly written proof of this trivial claim if the correct proof is so easy to be made. $\endgroup$ – Luboš Motl Jun 2 '16 at 16:22
1
$\begingroup$

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 ...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.