# Wick's theorem

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1. Feb 18, 2017

### gasar8

Hello, I have got a question regarding Wick contractions.
At lectures, we wrote that only $a_i a_j^{\dagger}$ contracted gives Kronecker delta $\delta_{ij}$, other creation anihillation combination of operators gives just 0.

But, when we did an exercise, we computed in Fermi sea:
$$\langle \phi_0|a_k^{\dagger} a_p^{\dagger} a_k a_p |\phi_0\rangle.$$
I understand that only fully contracted terms survive, so we contracted $a_k^{\dagger} a_k$, $a_p^{\dagger} a_p$ and $a_k^{\dagger} a_p$, $a_p^{\dagger} a_k$ and got a non-zero value. Why is that, if only the $a_i a_j^{\dagger}$ gives delta?
Is it because of the Fermi sea, because we got some particle numbers $n-s$? Or do we need to use commutation relations and change it so, that the dagger is at right? In this case, I assume:
$$\langle \phi_0|a_k^{\dagger} a_k |\phi_0\rangle = \langle \phi_0|(a_k a_k^{\dagger}-1) |\phi_0\rangle = \langle \phi_0|a_k a_k^{\dagger} |\phi_0\rangle-1=0.$$
But this is not what we get. :(

2. Feb 23, 2017