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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.

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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:

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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 $\beta$.

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  • $\begingroup$ $m$ could and is fact does take half-integer values. $\endgroup$ – ZeroTheHero Sep 19 at 14:23
  • $\begingroup$ I just meant for orbital angular moment, that's only for spin right? $\endgroup$ – Alex Gower Sep 19 at 14:26
  • $\begingroup$ The point is precisely to prove that $\ell$ takes integer values so if you assume this you’ve proved nothing. See Gatland, I.R., 2006. Integer versus half-integer angular momentum. American journal of physics, 74(3), pp.191-192. $\endgroup$ – ZeroTheHero Sep 19 at 14:29
  • $\begingroup$ Oh sorry I was assuming we have 2$\pi$ invariance for orbital momentum eigenstates, but I guess if you assume 4$\pi$ invariance like you do with spin then there is the extra question of why orbital angular momentum does not permit half integers. Could you write a summary answer for why this is? $\endgroup$ – Alex Gower Sep 19 at 15:00
  • $\begingroup$ actually I can't since the question is closed. $\endgroup$ – ZeroTheHero Sep 19 at 15:10

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