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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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Let $\theta$ denote a Grassmann-odd variable.

  1. If an integral $$I~:=~\int\! d\theta\tag{1}$$ on the algebra ${\cal A} $ of superfunctions $f(\theta)=\theta a + b$ should be

    • a (graded) linear operation,

    • translation invariant $$\int\! d\theta ~f(\theta+\theta_0) ~=~\int\! d\theta~f(\theta),$$

    • and the output $\int\! d\theta~ f(\theta)$ should not depend on the integration variable $\theta$ (apart from $a$ and $b$),

    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 $\partial / \partial\theta$ !

  2. If a definite integral $$I(\theta_1,\theta_2)~\stackrel{?}{:=}~\int_{\theta_1}^{\theta_2}\! d\theta\tag{2}$$ on the algebra ${\cal A} $ of superfunctions $f(\theta)=\theta a + b$ should be

    • a (graded) linear operation,

    • translation invariant $$\int_{\theta_1+\theta_0}^{\theta_2+\theta_0}\! d\theta ~f(\theta+\theta_0) ~=~ \int_{\theta_1}^{\theta_2}\! d\theta~f(\theta),$$

    • the output $\int_{\theta_1}^{\theta_2}\! d\theta~ f(\theta)$ should only depend on the limits $\theta_1$ and $\theta_2$ (apart from $a$ and $b$),

    • and vanishes for closed contours $$\begin{align}\int_{\theta_1}^{\theta_1}\! d\theta~f(\theta)~=~&0, \tag{i}\cr \left( \int_{\theta_1}^{\theta_2}+\int_{\theta_2}^{\theta_1}\right) d\theta~f(\theta)~=~&0,\tag{ii}\cr \left( \int_{\theta_1}^{\theta_2}+\int_{\theta_2}^{\theta_3}+\int_{\theta_3}^{\theta_1}\right) d\theta~f(\theta)~=~&0,\tag{iii} \end{align}$$

    then it is easy to check that the definite integral $$\begin{align}I(\theta_1,\theta_2)~\stackrel{?}{:=}~&\int_{\theta_1}^{\theta_2}\! d\theta\tag{2}\cr ~~~\propto~&~~ (\theta_2-\theta_1)\int\! d\theta\tag{2'}\end{align}$$ is proportional to $(\theta_2-\theta_1)$ times the usual Berezin integration $\int\! d\theta$.

    Sketched proof of eq. (2'): $$\begin{align}\int_{\theta_1}^{\theta_2}\! d\theta~ f(\theta)~=~&P(\theta_1,\theta_2)\cr ~=~&A\theta_1\theta_2+B_1\theta_1+B_2\theta_2+C\end{align}$$ must be a second-order polynomial in $\theta_1$ and $\theta_2$ (where the coefficients $A,B_1,B_2,C$ depend linearly on $a,b$). The condition (i) implies $B_1+B_2=0$ and $C=0$, while (iii) implies $A=0$. Translation invariance shows that the integral cannot depend on $b$. $\Box$

    This definite integral (2') is manifestly soul-valued, and hence not very useful. It also violates the naive expectation that the definite integration (2) should reverse Grassmann-parity. Moreover, the rule for integration by substitution $\theta^{\prime}=t\theta$ is non-standard: $$ \int_{\theta^{\prime}_1=t\theta_1}^{\theta^{\prime}_2=t\theta_2}\! d\theta^{\prime}~f(\theta^{\prime})~~=~~\int_{\theta_1}^{\theta_2}\! d\theta~f(t\theta),$$ i.e. there is no Jacobian factor! For this and other reasons, the definite integral (2) is usually considered to be ill-defined.

See also this related Math.SE post.

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  • $\begingroup$ 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? $\endgroup$
    – F R
    Oct 16, 2011 at 16:05
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Jan 29, 2018 at 20:31

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