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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?

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As pointed out, the cross sections for certain processes diverge in the IR. However, we know from everyday life that measurements don't diverge. In other words in any actual experiment the number of photons is finite. While physically very obvious, it is apriori unclear how QFT is consistent with this observation. Based on this observation, one might naively claim that QFT breaks down in this regime. However, this is not the case.

The solution (as you mentioned above) is we need to make sure we calculate observables to a consistent order in the coupling. In any physical measurement one can't measure "particles in the final state". Instead what we can measure is energy deposited on a detector. If the particles are energetic enough one can differentiate these in the detector and infer the "particles in the final state". These energy measurements are what we know from experience that they are finite.

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.

Also regarding your second question about we are concerned about cross-section divergences as opposed to S-matrix divergences. The answer is simple. S-matrix elements aren't observable, and hence unphysical. All we know from experience is that cross-sections are finite.

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  • $\begingroup$ 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? $\endgroup$ – aQuestion Apr 25 '15 at 20:45
  • $\begingroup$ 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. $\endgroup$ – JeffDror May 1 '15 at 10:11

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