Proof that total orbital angular momentum square can only take integer values of $\hbar^2$

Question

I am comfortable with the argument that in order for the wavefunction to be single valued/2$\pi$ invariant this means that $L_z$ must be an integer value of $\hbar$.

$$U(2\pi e_z)=e^{-(2\pi i/\hbar)\hat L_z}=1$$

However I don't know how it follows from this that $|L|^2 = \beta \hbar^2$ where $\beta$ in an integer. $L_x$ and $L_y$ are indetermined so it seems fishy to say 'if' they were measured they would have integer $\hbar$ values and so $L_x^2 + L_y^2 + L_z^2$ would be an integer $\hbar^"$ value, since we can never actually measure all of these at once.

Answer 1

Could it be that $L_z=m\hbar^2$ (for integer $m$) alone does not prove that $|L|^2 = \beta \hbar^2$ (for integer $\beta$), but instead:

Once you prove that $m^2 \leq \beta$ through this:

we therefore know $m$ will have some maximum possible (integer) value.

We can then use the definition of the angular momentum ladder operator to show that (for instance)

$L_{+}|\beta,m_{max}> = \hbar \sqrt{\beta -m_{max}(m_{max}+1)}|\beta,m_{max}+1>$

(as of yet the definition of $\beta$ in the ladder operator does not require $\beta$ to be an integer).

However since we require that $|\beta,m_{max}+1>=0$ (so that there will be some maximum value of $m$ and you cannot arbitrarily apply the raising operator). This therefore means that $\beta = m_{max}(m_{max}+1)$ and since $m_{max}$ must be an integer, so must $\beta$.