# Why IR divergences cancel by cross sections of next-to-leading diagrams?

I was reading QFT & Standard Model by Schwartz, Chapter 20 which is about IR divergences. He says that

IR divergences only cancel cross sections for processes involving different initial or final states are combined ...... Although the cross section for the $2\rightarrow2$ process is IR divergent at order $e{_R}^{4}$, as is the cross section for the related $2\rightarrow3$ process (like $e^{+}e^{-}\rightarrow\mu^{+}\mu^{-} \gamma$), their sum is IR finite.

I'd like to know how we know the next-to-leading diagrams' cross sections cancel IR divergences? Also, I'd be appreciate if someone could mention why we need cross sections for cancellation, indeed? Why S matrix not enough?

However, if the energies of one or more of these particles is very small they won't be seen in a detector. In this case these processes will appear as processes with fewer outgoing particles. Thus the right way to calculate the cross section of seeing a signal in a detector for a $2\rightarrow 2$ process will take the schematic form, $$\sigma \sim \int d\sigma_ {2 \rightarrow 2} + \int _{ E_3 < \mbox{detector res}} d\sigma_ {2 \rightarrow 3}$$ Thus we don't need to worry about if the $2 \rightarrow 2$ process diverges. As long as the total cross section is finite we can conclude that we are getting sensible results that can be consistent with our observations.
• Thanks for the answer JeffDror. Can I say IR scale do not diverge if it is observable, that"s why we solve the cross sections. However, these final states $2\rightarrow3$ exists but negligible since they are too small. But on the other hand not negligible since they cancel IR divergences? – aQuestion Apr 25 '15 at 20:45
• I don't agree that $2\rightarrow 3$ processes are necessarily negligible. In particular, due to a different divergence, a collinear divergence, they are actually very likely. This is why particles always come out showered at any particle accelerator. However, in the end as long as we consider some cone around each final particle, this is easy enough to deal with. – JeffDror May 1 '15 at 10:11