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Suppose we have a Hamiltonian operator $\hat{H}$ and another operator $\hat{A}$ such that $[\hat{H},\hat{A}]=0$. Then, if the spectrum of $\hat{H}$ is non-degenerate, from my understanding the eigenvectors of $\hat{H}$ are also eigenvectors of $\hat{A}$. In the case where $\hat{H}$ admits some degeneracy I don't believe this is true. From my understanding there still exists eigenvectors shared by both operators, but a change of basis may be required.

How may one find eigenvectors that diagonalize $\hat{H}$ and $\hat{A}$ simultaneously if $\hat{H}$ admits some degeneracy?

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Any linear combination of degenerate eigenvectors is also an eigenvector, corresponding to the same value. Thus, we can first find the eigenvectors of $H$ and then look for eigenvectors of $A$ as linear combinations of the eigenvectors of $H$ corresponding to the same eigenvalue.

In other words, if $H|n,\nu\rangle=\epsilon_n|n,\nu\rangle$, then $$ HA|n,\nu\rangle=AH|n,\nu\rangle=A\epsilon_n|n,\nu\rangle=\epsilon_nA|n,\nu\rangle $$ that is $A|n,\nu\rangle$ is an eigenvector of $H$, corresponding to the same eigenvalue $\epsilon_n$. If the spectrum is non-degenerate, then $A|n,\nu\rangle$ and $|n,\nu\rangle$ are the same state - they differ only by a constant factor, which is the eigenvalue of $A$: $$ A|n,\nu\rangle=a_{n,\nu}|n,\nu\rangle. $$ However, if the spectrum of $H$ is degenerate, then $$ A|n,\nu\rangle=\sum_{\mu}c_{n,\mu}|n,\mu\rangle. $$ We thus end up with diagonalizing matrices of the size equal to the degeneracy of each degenerate state - this is, of course, easier then solving the ad-hoc eigenstate problem for $A$.

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