Q:

Operator Commutativity $\hat{L}_{z}$; $\hat{p}^{2}$; $[\hat{p}^{2}, \hat{L}]$

Define: 
$$\hat{p}^{2}= \hat{p}^{2}_{x}+\hat{p}^{2}_{y}+\hat{p}^{2}_{z}= -\hbar^{2} \Delta$$
$$\hat{L}_{z}= xp_{y}-yp_{x}= -i\hbar \big(x\frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \big)$$
consider the Hamiltonian in three dimensions with a potential that depends only on the distance from the origin:
$$\hat{H} = \frac{\hat{p}^{2}}{2m} + \hat{V}(r), r=(x^{2}+y^{2}+z^{2})^{1/2}$$ 
Since: $ \hat{p}^{2}\hat{L}_{z} = -\hbar^{2}\Delta\big(-i\hbar(x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x})\big) = i\hbar^{3}(\frac{\partial ^{2}}{\partial x} + \frac{\partial ^{2}}{\partial y} + \frac{\partial ^{2}}{\partial z})(x\frac{\partial}{\partial y} - y \frac{\partial}{\partial x}) = 0 $ and $\hat{L}_{z}\hat{p}^{2}=(-i\hbar (x\frac{\partial}{\partial y}- y\frac{\partial}{\partial x})(-\hbar^{2}\Delta)=i\hbar^{3}(x\frac{\partial}{\partial y} - y \frac{\partial }{\partial x})(\frac{\partial ^{2}}{\partial x} + \frac{\partial^{2}}{\partial y} + \frac{\partial^{2}}{\partial z}) = 0$
$$[\hat{p}^{2},\hat{L}_{z}] = \hat{p}^{2}\hat{L}_{z} - \hat{L}_{z}\hat{p}^{2} = 0-0 = 0$$
The operator $\hat{L}_{z}$ commutes with $\hat{p}^{2}$.
And $\hat{p}^{2} \hat{V} = -\hbar^{2}(\frac{\partial^{2} \hat{V}}{\partial x} + \frac{\partial^{2} \hat{V}}{\partial y} + \frac{\partial^{2} \hat{V}}{\partial z})$ together ith $\hat{V} \hat{p}^{2} = \hat{V}(-\hbar^{2}\Delta) = -\hbar^{2}(\frac{\partial^{2} \hat{V}}{\partial x} + \frac{\partial^{2} \hat{V}}{\partial y} + \frac{\partial ^{2} \hat{V}}{\partial z})$
$$[\hat{p}^{2},\hat{V}]=\hat{p}^{2}\hat{V}- \hat{V}\hat{p}^{2} = 0$$
 Edit : I tried showing that $\hat{L}_{z}$ commutes with $\hat{V}$ using Norberts method: $(\hat{V}\hat{L}_{z})(f) = V(-i\hbar(x\frac{\partial (f)}{\partial y} - y \frac{\partial (f)}{\partial x}))$ and $(\hat{L}_{z}\hat{V})(f)= \hat{L}_{z}(V(r)(f)) =-i\hbar(x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x})(V(r)f) = -i\hbar((f)x\frac{\partial V}{\partial y} + (V)x\frac{\partial f}{\partial y}-(V)y\frac{\partial (f)}{\partial x} - (f)y \frac{\partial (V)}{\partial x})$  Why are they not the same?
I have never worked with operators before, so can you tell me if this is alright? I will greatly appreciate your advice. 

A:

You are considering operators, which are by definition acts on functions. If you want to compare two operators you must compare results of applying them to all allowed functions. Assume that this operators defined on space of smooth functions in $\mathbb{R}^3$. Then
for all smooth function $f$ we have
$$
(\hat{p}^2 \hat{V})(f)=\hat{p}^2(Vf)=-\hbar^2\Delta(Vf)=
-\hbar^{2}\left(\frac{\partial^{2} (Vf)}{\partial x^2} + \frac{\partial^{2} (Vf)}{\partial y^2} + \frac{\partial ^{2} (Vf)}{\partial z^2}\right)=
$$
$$
-\hbar^{2}\left(f\frac{\partial^{2} V}{\partial x^2} + f\frac{\partial^{2} V}{\partial y^2} + f\frac{\partial ^{2} V}{\partial z^2}+V\frac{\partial^{2} f}{\partial x^2} + V\frac{\partial^{2} f}{\partial y^2} + V\frac{\partial ^{2} f}{\partial z^2}+2\frac{\partial f}{\partial x}\frac{\partial V}{\partial x}+2\frac{\partial f}{\partial y}\frac{\partial V}{\partial y}+2\frac{\partial f}{\partial z}\frac{\partial V}{\partial z}\right)=
$$
$$
-\hbar^{2}\left(f\Delta V+V\Delta f+2\langle\nabla f,\nabla V\rangle\right)
$$
On the other hand
$$
(\hat{V}\hat{p}^2)(f)=\hat{V}(\hat{p}^2(f))=\hat{V}\left(-\hbar^{2}\left(\frac{\partial^{2} (f)}{\partial x^2} + \frac{\partial^{2} (f)}{\partial y^2} + \frac{\partial ^{2} (f)}{\partial z^2}\right)\right)=-\hbar^{2}V\Delta f
$$
Thus we see that $\hat{V}\hat{p}^2\neq\hat{p}^2\hat{V}$

