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  Product of Grassmann numbers

+ 2 like - 0 dislike

Let $\bar\theta$,$\theta$ be two Grassmann numbers. Then their product is a commuting number. If you were to integrate a function of the two $f(\bar\theta \theta)$ over $\bar\theta$,$\theta$ would it be possible to make a substitution $a=\bar\theta \theta$ and integrate over $a$? What kind of integral would that be?

This post imported from StackExchange Physics at 2014-03-05 14:51 (UCT), posted by SE-user TeeJay
asked Nov 20, 2013 in Theoretical Physics by TeeJay (20 points) [ no revision ]
A question for MathSE possibly?

This post imported from StackExchange Physics at 2014-03-05 14:51 (UCT), posted by SE-user John Rennie
Related question by OP: physics.stackexchange.com/q/88312/2451

This post imported from StackExchange Physics at 2014-03-05 14:51 (UCT), posted by SE-user Qmechanic

1 Answer

+ 2 like - 0 dislike

Let's say $f$ admits a taylor series $f(\bar\theta \theta) = A + B \bar\theta \theta + C\bar\theta \theta \bar\theta \theta + \dots$. Now, $\bar\theta \theta \bar\theta \theta = -\bar\theta^2 \theta^2 = 0$, etc., so our function terminates at the linear term. Furthermore, the integral of $f$ over $d \bar\theta\, d\theta$, by the rules of Berezin integration, is precisely equal to B.

Btw, the reason that integration is defined in this way is so that $\int\!d\theta\, \frac{\partial}{\partial\theta} f(\theta) = 0$, which is something you might expect for "well-behaved" functions.

Getting back to your question, in physics, the way you would "integrate" over $\bar\theta\theta$ in a path integral is to define $\bar\psi \psi = e^{i\phi}$ and then integrate over $\phi$. This is called Bosonization, because it expresses the dynamics of fermions in terms of bosons. It isn't always possible, afaik.

This post imported from StackExchange Physics at 2014-03-05 14:51 (UCT), posted by SE-user lionelbrits
answered Nov 20, 2013 by lionelbrits (110 points) [ no revision ]

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