# Infrared divergence in scattering via annihilation

In textbooks when discussing radiative corrections to QED scattering they normally only consider the loops corresponding to the vertex function and corrections to the electron and photon propagators, and also the diagrams for emission of soft photons from external legs (see e.g. Chapter 6, diagram 6.1 in Peskin and Schroeder).

But there are additional loops, for instance electron positron scattering via intermediate two photon annihilation

This diagram has no ultraviolet divergence, but as one of the virtual photon momenta $k\rightarrow 0$ the electron propagators go on shell and there is an infrared divergence.

My question is how is this divergence resolved? My thought is that the electron propagators are modified by the field of the external legs in higher order diagrams, and that shifts the pole by something on the order of a binding energy.

But is this interpretation correct and does it have anything to do with the Lamb shift? How is this divergence dealt with in practice in calculations?

• 1. IR divergences are resolved the same way any divergences are resolved: you regularize it, analyze the regularized expression, make corrections, take off the regularizer after corrections are made. 2. IR divergences are canceled by radiative diagrams: the ones with arbitrary low-energy photons emitted by the external legs. These correspond physically to the following: we can never be sure that a small portion of energy haven't escaped our experimental setup in the form of an undetected small-energy photon in the out state. – Prof. Legolasov Apr 11 '17 at 0:28
• Yes I'm familiar with the interpretation where the low energy photons on the external legs cancel the IR divergences in the vertex function and electron self energy. It is discussed for instance in Chapter 6 of Peskin and Schroeder.Does this IR divergence cancel in the same way? I don't think so, but I'd be happy to be shown wrong. – octonion Apr 11 '17 at 1:12
• I looked again at the general proof of the cancellation of IR divergences and I think you are right. For the scattering case this IR divergence is also cancelled by the soft photons from the external legs. I had this question because it looked like those external legs were already canceling the divergences from the vertex function, but in the full 4-point function there are additional external legs too. Also if this diagram is used as a correction in bound state problems the IR divergence has a completely different interpretation so I thought the scattering also had an unusual interpretation. – octonion Apr 11 '17 at 9:52