Replacing an operator with its expectation value

Question

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...):

Answer 1

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 $L_z$.