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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?

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    $\begingroup$ You are confusing numbers with operators. The operators ${\hat \psi}_\alpha$ satisfy $\{ {\hat \psi}_\alpha , {\hat \psi}_\beta^\dagger \} = \delta_{\alpha\beta}$. The Grassmannian numbers $\psi_\alpha$ anti-commute. $\endgroup$
    – Prahar
    Commented Mar 21 at 14:29
  • $\begingroup$ Related: physics.stackexchange.com/q/232382/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Mar 22 at 0:01
  • $\begingroup$ @Qmechanic But the relationship for classical $\{\psi_\alpha,\psi^\dagger_\beta\}$ in your answer seems not consistent with the answer below this question. What have I missed? $\endgroup$
    – Errorbar
    Commented Mar 22 at 4:35
  • $\begingroup$ Be aware that the notation $\{\cdot,\cdot\}$ denotes an anticommutator for some authors while a super-Poisson bracket for others. $\endgroup$
    – Qmechanic ♦
    Commented Mar 22 at 5:08

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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.

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  • $\begingroup$ I haven't got it. I don't know how to verify if an operator is quantum. Does it mean second quantization? And if the first two relationships are remain satisfied for the quantum operatots, doesn't it mean they are still Grassmann numbers (stasfying anticommute relationship). $\endgroup$
    – Errorbar
    Commented Mar 21 at 14:25
  • $\begingroup$ "How to verify if an operator is quantum": when they don't (anti-)commute, such as $\{\Psi_\alpha,\Psi^\dagger_\beta\}=\delta_{\alpha \beta}$. $\endgroup$
    – MadMax
    Commented Mar 21 at 14:28
  • $\begingroup$ So do you mean in the canonical quantization, they are promoted to operators; and in the path integral quantization the are just numbers? Can I say that if I see a Hamiltonian then it's a canonical quantization, and if I see an action it's path integral quantization? $\endgroup$
    – Errorbar
    Commented Apr 1 at 8:25

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