Consider the path integral computation for the partition function: $$Z=Tr\space [exp(-\beta H)]=\int DxD\bar{\psi}D\psi|_{P} exp(-S_E)\space(1),$$$$Z=Tr\space [\exp(-\beta H)]=\int_{AP} D\bar{\psi}D\psi ~Dx~\exp(-S_E)\tag{10.125},$$ and that for the Index (Mirror Symmetry, (10.125) and (10.126)): $$Tr[(-1)^Fexp(-\beta H)]=\int DxD\bar{\psi}D\psi|_{AP} exp(-S_E)\space (2).$$$$Tr[(-1)^F\exp(-\beta H)]=\int_PD\bar{\psi}D\psi~ Dx~\exp(-S_E) .\tag{10.126}$$ In evaluating these path integrals, it is indicated that we choose in
(110.125): periodic boundary conditions for both bosons and fermions, and inbut anti-periodic boundary conditions for the fermions.
(210.126): periodic boundary conditions for both bosons, but anti-periodic boundary conditions for the and fermions.
What is understandable is the fact that when I try to cast out these traces in path integral representation, upon using the position basis, the periodicity of the bosonic paths becomes evident. Since that's the only way to obtain the path integral representation. When I try to do the same using the fermionic coherent state basis, the fermionic paths must come out antiperiodic/ periodic in (110.125)/(210.126) in order for the path integral representation to be realized.
What is not clear is the explanation given in the book Mirror Symmetry:, p. 191:
"The fact that inserting $(-1)^F$ operator corresponds to changing the boundary conditions on fermions is clear from and follows from the fact that fermions anti-commute with $(-1)^F$. So before the trace is taken, the fermions are multipled with an extra minus sign. What is not completely obvious is that without the insertion of $(-1)^F$, the fermions have anti-periodic boundary condition along the circle..."
Need a rudimentary sort of help in beginning to get going and through with this argument.
References:
- K. Hori, S. Katz, A. Klemm, R. Pandharipande, R. Thomas, C. Vafa, R. Vakil, and E. Zaslow, Mirror Symmetry, 2003. The PDF file is available here.
