# Simple question about anticommuting numbers

1. Apr 29, 2009

### pellman

... I hope.

I wasn't sure which math forum to put this in to 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?

2. Apr 29, 2009

### samalkhaiat

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

and

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

sam

3. Apr 29, 2009

### pellman

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

$$\int d \theta = 0$$

?

4. Apr 29, 2009

### 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.

5. May 2, 2009

Thank you.