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I Lagrangian and Feynman diagrams

  1. May 17, 2016 #1
    Hello,

    Consider the the following Lagrangian of the $\phi ^4$ theory:
    $$\begin{align*} \mathcal{L} = \frac{1}{2} [\partial ^{\mu} \phi \partial _{\mu} \phi - m^2 \phi ^2] - \frac{\lambda}{4!} \phi ^4 \end{align*}$$

    Now I'm interested in Feynman diagrams.

    1. The second term gives the propagator an the third a vertex but what about the first term $$\frac{1}{2} [\partial ^{\mu} \phi \partial _{\mu} \phi]~?$$

    2. How does this kinetic term looks like in the feynman diagram formalism?

    3. I also don't understand how to renormalize this first term (maybe this question is solved if I know how this term looks like in the sense of feynman diagrams).

    Thank you.
     
  2. jcsd
  3. May 17, 2016 #2

    Demystifier

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    The second term does not give the propagator. The first and second terms together do.
     
  4. May 17, 2016 #3
    I can rewrite the Lagrangian in a form where a new field $\sigma$ appears and for this field exists no kinetic term. I thougth this means that there are no external lines for this field. So is there a link between this kinetic term and the external lines?

    And how is it possible that we obtain
    $$ \mathcal{L} = \frac{1}{2} (1+ \delta Z)(\partial_{\mu} \phi)^2 + \frac{1}{2}(m^2 + \delta m^2) \phi^2 +...$$
    for the renormalization of the kinetic term if it is "bounded" in the propagator? More precise shouldn't the counterterm Z be part of . So what are the diagrams for Z?
     
  5. May 17, 2016 #4

    vanhees71

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    the counter term comes from self-energy diagrams, i.e., diagrams with two external lines. Power counting tells you that this is quadratically divergent. Due to Lorentz invariance, it can only depend on ##p^2## (with ##p## being the four momentum of the external lines) the counter term thus is of the form ##A p^2-Bm^2## with ##A## and ##B## dimensionless quantities. This tells you that the self-energy diagrams' counter terms contribute to ##\delta Z## and ##\delta m^2##. Your 2nd term should have a minus sign in front, by the way, i.e., it's ##...-1/2 (m^2+\delta m^2)\phi^2##.
     
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