2
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.