Find the expectation value of angular momentum $L_z$ of a wave function in energy eigenstates

Question

$\psi_{nlm}(r,\theta,\varphi,t)$ as expansion in energy eigenstates:

$$\psi(r,\theta,\varphi,t) =\sum_{n=1}^{\infty}\sum_{l=0}^{n-1}\sum_{m=-1}^{l}c_{nlm}ψ_{nlm} (r,θ,ϕ) \exp \left(-\frac{iE_n t}{\hbar}\right)$$

Here the $c_{nlm}$ are complex constants.

My derivation is as follows:

However, I have some queries about the derivation.

On the third line we have: $\hat{L}_z\psi_{n_*l_*m_*}$, which is the angular momentum operator acting on a energy eigenstate. I know that the eigenfunctions of $\hat{L}_z$ are the spherical harmonics $Y_{lm}$, and the realtionship is $$L_{z}|l,m\rangle=m\hbar |l,m\rangle.$$

However, in the above we have the total hydrogen atom electron energy eigenfunction which is $\psi_{nlm}(r,\theta,\varphi)=R_{nl}(r)Y_{lm}(\theta,\varphi)$, so can we consider the $R_{nl}$ as just a constant? Is the eigenfunction condition $$L_{z}|n,l,m\rangle=m\hbar |n,l,m\rangle$$ still valid?

Answer 1

We have that the operator $L_z$ in the position representation you are working with is simply given by

$L_z=-i\hbar\frac{\partial}{\partial\phi}$.

Hence you are correct, since $R_{nm}(r)$ does not depend on $\phi$, the presence of $R_{nm}$ does not matter here for the last equation.