# Diagonalizing Operators Simultaneously [duplicate]

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?

• Commented Jun 16, 2022 at 12:09

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 $$A$$.