# 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!

• 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... Jun 7, 2022 at 0:40
• 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.... Jun 7, 2022 at 0:59
• Can you give me more details on how to generalize those operators? Is there a textbook that has this kind of exercise? Thanks! Jun 7, 2022 at 1:53
• Yes, the anticommutator is $\{f_i,f_j^\dagger\} =\delta_{ij}$, and 0 for other cases. Jun 7, 2022 at 4:39
• Does this answer your question? What is difference between fermions and spins?. Also closely related: explicit representation of creation/annihilation operators... Jun 7, 2022 at 13:39

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.