# 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.

• As a side note: Never ever in physics you should try to do something harder first e.g. clearly calculating the commutator $[L_x,V]$ is much easier then calculating $[L_x^2,V]$. In other words, you should always try to be 'lazy' first :) – Gonenc Feb 28 '18 at 19:51

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.

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 $V(r)$.

• Nice. The first part is what I was thinking of when I made my comment about $[L^2, P^2]$, but using $r^2 = x_l x_l$ is quite clever. – Mike Feb 28 '18 at 20:36