If I consider operators $L^2$ and $L_z$ in spherical coordinates where $L^2$ is the angular momentum squared operator and $L_z$ is the $z-axis$ component of the angular momentum, is a function like $\psi=f(r)Y_l^m(\theta,\phi)$, where $Y_l^m$ denotes spherical harmonics, an eigenfunction of $L^2$ and $L_z$?
So that: $$L^2f(r)Y_l^m(\theta,\phi)=\hbar^2l(l+1)f(r)Y_l^m(\theta,\phi)$$ and $$L_zf(r)Y_l^m(\theta,\phi)=\hbar mf(r)Y_l^m(\theta,\phi)$$ and then the common term $f(r)$ could simplify?
Best Answer
The answer is yes, and I claim that this is "obvious", which I will clarify in a moment. Numbered equations will refer to the third edition of Sakurai's Modern Quantum Mechanics (Cambridge University Press, 2021). I've found it a nice intermediate quantum book.
(By the way, you don't have to consider $L^2$ and $L_z$ in any particular coordinate basis for your claim to be true. Looking at their action in spherical coordinates is where the "obvious" comes from.)
Let's start by writing out the actions. For some state $\lvert\alpha\rangle$, $L_z$ acts on it like $$\begin{equation} \langle r, \theta, \phi \vert L_z \vert \alpha \rangle = - i \hbar \frac{\partial}{\partial \phi} \langle r, \theta, \phi \vert \alpha \rangle, \tag{3.218}\end{equation}$$ (we write $\langle r, \theta, \phi\rvert$ to make explicit that we're finding a matrix element in the spherical-coordinate basis). Similarly, the action of $L^2$ in the same basis is $$\begin{equation} \langle r, \theta, \phi \vert L^2 \vert \alpha \rangle = -h^2 \left[ \frac{1}{\sin^2 \theta} \frac{\partial^2}{\partial \phi^2} + \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta \frac{\partial}{\partial \theta} \right) \right] \langle r, \theta, \phi \vert \alpha \rangle. \tag{3.224}\end{equation}$$
You'll notice that $r$ appears in neither of these actions: any function $f(r)$ is treated like a constant by both the $L_z$ and $L^2$ operators, and you can treat it as such in their manipulations.
Finally, we happen to know (see section 3.6.2 of Sakurai) that the spherical harmonics are, in fact, eigenfunctions of $L_z$ and $L^2$. (So is the Hamiltonian of a spinless particle in a spherically symmetric potential, in fact.) So without worrying about the details of what the eigenvalue is, just denoting it with the subscript of the operator in question, $$\begin{equation} L_z \lvert Y_l^m \rangle = \lambda_{L_z} \lvert Y_l^m \rangle, \quad L^2 \lvert Y_l^m \rangle = \lambda_{L^2} \lvert Y_l^m \rangle; \end{equation}$$ because of the aforementioned lack of dependence on the radius $r$ $$\begin{equation} L_z \lvert f(r) Y_l^m \rangle = \lambda_{L_z} f(r) \lvert Y_l^m \rangle, \quad L^2 \lvert f(r) Y_l^m \rangle = \lambda_{L^2} f(r) \lvert Y_l^m \rangle \end{equation}$$ just as you thought.