# Fermion commutation with two quantum numbers

I am trying to understand the second quantization formalism. Let's say we have a system of fermions (e.g. electrons) with spin in an array of quantum dots. The creation and annihilation operators $$c^{\dagger}_{i\sigma}$$ and $$c_{i\sigma}$$, where $$i$$ and $$\sigma$$ indicate the index of the dot and the spin respectively, obey the caconical fermion commutation rules.

\begin{align} c^{\dagger}_{i\alpha}c_{j\beta}+c_{j\beta}c^{\dagger}_{i\alpha}=\delta_{i,j}\delta_{\alpha,\beta}, \hspace{5mm}c^{\dagger}_{i\alpha}c^{\dagger}_{i\alpha}=0, \hspace{5mm} c^{\dagger}_{i\alpha}c_{j\beta}^{\dagger}=-c_{j\beta}^{\dagger}c^{\dagger}_{i\alpha} \end{align}

I will be looking in the case of just 2 fermions, i=1,2 and σ=↑,↓. So both dots contain 1 fermion (1,1) or either of the dots contains 2 fermions (0,2) and (2,0).

I can construct the following state $$c^{\dagger}_{1\downarrow}c^{\dagger}_{2\uparrow}|0,0\rangle=|\downarrow,\uparrow\rangle$$. The third commutation rule tells us that if we construct the state $$c^{\dagger}_{2\uparrow}c^{\dagger}_{1\downarrow}|0,0\rangle$$, we need to place a minus sign due to exchange of variable indices. However since we exchanged 2 variable indices $$i$$ and $$\sigma$$ do we have to add two minus signs? i.e. if $$c^{\dagger}_{1\downarrow}c^{\dagger}_{2\uparrow}|0,0\rangle=|\downarrow,\uparrow\rangle$$ holds then is $$c^{\dagger}_{2\uparrow}c^{\dagger}_{1\downarrow}|0,0\rangle=-|\downarrow,\uparrow\rangle$$ or $$c^{\dagger}_{2\uparrow}c^{\dagger}_{1\downarrow}|0,0\rangle=--|\downarrow,\uparrow\rangle=|\downarrow,\uparrow\rangle$$?