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 $1/2$.


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