- #1

pellman

- 684

- 5

I'm not seeing this.

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

[tex]\int 1 d\psi=0[/tex]

[tex]\int \psi d\psi=1[/tex]

Ok. So let's check.

[tex]\int\delta(\psi-\psi')f(\psi)d\psi[/tex]

[tex]=\int(\psi-\psi')f(\psi)d\psi[/tex]

[tex]=\int(\psi-\psi')(a+b\psi)d\psi[/tex]

[tex]=a\int \psi d\psi + 0 -a\psi'\int d\psi-b\psi'\int\psi d\psi[/tex]

[tex]=a - 0 -b\psi'[/tex]

[tex]=f(-\psi')[/tex]

[tex]\neq f(\psi')[/tex]

Are one of my assumptions wrong?