# Commutator of $V(\hat{\vec{r}})$ and $\hat{L_z}$

Can someone explain to me why this is true?

to me I see that $$x[\hat{p_y},V(r)]- y[\hat{p_x},V(r)]$$

$$=x(\hat{p_y}V(r)- V(r)\hat{p_y} )- y (\hat{p_x}V(r)+ \hat{p_x}V(r)).$$

The only way this can equal what is in the image is if $$V(r)\hat{p_y} = 0$$ and $$V(r)\hat{p_x}=0$$ but why would that be ?

• Do you understand that $[\partial_y ,V(r)]= \partial_y V(r)$ in operator calculus notation? Commented Apr 24, 2022 at 0:11
• isn't it $\partial_y V(r) - V(r) \partial_y$ ? since this is applied to a state like $\psi$ ? Commented Apr 24, 2022 at 0:13
• No! The first partial is supposed to act on everything to its right. Apply the chain rule to get $\partial_y V(r) \psi(r)=( \partial_y V(r)) \psi(r)+ V(r) \partial_y \psi(r)$, etc. You need the associative string of linear operations' notation.... Heaviside! Commented Apr 24, 2022 at 0:16

As @CosmasZachos' comment already pointed out, the equation $$\left[p_y,V(\vec{r})\right]= -i\hbar \frac{\partial V(\vec{r})}{\partial y}$$ is an operator equation. To prove such an operator equation you need to show that the operators on the left and on the right side yield the same result when operating on any arbitrary wave function $$\psi(\vec{r})$$. So let us do this: \begin{align} \left[p_y,V(\vec{r})\right]\psi(\vec{r}) &=-i\hbar\left[\frac{\partial}{\partial y},V(\vec{r})\right]\psi(\vec{r}) \\ &=-i\hbar\left(\frac{\partial}{\partial y}V(\vec{r}) -V(\vec{r})\frac{\partial}{\partial y}\right)\psi(\vec{r}) \\ &=-i\hbar\left(\frac{\partial}{\partial y}(V(\vec{r})\psi(\vec{r})) -V(\vec{r})\frac{\partial}{\partial y}\psi(\vec{r}) \right) \\ &=-i\hbar\left(\frac{\partial V(\vec{r})}{\partial y}\psi(\vec{r}) +V(\vec{r})\frac{\partial \psi(\vec{r})}{\partial y} -V(\vec{r})\frac{\partial\psi(\vec{r})}{\partial y} \right) \\ &=-i\hbar\frac{\partial V(\vec{r})}{\partial y}\psi(\vec{r}) \end{align} Since this equation holds for every $$\psi(\vec{r})$$ both operators acting on $$\psi(\vec{r})$$ must be equal. $$\left[p_y,V(\vec{r})\right] =-i\hbar\frac{\partial V(\vec{r})}{\partial y}$$