Equation for the simultaneous eigenfunction of three operators in spherical coordinates

Question

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.

Answer 1

I will try to answer your many questions.

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.
3. Yes, the factorized eigenfunctions form a complete set. This follows from the spectral theorem.
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.