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I can unterstand why because the integration over Grassman variables has to be translational invariant too, one has

$$ \int d\theta = 0 $$

and

$$ \int d\theta \theta = 1 $$

but I dont see where the rule for this double integration

$$ \int d^2 \theta \bar{\theta}\theta = -2i $$

comes from.

So can somebody explain to me how this is motivated and/or derived?

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  • $\begingroup$ Is the last integral supposed to read something like $\int d^2\!\theta \, \bar{\theta}\theta$? $\endgroup$ – Olof Apr 14 '13 at 12:25
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As with anything that has to do with supersymmetry the details will be dependent on your exact conventions, but we can obtain the result as follows:

Assume we have two Grassman variables $\theta_1$ and $\theta_2$. By applying your first formula twice we find $$\int d\theta_1 d\theta_2 \, \theta_2 \theta_1 = 1$$ Now combine these into $$\theta = \theta_1 + i\theta_2 \qquad \text{and} \qquad \bar{\theta}=\theta_1-i\theta_2.$$ We then have $$\bar{\theta} \theta = - 2i\theta_2\theta_1$$ and hence $$\int d\theta_1 d\theta_2 \bar{\theta} \theta = - 2i$$ which is exactly your second integral, if we identify the measure $$d^2\theta = d\theta_1 d\theta_2.$$

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