# Another Question About Angular Momentum in QM

1. Feb 23, 2013

### Bobbo Snap

1. The problem statement, all variables and given/known data
Show that the Hamiltonian $H = (p^2/2m)+V$ commutes with all three components of L, provided that V depends only on r.

2. Relevant equations
In previous parts of the problem, I've worked out the following relations:
$$[L_z,x] = i\hbar y, \quad [L_z,y] = -i\hbar x, \quad [L_z, z] = 0,\\ [L_z, p_x] = i\hbar p_y, \quad [L_z, p_y] = -i\hbar p_x, \quad [L_z, p_z] = 0,\\ [L_z, L_x] = i \hbar L_y, \quad [L_z, r^2] = xy(xp_x - yp_y), \quad [L_z, p^2] = (xp_x -yp_y)p_xp_y$$

where
$$p_i = \frac{\hbar}{i}\frac{\partial}{\partial i}, \quad r^2 = x^2 + y^2 + z^2,\\ p^2 = p_x^2 + p_y^2 + p_z^2, \quad L_z = yp_z - zp_y$$

3. The attempt at a solution
I started with $[L_z, H]$, expanded it, cancelled a bunch of stuff and wound up with,
$$[L_z, H] = \frac{1}{2m}(xp_yp_x^2 - yp_xp_y^2) + xp_yV - yp_xV + Vxp_y - Vyp_X.$$
Assuming I did everything right, I need this equation to equal 0 to show $L_z$ commutes with $H$ and the problem suggests that this can only be the case if $V$ is a function of $r$, but I don't see how to get there. Any help would be appreciated.

Last edited: Feb 23, 2013
2. Feb 23, 2013

### Bobbo Snap

Ok, so I can simplify that last equation to get
$$[L_z, H] = \frac{1}{2m}[L_z, p^2] + [L_z, V] =\frac{i\hbar}{2m}(p_y^2 -p_x^2) + [L_z, V]$$

How does this imply $V$ must be a function of $r$?
(I guess I should have seen this from the form of the Hamiltonian).

Last edited: Feb 23, 2013