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Is this true that for fermion statistical systems

in the thermal phase, with Euclidean time, $$ \beta=1/T=t_E $$

the Euclidean time will be chosen to be anti-periodic for fermion boundary condition?

Why is that? How to determine the fermion boundary condition for a thermal compact circle?

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  • $\begingroup$ A simple way to see this is to consider the fermion 2-point function on a circle, $\langle \bar \psi(t) \psi(0) \rangle$. Send $t\to t +\beta$, which corresponds to moving the fermion in a loop around the circle returning to its initial location. In doing so it crossed the other fermion and picked up a minus sign. To get periodic fermions, one must include $(-)^F$ in the trace, which cancels this sign. $\endgroup$ – Elliot Schneider Mar 3 '18 at 3:38

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