# Fermionic coherent state in Fock representation

The notes I follow define a Fermionic coherent state $$|c\rangle$$ as $$\begin{equation} \hat{c}|c\rangle=c|c\rangle \end{equation}$$ where $$\hat{c}$$ is the Fermionic annihilation operator and $$c$$ is a Grassman number. Then the notes give its explicit form in the Fock occupation space $$\begin{equation} |c\rangle=e^{\hat{c}^{\dagger}c}|0\rangle \end{equation}$$ Is it really a good form of it? The following calculation seems to contradict the definition : $$\begin{equation} e^{\hat{c}^{\dagger}c}|0\rangle=|0\rangle+\hat{c}^{\dagger}c|0\rangle=|0\rangle-c\hat{c}^{\dagger}|0\rangle=|0\rangle-c|1\rangle \end{equation}$$ $$\begin{equation} \hat{c}|c\rangle=\hat{c}\left(|0\rangle-c|1\rangle\right)=0+c\hat{c}|1\rangle=c|0\rangle \end{equation}$$ In the first line, I expanded the exponent using the fact that the square of a Grassman number is 0 or the Pauli exclusion principle (I can't create two fermions in a place). Then I simply acted by the annihilation operator on that state. It doesn't look like an eigenvalue. Is that representation in the Fock space wrong or do I make a mistake somewhere?

$$\newcommand{\ket}{|#1\rangle}$$Your definitions and calculation are correct. To prove that $$\hat{c}\ket{c}=c\ket{c}$$ from where you left you have to use the fact that Grassman numbers verify $$c^2=0$$: $$\hat{c}\ket{c} = c\ket{0} = c\ket{0} - c^2\ket{1} = c(\ket0 - c\ket1) = c\ket c$$