Why two different spinors are Grassmann quantities? In Rydberg Quantum Field Theory page 441 (this edition, unfortunately page 441 is not in the link) it says 

If $\xi$ and $\eta$ are Majorana spinors [...] and since $\xi$ and $\eta$ are Grassmann quantities, $$ \xi_i^*\eta_j^* = -\eta_j^*\xi_i^* $$

I know that Grassmann quantities obey the Grassmann algebra, i.e. $\{\xi_i,\eta_j\} = 0$. But why two different spinor fields obey this algebra? 
As far as I understood two spinor fields $\xi$ and $\eta$ obey the anticommutation relations between each and its conjugate, $\{\xi_i(\vec{x},t),\xi_j(\vec{y},t)\}=\delta_{ij}\delta^3(\vec{x}-\vec{y})$ and the same for $\eta$. But in this case we have two different fields $\xi$ and $\eta$ which are not the conjugate of each other. So why do they have to satisfy this Grassmann algebra?
For example, in QED there are two fields $\psi$ and $A^\mu$ and each has its commutation/anticommutation relation with its own conjugate (in the case of $A^\mu$ with itself). But there are not any relations between $\psi$ (or $\bar{\psi}$) and $A^\mu$, they are different fields. I mean $[\psi,A^\mu] = \{\psi,A^\mu\} = 0$ and the same for $\bar{\psi}$.
 A: OP's title question (v3) reads:

Why two different spinors are Grassmann quantities?

We interpret this as asking why independent spinors anticommute? Here is one line of reasoning: 


*

*A general supernumber $z=z_0+z_1$ has a Grassmann-even and a Grassmann-odd component, $z_0$ and $z_1$, respectively.

*$z_0$ and $z_1$ are supernumbers of definite Grassmann parity, $|z_0|=0$ and $|z_1|=1$, respectively.

*A Grassmann-odd supernumber $z$ with $|z|=1$ squares to zero: $z^2=0$.

*A sum $z+w$ of two Grassmann-odd supernumbers $z$ and $w$ is still a Grassmann-odd supernumber. Hence it squares to zero
$$ 0~=~(z+w)^2~=~\underbrace{z^2}_{=0} + zw + wz +\underbrace{w^2}_{=0}~=~ zw + wz. $$
Hence two Grassmann-odd supernumbers $z$ and $w$ anticommute, cf. OP's question.

*More generally, two supernumbers $z$ and $w$ of definite Grassmann parity supercommute:
$$[z,w]_S~:=~zw-(-)^{|z||w|}wz~=~0,$$
where $[\cdot,\cdot]_S$ denotes the supercommutator. In other words, the supercommutator $[z,w]_S=0$ is zero. 

*When we replace the supernumbers with two superoperators $\hat{z}$ and $\hat{w}$ of definite Grassmann parity, i.e. go from classical to quantum theory, the supercommutator 
$$ [\hat{z},\hat{w}]_S~=~{\cal O}(\hbar)$$
can acquire non-zero terms at first (and higher) orders in Planck's constant $\hbar$.  
