# Lagrangian and Feynman diagrams

• I
• Neutrinos02
In summary, the Lagrangian for the $\phi ^4$ theory contains three terms, with the first term representing the kinetic term, the second term representing the propagator, and the third term representing the vertex. The kinetic term can be rewritten in a form where a new field $\sigma$ appears, but this field has no kinetic term and thus no external lines. The counter term for the kinetic term comes from self-energy diagrams with two external lines and contributes to the renormalization of the kinetic term, represented by the terms ##\delta Z## and ##\delta m^2##.

#### Neutrinos02

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.

Neutrinos02 said:
1. The second term gives the propagator an the third a vertex but what about the first term $$\frac{1}{2} [\partial ^{\mu} \phi \partial _{\mu} \phi]~?$$
The second term does not give the propagator. The first and second terms together do.

• bhobba
Demystifier said:
The second term does not give the propagator. The first and second terms together do.

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?

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##.

• bhobba