# Equation for the simultaneous eigenfunction of three operators in spherical coordinates

If I'm considering the three operators $$H,L^2,L_z$$ with the condition $$[H,L^2]=[H,L_z]=[L^2,L_z]=0$$, I can find a complete set of simultaneous eigenfunctions. If I study this problem in spherical coordinates and the Hamiltonian is a "center" Hamiltonian $$H=\frac{p^2}{2m}+V(r)$$ the three operators are differential operator. My questions arise from this sentence I've read in "Landau-Lifshitz Quantum Mechanics":

We shall consider stationary states in which $$L^2$$ and $$L_z$$ have definite values. In other words, we shall seek the common eigenfunctions of the operators $$H$$, $$L^2$$ and $$L_z$$. [...] We thus seek solutions of equation (...) in the form $$\psi=R(r)Y_l^m(\theta,\phi)$$.

I understand that we CAN search simultaneous eigenfunctions of the three operators since they commute but:

1. Must their simultaneous eigenfunctions be equal to $$\psi(r,\theta,\phi)=R(r)Y_l^m(\theta,\phi)$$?

2. In case answer 1 is yes, does this come from the fact the three operators are, in spherical coordinates, differential operator?

3. Do these factorized eigenfunctions form the complete set?

4. Could there be other simultaneous eigenfunctions of a different form than these?

5. Do we search for stationary states in which $$L^2$$ and $$L_z$$ have definite values just because it is convenient?

Sorry for too many questions but with numbers it is clearer to understand the question for you and the answers for me.

1. Yes, for central potentials $$V(r)$$ there need to be simultaneous eigenfunctions which are factorizable like this.
2. This comes from the fact that the Hamiltonian can be separated into a sum $$H=\frac{1}{2m}\left(p_r^2+\frac{L^2}{r^2}\right)+V(r)$$ where the operator $$p_r$$ does not contain any angular coordinates or derivatives.
4. In principle there might be another set of simultaneous eigenfunctions, at least for some special potentials $$V(r)$$. For example: The Kepler problem (with $$V(r)\propto \frac{1}{r}$$) is known to be separable also in parabolic coordinates. But why would we care about this?
5. Yes, we do it because of convenience, and because central potentials ($$V$$ depending only on $$r$$) are an important special case.
• @Luigid Yes, it is this form of $H$ which makes the TISE solvable by separation of variables. This leads to solutions with a $r$-dependent and a $(\theta,\phi)$-dependent factor. May 19 at 22:17
• So let me try: Landau is saying that the central Hamiltonian commutes with those two operators and since it is good to work with simultaneous eigenstates of the Hamiltonian and other operators we want to find these simultaneous eigenstates. The "shape" of the TISE suggest us to factorize in some ways the eigenfunction. SO we are searching a factorized simultaneous eigenfunction and this leads to search a $\psi=R(r)Y_l^m(\theta,\phi)$ because it is factorized, and helps us to simply solve TISE, AND it is easy to verify that these wavefunctions are simultaneous wavefunctions. Is this right? May 19 at 22:35
• @Luigid Just put this $\psi(r,\theta,\phi)$ into the TISE. Then you will get one differential equation for $R(r)$ and one differential equation for $Y(\theta,\phi)$. May 19 at 22:47