# Fermionic path integral with boundary

Given a path integral:

$$K(\eta,\xi) = \int\limits_{\psi(0)=\eta}^{\psi(1)=\xi} e^{\int_0^1\dot{\psi}(t)\psi(t) dt} D\psi\tag{1}$$

where $$\psi(t)$$ are a real Grassmann fields.

I get two answers depending on whether the path is split into an odd or even number of parts in time. The answer is either Grassmann-odd or Grassmann-even:

$$K^1(\eta,\xi) = \eta - \xi\tag{2}$$ $$K^0(\eta,\xi) = 1 + \eta \xi\tag{3}$$

which satisfy:

$$\int K^a(A,B)K^b(B,C) dB = K^{(a+b+1) (mod\ 2)}(A,C). \tag{4}$$

How are we to understand these two solutions? Should we always split time into an even number of intervals and so just go with the first solution?

I can't find anywhere in the literature which deals with fermion integrals with boundary, even though there is lots on Bosonic path integrals with boundary conditions.

OP does not mention which physical system they have in mind. But one thing is for sure: To produce a non-trivial physical quantity that could be measured in a detector, all Grassmann-variables should be integrated out, including the boundary variables $$\xi$$ and $$\eta$$.
1. On one hand, it is conventionally assumed that the discretization of a path integral contains an even number of integrations. In this case we can directly integrate $$\int\!d\xi~d\eta~K_0(\eta,\xi)=1$$ and get a non-trivial quantity.
2. On the other hand, with an odd number of integrations, $$\int\!d\xi~d\eta~K_1(\eta,\xi)\equiv0$$, so $$\int\!d\xi~d\eta~K_1(\eta,\xi)M(\eta,\xi)$$ should as a minimum be prepared with a non-trivial measure factor $$M(\eta,\xi)$$ to make sense. This is reminiscence of the ghost number anomaly in string theory [1] so we will stop short of ruling out this somewhat wierd case.