Hi,

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

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

$\displaystyle \theta_i $ are complex Grassmann Numbers.

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

A is a hermitian matrix.

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

$ I got:

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

Expansion

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

gives

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


Now I don't know, how to go on. I know that $\displaystyle \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