Matrix form of fermionic creation and annihilation operators in two-level system I'm trying to find the matrix form of fermionic creation and annihilation operators in two-level systems from this text. I understand that for one site, the operators take the form:
$$
f_{0}=\left(\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right), \quad f_{0}^{\dagger}=\left(\begin{array}{ll}
0 & 0 \\
1 & 0
\end{array}\right),
$$
where
$$
\begin{aligned}
f_{0}|1\rangle &=|0\rangle = \left(\begin{array}{l}
1 \\
0
\end{array}\right), & f_{0}|0\rangle=0 \\
f_{0}^{\dagger}|1\rangle &=0, & f_{0}^{\dagger}|0\rangle=|1\rangle=\left(\begin{array}{ll}
0 \\
1
\end{array}\right)
\end{aligned}
$$
For two sites, I was able to deduce
$$
f_{0}^{\dagger}=\left(\begin{array}{l11l}
0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
\end{array}\right), \quad f_{0}=\left(\begin{array}{l11l}
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
\end{array}\right),
f_{1}^{\dagger}=\left(\begin{array}{l11l}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
\end{array}\right), \quad f_{1}=\left(\begin{array}{l11l}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & -1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{array}\right),
$$
which allow for these rules as indicated by the text
$$
\begin{aligned}
f_{0}^{\dagger}|0,0\rangle &=|1,0\rangle ; \quad f_{0}^{\dagger}|1,0\rangle=0 \\
f_{0}|1,0\rangle &=|0,0\rangle ; \quad f_{0}|0,0\rangle=0 \\
f_{0}|0,1\rangle &=0 ; \quad f_{0}^{\dagger}|1,1\rangle=0 \\
f_{1}^{\dagger}|0,0\rangle &=|0,1\rangle ; \quad f_{1}|0,0\rangle=f_{1}|1,0\rangle=0 \\
f_{1}^{\dagger}|1,0\rangle &=-|1,1\rangle ; \quad f_{1}|0,1\rangle=|0,0\rangle \\
f_{1}^{\dagger}|0,1\rangle &=f_{1}^{\dagger}|1,1\rangle=0 ; \quad f_{1}|1,1\rangle=-|1,0\rangle\\
f_{0}^{\dagger}|0,1\rangle &=|1,1\rangle ; \quad f_{0}|1,1\rangle=|0,1\rangle
\end{aligned}
$$
where $|0,0\rangle =\left(\begin{array}{l}
1 \\
0 \\
0 \\
0 \\
\end{array}\right), |1,0\rangle = \left(\begin{array}{l}
0 \\
1 \\
0 \\
0 \\
\end{array}\right), |0,1\rangle = \left(\begin{array}{l}
0 \\
0 \\
1 \\
0 \\
\end{array}\right), |1,1\rangle = \left(\begin{array}{l}
0 \\
0 \\
0 \\
1 \\
\end{array}\right)$.
My question is: am I thinking about this the right way? And what is the general formula of the operators for when there are $n$ sites instead? Is there some material that discusses this? Thank you!
 A: Briefly: you have to order your sites and add a string $\eta_{\alpha}$ of operators in front of the creation and annihilation operators
$$
 \overline{f}_{\alpha}=\eta_{\alpha}f_{\alpha}, \qquad \overline{f}_{\alpha}^{\dagger}=\eta_{\alpha}f_{\alpha}^{\dagger}, \qquad \eta_{\alpha}=\prod_{\beta=1}^{\alpha-1}\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}_{\beta}
$$
The point is that your single site operators $f_{\alpha}=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}_{\alpha}$ and $f_{\alpha}^{\dagger}$, obey the right anticommutation rules on the site $\alpha$, but they commute of the on different sites.
You can see that $\overline{f}_{\alpha}$ and $\overline{f}_{\alpha}^{\dagger}$ , thanks to the string $\eta_{\alpha}$ we have attached to them, obey the right anticommutation relations
$$
\{\overline{f}_{\alpha}^{\dagger}, \overline{f}_{\beta}\} = \delta_{\alpha\beta} \qquad  \{\overline{f}_{\alpha}, \overline{f}_{\beta}\}=0
$$
This implies that when we costruct a state from the vacuum
$$
|\alpha,\beta,\gamma\rangle = \overline{f}_{\alpha}^{\dagger}\overline{f}_{\beta}^{\dagger}\overline{f}_{\gamma}^{\dagger}|0\rangle
$$
this is antisymmetric under exchange of two indices: that is what we want from a fermionic state.
