I have the two following operators: $$ L_z=-i\hbar\frac{\partial}{\partial\phi} \\ L^2=-\hbar^2 \left(\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\sin\theta\frac{\partial}{\partial\theta} +\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2} \right)$$
As they commute (and are diagonalizable) we are looking for common eigenvectors $\psi_{m,l} $
$\psi_{m,l}$ being an eigenvector of $L_z$ yields to: $$L_z \psi_{m,l} = \hbar m\psi_{m,l} $$ $\psi_{m,l}$ being an eigenvector of $L^2$ yields to: $$L^2\psi_{m,l}=\hbar^2 \lambda \psi_{m,l} $$
where $\lambda$ is a positive eigenvalue (positive because $L^2$ can be written as ${L_x}^2 + {L_y} ^2 +{L_z}^2$) that can be written as $l(l+1)$ (because $f: l->l(l+1)$ establishes a bijection between $[0,+\infty[ $and $[0,+\infty[$)
My question is why the so-called eigenfunction $\psi_{m,l} (\theta, \phi)$ only depend on $\theta$ and $\phi$ and not on $r$?
In other words, why can't we write $\psi_{m,l} (r,\theta, \phi)$ ?
PS: $\hbar$ and $\hbar ^2$ are only there for dimensions (physical dimensions, like length, time..) purposes.
PS2: Both operators are diagonalizable because they are hermitian. Their hermiticity implies normality, and normality implies unitarily diagonalizable.
PS3: I have to say I don't know at all how to transpose those expressions of $L_z$, $L^2$ into matrices , maybe writting them as matrices would solve my question.. But I don't know
Given the eigenfunctions $\psi_{m,l}(\theta,\phi)$, because $L_z$ and $L^2$ have no radial dependency, you have the freedom to transform $$\psi_{m,l}(\theta,\phi)\rightarrow \psi_{m,l}(\theta,\phi,r)=f(r)\cdot \psi_{m,l}(\theta,\phi)$$ for some function $f$, and $\psi_{m,l}(\theta,\phi,r)$ will still be an eigenfunction of both operators. But we have the freedom to choose $f(r)=1$, which turns out to be the most convenient choice in situations where $L_z$ and $L^2$ appear (e.g. the eigenstates of hydrogen-like atoms).