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
$$
\rm{Tr} (−1)^F \exp(−\beta H).
$$

*

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


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


*What are the ground state sectors (bosonic vs fermionic) for the above boundary conditions?
 A: 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 $c(\beta) = c(0)$
