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I was solving a homework problem where the question gives the representation of two operators in matrix form, in some arbitrary set of basis vectors. It then asks to find the simultaneous eigenstates of the two operators (they commute). Finally, it asks if specifying the eigenvalues uniquely specifies the eigenstates. I first found the eigenvalues of each operator by writing out the characteristic equation for each matrix - I assume the eigenvalues should not change when you change basis set. The spectrum of each operator has degeneracies. Then, I wrote down constraints on each each component of the eigenvector and guessed a set of simultaneous eigen states. From the constraints, I showed that there exists a unique choice of simultaneous eigen states for these two operators.

My question is are there situations in which the choice of simultaneous eigenstates is not unique. Or must they never be unique - in which case my conclusion was wrong?

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    $\begingroup$ Take $A={\bf 1}$ and $B=2{\bf 1}$. Then a basis for simultaneous eigenspaces is not unique. Related: physics.stackexchange.com/q/361341/2451 , physics.stackexchange.com/q/142548/2451 and links therein. $\endgroup$ – Qmechanic Jan 24 '18 at 22:01
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    $\begingroup$ Right. It's enough for your commuting operators to share an eigenspace of dimension higher than 1 to see counterexamples in which your common eigenvectors are not unique. And that be accomplished with ease, as @Qmechanic points out. $\endgroup$ – secavara Jan 24 '18 at 22:08
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Yes. A very simple examples would be $L^2$ and $L_z^2$. The $\vert \ell m\rangle$ states $\vert 1,1\rangle$ and $\vert 1,-1\rangle$ are both eigenvectors of the two operators, but so is the arbitrary combination $$ \vert\psi(\theta,\phi)\rangle =\cos\theta \vert 1,1\rangle + e^{i\phi}\sin\theta \vert 1,-1\rangle $$ given in terms of the continuous parameters $\theta$ and $\phi$.

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