Boundary conditions on bosons and fermions in computing Partition function/Index Path integrals

Question

Consider the path integral computation for the partition function: $$Z=Tr\space [\exp(-\beta H)]=\int_{AP} D\bar{\psi}D\psi ~Dx~\exp(-S_E)\tag{10.125},$$ and that for the Index (Mirror Symmetry, (10.125) and (10.126)): $$Tr[(-1)^F\exp(-\beta H)]=\int_PD\bar{\psi}D\psi~ Dx~\exp(-S_E) .\tag{10.126}$$ In evaluating these path integrals, it is indicated that we choose in

(10.125): periodic boundary conditions for bosons, but anti-periodic boundary conditions for the fermions.

(10.126): periodic boundary conditions for both bosons 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 (10.125)/(10.126) 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:

1. 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.

Answer 1

Note periodic boundary condition corresponds to $$\int d\psi \langle \psi |e^{-\beta H}|\psi\rangle=\text{Tr}((-1)^Fe^{-\beta H}).$$ This can be seen by letting $|\psi\rangle=|0\rangle+|1\rangle\psi$, so that $\hat{\psi}|\psi\rangle=|\psi\rangle\psi$, where I use the convention $\hat{\psi}$ acts as lowering operator. Note $\langle\psi'|\psi\rangle=\psi'-\psi$, the Grassman version of delta function. That indicates $\langle\psi'|0\rangle=\psi'$ and $\langle\psi'|1\rangle=-1$. So $$\int d\psi\langle\psi|e^{-\beta H}|\psi\rangle=\langle0|e^{-\beta H}|0\rangle-\langle1|e^{-\beta H}|1\rangle=\text{Tr}((-1)^Fe^{-\beta H}).$$

What $(-1)^F$ does in operator language is to change $|\psi\rangle$ to $|-\psi\rangle$, so that $\hat{\psi}|-\psi\rangle=-|-\psi\rangle\psi$. This corrects the minus sign in $\langle1|e^{-\beta H}|1\rangle$, and therefore anti periodic boundary condition corresponds to

$$\int d\psi \langle \psi |e^{-\beta H}|-\psi\rangle=\text{Tr}(e^{-\beta H}).$$

By the way, from path integral point of view one can impose either periodic boundary condition or anti periodic boundary condition for fermions. Depending on the symmetries of the theory, one might even be able to impose $\psi(t+L)=e^{i\alpha}\psi(t)$.

A good reference on this issue is Polchinski Volume I Appendix A.