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

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.

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 $$\beta$$.

• $m$ could and is fact does take half-integer values. – ZeroTheHero Sep 19 at 14:23
• I just meant for orbital angular moment, that's only for spin right? – Alex Gower Sep 19 at 14:26
• 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. – ZeroTheHero Sep 19 at 14:29
• 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? – Alex Gower Sep 19 at 15:00
• actually I can't since the question is closed. – ZeroTheHero Sep 19 at 15:10