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LSZ formula help!

  1. Feb 9, 2008 #1
    I need some help getting started with calculating a two to two scattering amplitude (tree approximation) in the field theory with lagrangian:
    g[tex]^{2}[/tex]L = ([tex]\partial[/tex][tex]\phi[/tex])[tex]^{2}[/tex] - V([tex]\phi[/tex])
    where V is a polynomial in [tex]\phi[/tex]. That is,
    V = [tex]\Sigma[/tex]v[tex]_{n}[/tex][tex]\phi^{n}[/tex]

    I am trying to calculate this using techniques of path integration and the lsz formula. I know the answer depends only on v[tex]_{2}[/tex], v[tex]_{3}[/tex], and v[tex]_{4}[/tex] but I don't understand why. When I tried calculating the 4-point green's function I got it depending only on v[tex]_{4}[/tex] so I must be doing something wrong, but I don't know what.
  2. jcsd
  3. Feb 10, 2008 #2


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    Can you post some of your steps?

    Yes, there is a four point amplitude that comes from the V_4 term. But you should also get diagrams where there is an internal line connecting two three point vertex (like the tree level of the electron-positron scattering in QED). These will contain v_3 and v_2 (v_2 appearing in the propagator since it's essentially the mass of the scalr field, right?).

    There are many many steps going from the lagrangian to an amplitude. Are you stuck with the LSZ part or in evaluating the amplitudes using path integrals? It's not clear where you are stuck, exactly.
  4. Feb 10, 2008 #3
    I actually figured it out. My mistake was that I was only going to first order. Like you said, you get a v_3 contribution from the second order term where you have an internal propagator. Thanks for the help.
  5. Feb 10, 2008 #4


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    Good. I almost mentioned that in my post because the diagrams I described are clearly of order (v_3)^2 so they are of second order in the coupling constant but teh diagrams are still tree level. So I almost mentioned the fact that the v_4 contribution and the v_3 contributions were of different order in the coupling constant expansion. But I decided to wait and hear back from you first.

    Good for you.
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