# [QED] Infrared divergences

1. Oct 16, 2012

### eoghan

Hi!

In e+/e- to mu+/mu- scattering, if the initial electron/positron radiates a photon, then I have an IR divergence. How do I treat it?

2. Oct 16, 2012

### Einj

Usually when you find an IR divergence it means that you didn't considered all the diagrams to a given order. For example, in the ee->μμ scattering you have the tree level diagram that is of order $M_1=O(e^2)$ (where M is the related matrix element). If you want to add the diagram in which the initial electron emits a photon then you are adding a diagram of order $M_2=O(e^3)$. Then, considering that you will have to take the square of the matrix elements your leading order will be $|M_2|^2=O(e^6)$.
To resolve the IR divergence you have to add, for example, the diagrams in which the one of the initial (or final) particles emits and then re-absorb this photon (that is a renormalization diagram). This new diagram lead to a matrix element $M_3=O(e^4)$ and so it could seem to be of higer order. But, since they all compete to the same phenomenon you will have to take the square of the sum of the diagrams and thus you obtain:

$$|M_1+M_2+M_3|^2=|M_1|^2+|M_2|^2+|M_3|^2+2Re(M_1^*M_2)+2Re(M_1^*M_3)+2Re(M_2^*M_3)=O(e^4)+O(e^6)+O(e^8)+O(e^5)+O(e^2)+O(e^7)$$

so, as you can see you have to include also the second mixed term in order to consider all the diagrams to the same order.

What you usually do is to give a small mass to the photon (in order to avoid divergences during calculations) and consider all these contribution. And the end of the calculation you can set the photon mass to zero.