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?


1 Answer 1


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 $c(\beta) = c(0)$

  • $\begingroup$ 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. $\endgroup$
    – WillG
    Commented Jun 28, 2023 at 15:46

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