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

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


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

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 1


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.

  • $\begingroup$ Many thanks! So we can fix $\xi(0), \bar{\xi}(0), \xi(1)$ and $\bar{\xi}(1)$ all at the same time? $\endgroup$
    – user35319
    Feb 18, 2020 at 12:35

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