Proof that commuting operators have a shared base of eigenfunctions

[jakotaco]

I have been told that if we have two operators, A and B, such that AB = BA then this is equivalent with that A and B have a common base of eigenfunctions.

However, the proof given was made under the assumption that the operators had a non-degenerate spectrum. Now I understand that one rather wants to give a short proof under some simplified condition and hope people will take your word that it applies without those simplifications. But in quantum mechanics, it is not as if degeneracy is uncommon.

So, does anyone know how to prove it more generally? Or maybe an explanation why the assumption isn't as bad as I though?

 

[strangerep]

I have been told that if we have two operators, A and B, such that AB = BA then this is equivalent with that A and B have a common base of eigenfunctions.

However, the proof given was made under the assumption that the operators had a non-degenerate spectrum. Now I understand that one rather wants to give a short proof under some simplified condition and hope people will take your word that it applies without those simplifications. But in quantum mechanics, it is not as if degeneracy is uncommon.

So, does anyone know how to prove it more generally? Or maybe an explanation why the assumption isn't as bad as I though?

If you search back through this forum for recent weeks, you'll find another thread
where I gave a proof - and got the same objection about degeneracy. Someone else
added the extra detail.

Personally, I don't think it's all that bad, since one can regard the degenerate
eigenspaces as vectors if you think more abstractly in terms of equivalence classes.
Others may find this a bit distasteful, however.

 

[jakotaco]

ok, thanks I found this thread by browsing your post history: https://www.physicsforums.com/showthread.php?t=417407

This is the step I am unsure about

$$H (A |n\rangle) ~=~ \lambda_n (A |n\rangle) ~.$$

Therefore, $(A |n\rangle)$ is an eigenvector of H with eigenvalue $\lambda_n$
and hence is proportional to $|n\rangle$. (This follows because all the
eigenvectors are mutually orthogonal.) So we have:

$$A |n\rangle ~\propto~ |n\rangle ~.$$

But I will try giving it some more thought, reading some of the similar threads on this forum and see if it comes more clearly.

 

[arkajad]

One way of doing it is by spectral decomposition:

$$A=\sum a_n P_n$$

$$B=\sum b_n Q_n$$

where $$a_n,b_n$$ are eigenvalues and $$P_n,Q_n$$ are projection operators on eigenspaces of A and B resp. Then it is an easy exercise to show that A and B commute if an only if all $$P_n$$ and $$Q_m$$ commute. If so, then

$$P_nQ_m$$ are also projections and we have orthogonal sum

$$\sum_{m,n}P_nQ_m=I$$.

You then take any orthonormal basis from vectors contained in $$P_nQ_m$$ subspaces and you are done one way. The other way is easy.