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.


  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.
  • 1
    $\begingroup$ Think of the path integral as a transition amplitude. You start with a state $|\psi\rangle$, evolve it forward in Euclidean time $\beta$, and produce the amplitude $\langle \psi'|e^{-\beta H} |\psi\rangle$. $\endgroup$ Jun 4, 2018 at 13:30
  • $\begingroup$ The trace instructs you to glue the two ends together and sum over all states, $\mathrm{tr}(e^{-\beta H}) = \sum_\psi \langle \psi | e^{-\beta H}| \psi \rangle$, corresponding to the path integral on a Euclidean time circle. The fermions are anti-periodic in the partition function because they are anti-commuting. $\endgroup$ Jun 4, 2018 at 13:30
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    $\begingroup$ Inserting $(-)^F$ inside the trace tells you multiply by $(-)^F$ before gluing the ends together, $\mathrm{tr}((-)^F e^{-\beta H}) = \sum_\psi \langle \psi | (-)^F e^{-\beta H} |\psi \rangle$. Now you evolve $|\psi\rangle$ forward in time by $\beta$, and then multiply it by $-1$ before gluing the amplitude, cancelling the earlier sign. $\endgroup$ Jun 4, 2018 at 13:31
  • $\begingroup$ @user81003 Thanks for the reply. Lets take it from here, what do you mean when you say "...the fermions are antiperiodic in the partition function because they are anti commuting". What fermions are we talking about? In the expression for the trace, what is visible is an evolution operator sandwiched between quantum states...where are the fermions? $\endgroup$
    – Kong
    Jun 4, 2018 at 14:59
  • $\begingroup$ @Kong The theory has $\mathrm{Z}_2$ symmetry $(-)^F = \pm 1$. By fermion I mean the states with $(-)^F = -1$. $\endgroup$ Jun 4, 2018 at 17:42

1 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.


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