A How can infrared divergences in the fermion propagator be cured in QED?

[Jamister]

TL;DR Summary

how to cure infrared divergences in fermion propagator in QED?

Summary: how to cure infrared divergences in fermion propagator in QED?

In calculating the fermion propagator in QED, we identify Ultraviolet and Infrared divergences. the Ultraviolet divergences solved by regularization, but I don't understand how to treat the Infrared divergences. Infrared divergences also appear in the QED vertex, but the solution there to Infrared divergences is by soft photons. How is it done in the fermion propagator?
This is the result of the 1PI diagram of the fermion propagator to leading order:

mu is the mass of the photon.

 

[vanhees71]

The IR divergency is canceled by considering the appropriate contributions to the electron-to-electron transition amplitude plus two (un-noticed) real photons. This is a very general feature, first established by Bloch and Nordsieck and later extended to non-Abelian gauge theories by Kinoshita, Lee, and Nauenberg. The standard treatment within usual perturbation theory can be found in

Weinberg, Quantum Theory of Fields, vol. 1 (Chpt. 13).

A very illuminating paper,

P. Kulish, L. Faddeev, Asymptotic conditions and infrared divergences in quantum
electrodynamics, Theor. Math. Phys. 4 (1970) 745.
http://dx.doi.org/10.1007/BF01066485
showing that the physical reason for the IR properties is that due to the long-ranged Coulomb force due to the masslessness of the photons the naive plane-wave single-particle states used as asymptotic-free states are not the right ones. Rather you have to dress the "free" bare particles with a cloud of (real and virtual) photons around them. This cloud cancels the IR divergences. These modified asymptotic free states are sometimes called "infraparticles".