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Wick's theorem

  1. Feb 18, 2017 #1
    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. :(
     
  2. jcsd
  3. Feb 23, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
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