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Homework Help: Operators: how to get rid of the two extra terms?

  1. Aug 29, 2012 #1
    I'm trying to work out equation 2.45 in these notes:


    Anyway [tex]\phi(\vec{x})[/tex] and [tex]\pi(\vec{x})[/tex] are given in equations 2.18 and 2.19

    When I multiply out [itex]\pi(\vec{x}) \vec{\div} \phi(\vec{x})[/itex] I get four terms.
    After normal ordering I can combine two of them to get the [tex]a_{\vec{p}}^\dagger a_{\vec{p}}[/tex] term I am meant to but I also have a [tex]a_{\vec{-p}} a_{\vec{p}}[/tex] and a [tex]a_{\vec{-p}}^\dagger a_{\vec{p}}^\dagger[/tex] term that I cannot get rid of.

    Does anyone know how to get rid of the two extra terms?

  2. jcsd
  3. Aug 30, 2012 #2
    Re: Operators

    When we use an operator like this, we do so by taking its expectation value against some state.

    [tex]\langle \psi | P | \psi \rangle[/tex]

    But in this formalism, states are constructed out of momentum eigenstates.

    [tex]|\psi\rangle = a^\dagger_p|0\rangle[/tex]

    That means that terms like [itex]a^\dagger_p a^\dagger_{-p}[/itex] in the momentum operator will lead to things like

    [tex]\langle 0 | a_p a^\dagger_p a^\dagger_{-p} a^\dagger_p|0\rangle[/tex]

    Terms like this will always vanish, because there aren't enough [itex]a_p[/itex]'s to annihilate all of the [itex]a^\dagger_p[/itex]'s (and there aren't any at all to annihilate the [itex]a^\dagger_{-p}[/itex]'s), so some of them will hit the vacuum on the left and vanish. A similar argument holds for terms like [itex]a_p a_p[/itex].

    I'm not sure whether that's a good answer, because you could say the same thing about the Hamiltonian too, and there those terms explicitly cancel without needing this argument, but I know that when you use it as an expectation value like this, those terms won't contribute.
    Last edited: Aug 30, 2012
  4. Aug 30, 2012 #3
    Re: Operators

    so basically they've just looked ahead and said that those terms won't be relevant?

    i agree that they won't be if you take an expectation value it just bugs me because what he's written as an equality doesn't appear to actually be an equality!
  5. Aug 30, 2012 #4
    I think Chopin's argument fails because the general state in Fock space is a superposition of 0-particle states, 1-particle states, 2-particle states, etc etc. It might very well survive after we hit it with 2 annihilation operators. :frown:

    However if you do the x-integration carefully, I think you might find that the resulting integrand is odd in p, and so the whole thing does vanish anyway. If memory serves me correctly.
  6. Aug 30, 2012 #5
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