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

$$\psi_{nlm}(r,\theta,\varphi,t)$$ 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?

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.