Showing an identity involving Grassmann variables

[JD_PM]

Homework Statement

Given a collection of independent Grassmann variables $\{\theta_i,\bar{\theta}_i, \eta_i, \bar{\eta}_i \}$, where $i \in \{ 1, ..., n\}$, and an invertible $n \times n$ matrix $B_{ij}$.

We want to show that the following identity is true

\begin{equation*}
Z[B, \eta, \bar{\eta}] = \int \prod_{i,j=1}^n d\bar{\theta}_i d \theta_j \exp\left(-\bar{\theta}_i B_{ij} \theta_j + \bar{\eta}_i \theta_j + \bar{\theta}_i \eta_j \right) = \det(B) \exp(\bar{\eta}_i(B^{-1})_{ij}\eta_j)
\end{equation*}

Relevant Equations

Please see below

All we should need for this problem are the basic rules for the Grassmann algebra

\begin{equation*}
\{ \theta_i, \theta_j\} = 0, \quad \theta^2_i=0
\end{equation*}

\begin{equation*}
\int d\theta_i = 0, \quad \int d\theta_i \ \theta_i = 1
\end{equation*}

Starting from left to right

Given that the RHS exponential term has no linear Grassmann terms, we introduce a change of variables such that the resulting integration yields no linear Grassmann terms

\begin{equation*}
\theta := -B^{-1} \theta' + B^{-1} \eta, \quad
\bar{\theta} := \bar{\theta}' + \bar{\eta} B^{-1}
\end{equation*}

Working out the exponential argument we get

\begin{align*}
-\bar{\theta}_i B_{ij} \theta_j + \bar{\eta}_i \theta_j + \bar{\theta}_i \eta_j &= -( \bar{\theta}_i' + \bar{\eta}_i(B^{-1})_{ij}) B_{ij} (-(B^{-1})_{ij} \theta_j' + (B^{-1})_{ij} \eta_j) \\
&+ \bar{\eta}_i (-(B^{-1})_{ij} \theta_j' + (B^{-1})_{ij} \eta_j) + (\bar{\theta}_i' + \bar{\eta}_i (B^{-1})_{ij}) \eta_j \\
&= \bar{\theta}_i'\theta_j' - \bar{\eta}_i (B^{-1})_{ij} \eta_j -\bar{\theta}_i' \eta_j + \bar{\eta}_i (B^{-1})_{ij} \theta_j' - \bar{\eta}_i(B^{-1})_{ij}\theta_j' \\
&+ \bar{\eta}_i (B^{-1})_{ij} \eta_j + \bar{\theta}_i'\eta_j + \bar{\eta}_i (B^{-1})_{ij} \eta_j \\
&= \bar{\theta}_i'\theta_j' + \bar{\eta}_i (B^{-1})_{ij} \eta_j
\end{align*}

Hence we get

\begin{align*}
Z[B, \eta, \bar{\eta}] &= \int \prod_{i,j=1}^n d\bar{\theta}_i' d \theta_j' \exp(\bar{\theta}_i'\theta_j' + \bar{\eta}_i (B^{-1})_{ij} \eta_j) \\
&= \exp(\bar{\eta}_i (B^{-1})_{ij} \eta_j) \int \prod_{i,j=1}^n d\bar{\theta}_i' d \theta_j' \exp(\bar{\theta}_i'\theta_j') \\
&\approx \exp(\bar{\eta}_i (B^{-1})_{ij} \eta_j) \underbrace{\int \prod_{i,j=1}^n d\bar{\theta}_i' d \theta_j' (1+ \bar{\theta}_i'\theta_j')}_{=1}
\end{align*}

Using the fact that the partition function $Z \propto 1/ \det(B)$

We get that


\begin{equation*}
Z[B, \eta, \bar{\eta}] = \int \prod_{i,j=1}^n d\bar{\theta}_i d \theta_j \exp\left(-\bar{\theta}_i B_{ij} \theta_j + \bar{\eta}_i \theta_j + \bar{\theta}_i \eta_j \right) \approx \det(B) \exp(\bar{\eta}_i(B^{-1})_{ij}\eta_j)
\end{equation*}

So my issue is that I could only get to the RHS working up to first order. Is the above OK so far? If yes, how could show the equality in a more general way?

Thank you!