I have some insight on the first question. For the ordinary partition function:
Tr e−βH=∑n⟨n|e−βH|n⟩
Then one inserts the fermionic path integral:
∫dc dc∗e−cc∗∑n⟨n|c⟩⟨c|e−βH|n⟩=∫dc dc∗e−cc∗∑n⟨−c|e−βH|n⟩⟨n|c⟩=∫dc dc∗e−cc∗⟨−c|e−βH|c⟩
Where one has used :
⟨n|c⟩⟨c|n⟩=−⟨c|n⟩⟨n|c⟩.
Then one inserts resolution of identity N times (here cN=−c0):
∫∏dcndc∗ne−∑ncnc∗n⟨cN|e−ΔτH|cN−1⟩…⟨c1|e−ΔτH|c0⟩
Where τ=β/N. In the continuum limit one gets the path integral after:
ΔτN∑n…→∫β0dτ…∏ndcndc∗n→DcDc∗cn−cn−1Δτ→∂τ
Z=∫DcDc∗e−∫β0dτ L(c,c∗)
With the boundary condition c(β)=−c(0).
For the Witten index:
Tr (−1)Fe−βH
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:
⟨c|e−βH|n⟩⟨n|c⟩
And after the insertions of identity the path integral with c(β)=c(0)
This post imported from StackExchange Physics at 2020-12-12 20:05 (UTC), posted by SE-user spiridon_the_sun_rotator