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!

 

[mighty2000]

Hello,

Thank you for your response. It seems like you have made some progress in simplifying the exponential argument and integrating over the Grassmann variables. However, there are a few things that need to be clarified in order to fully answer your question.

Firstly, it would be helpful to know what problem you are trying to solve using the Grassmann algebra. This would give us a better understanding of the context and the specific rules that need to be applied.

Secondly, it seems like you have made a mistake in the change of variables that you introduced. The RHS exponential term should contain a linear Grassmann term, so the change of variables should not eliminate it completely. Additionally, the RHS exponential term should be independent of the Grassmann variables, so introducing an extra term like $\bar{\eta}_i \theta_j$ would not be allowed.

Finally, in order to show the equality in a more general way, you would need to show that the integration over the Grassmann variables yields a determinant of the matrix B. This can be done by expanding the determinant using the Leibniz formula and then using the properties of the Grassmann variables to simplify the terms.

I hope this helps. If you provide more information about the problem and clarify the steps in your solution, I would be happy to provide further assistance. Keep up the good work!