Grassmann Integration: Clarifying Notation in "hep-th/0108200

[Korybut]

Hi, everyone!

I am trying to understand notation of this textbook http://arxiv.org/abs/hep-th/0108200

page 8, formulas 2.1.4 and 2.1.5

$$\int d \theta_\alpha \theta^\beta=\delta_\alpha^\beta$$

this could be found in any textbook the weird that from the above formula follows

$$\int d^2 \theta \; \theta^2=-1$$

I know what θ² means, but what is d²θ I could hardly guess. According to standard Berezin definition there should be $i$ in the r.h.s. of the last formula

Please help to clarify this

Best wishes
Korybut Anatoly

 

[MathematicalPhysicist]

Grassman integration is the same as Grassman's derivation, so unless I am mistaken it should be 2 and not -1.

Think of it as: d^2(\theta^2)/d^2 \theta

I leave this to the experts.

 

[Avodyne]

I'm not a fan of the notation in this text. But in general, for a Grassmann variable with two components like $\theta$, we have $d^2\theta=d\theta_1 d\theta_2$. This is like writing $d^3x=dx_1 dx_2 dx_3$ for a vector $\vec x$.

Then, if $\theta^2 =\theta_1\theta_2$ (which is true in everybody's convention up to some factor like $-1$ or $i$), we have
\int d^2\theta\,\theta^2 = \int d\theta_1d\theta_2\,\theta_1\theta_2=-\int d\theta_2d\theta_1\theta_1\theta_2=-\int d\theta_2\,\theta_2=-1.
Edit: I'm using the convention $\int d\theta_\alpha\,\theta_\beta=\delta_{\alpha\beta}$, which differs from this text.

 

[Korybut]

I'm not a fan of the notation in this text. But in general, for a Grassmann variable with two components like $\theta$, we have $d^2\theta=d\theta_1 d\theta_2$. This is like writing $d^3x=dx_1 dx_2 dx_3$ for a vector $\vec x$.

Then, if $\theta^2 =\theta_1\theta_2$ (which is true in everybody's convention up to some factor like $-1$ or $i$), we have
\int d^2\theta\,\theta^2 = \int d\theta_1d\theta_2\,\theta_1\theta_2=-\int d\theta_2d\theta_1\theta_1\theta_2=-\int d\theta_2\,\theta_2=-1.
Edit: I'm using the convention $\int d\theta_\alpha\,\theta_\beta=\delta_{\alpha\beta}$, which differs from this text.

I know that $\theta$ anticommute, but how one deduce that $d\theta$ obey the same rule

 

[julian]

\int d \theta_1 d \theta_2 \theta_1 \theta_2 = - \int d \theta_1 d \theta_2 \theta_2 \theta_1 = - \int d \theta_1 \theta_1 = -1 = - \int d \theta_2 d \theta_1 \theta_1 \theta_2

and

\int d \theta_1 d \theta_2 \theta_2 \theta_1 = 1 = \int d \theta_2 d \theta_1 \theta_1 \theta_2 = - \int d \theta_2 d \theta_1 \theta_2 \theta_1.

 

[julian]

So I've proved

\int d \theta_1 d \theta_2 ( \theta_1 \theta_2) = - \int d \theta_2 d \theta_1 (\theta_1 \theta_2)

This obviously implies

\int d \theta_1 d \theta_2 ( \theta_2 \theta_1) = - \int d \theta_2 d \theta_1 (\theta_2 \theta_1).

So d \theta_1 d \theta_2 = - d \theta_2 d \theta_1