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Infrared Divergences in Gauge Theories

+ 3 like - 0 dislike
279 views

Hi,
There are some confusions regarding infrared (IR) divergences in gauge theory:

  • What is the primary reason for the appearance of IR divergences in gauge theory? Anything other than the existence of massless particles in the spectrum of the theory. 
     
  • Why there is no IR divergence for massive theories? The propagator is given by:
    $$K(p)\propto\frac{1}{p^2-m^2}$$
    Therefore, it is expected that the internal momentum integration crosses the pole given by mass. If $i\epsilon$ prescription is used to avoid this, why does it can not be used in the case of massless theories with propagator proportional to $\frac{1}{p^2}$?
     
  • Do IR divergences appear only at the loop-level?
     
  • The S-matrix of a QFT is defined by assuming that the interactions die off at spatial infinity. But in a theory with massless particles, this can not be assumed. Then how should we interpret the results of scattering amplitude computations in gauge theory?
asked Sep 3 in Theoretical Physics by SI1989 (85 points) [ revision history ]
recategorized Sep 3 by Dilaton

In QM we calculate the probabilities of events. Now imagine a compound target consisting of many interacting "particles" and thus having many internal degrees of freedom. When you hit one of these particles with a projectile, the target starts moving as a whole and generally gets "excited" too. The latter means a change of the initial state of internal motion of its constituents. If there is a non-zero threshold for exciting the target, then there is a probability of not exciting the internal motion while hitting one of its constituents. In other words, there is a non-zero probability to "push" the system "elastically". In particular, for the transferred energy $\Delta E\le E_{Threshold}$ there are no excitations in the final state, i.e., there are no inelastic channels open - they are inaccessible due to lack of transferred energy. The system gets the transferred energy as whole. So the probability of elastic scattering to any angle is unity.

But If there is no threshold of exciting the target internal motion, then the probability of elastic scattering is always zero - you cannot push a system acting on one of its parts and not change the target internal "order". The energy of exciting may be very small, but we calculate probability, not energy.

Now let us imagine that in our calculations of transition probabilities we threat the interaction of the projectile with one of the target constituents perturbatively. For example, we use the first Born approximation. If the state of the target is known exactly, then the probability of elastic scattering will be immediately obtained as zero. But, but! If the interaction of target particles is neglected in the first Born approximation (it is postponed to next perturbation orders), then you can get a unity probability for elastic scattering.

Thus, in perturbative calculations of probabilities we may occasionally start from unity, which is very far from the exact result - zero. No wonder the perturbation series may diverge in higher orders since the initial approximation for elastic amplitude was chosen too bad. This is the true reason of IR divergences- you cannot push a charge without automatically exciting electromagnetic (or other massless gauge) degrees of freedom coupled to the charge in the exact solution, but you start unfortunately from too inexact initial solution where the charge and soft modes are decoupled.


What is done in QFT in the end is taking into account the coupling to soft target modes perturbatively in all orders (i.e., exactly) and obtaining a non-zero results for inclusive transitions. Inclusive cross sections are different from zero and slightly depend on how much energy is allocated to soft modes. This is the only correct result while dealing with "soft" targets.

@VladimirKalitvianski

Thank you for the nice comment.

2 Answers

+ 3 like - 0 dislike

There are a bunch of questions here, so I will just do my best to answer them as efficiently as possible.

First of all, one can integrate a propagator to a force law. A massive propagator $1/(p^2+m^2)$ induces a force law $e^{-rm}$, in natural units, while a propagator $1/p^2$ induces a force law of Coulomb type, eg. $1/r^2$ in 3+1D. Notice that the first has no small r divergence while the second does. This is the basic statement, though it is modified in many interesting ways beyond tree level.

In formulating scattering problems in gauge theory, one must be quite careful about how to specify boundary conditions of the gauge field at infinity. Peskin and Schroeder have a great description of this, but there are also some new perspectives coming from Andy Strominger and his collaborators. See, for example, these lecture notes (there are also videos on youtube).

Hope this helps!

answered Sep 9 by Ryan Thorngren (1,605 points) [ revision history ]

@RyanThorngren

Thank you for the answer. 

By your last point about boundary conditions, do you mean that the S-matrix can be defined but we need to specify a specific boundary condition for gauge field at infinity? Also, can you please specify the relevant sections of Peskin & Schroeder?
 

Photon plane waves are themselves a boundary condition for the gauge field at infinity, but we will need to account for radiative corrections to define the S-matrix. I don't have my PS on hand, but it's in there, just read the whole thing! :P

+ 3 like - 0 dislike

Massive theories give rise to a partially discrete spectrum in the Kallen-Lehmann formula, corresponding in Haag-Ruelle scattering theory to ordinary particles. Massless theories produce only a continuous spectrum, corresponding in a generalized Haag-Ruelle scattering theory to infraparticles. This means that the infraparticles belong to end points of branch cuts (in the complex scaled spectrum) rather than to poles. This changes the whole physics and produces the IR problem (divergences due to treating the branch cuts as if they were poles).

In particular, in a theory with massless states the asymptotic states are no longer plane wave states but coherent states of some sort with many more degrees of freedom. All this is not very well understood. 

answered Sep 16 by Arnold Neumaier (12,355 points) [ no revision ]

Thank you for the very nice answer. I am very interested to know about the issue on a more fundamental level. Is there any reference on these topics?

Also, you can see that in the comments of the above answer, Ryan mentions that photon plane wave (together with radiative corrections) provides a boundary condition for the gauge field. However, you are mentioning that the boundary conditions are given by coherent states. These two seems to be in contradiction. Am I correct?

@SI1989: I think there is no contradiction here: Coherent states contain infinite number of (soft) photons or plane waves, that's it. They all are included (at least) in the final state after scattering.

@SI1989: One must take linear combinations of plane waves (giving coherent states) in order to have in-and out states with a finite (i,.e.., nonzero) S-matrix contribution.

The standard reference on coherent states and the IR problem is 

P. Kulish and L. Faddeev, Asymptotic conditions and infrared divergences in quantum electrodynamics, Theor. Math. Phys., 4 (1970), p. 745.
 

For infraparticles, see, e.g., https://arxiv.org/abs/0709.2493.

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