What is the possible potential?

Question

The professor asked us to do this one:

"..Determine all potentials $V(r,\theta, \phi)$ for which it is possible for find solutions of the time independent Schrodinger equation which are also eigenfunctions of the operator $L_{z}$."

I try to solve this problem by assuming separation of variables, and I get $$\Phi(r,\theta,\phi)=R(r)F(\theta)e^{im_{l}\phi}$$ Unfortunately I do not know what to do next for this - should I put it into the formula $H\Phi=E\Phi$ in radial form to see whether it works? The professor give us the hint that the commutator $$[L_{z},V(r,\theta,\phi)]=0$$ I do not know how to use this relationship.

Answer 1

Try writing out $[L_z,V(r,\theta,\phi)]=0$. In other words, V has to be a function such that:
$L_zV \psi = VL_z\psi$
for absolutely any function $\psi$. Does that tell you anything about V?

Answer 2

I'm not going to answer your homework problem. Instead I'm going to construct a counter example to the hint $[L_{z},V]=0$, which is wrong.

Consider first, for simplicity, a spherically symmetric potential $V(r)$ and a product solution

$\Psi(r,\theta,\phi)=R(r)\Theta(\theta)e^{im\phi}$

to the time independent Schroedinger equation $(H-E)\Psi=0$, with Hamiltonian $H:=T+V$, with kinetic term $T:=-\frac{\hbar^2}{2m}\Delta$, and with norm $0< ||\Psi||< \infty$.

In fact, you can think of a Hydrogen-like atom, where $V(r)$ is the Coulomb potential, $R(r)$ is a Laguerre polynomial, and $\Theta(\theta)$ is an associated Legendre polynomial. If the principal quantum number $n\neq 1$ is not one, then $R(r)$ will have a radial node, i.e., there exists a radius $r_{*}>0$, such that $R(r_{*})=0$.

Now construct a new potential of the form

$\tilde{V}(r,\theta,\phi)=V(r)+\delta(r-r_{*})W(\theta,\phi)$.

The solution $\Psi$ from before will also solve the new problem $(\tilde{H}-E)\Psi=0$, where $\tilde{H}:=T+\tilde{V}$, since $R(r)\delta(r-r_{*})=0$. But $L_z$ may not commute with $\tilde{V}$.