Understanding Wick Contractions in Fermi Sea

[gasar8]

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

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

 

[BigJDubs]

The reason why other creation-annihilation combinations of operators give non-zero values in this particular example is because of the Pauli Exclusion Principle. According to the Pauli Exclusion Principle, no two fermions can occupy the same quantum state at the same time. This means that if you have a combination of creation and annihilation operators that would result in two particles occupying the same quantum state, then it would be zero due to the Pauli Exclusion Principle. In your example, the a_k^{\dagger} a_p^{\dagger} a_k a_p combination would create two particles in the same state, so it is zero. However, the a_k^{\dagger} a_k and a_p^{\dagger} a_p combinations do not violate the Pauli Exclusion Principle, so they do not equal zero.