- #1

JD_PM

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- 158

- 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!

\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!