Showing an identity involving Grassmann variables

Keep up the good work!In summary, for the problem being solved using Grassmann algebra, the basic rules are needed, including the anti-commutation and integration rules. A change of variables is introduced to simplify the exponential argument, but it seems to have a mistake as it eliminates the linear Grassmann term and introduces an extra term. To show the equality in a more general way, the integration over the Grassmann variables needs to be expanded using the Leibniz formula and simplified using the properties of the Grassmann variables. More information and clarification of the steps in the solution is needed to provide further assistance.
  • #1
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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)##

kdolzekdekdkeode.png


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! :biggrin:
 
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  • #2


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!
 

Related to Showing an identity involving Grassmann variables

1. What are Grassmann variables?

Grassmann variables are mathematical objects that are used in the field of algebraic geometry and theoretical physics. They are anticommuting variables, meaning that they do not follow the usual rules of multiplication and instead satisfy the Grassmann algebra. They are often used to describe fermionic particles in quantum field theory.

2. How are Grassmann variables used in showing an identity?

Grassmann variables are used in showing an identity by manipulating them using the rules of Grassmann algebra. These rules involve anticommutativity, distributivity, and linearity. By carefully applying these rules, one can transform the expression on one side of the identity to match the expression on the other side.

3. What is the Grassmann algebra?

The Grassmann algebra is a mathematical structure that extends the usual algebra of real or complex numbers. It is based on the anticommutativity property of Grassmann variables, where the product of two variables is equal to the negative of the product of the same variables in reverse order. This algebra is used in many areas of mathematics and physics, including differential geometry and quantum mechanics.

4. What are some common identities involving Grassmann variables?

Some common identities involving Grassmann variables include the binomial theorem, the Jacobi identity, and the Grassmann Fourier transform. These identities are useful in various areas of mathematics and physics, such as quantum field theory and supersymmetry.

5. How can I learn more about Grassmann variables and their applications?

There are many resources available for learning about Grassmann variables and their applications. Some recommended books include "Supersymmetry and Supergravity" by Peter West and "Grassmannians, Moduli Spaces and Vector Bundles" by Christian Okonek, Michael Schneider, and Heinz Spindler. Additionally, online lectures and tutorials are available on websites such as YouTube and arXiv.org.

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