Can a spherical function with only $r$-dependence be an eigenstate of $L^2$ and $L_z$?

Question

My guess is no, because the equations for $L^2$ and $L_z$ have no $r$-dependence:

$$L^2 f_m^\ell ( \theta , \phi ) = \hbar^2 \ell(\ell+1) f_m^\ell ( \theta , \phi )$$

$$L_z f_m^\ell ( \theta , \phi )= \hbar m f_m^\ell ( \theta , \phi )$$

Is this the correct way of thinking about it? Or would such a function still be an eigenstate with eigenvalues both = 0?

Answer 1

These are differential operators, so a function of $r$ only is effectively a constant, corresponding to the zero eigenvalues of both $L^2$ and $L_z$.