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  How to get classical action for fermion's path integral ?

+ 1 like - 0 dislike

 Consider our Hamiltonian is

$$ H= E_1 a^\dagger_1 a_1 + E_2 a^\dagger_2 a_2 + T_c(a^\dagger_2 a_1 + a^\dagger_1 a_2)$$

Explicitly, The fermion coherent-state path integral for the forward transition amplitude is given by

$$ \langle \zeta_f| U(t-t_0)| \zeta_i \rangle  = \int D[\zeta^*, \zeta]  \exp (iS_c[\zeta^*, \zeta])  -------(A) $$

Here $\zeta^*, \zeta$ are Grassmann variables characterizing fermion coherent states. The $S_c$ is defined as

$$ S_c[\zeta^*, \zeta] = \sum_{i=1,2}\left\lbrace \frac{-i}{2} \left[\zeta^*_{if} \zeta_i(t) + \zeta^*_i(t_0)\zeta_{i0} \right]  \qquad \\ + \int_{t_0}^{t} d\tau \left[ \frac{i}{2} (\zeta^*_i\dot{\zeta}_i - \dot{\zeta}^*_i \zeta_i )- \left[ E_i \zeta^*_i \zeta_i + T_c \zeta^*_i \zeta_{i'}\right] \right] \right\rbrace $$    

In Eq.(A), the path integral $D[\zeta^*, \zeta]$ integrates over all paths $\zeta_i(\tau)$ and $\zeta^*_i(\tau)$  bounded by $\zeta_i(t_0)=\zeta_{i0}$ and $\zeta^*_i(t_0)=\zeta^*_{if}$, with $i \neq i'$. 

I have no background in Path integral calculations. I don't know how can i get above expression for $S_c$.  Any help ?  

These expressions are taken from https://journals.aps.org/prb/pdf/10.1103/PhysRevB.78.235311

asked Oct 31, 2019 in Theoretical Physics by Khan [ no revision ]
recategorized Oct 31, 2019 by Dilaton

Then you should read something to get the missing background first. An online source is, for example, A. Rogers, Fermionic Path Integration and Grassmann Brownian Motion

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