# Delta function for grassmann numbers?

1. Dec 19, 2009

### pellman

Claim: if $$\psi$$ is a variable grassmann number, then $$\delta(\psi)=\psi$$ is a Dirac delta function for integrals over $$\psi$$.

I'm not seeing this.

A general function of a grassmann number can be written $$f(\psi)=a+b\psi$$ because anti-commutativity requires $$\psi^2=0$$. According to the wikipedia article (which doesn't elaborate on why), integrals satisfy

$$\int 1 d\psi=0$$
$$\int \psi d\psi=1$$

Ok. So let's check.

$$\int\delta(\psi-\psi')f(\psi)d\psi$$
$$=\int(\psi-\psi')f(\psi)d\psi$$
$$=\int(\psi-\psi')(a+b\psi)d\psi$$
$$=a\int \psi d\psi + 0 -a\psi'\int d\psi-b\psi'\int\psi d\psi$$
$$=a - 0 -b\psi'$$
$$=f(-\psi')$$
$$\neq f(\psi')$$

Are one of my assumptions wrong?

2. Dec 19, 2009

### Hurkyl

Staff Emeritus
The definition of $\delta$ is $\int_{-\infty}^{+\infty} \delta(x) f(x) \, dx = f(0)$.

Does that change of variable make sense for such integrals? Does it make sense for f?

3. Dec 19, 2009

### pellman

I don't know, Hurkyl. I'm not even sure what integration itself means here. I am just following the steps in a formal sort of way. Warren Siegel on page 48 of Fields has that $$\int d\psi'(\psi'-\psi)f(\psi')=f(\psi)$$ (the prime-unprimed reverse of what I have above.) But I get $$f(-\psi)$$ .

Never mind if we can call it the delta function. Siegel himself puts "anti-commuting delta function" in quotes.

4. Dec 19, 2009

### Hurkyl

Staff Emeritus
I remember trying to read through that book; I stopped after finding that section extremely off-putting.

Anyways, one thing Siegel says in that section is if we write $f(\psi) = a + b \psi$, then a and b aren't necessarily complex scalars -- they can be commuting numbers or anticommuting numbers or whatever.

Assuming that we really do have
$$\int d\psi (a + b \psi) = b$$
whenever a and b don't involve psi, then if I take special care not to pass any variable through another I get

$$\int d\psi' (\psi' - \psi) f(\psi') = -\psi b + \int d\psi' \psi' a$$

which can be rearranged to

$$= f(\psi) - \{\psi, b\} + \int d\psi' [\psi', a]$$

of course, it can be rearranged into other things -- such as

$$= f(-\psi) - [\psi, b] + \int d\psi' [\psi', a]$$

Last edited: Dec 19, 2009
5. Dec 20, 2009

### pellman

I think I'm with you about that section. I'll just let it go and press on.

Thanks!!