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}<br /> <br />
I got:

<br />
\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