# 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