= Solution
The exponential acts on the <Fock vacuum>: $|\psi\rangle=e^{\zeta\sum_n\psi_na_n^\dagger}|0\rangle$. For bosons, $\psi_n$ are commuting <complex numbers>. Since $[a_n,\sum_m\psi_ma_m^\dagger]=\psi_n$, commuting $a_n$ through the exponential gives \b[$a_n|\psi\rangle=\psi_n|\psi\rangle$]. These are <unnormalized bosonic coherent states>.
For fermions, the labels are independent odd <Grassmann variables>, which anticommute with one another and with the fermionic operators. For one mode, the <fermionic coherent state> is $|\psi\rangle=(1-\psi a^\dagger)|0\rangle$. The <canonical anticommutation relations> give $a(-\psi a^\dagger)|0\rangle=\psi|0\rangle=\psi|\psi\rangle$, using $\psi^2=0$. The even factors $1-\psi_na_n^\dagger$ commute between modes, so the same argument applies to every $n$. A Grassmann eigenvalue is a formal extension of the state space, not an ordinary complex eigenvalue of the nilpotent annihilation operator.
With dual state $\langle\psi|=\langle0|e^{\zeta\sum_na_n\bar\psi_n}$, both cases obey $\langle\psi|\psi'\rangle=e^{\sum_n\bar\psi_n\psi_n'}$.
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