*The definition of the Hilbert space $\mathcal{H}$ is part of the specification of the quantum system. Hence, if $\mathcal{H}$ is the space of square-integrable functions on the circle, then $\mathcal{H}$ contains*functions

*, hence $e^{ i \alpha}$ should be equal to 1 because otherwise $\psi$ is not a function.*

A second answer is as follows. Suppose I leave the exact definition of $\mathcal{H}$ open for now, but impose the condition \eqref{eq:20170129b}. Then $\eta$ defined by \begin{equation}\label{eq:20170129c} \eta(t,x) = e^{ - i \alpha \dfrac{x}{2 \pi}} \psi(t,x) \end{equation} is periodic, therefore $\eta$ is a function on the circle. Furthermore $\eta$ satisfies \begin{equation}\label{eq:20170129d} i \frac{\partial \eta}{\partial t} = - \frac{1}{2 m} \left( \partial_x + i \frac{\alpha}{2 \pi} \right)^2\eta \end{equation} This is the SchrÃ¶dinger equation of a particle on a circle with constant vector potential $A = \frac{\alpha}{2 \pi} dx$. Therefore, the second answer to the puzzle is

*Yes, one can impose the condition \eqref{eq:20170129b}. The Hilbert space is then not the space of square integrable functions [2]. Hence the physical system is different, namely, it is a particle on a circle with Wilson loop.*

I was motivated to write this post when I was reading the famous article by Dirac about magnetic monopoles [3] and section 18.3 in [4].

References and comments

[1] For example, exercise 2.43 in Griffiths, Introduction to Quantum Mechanics

[2] The Hilbert space is the set of square integrable sections of some kind of bundle over the circle.

[3] A.M. Dirac "Quantised Singularities in the Electromagnetic Field", 1931. Journal Site, pdf file

[4] A First Course in String Theory, Zwiebach, 2009

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