How Does Anti-Commutativity Affect Grassmann Number Integration?

[pellman]

... I hope.

I wasn't sure which math forum to put this into get an answer, but since the application is quantum, I figured this forum would be better.

http://en.wikipedia.org/wiki/Grassmann_number

We see here that Grassman numbers have the property

\int [ \frac{\partial}{\partial\theta}f(\theta)]d\theta=0

I don't see it. Suppose f(\theta)=a + b\theta.

Then \frac{\partial}{\partial\theta}f(\theta)=b

And so

\int [ \frac{\partial}{\partial\theta}f(\theta)]d\theta=b\theta + constant

right? At least that is what we would get with regular numbers. How does the anti-commutativity affect that?

 

[samalkhaiat]

We see here that Grassman numbers have the property

\int [ \frac{\partial}{\partial\theta}f(\theta)]d\theta=0

I don't see it. Suppose f(\theta)=a + b\theta.

Then \frac{\partial}{\partial\theta}f(\theta)=b

And so

\int [ \frac{\partial}{\partial\theta}f(\theta)]d\theta=b\theta + constant

right? At least that is what we would get with regular numbers. How does the anti-commutativity affect that?

\int d \theta = 0, \ \ \int d \theta \ \theta = 1

and

\int d \theta f(\theta) = \frac{d}{d \theta} f(\theta) = b


sam

 

[pellman]

\int d \theta = 0, \ \ \int d \theta \ \theta = 1

and

\int d \theta f(\theta) = \frac{d}{d \theta} f(\theta) = b


sam

Well, now I am totally confused. Considering a single anti-commuting variable theta, the defining property is \theta^2=0 right? Is there anything else? If not, how does that get us

\int d \theta = 0

?

 

[Avodyne]

\int d\theta = 0 and \int d\theta\,\theta = 1 are just definitions. They are motivated by the following considerations. An integral over \theta is supposed to be an analog of a definite integral over a real variable x from -\infty to +\infty. So consider I=\int_{-\infty}^{+\infty}dx\,f(x). Assume the integral coverges. One key property is linearity: if we multiply f(x) by a constant c, the result is cI. Another is translation invariance: if we replace f(x) with f(x+a), the result is still I.

Now, for a Grassmann variable, the most general function is f(\theta)=a+b\theta. Let I=\int d\theta\,f(\theta). If we want I to be both linear and translation invariant, we must define I=b, up to a possible numerical multiplicative constant.

 

[pellman]

Thank you.