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

Question

 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

Comments

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