# Commutator of $\hat {L}_x$ and $\hat{V}(\hat{r})$ [duplicate]

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Consider the angular momentum operator $$\hat{L_x}=\hat y\hat{p}_z-\hat{z}\hat{p}_y$$ and the potential operator $$\hat{V}$$ where the potential $$\hat{V}=\hat{V}(\hat{r})$$ is spherically symmetric.

It can be shown that $$[\hat {L}_x, \hat {V}] = \hat y [\hat {p}_z, \hat V]-\hat z [\hat {p}_y, \hat V].$$ I tried showing this myself and found that this is only true if $$[\hat y, \hat V]=0$$ and $$[\hat z, \hat V]=0.$$

Why are the last two equations equal to zero?

## marked as duplicate by Kyle Kanos, ahemmetter, Cosmas Zachos, A.V.S., ZeroTheHeroDec 27 '18 at 15:00

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## 1 Answer

Because the operator $$\hat{\boldsymbol{r}}$$ is just

$$\hat{\boldsymbol{r}}\psi=\boldsymbol{r}\psi$$

and similarly

$$\hat{V}(\hat{\boldsymbol{r}})\psi=V(\boldsymbol{r})\psi$$

i.e. simply multiplication with a real number. And as you know, real numbers are commutative.