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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.

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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?

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Hi, so the hint is if $\Phi(r,\theta,\phi)$ is an eigenfunction of $L_{z}$, then $V(r,\theta,\phi)$ is also an eigenfunction of $L_{z}$? Is this necessarily true? –  Kerry Mar 18 '11 at 4:49
    
@user2597; It's not necessarily true, but it's close. When two operators commute, it's possible to select a complete set of eigenfunctions that are eigenfunctions of both. So you can assume that $\Phi$ is an eigenfunction of both. [Other language: Another way of saying $[L_z,V]=0$ is that $L_z$ and $V$ are compatible observables. And another way of saying this is that they form part of a complete set of commuting observables.] –  Carl Brannen Mar 18 '11 at 6:17
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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 $\tilde{V}$.

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You probably mean \nabla and not \Delta in the expression for kinetic energy. And probably want it squared, as well. –  dmckee Mar 18 '11 at 15:25
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The Laplacian $\Delta:=\nabla^2$. –  Qmechanic Mar 18 '11 at 15:29
    
Ah, so. That usage is far enough from universal that it is probably worth saying so. I've never liked it myself, because I want my second order operator to look second order. –  dmckee Mar 18 '11 at 15:43
    
Dear @dmckee. Thanks for pointing it out and help improving the answer. If $\Delta$ was unclear to you, I'm sure it was unclear to others as well. –  Qmechanic Mar 18 '11 at 16:00
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Dear @Marek: We may debate forever what is natural and what is unnatural, but Nature always makes her own choices, some of which only the future will tell. –  Qmechanic Mar 18 '11 at 19:53
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