Boundary conditions of fermionic coherent states path integral

Given the algebra of a fermionic oscillator

$$\{\hat{a},\hat{a}^\dagger \}=1\,, \qquad \hat{a}^2=(\hat{a}^\dagger)^2=0,$$

with coherent states $$\hat{a}|\xi\rangle=\xi|\xi\rangle$$, let's consider the transition amplitude between coherent states $$|\eta\rangle$$ and $$\langle\bar{\lambda}|$$ with hamiltonian $$\hat{H}$$ is given by

$$\langle\bar{\lambda}|e^{-i\hat{H}}|\eta\rangle = \int_{\xi(0)=\eta}^{\bar{\xi}(1)=\bar{\lambda}} D\bar{\xi}D\xi e^{iS[\bar{\xi},\xi]}$$

for

$$S = i\int_0^1 d\tau \, \bar{\xi}\dot{\xi}(\tau)-H(\bar{\xi},\xi)-i\bar{\xi}\xi(1).$$

Now my question is: do the boundary conditions automatically imply $$\xi(1)=\lambda$$ and $$\bar{\xi}(0)=\bar{\eta}$$? If not, does that mean that the integral involves all possible boundary conditions $$\xi(1)$$ and $$\bar{\xi}(0)$$?

1 Answer

Notation in this answer: In this answer, let $$z,z^{\ast}$$ denote two independent complex Grassmann-odd numbers. Let $$\overline{z}$$ denote the complex conjugate of $$z$$.

With this notation OP's Grassmann-odd/fermionic coherent state path integral reads

$$\langle\lambda^{\ast}|e^{-i\hat{H}}|\eta\rangle ~=~ \int_{\xi(0)=\eta}^{\bar{\xi}(1)=\lambda^{\ast}} \!{\cal D}\bar{\xi}~{\cal D}\xi~ e^{iS[\bar{\xi},\xi]}.$$

In particular, the complex conjugate $$\bar{\xi}(0)~=~\bar{\eta} \qquad\text{and}\qquad \xi(1)~=~\bar{\lambda}^{\ast}$$ of the boundary conditions are also satisfied in the path integral, cf. OP's specific question.

Nevertheless, a hallmark feature of coherent state path integrals should probably be stressed: For generic boundary conditions, there don't exist classical paths! The same situation happens for Grassmann-even/bosonic coherent state path integrals. It is related to the overcompleteness of the coherent states, cf. this related post.

• Many thanks! So we can fix $\xi(0), \bar{\xi}(0), \xi(1)$ and $\bar{\xi}(1)$ all at the same time? – user35319 Feb 18 at 12:35