quantum mechanics - Matrix representation for fermionic annihilation operator

My guess it should look something like this:

$ c_\sigma = (\left|0\right>\left<\uparrow\right|+\left|\downarrow\right>\left<\downarrow\uparrow\right|)\delta_{\sigma,\uparrow}+(\left|0\right>\left<\downarrow\right|+\left|\uparrow\right>\left<\downarrow\uparrow\right|)\delta_{\sigma,\downarrow}$

where $\delta$ is a Kronecker delta and states $\left|0\right>,\left|\downarrow\right>,\left|\uparrow\right>,\left|\downarrow\uparrow\right>$ are orthonormal.

Now it behaves like annihilation operator

$c_{\downarrow}\left|0\right>=\left|0\right>, c_{\uparrow}\left|0\right>=\left|0\right>$

$c_{\downarrow}\left|\downarrow\right>=\left|0\right>, c_{\uparrow}\left|\downarrow\right>=\left|0\right>$

$c_{\downarrow}\left|\downarrow\uparrow\right>=\left|\uparrow\right>, c_{\uparrow}\left|\downarrow\uparrow\right>=\left|\downarrow\right>$

but anticommutator for example $[c_{\uparrow},c_{\downarrow}]_+$ isn't zero.

Is it possible to define it like that (in terms of basis states)?

Answer

Main point: You should allow the possibility of sign factors appearing into the definition of the Hilbert space representation of fermionic operators, cf. fermionic Fock space.

In more detail, consider the CAR algebra

$$\tag{1} \{c_{\sigma}, c_{\tau}\}_+~=~0, \qquad \{c_{\sigma}, c^{\dagger}_{\tau}\}_+~=~\delta_{\sigma,\tau} {\bf 1}, \qquad\{c^{\dagger}_{\sigma}, c^{\dagger}_{\tau}\}_+~=~0, \qquad \sigma,\tau\in \{\uparrow,\downarrow\}. $$

Next define

$$\tag{2} c_{\sigma}\left|0\right>~:=~0, \qquad \left|\sigma\right>~:=~ c^{\dagger}_{\sigma}\left|0\right>, \qquad \left|\sigma\tau\right>~:=~ c^{\dagger}_{\sigma}\left|\tau\right>, \qquad \sigma,\tau\in \{\uparrow,\downarrow\}.$$

Note that these definitions imply that

$$\tag{3} \left|\sigma\tau\right> ~=~ -\left|\tau\sigma\right>, \qquad \sigma,\tau\in \{\uparrow,\downarrow\}.$$

In particular

$$\tag{4} \left|\sigma\sigma\right> ~=~ 0, \qquad \sigma\in \{\uparrow,\downarrow\}. $$