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.
 A: Mike has already answered the question but I think there is another neat answer, in which you can use Cartesian coordinates. Notice that (I won't use hats for operators and repeated indices are summed over!) $L_i = \epsilon_{ijk} x_j p_k$ as an operator hence we have:
$$[L_i, x_l] = \epsilon_{ijk}[x_jp_k,x_l] = \epsilon_{ijk}x_j[p_k,x_l] = -i\hbar \epsilon_{ijk} x_j \delta_{kl} = -i\hbar \epsilon_{ijl}x_j = i\hbar \epsilon_{ilj} x_j$$
where in the second step I used the commutator identity $[A,BC]=B[A,C]+[A,B]C$ and in the last step swapped $j$ and $l$. Similarly we have (maybe upto a sign):
$$[L_i,p_l] = i\hbar \epsilon_{ilj} x_j $$
Hence we get with $r^2=x_lx_l$:
$$[L_i,r^2]=[L_i,x_lx_l] = 2i\hbar \epsilon_{ilj}x_jx_l = 0 = [L_i, p^2]$$
notice that this corresponds to the intuition that the length of the vector $x$ and $p$ doesn't change under rotation. Lastly, we only need to observe that any function $V(|x|)$ can be written as a function $V(r^2)$ and thus:
$$[L_i, V(r^2)]=0=[L_i, V(|x|)]$$
Note that we have thus proven something much stronger $[H,L_i]=0$ if the potential is only dependent on the radius. 
One a side note: the classical analogous of this problem is to show that the angular momentum vector is conserved if in a central potential $V(r)$.
A: 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.
