# How to draw self-energy diagrams in phi^4 theory

1. Mar 20, 2015

Okay guys so I the lagrangian

$\mathcal{L}=\dfrac{1}{2}\left( \partial_{\mu}\phi\partial^{\mu}\phi-m^{2}\phi^{2} \right)+\dfrac{\lambda}{4!}:\phi(x)^{4}:$

where $\phi(x)$ is a real scalar field.

I want to know how you can draw the self-energy diagrams at order $\lambda$ and $\lambda^{2}$ for a 2 => 2 scattering system. How do you go about doing this just by looking at the Lagrangian?

Thanks guys....

2. Mar 20, 2015

### Einj

First of all, what do you mean by self-energy? Usually one calls self-energy the higher-order corrections to the single particle propagator, i.e. a 1 -> 1 process. Here you have 2 initial particle and 2 final one. What do you define as self-energy?
However, in the most general way possible, since your interaction term is $\frac{\lambda}{4!}\phi^4$, this means that this theory admits vertices with 4 legs (shaped like an X, if you want). Each X counts as $\lambda$.
In your case the basics 2 -> 2 process simply happens when you have one particle at each leg of the X. This is a order $\lambda$ diagram, the only allowed for the considered scattering. If you want to go to order $\lambda^2$ then you have two possibilities:

1) Take the order $\lambda$ diagram just mentioned and you add a "bubble" to one of the external leg. Meaning, that one of the legs emits and reabsorbs a particle at the same point (you essentially join two of the legs of your X together). You can do this for each of the four legs. This is what I would call the self-energy correction to your tree level 2 -> 2 scattering.
2) You can also take two X and join two of their legs together, so that you end up with a bubble with 4 total external legs (two coming from the same point on one side and two coming from the same point on the other side). This is again order $\lambda^2$ but it doesn't strictly contain any self-energy.

Is this what you were looking for?

3. Mar 23, 2015

### vanhees71

First of all you have to correct the sign of your interaction term, because otherwise there wouldn't be a stable ground state and all calculations in perturbation theory obsolete!