# I Lagrangian and Feynman diagrams

1. May 17, 2016

### 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.

2. May 17, 2016

### Demystifier

The second term does not give the propagator. The first and second terms together do.

3. May 17, 2016

### Neutrinos02

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?

4. May 17, 2016

### vanhees71

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 $...-1/2 (m^2+\delta m^2)\phi^2$.