Second quantization canonical commutation relation : $\{c_\alpha,c_\beta^\dagger\}=\delta_{\alpha,\beta}$ a counter example? Suppose two different states $\alpha$ and $\beta$ of some system of fermions such that each state only allows zero or one particle.
The canonical commutation relation $\{c_\alpha,c_\beta^\dagger\}=\delta_{\alpha,\beta}$ is supposed to hold for any state, however if we suppose $\alpha\neq\beta$ the state $|1_\alpha, 0_\beta \rangle$ seems to be a counter example as $\{c_\alpha,c_\beta^\dagger\}|1_\alpha, 0_\beta \rangle=c_\alpha c_\beta^\dagger|1_\alpha, 0_\beta \rangle + c_\beta^\dagger c_\alpha|1_\alpha, 0_\beta \rangle=2|0_\alpha, 1_\beta \rangle \neq 0$. 
Could someone point to the problem in my reasonning ?
 A: As @Javier suggested in his comment. this can be seen easily if $|1_{\alpha},0_{\beta}⟩ $ is written in the form $ c_{a}^{\dagger} |0⟩ $. 
{$c_{\alpha} ,c^{\dagger}_{\beta}$} $ c_{a}^{\dagger} |0⟩ $ = $c_{\alpha} c^{\dagger}_{\beta} c_{a}^{\dagger} |0⟩ + c^{\dagger}_{\beta} c_{\alpha}  c_{a}^{\dagger} |0⟩ $ 
$ = (\delta_{\alpha \beta} - c^{\dagger}_{\beta} c_{\alpha}) c_{a}^{\dagger} |0⟩ + c^{\dagger}_{\beta} c_{\alpha}  c_{a}^{\dagger} |0⟩ $
$ = \delta_{\alpha \beta} c_{a}^{\dagger} |0⟩  = 0 $ (if $\alpha \neq \beta$) 
The method you have used in the question gives a wrong result because $ c^{\dagger}_{\beta} c_{\alpha}  |1_{\alpha} 0_{\beta}⟩ = -|0_{\alpha} 1_{\beta}⟩ $ and not $|0_{\alpha} 1_{\beta}⟩$. This is because the beta state comes after the alpha state in the ket and hence the creation operator associated with beta state has to pass over the alpha state before acting on the beta state.
In general,
$ c_{x}^{\dagger} |n_{1},n_{2},n_{3},...n_{x},n_{x+1},...⟩ = $ i) 0 if $n_{x} = 1$ or ii) $(-1)^{(1-n_{1}) + (1-n_{2}) + .... + (1-n_{x-1})} $ if $ n_{x} = 0$.
$ c_{x}|n_{1},n_{2},n_{3},...n_{x},n_{x+1},...⟩ = $ i) 0 if $n_{x} = 0$ or ii) $(-1)^{n_{1}+n_{2}+....+n_{x-1}} $ if $ n_{x} = 1$.
