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How to draw self-energy diagrams in phi^4 theory

  1. Mar 20, 2015 #1
    Okay guys so I the lagrangian

    [itex]\mathcal{L}=\dfrac{1}{2}\left( \partial_{\mu}\phi\partial^{\mu}\phi-m^{2}\phi^{2} \right)+\dfrac{\lambda}{4!}:\phi(x)^{4}:[/itex]

    where [itex]\phi(x)[/itex] is a real scalar field.

    I want to know how you can draw the self-energy diagrams at order [itex]\lambda[/itex] and [itex]\lambda^{2}[/itex] for a 2 => 2 scattering system. How do you go about doing this just by looking at the Lagrangian?

    Thanks guys....
  2. jcsd
  3. Mar 20, 2015 #2
    First of all, what do you mean by self-energy? Usually one calls self-energy the higher-order corrections to the single particle propagator, i.e. a 1 -> 1 process. Here you have 2 initial particle and 2 final one. What do you define as self-energy?
    However, in the most general way possible, since your interaction term is ##\frac{\lambda}{4!}\phi^4##, this means that this theory admits vertices with 4 legs (shaped like an X, if you want). Each X counts as ##\lambda##.
    In your case the basics 2 -> 2 process simply happens when you have one particle at each leg of the X. This is a order ##\lambda## diagram, the only allowed for the considered scattering. If you want to go to order ##\lambda^2## then you have two possibilities:

    1) Take the order ##\lambda## diagram just mentioned and you add a "bubble" to one of the external leg. Meaning, that one of the legs emits and reabsorbs a particle at the same point (you essentially join two of the legs of your X together). You can do this for each of the four legs. This is what I would call the self-energy correction to your tree level 2 -> 2 scattering.
    2) You can also take two X and join two of their legs together, so that you end up with a bubble with 4 total external legs (two coming from the same point on one side and two coming from the same point on the other side). This is again order ##\lambda^2## but it doesn't strictly contain any self-energy.

    Is this what you were looking for?
  4. Mar 23, 2015 #3


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    First of all you have to correct the sign of your interaction term, because otherwise there wouldn't be a stable ground state and all calculations in perturbation theory obsolete!
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