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I've got a pretty simple question that's been bothering me for a while.

This is the issue. Suppose we have a superconductor with phase $\phi$, and we change this to $\phi+2\pi$. One would think that the wavefuction would have to turn back into itself, but apparently, if the parity (even or oddness of the # of fermions) is odd, then the wavefunction is actually antiperiodic in $\phi$. Is there a simple physical intuition for why this is the case?

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Someone might give a more precise answer in terms of group theory but I'll give it a go anyway ; feel free to edit my post.

Instead of considering the case of an odd number of fermions, one can consider just a single spin $1/2$ - fermion to discuss $2 \pi$ rotations. Spin $1/2$ is a representation of dimension 2 of the rotation group, which is called the spinor representation. Being a representation of even dimension, it is an unfaithful representation of rotations, which means that it will not behave in a "usual way" regarding rotations of space. This is the mathematical reason behind the fundamental quantum nature of spins, and why there is no classical equivalent to them.

There is a famous qualitative explanation to the nature of spins called the Dirac's belt trick which explains why a rotation of $2 \pi$ is not enough to recover the initial state for a spin - $1/2$ (the wavefunction picks up a minus sign), instead one needs a rotation of $4 \pi$.

Representations of odd dimension such as the vector representation, of dimension 3 (hence representing objects of spin 1), are faithful representations of the rotation group as they behave in the "usual" way towards rotations.

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