Question regarding commutator of $\hat{p}$ and $V(\hat{r})$

Question

I recently did an exercise for which I had to calculate $[\hat{H}, \hat{p}] = [V(\hat{r}), \hat{p}]$ with $- \nabla V(r) = F(r)$ and calculated it the following way: $[\hat{H}, \hat{p}]\psi(r) = [V(\hat{r}), \hat{p}] \psi(r) = (V(\hat{r})\hat{p} - \hat{p}V(\hat{r})) \psi(r) = (-i \hbar V(\hat{r}) \nabla + i \hbar \nabla V(\hat{r})) \psi(r)$ and I propose that this equals $$-i \hbar(\nabla V(\hat{r})) \psi(r)$$ but that would mean using a "product rule" on $V$ as well even though $V$ is an operator. Am I wrong in assuming this?

Answer 1

You're dealing with operators in position-representation and you have to evaluate expressions like $$\hat{A} \psi(x) \equiv \langle x|\hat{A}|\psi\rangle \quad ,$$ with $\psi(x) \equiv \langle x|\psi\rangle$, for some operator $\hat{A}$. This can be rewritten to $$\int \mathrm{d}x^\prime\, \langle x|\hat{A}|x^\prime \rangle \psi(x^\prime) \quad . $$ If we assume that for $\hat{V}$ it holds that $ \langle x|\hat{V}|x^\prime\rangle = V(x) \,\delta(x-x^\prime)$ (i.e. it is diagonal in position-representation), then it follows that $$\hat{V}\psi(x) = V(x)\, \psi(x) \quad .$$

So yes, it makes sense to use a product rule here.