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.


1 Answer 1


OP's formulas are correct (possibly modulo sign conventions). Boundary values appear e.g. in a fermionic coherent state path integral, see e.g. this Phys.SE post.

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.


  1. R. Blumenhagen, D. Lust & S. Theisen, Basic Concepts of String Theory, 2012; p. 116 eq. (5.49).

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