[SOLVED] Proof of Gaussian Integral Identity with Grassmann Numbers

• Feb 21st 2009, 09:40 AM
Kai.Grohs
[SOLVED] Proof of Gaussian Integral Identity with Grassmann Numbers
Hi,

I would like to show explicitly the following identity for N = 2

$\int d \theta^-_1 d \theta_1 ... d \theta^-_N d \theta_N e^{-\theta^- A \theta} = det A$

$\theta_i$ are complex Grassmann Numbers.

$\theta^-_i$ are the complex conjugated of $\theta_i$.

A is a hermitian matrix.

With $\int \prod d \theta^-_i d \theta_i e^{- \theta^-_i A_{ij} \theta_j}

$
I got:

$
\int d \theta^-_1 d \theta_1 d \theta^-_2 d \theta_2 e^{- (\theta^-_1 A_{11} \theta_1 + \theta^-_2 A_{21} \theta_1 + \theta^-_1 A_{12} \theta_2 + \theta^-_2 A_{22} \theta_2)}$

Expansion

$e^x = \sum_{k=1}^\infty \frac{x^k}{k!}$

gives

$\int d \theta^-_1 d \theta_1 d \theta^-_2 d \theta_2 (1 + (-\theta^-_1 A_{11} \theta_1 - \theta^-_2 A_{21} \theta_1 - \theta^-_1 A_{12} \theta_2 - \theta^-_2 A_{22} \theta_2)$
$+ \frac{1}{2} (-\theta^-_1 A_{11} \theta_1 - \theta^-_2 A_{21} \theta_1 - \theta^-_1 A_{12} \theta_2 - \theta^-_2 A_{22} \theta_2)^2 )$

Now I don't know, how to go on. I know that $\theta^2 = 0$ and the integral over 1 vanishes.

But this doesn't help me enough to continue the proof.

Could anybody help me with this?

Regards,
Kai