How to Show \(\psi_{i}\ket{\xi} = \xi_{i}\ket{\xi}\)?

[thatboi]

I came across the following formula (2.68) in di Francesco's CFT book for a fermionic coherent state:
$$\ket{\xi} = e^{\psi^{\dagger}T\xi}\ket{0}$$
where$\ket{\xi} = \ket{\xi_{1},...,\xi_{n}}$, $\xi_{i}$ is a Grassman number, $T$ is some invertible matrix, and $\psi^{\dagger}$ is the fermion creation operator. I want to show that $$\psi_{i}\ket{\xi} = \xi_{i}\ket{\xi}$$ and am struggling to do so. I tried considering just the simplest case where $T$ is a 2x2 matrix. Then we would have (take $i=1$ as an example):
$$\psi_{1}\ket{\xi} =\left(1+T_{2,1}\xi_{1}\psi^{\dagger}_{2}\right)\left(1+T_{2,2}\xi_{2}\psi^{\dagger}_{2}\right) \psi_{1}\left(1+T_{1,1}\xi_{1}\psi^{\dagger}_{1}\right)\left(1+T_{1,2}\xi_{2}\psi^{\dagger}_{1}\right)\ket{0}$$
where I have used the fact that $e^{\eta} = 1+\eta$ for $\eta$ a Grassman number. Further expanding then gives:
$$\psi_{1}\left(1+T_{1,1}\xi_{1}\psi^{\dagger}_{1} + T_{1,2}\xi_{2}\psi^{\dagger}_{1} + T_{1,1}T_{1,2}\xi_{1}\xi_{2}\psi^{\dagger}_{1}\psi^{\dagger}_{1}\right)\ket{0}$$
Now the last term is automatically $0$ by Pauli exclusion, so passing the $\psi_{1}$ through results in:
$$\left(-T_{1,1}\xi_{1}-T_{1,2}\xi_{2}\right)\ket{0}$$
where the additional minus sign is because $\{\psi,\xi\}=0$.
I am not sure how to proceed from here. Any advice appreciated.

 

[Demystifier]

You missed one minus sign in the third equation, so your final equation should have plus signs instead of the minus ones. To proceed, now compute $\xi_1|\xi\rangle$ in a similar way.

 

[thatboi]

You missed one minus sign in the third equation, so your final equation should have plus signs instead of the minus ones. To proceed, now compute $\xi_1|\xi\rangle$ in a similar way.

Do you mean when I pass the $\psi_{1}$ through the $\psi^{\dagger}_{2}$ terms? I don't see why a minus sign would be picked up ($\psi^{\dagger}_{1}$ should commute with both $\xi_{1}\psi^{\dagger}_{2}$ and $\xi_{2}\psi^{\dagger}_{2}$).

 

[Demystifier]

I don't see why a minus sign would be picked up ($\psi^{\dagger}_{1}$ should commute with both $\xi_{1}\psi^{\dagger}_{2}$ and $\xi_{2}\psi^{\dagger}_{2}$).

You are right, my bad.

 

[thatboi]

You are right, my bad.

Haha no worries, I can't edit the original post now, so I'll proceed based off what you said, it's a bit of a mess so let me know if you see anything suspicious.
$$\begin{equation}
\begin{split}
\xi_{1}\ket{\xi} &= \xi_{1}\left(1+T_{2,1}\xi_{1}\psi^{\dagger}_{2}\right)\left(1+T_{2,2}\xi_{2}\psi^{\dagger}_{2}\right) \left(1+T_{1,1}\xi_{1}\psi^{\dagger}_{1}\right)\left(1+T_{1,2}\xi_{2}\psi^{\dagger}_{1}\right)\ket{0}\\
&= \xi_{1}\left(1+T_{2,1}\xi_{1}\psi^{\dagger}_{2} + T_{2,2}\xi_{2}\psi^{\dagger}_{2}\right)\left(1+T_{1,1}\xi_{1}\psi^{\dagger}_{1}+T_{1,2}\xi_{2}\psi^{\dagger}_{1}\right)\ket{0}\\
&= \left(\xi_{1}+T_{2,2}\xi_{1}\xi_{2}\psi^{\dagger}_{2}\right)\left(1+T_{1,1}\xi_{1}\psi^{\dagger}_{1}+T_{1,2}\xi_{2}\psi^{\dagger}_{1}\right)\ket{0}\\
&=\left(\xi_{1}+T_{1,2}\xi_{1}\xi_{2}\psi^{\dagger}_{1} + T_{2,2}\xi_{1}\xi_{2}\psi^{\dagger}_{2}\right)\ket{0}
\end{split}
\end{equation}$$
In the second line I dropped the $\psi^{\dagger}_{i}\psi^{\dagger}_{i}$ terms due to Pauli exclusion again.
This still doesn't really look like what I had in my original post so I'm stuck.