# Quantisation of $z$-angular momentum eigenvalues

Consider the eigenvalue equation for the $$\hat{l}_z$$ angular momentum operator: $$\hat{l}_zY_{lm_l}(\theta,\phi)=m\hbar Y_{lm_l}(\theta,\phi)$$ with separable solution $$Y_{lm_l}(\theta,\phi)=\Theta_{lm_l}(\phi)\exp(im\phi)$$

I am unsure about the following passage from Gasiorowicz's Quantum Physics:

It is sometimes argued that since […] a transformation $$\phi\to\phi+2\pi$$ leaves the system invariant, it is necessary that $$\exp(2\pi im)=1$$, so that $$m$$ is an integer.

This is not quite correct, since the quantities that enter into physical observables are of the type $$\int_0^{2\pi}d\phi\Psi_1^*(\phi)A\Psi_2(\phi)$$, with wavefunctions $$\Psi(\phi)$$ of the form $$\Psi(\phi)=\sum_{m=-\infty}^\infty C_m\exp(im\phi).$$

If we require that these arbitrary wavepackets do not change under the transformation $$\phi\to\phi+2\pi$$, then we are led to the conclusion that the most general allowed values of $$m$$ are $$m=c+$$ integer, where $$c$$ is a constant.

I recognise the integral in the passage as the expectation value of an observable $$A$$. I understand why the boundary conditions imply $$m$$ must be an integer; what I don't understand is how/why is $$m$$ allowed, in principle, to be equal to an integer plus a constant. While it may seem an irrelevant issue, I suspect that when $$c=1/2$$ then $$m$$ is half-integer and therefore an eigenvalue of spin, so I would like to understand exactly where this constant factor comes in. If I just plug in $$m+$$ constant in the equation and then I transform from $$\phi$$ to $$\phi+2\pi$$ I seem to break the boundary condition for single-valuedness.

Wavefunctions are not observable and and only unique up to a complex phase (i.e. $$e^{ic \phi}$$), since any complex phase will be equal to one if multiplied with its conjugate. In other words, you can multiply your wavefunction with any $$e^{ic \phi}$$, and it's still considered the same wavefunction! That said, $$c$$ can be any numbers, not necessary $$1/2$$.