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 $$a^\dag$$?