- #1
thatboi
- 133
- 18
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.
$$\ket{\xi} = e^{\psi^{\dagger}T\xi}\ket{0}$$
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.