# Boundary (Anti)Periodic conditions and fermion partition functions

The path integral with antiperiodic fermions (Neveu-Schwarz spin structure) on a circle of circumference $$\beta$$, in a theory with Hamiltonian $$H$$, has partition function $$\rm{Tr} \exp(−\beta H)$$ The path integral with periodic fermions (Ramond spin structure) has partition function $$\rm{Tr} (−1)^F \exp(−\beta H).$$

1. Why antiperiodic and periodic boundary conditions give these two partition functions? Intuitions and derivations?

2. How are the two boundary conditions related to spin structures? (Ramond and Neveu-Schwarz spin structures)

3. What are the ground state sectors (bosonic vs fermionic) for the above boundary conditions?

I have some insight on the first question. For the ordinary partition function: $$\text{Tr} \ e^{-\beta H} = \sum_n \langle n | e^{-\beta H} | n \rangle$$ Then one inserts the fermionic path integral: $$\int dc \ dc^{*} e^{-c c^{*}}\sum_n \langle n| c \rangle \langle c | e^{-\beta H} | n \rangle = \int dc \ dc^{*} e^{-c c^{*}} \sum_n \langle -c | e^{-\beta H} | n \rangle \langle n| c \rangle = \int dc \ dc^{*} e^{-c c^{*}} \langle -c | e^{-\beta H} | c \rangle$$ Where one has used : $$\langle n | c \rangle \langle c | n \rangle = -\langle c | n \rangle \langle n | c \rangle$$.
Then one inserts resolution of identity $$N$$ times (here $$c_N = -c_0$$): $$\int \prod dc_n dc_n^{*} e^{-\sum_n c_n c_n^{*}} \langle c_N | e^{-\Delta \tau H} | c_{N-1} \rangle \ldots \langle c_1 | e^{-\Delta \tau H} | c_{0} \rangle$$ Where $$\tau = \beta / N$$. In the continuum limit one gets the path integral after: $$\Delta \tau \sum_n^{N} \ldots \rightarrow \int_0^{\beta} d \tau \ldots \qquad \prod_n dc_n dc_n^{*} \rightarrow \mathcal{D} c \mathcal{D} c^{*} \qquad \frac{c_n - c_{n-1}}{\Delta \tau} \rightarrow \partial_\tau$$ $$Z = \int \mathcal{D} c \mathcal{D} c^{*} e^{-\int_0^{\beta} d \tau \ L (c, c^{*})}$$ With the boundary condition $$c(\beta) = - c(0)$$.
For the Witten index: $$\text{Tr} \ (-1)^{F} e^{-\beta H}$$ One uses the anticommutation of $$(-1)^{F}$$ with fermions: $$\{(-1)^{F}, c\} = 0$$ Proceeding as above, but and evaluating $$(-1)^{F}$$ on a particular state, one gets now: $$\langle c | e^{-\beta H} | n \rangle \langle n| c \rangle$$ And after the insertions of identity the path integral with $$c(\beta) = c(0)$$
• I'm a little confused by the identity $\langle n|c\rangle \langle c|n\rangle = -\langle c|n\rangle \langle n|c\rangle$. Each of $\langle \cdot | \cdot \rangle$ is just a complex number resulting from the inner product on a Hilbert space, so shouldn't they commute? I thought the $|c\rangle$ were the Grassmann numbers. Commented Jun 28, 2023 at 15:46