# Commutators and implications of them not equalling 0

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?

• Can a general linear map have a nonzero kernel? Nov 18, 2020 at 16:49
• @NDewolf Well, you've got $T(u) = 0_V$ but the kernel contains a set of vectors Nov 18, 2020 at 16:58
• 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. Nov 18, 2020 at 17:19
• There is no such thing as $\psi=0$. Nov 18, 2020 at 18:32

## 1 Answer

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 $$[L_x, L_y] |\psi\rangle = 0$$.