Delta function for grassmann numbers?

[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?

 

[Hurkyl]

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?

 

[pellman]

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?

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.

 

[Hurkyl]

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]

 

[pellman]

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

Thanks!