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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!

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    $\begingroup$ I feel it is terrible , terrible practice to use the same symbols, e.g. $f_0$ for both 2x2 and 4x4 matrices, as you do. Personally, I would not label your states your way, but I'd use, instead, $|a,b\rangle\equiv |a\rangle \otimes |b\rangle$ , in the "right is a block of left" convention. You must simply review your direct products. This looks like a homework problem... $\endgroup$ Jun 7, 2022 at 0:40
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    $\begingroup$ Despite your reference, for just one oscillator, your don't need a subscript. In any case, in your left-into-right convention, your seem to have the right idea. Have you checked all anticommutators? Like $\{ f_0, f_1\}$? hmmm.... $\endgroup$ Jun 7, 2022 at 0:59
  • $\begingroup$ Can you give me more details on how to generalize those operators? Is there a textbook that has this kind of exercise? Thanks! $\endgroup$
    – Kim Dong
    Jun 7, 2022 at 1:53
  • $\begingroup$ Yes, the anticommutator is $\{f_i,f_j^\dagger\} =\delta_{ij}$, and 0 for other cases. $\endgroup$
    – Kim Dong
    Jun 7, 2022 at 4:39
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    $\begingroup$ Does this answer your question? What is difference between fermions and spins?. Also closely related: explicit representation of creation/annihilation operators... $\endgroup$ Jun 7, 2022 at 13:39

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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.

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