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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)?

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Please, define your notation, and possibly restate the question to get the best response. –  Peter Kravchuk Apr 28 '13 at 23:08
    
In particular, $\sigma_\pm=\sigma_1\pm i\sigma_2$ is a good pair of fermionic creation/annihilation operators. But I guess that it is not what you want, so please work on your question. –  Peter Kravchuk Apr 28 '13 at 23:11
    
@PeterKravchuk I hope now it's more clear. –  swish Apr 28 '13 at 23:25
    
Thanks. A way better. –  Peter Kravchuk Apr 28 '13 at 23:51
    
You will also encounter another problem with $[c_\downarrow,c_\downarrow]_+$. If you do, consider acting on $|\downarrow\rangle$ with this anticommutator. In addition to sign issue pointed out by Qmechanic, you have another inaccuracy. –  Peter Kravchuk Apr 29 '13 at 0:05
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1 Answer

up vote 2 down vote accepted

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

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

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

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

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By the way, in a comment to swish previous question I mentioned that creation/annihilation operators for different fields may commute, in principle. This corresponds to taking the Fock space to be the ordinary tensor product, while it looks like the natural way is to take it as a graded tensor product. I wonder, at what point the bosonic tensor product turns ugly? I think that for spin projections it is near the rotational invariance, but what about two interacting but different fields? –  Peter Kravchuk Apr 29 '13 at 0:02
    
So I should change my states to |n>*|spin> in order to make proper operators? –  swish Apr 29 '13 at 10:36
    
I updated the answer. –  Qmechanic May 1 '13 at 19:04
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