Do Grassmann Numbers Commute With Fermionic Operators?

[pellman]

Homework Statement

Warren Siegel, Fields, ex. IA2.3(b)

Define the eigenstate of the fermionic annihilation operator as a|\zeta\rangle=\zeta |\zeta\rangle. \zeta is a Grassmann (anti-commuting) number.

Show that

a^\dag|\zeta\rangle=-\frac{\partial}{\partial\zeta}|\zeta\rangle.

Homework Equations

\{a,a^\dag\}=aa^\dag + a^\dag a = 1
\{a^\dag\,a^\dag\}=0
\{a,a\}=0

The Attempt at a Solution

For small \Delta\zeta we have

|\zeta+\Delta\zeta\rangle = |\zeta\rangle+\Delta\zeta\frac{\partial}{\partial\zeta}|\zeta\rangle

(Actually this is exact since \Delta\zeta^2=0.)

so we could show first that

|\zeta+\Delta\zeta\rangle = |\zeta\rangle-\Delta\zeta a^\dag|\zeta\rangle

That is as far as I have gotten. And it is not clear to me.. do the anti-commuting c-numbers \zeta commute with the operator a^\dag?

 

[pellman]

Actually, I think I about have this one. But it hinges on .. do the anti-commuting c-numbers \zeta commute or anti-commute with the operator a^\dag?