I'm struggling to understand the commutator theory for quantum mechanics. I know there's the proof to do with $[P,Q]=0$ therefore there is a set of simultaneous eigenstates for $P$ and $Q$. However, if $[P,Q] \neq 0$, does it also mean that then for all $|\psi\rangle \neq 0$, we have $[P,Q]|\psi\rangle \neq 0$? Or there is a way for it to equal 0?

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    $\begingroup$ Can a general linear map have a nonzero kernel? $\endgroup$
    – NDewolf
    Nov 18, 2020 at 16:49
  • $\begingroup$ @NDewolf Well, you've got $T(u) = 0_V$ but the kernel contains a set of vectors $\endgroup$ Nov 18, 2020 at 16:58
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    $\begingroup$ Two non-commuting operators can share an eigenvector. They just can't share all of them. If $|\psi\rangle$ is an eigenvector, then the commutator acting on this vector will give zero. I provide some clarification here. $\endgroup$
    – march
    Nov 18, 2020 at 17:19
  • $\begingroup$ There is no such thing as $\psi=0$. $\endgroup$
    – my2cts
    Nov 18, 2020 at 18:32

1 Answer 1


As a counterexample: in general, the operators $\hat{L}_x$ and $\hat{L}_y$ do not commute, since $[\hat{L}_x, \hat{L}_y] = i \hbar \hat{L}_z$. However, for any state $|\psi\rangle \neq 0$ which is an eigenvector of $\hat{L}_z$ with $L_z = 0$, we have $[L_x, L_y] |\psi\rangle = 0$.


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