quantum mechanics - Angular momentum squared and Hamiltonian?

I'm trying to show, for the Hamiltonian $H = \vec{P}^2/2m + V(\vec{X})$, that $[\vec{L}^2,H]=0$ if $V(\vec{X}) = V(|\vec{X}|)$, and I pretty much almost have it, there's just one thing I'm getting messed up on.

So what I did was: $$ [\vec{L}^2,H] = [L_x^2 + L_y^2 + L_z^2, \frac{\vec{P}^2}{2m} + V(\vec{X})] = [L_x^2 + L_y^2 + L_z^2, \frac{\vec{P}^2}{2m} + V(|\vec{X}|) ] $$ $$ = [L_x^2,\frac{\vec{P}^2}{2m} + V(|\vec{X}|)] + [L_y^2,\frac{\vec{P}^2}{2m} + V(|\vec{X}|)] + [L_z^2,\frac{\vec{P}^2}{2m} + V(|\vec{X}|)] $$ Looking at the $L_x^2$ component: $$ \rightarrow [L_x^2,\frac{\vec{P}^2}{2m} + V(|\vec{X}|)] = \frac{1}{2m}[L_x^2,P^2] + [L_x^2,V(|\vec{X}|)] $$ $$ = \frac{1}{2m}[L_x^2,P_x^2+P_y^2+P_z^2] + [L_x^2,V(|\vec{X}|)] $$ $$ = \frac{1}{2m} \bigg( [L_x^2,P_x^2]+[L_x^2,P_y^2]+[L_x^2,P_z^2] \bigg) + [L_x^2,V(|\vec{X}|)] $$

Now for the momentum parts, that's a lot of tedious work that I won't type up, but at the end of the day, I get that $$ [L_x^2,P_x^2]=[L_x^2,P_y^2]=[L_x^2,P_z^2]=0 $$ and similarly for $[L_y^2,P^2]=0$ and $[L_z^2,P^2]=0$. My problem is this, for this to work out, I need $$ [L_x^2,V(|\vec{X}|)] = 0 $$ and similarly $$ [L_y^2,V(|\vec{X}|)]=[L_z^2,V(|\vec{X}|)]=0 $$ but I don't really understand why that would be the case? Any insight would be appreciated.

Answer

Remember that $L_x$, $L_y$, and $L_z$ are generators of rotation about the $x$, $y$, and $z$ axes, respectively. But $V(\vec{X})=V(|\vec{X}|)$ says that your potential is invariant under rotations. So on those grounds, physically you would expect any of these angular-momentum operators to commute with such a potential operator $V$.

Mathematically, I would say that the easiest way to see this is to use the angular momentum operator in spherical coordinates. More specifically, I would say that you should probably back up and just prove $[L^2, V(|\vec{X}|)] = 0$ by expanding $L^2$ as $\frac{1}{2}(L_{+}L_{-} + L_{-}L_{+}) + L_{z}^2$. Note that those operators have no $\partial / \partial_r$ term, so they commute with $V(r)$.

I'm guessing you did all your work so far using a Cartesian basis. That's valid, but it often makes sense to look around for other ways of doing a problem. And changing coordinates is one of the first things you should think about.