# Is it possible for an operator to have only one eigenvalue in this case? - in need of a proof

First of all i have to state that i am a newcommer to spectral theory so please take it easy on me :). On lectures our professor derived 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}

which represents a connection between a general wavefunction $\psi(r,\varphi,\vartheta)$ and a wavefunction $\psi(r,0,\vartheta)$ at $\varphi=0$ where $\vartheta = const.$. Equation above is written in spherical coordinate system:

Our professor said that we can replace an operator $\hat{L}_z$ with its expectation value - he denoted it just $L_z$.

Q1: Does this mean that an operator $\hat{L}_z$ has only one possible eigen equation? And only one possible eigenvalue? Lets denote the later $L_z$ (no hat).

Q2: Is there a way to prove this? I mean we can state that $\psi(r,\varphi,\vartheta) = \psi(r,0,\vartheta)$ at $\varphi=0$. And if we take Eulers identity into consideration we notice that first equation only holds if:

\begin{align} \hat{L}_z\frac{i}{\hbar}\varphi = 2\pi i \end{align}

or in more genaral if:

\begin{align} \hat{L}_z\frac{i}{\hbar}\varphi = n\,2\pi i \qquad n=\pm1,\pm2,\pm3\dots \end{align}

Is there any way to conclude from this that operator $\hat{L}_z$ has only one eigenvalue named $L_z$?

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