1
$\begingroup$

Let $(\theta,A\theta)=\theta_i A_{ij}\theta_j$ where $A$ is some $(2\times2)$ antisymmetric matrix.

I want to generalize the following

$$I(A) =\int d\theta_1d\theta_2~ \exp\Bigg[\frac{1}{2}(\theta,A\theta)\Bigg]=\int d\theta_1d\theta_2~ (1+\theta_1\theta_2A_{12}) = A_{12}=\sqrt{\det A}$$

to the $n$-tuple case.

Let now $$A:=\begin{bmatrix} 0 & 1 & \; & \; \\ \;-1 & 0 & & \; \\ \; & \; & 0 & 1 \\ \; & \; & -1 & 0 \\ \; & \; & \; & \, &\ddots \\ \end{bmatrix}.$$

I evaluate, I get the following $$I(A) = \int d\theta_n\dots d\theta_1\,\exp\Bigg[\frac{1}{2}(\theta,A\theta)\Bigg]\\ = \int d\theta_n\dots d\theta_1\, (\theta_1\theta_2+\theta_3\theta_4+\cdots)\\=0$$

The answer should be $$I(A) = 1.$$

In the above I use (perhaps incorrectly?)

$$\int d\theta_n\dots d\theta_1\, \theta_n\dots \theta_1\, = 1 $$ and $$\int d\theta_n\dots d\theta_1= 0. $$

Where do I err?

EDIT: I think I know how to fix this: it is the last term in the expansion of the exponential the contributes. All other terms give zero (just like the one above). I will add the solution later.

$\endgroup$

2 Answers 2

2
$\begingroup$

The problem is that you are generalizing the real Gaussian integral to the Grassmannian case rather than the complex Gaussian integral $\int dz d\bar{z} e^{-(z, Az)}$, where $(z, A z) = \sum_{i,j} z_i A_{ij} \bar{z}_j$ and $d z d\bar{z} = dz_1 d\bar{z}_1 \ldots dz_n d\bar{z}_n$.

You can see that the determinant will not arise by looking at the $2$-dimensional case. That is, let $A$ be a $2 \times 2$ matrix with entries $A_{ij}$. Then $(\theta, A \theta) = \theta_1 A_{12} \theta_2 + \theta_2 A_{21} \theta_1$. The anti-commutativity kills the $A_{11}$ and $A_{22}$ that you need in the determinant of $A$.

To fix this, introduce "conjugate variables" $\bar{\theta}_1$ and $\bar{\theta}_2$. The bar doesn't denote any kind of operation on the $\theta_i$. The $\bar{\theta}_i$ are just new Grassmann variables. Then define \begin{equation} (\theta, A \bar{\theta}) = \theta_1 A_{11} \bar{\theta_1} + \theta_1 A_{12} \bar{\theta}_2 + \theta_2 A_{21} \bar{\theta}_1 + \theta_2 A_{22} \bar{\theta}_2. \end{equation} The terms $\theta_i A_{ii} \bar{\theta}_i$ no longer vanish because there are no relations between $\theta_i$ and $\bar{\theta}_i$. Now \begin{equation} \int d\theta_1 d\bar{\theta}_1 d \theta_2 d\bar{\theta}_2 e^{-(\theta, A \bar{\theta})} = \int d\theta_1 d\bar{\theta}_1 d \theta_2 d\bar{\theta}_2 \frac{1}{2} (\theta, A \bar{\theta})^2 \end{equation} since the first two terms $1$ and $-(\theta, A \bar{\theta})$ in the exponential vanish under integration and the higher-order terms $(\theta, A \bar{\theta})^n$ for $n \geq 3$ vanish by anti-commutativity.

Now a computation shows that $(\theta, A \bar{\theta})^2 = 2 \theta_1 \bar{\theta}_1 \theta_2 \bar{\theta}_2 \det A$, so \begin{equation} \int d\theta_1 d\bar{\theta}_1 d \theta_2 d\bar{\theta}_2 e^{-(\theta, A \bar{\theta})} = \det A. \end{equation}

Given Grassmann variables $\theta_i, \bar{\theta}_i$ for $i = 1, \ldots, n$, defining $(\theta, A \bar{\theta}) = \sum_{i,j=1}^n \theta_i A_{ij} \bar{\theta}_j$ yields $\int d\theta_1 d\bar{\theta}_1 \ldots d\theta_n d\bar{\theta}_n e^{-(\theta, A \bar{\theta})} = \det A$ by a similar computation (this is really an exercise in bookkeeping).

$\endgroup$
2
$\begingroup$

You are absolutely right in your edit. Only the highest order term of the expansion survives integration. This is because $\int d\theta_i=0$ so that we need a term where all grassmann variables are there ($\int d\theta_n...d\theta_1 \theta_1...\theta_n=1$).

I noticed that this post is really old but still felt like replying as this connects to subjects maters I am currently studying about majorana fermions in physics.

In general one can always write a "gaussian" integral over real grassmann variables as \begin{equation} \int d\theta_1d\theta_2...d\theta_N \exp\left\{-\frac{1}{2}\sum_{i,j=1}^N\theta_iA_{ij}\theta\right\}. \end{equation} Where $A$ is a $N\times N$ real skew-symmetric matrix. Now notice that such matrices have the following properties:

Assume $\vec{x}$ to be an eigenvector of $A$ than $A\vec{x}=\lambda\vec{x}$ and $(A\vec{x})^\dagger=(\lambda\vec{x})^\dagger \rightarrow \vec{x}^\dagger A^T=\lambda^*\vec{x}^\dagger$ using this

\begin{equation} \label{Im} \begin{split} \vec{x}^\dagger A^T\vec{x}&=\lambda^*\vec{x}^\dagger\vec{x}\\ \vec{x}^\dagger (-A)\vec{x}&=\lambda^*\vec{x}^\dagger\vec{x}\\ -\lambda\vec{x}^\dagger\vec{x}&=\lambda^*\vec{x}^\dagger\vec{x}\\ -\lambda&=\lambda^* \end{split} \end{equation}

so that

\begin{equation} \label{pairs} \begin{split} (A\vec{x})^*&=(\lambda\vec{x})^*\\ A\vec{x}^*&=\lambda^*\vec{x}^*\\ A\vec{x}^*&=-\lambda\vec{x}^*. \end{split} \end{equation}

We have found that $\lambda_i$ is pure imaginary and that all eigenvalues appear in $\pm\lambda$ pairs. Consequently $A$ has the following spectrum of eigenvalues

\begin{equation} \label{spec} \lambda(A)=\{\pm i\lambda_1, \pm i\lambda_2,...,\pm i\lambda_m\}\quad\quad\text{Where}\quad\lambda_i\in\mathbb{R} \end{equation}

As a side note observe that if $N$ is uneven that at least one of the eigenvalues must be zero. This is not the case for even $N$ where we have defined $(N=2m)$. Using the property that the determinant is so called similarity invariant (similar matrices give the same determinant) we can write

\begin{equation} \label{det} \det(A)=\prod_{i=1}^m\lambda_i^2 \end{equation}

Also note that such a matrix can be put in to block-diagonal form by special-orthogonal transformation $D=OAO^T$ (see wiki) to get \begin{equation} \begin{split} D&=diag\left(\left( \begin{matrix} 0 & \lambda_1 \\ -\lambda_1 & 0 \end{matrix} \right) , \left( \begin{matrix} 0 & \lambda_2 \\ -\lambda_2 & 0 \end{matrix} \right) ,..., \left( \begin{matrix} 0 & \lambda_m \\ -\lambda_m & 0 \end{matrix} \right)\right) \end{split} \end{equation}

We can now start looking at the integral. First lets show that the measure will be invariant under change of variables $O_{ij}\psi_j=\psi_i'$ since

\begin{equation} \begin{split} \prod_i^N\theta_i'&=\frac{1}{N!}\epsilon_{i_1,i_2,...,i_N}\theta_{i_1}'\theta_{i_2}'...\theta_{i_N}'\\ &=\frac{1}{N!}\epsilon_{i_1,i_2,...,i_N}O_{i_1,i_1'}\theta_{i_1'}O_{i_2,i_2'}\theta_{i_2'}...O_{i_N,i_N'}\theta_{i_N'}\\ &=\frac{1}{N!}\epsilon_{i_1,i_2,...,i_N}U_{i_1,i_1'}O_{i_2,i_2'}...O_{i_N,i_N'}\epsilon_{i_1',i_2',...,i_N'}\prod_i^N\theta_{i}\\ &=\det(O)\prod_i^N\theta_{i}\\ &=\prod_i^N\theta_{i} \end{split} \end{equation}

This means that (omitting the apostrophe's) we can write the integral in the following form and start to solve it

\begin{equation} \begin{split} &\int d\theta_1d\theta_2...d\theta_N e^{-\frac{1}{2}\theta_iD_{ij}\theta_j}\\ &=\int d\theta_1d\theta_2...d\theta_N \exp\left\{-\frac{1}{2}\left(\lambda_1\theta_1\theta_2-\lambda_1\theta_2\theta_1+\lambda_2\theta_3\theta_4-\lambda_2\theta_4\theta_3+...+\lambda_m\theta_{m-1}\theta_m-\lambda_m\theta_{2m}\theta_{2m-1}\right)\right\}\\ &=\int d\theta_1d\theta_2...d\theta_N \exp\left\{\lambda_1\theta_2\theta_1+\lambda_2\theta_4\theta_3+...+\lambda_m\theta_{2m}\theta_{2m-1}\right\}\\ &=\int d\theta_1d\theta_2...d\theta_N \exp\left\{\sum_{i=1}^m\lambda_i\theta_{2i}\theta_{2i-1}\right\}\\ &=\int d\theta_1d\theta_2...d\theta_N\prod_{i=1}^m\exp\left\{\lambda_i\theta_{2i}\theta_{2i-1}\right\}\\ &=\int d\theta_1d\theta_2...d\theta_N\prod_{i=1}^m(1+\lambda_i\theta_{2i}\theta_{2i-1})\\ &=\int d\theta_1d\theta_2...d\theta_N\left(\prod_{i=1}^m\lambda_i\right)\theta_2\theta_1\theta_4\theta_3...\theta_N\theta_{N-1}\\ &=\prod_{i=1}^m\lambda_i\\ &=\sqrt{\det(A)} \end{split} \end{equation}

Where we have used that the only the part of the product $(\prod_{i=1}^m(1+\lambda_i\theta_{i+1}\theta_i))$ that survives integration is where we multiply all the grassmann variables (All other terms will vanish because $\int d\theta=0$). At the end of the calculation I used that pairs of grassmann variables commute ([$\theta_2\theta_1,\theta_4\theta_3]=0$)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .