# Lagrangian and Feynman diagrams

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.

Demystifier
Gold Member
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
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?

vanhees71
• 