Replacing an operator with its expectation value While dealing with a circling particle in an spherical symetric potential our professor said that we can replace an operator of $z$ component of angular momentum $\hat{L}_z$ with the expectation value - he denoted it just $L_z$ - of the angular momentum if $L_z$ is constant. Why is that? 
So we first had this equation: 
\begin{align}
\underbrace{\psi (r,\varphi,\vartheta)}_{\llap{ \text{wave function in spherical coordinates}}} &= \exp\left[\hat{L}_z \frac{i}{\hbar}\, \varphi\right] \underbrace{\psi (r,0,\vartheta)}_{\rlap{\text{wave function in spherical coordinates at $\varphi=0$}}}
\end{align}
and we got this one (notice that there is no operator over an $L_z$): 
\begin{align}
\psi (r,\varphi,\vartheta) &= \exp\left[L_z \frac{i}{\hbar}\, \varphi\right] \psi (r,0,\vartheta)
\end{align}
Anyway here is the spherical coordinate system we ve been using all the time (the blue spherical aure is supposed to be a spherical potential...): 

 A: In this context, "$L_z$ is constant" means that the operator $\hat{L}_z$ has only one eigenvalue in the space of states under consideration, let's call it $\mathcal{H}$ - in other words, for any quantum state or wavefunction $\psi$ that could occur given the constraints of the problem ($\forall\psi\in\mathcal{H}$), $\psi$ is an eigenstate of $\hat{L}_z$ with the particular eigenvalue $L_z$.
You can find a basis of eigenvectors of $\hat{L}_z$ which span $\mathcal{H}$, and all associated eigenvalues are $L_z$. The spectral theorem says that, under certain conditions (which do apply here), an operator is completely determined by its eigenvalues and eigenvectors. So any other operator $\hat{L}_z'$ which satisfies
$$\hat{L}_z'\psi = L_z\psi\ \forall\ \psi\in\mathcal{H}$$
is equivalent to $\hat{L}_z$ as long as your state space is limited to $\mathcal{H}$. Basically, for this problem, all operators satisfying $\hat{L}_z'\psi = L_z\psi$ are interchangeable.
One such operator is multiplication by the constant $L_z$,
$$\hat{L}_z' = L_z$$
So you can replace the original operator $\hat{L}_z$ by the constant $L_z$.
