# What is the origin of infrared divergence?

1. Sep 5, 2011

### ndung200790

Please teach me this:
It seem that the ultraviolet divergence has origin of we unknow the physics at very small distance(very large momentum,then very small distance).So we must cut off the very large momentum by renormalization procedure.But I do not understand the origin of infrared divergence in Feynman diagrams.By the way,it seem that in order to cut off infrared divergence,the general procedure is to put a small mass for the massless particles.
Thank you very much in advance.

2. Sep 5, 2011

### vanhees71

The infrared divergence originates from using the wrong "asymptotic states" when massless particles (field quanta) like photons are involved.

The most simple example for this deficiency of a naive use of asympotic states is well-known also in non-relativistic quantum theory: For non-relativistic Coulomb scattering the asymptotic states are not plane waves (momentum eigenstates) but the Coulomb-scattering states, which have additional non-trivial phase factors.

A very nice treatment of IR divergences from this point of view is given in the paper (including the Coulomb-scattering example)

P. P. Kulish, L. D. Faddeev, Asympototic conditions and IR divergences in quantum electrodynamics, Theo. Math. Phys. 4, 745 (1970)
DOI: 10.1007/BF01066485

3. Sep 5, 2011

### Bill_K

Infrared divergences arise in two ways: (a) the emission probability of soft photons (bremsstrahlung) becomes infinite with increasing wavelength, and (b) the radiative correction from a virtual photon diverges when integrated down to zero energy. If we do perturbation theory in the simplest way, we keep (b) but not (a) and obtain an infinite result.

The solution is to realize that the energy of an outgoing electron can only be determined experimentally to within some limit ΔE, and that therefore the emission of soft photons with energy < ΔE must be included. When (a) and (b) are combined, the result is finite.