Confusion about whether a fermion field and its conjugate as an Grassmann number

Question

I'm confused about what "a Grassmann-odd number" really means and how does it apply to fermions.

In some text, it says that "if $\varepsilon \eta+\eta \varepsilon =0 $, then $\varepsilon $ and $\eta$ are Grassmann-odd numbers.

And in wiki, the Grassmann algebra are those whose generators satisfy $$\theta_1\theta_2+\theta_2\theta_1=0$$ and fermion space is one of the Grassmann algebras.

So, following these definitions, a fermion field is a Grassmann-odd number because $$\{\psi_\alpha,\psi_\beta\}=0$$ so is its conjugate. But $$\{\psi_\alpha,\psi^\dagger_\beta\}=\hbar\delta_{\alpha \beta}.$$ So why don't $\psi_\alpha$ and $\psi^\dagger_\beta$ anticommute if they are all Grassmann-odd number?

I suppose that if they are not belonging to the same Grassmann algebra. But if so, when the essay says "$\epsilon$ is a Grassmann-odd number" (e.g.Chapter II of an essay about supersymmetry), what does it mean?

Answer 1

It is wrong to claim that:

$$\{\psi_\alpha,\psi^\dagger_\beta\}=\delta_{\alpha \beta}$$

since Grassmann odd numbers should always anticommute with each other: $$\{\psi_\alpha,\psi^\dagger_\beta\}=0$$ Only after you promote (via quantization) the classical Grassmann odd numbers $\psi_\alpha$, $\psi^\dagger_\beta$ to the quantum fermionic operators $\Psi_\alpha$, $\Psi^\dagger_\beta$ then you have $$\{\Psi_\alpha,\Psi^\dagger_\beta\}=\delta_{\alpha \beta}$$ where $\Psi_\alpha$ and $\Psi^\dagger_\beta$ are NOT Grassmann odd numbers any more.