# 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 ?

• Briefly because I'm on mobile: fermion creation and destruction operators may act with a minus sign, depending on how far along the ket is the part they act on. Write your state as $c^\dagger_\alpha |0\rangle$ and you'll see why. – Javier Sep 6 '18 at 12:09

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 $n_{x} = 1$.

• Thank you for suggesting this way of seeing things, this is helpful. However, I am uncomfortable to accept this as an answer because I am still confused. In your development, you suppose and use the relation $\{c_\alpha,c_\beta^\dagger\}=\delta_{\alpha,\beta}$, while I am precisely casting doubt on it. The argument looks a little circular to me. Moreover, it does not explain what is wrong with the way of doing in the question. I would appreciate if you could give me more insight. – Mary Sep 6 '18 at 14:26
• I have edited the answer to include the reason as to why the method you took, gave the wrong result. Regarding your first query, you based your doubts on the relation citing this "counterexample". But since this is no longer a counterexample, why do you still have a doubt regarding the correctness of the anti-commutation relation ? – Hari Sep 6 '18 at 15:14
• Thanks, you clarified that very well. I was just still confused, that is not the case any more. – Mary Sep 6 '18 at 20:18