Fermionic partition function

Question

When we calculate the fermionic partition function, we use $\newcommand{\ket}[1]{| #1 \rangle} \newcommand{\bra}[1]{\langle #1 |} \newcommand{\braket}[1]{\langle #1 \rangle}$

$Z=tr(e^{-\beta H})=\int \bra{\bar \psi}e^{-\beta H}\ket{\psi} d\psi$

and in the bosonic case we use

$Z=tr(e^{-\beta H})=\int \bra{ \psi}e^{-\beta H}\ket{\psi} d\psi$

why $\braket{\bar \psi|\psi}=1$? Or following question, why is the resolution of identity given by $\int \ket {\bar \psi}\bra{\psi}d\psi e^{-\bar \psi \psi}d\bar \psi=1$ instead of $\int \ket { \psi}\bra{\psi}d\psi e^{-\bar \psi \psi}d\bar \psi=1$?

I understand that there is something to do with the fermionic statistics, but I am not sure how we see it here.

Answer 1

These are coherent states, you need to take a look on what is called "holomorphic representation of path integral". The coherent states which are the eigenvectors of the annihilation operator are not normalized. For the fermionic case you also need to use Grassmann variables.