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Is there a reason why $\int\! d\theta~\theta = 1$ for a Grassmann integral? Books give arguments for $\int\! d\theta = 0$ which I can follow, but not for the former one.

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1 Answer 1

If the integral $I:=\int d\theta$ on the algebra ${\cal A} $ of superfunctions $f(\theta)=\theta a + b$ should be

1) a (graded) linear operation,

2) translation invariant, i.e., $\int d\theta ~f(\theta+\theta') =\int d\theta~f(\theta)$,

3) and if the output $\int d\theta~ f(\theta)$ should not depend on the integration variable $\theta$,

then it is easy to check that the usual formulas for the Berezin integral is the only possibility up to an overall multiplicative normalization factor.

Interestingly, this means that Berezin integration $\int d\theta$ is the same as differentiation $\partial / \partial\theta$.

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Thank you for giving me a better answer than I've ever found. Condition (3) is what I was missing, and I assume that it is true since the Berezin integral is to be thought of as a definite integral, and finite at that? –  F R Oct 16 '11 at 16:05

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