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  Is QCD free from all divergences?

+ 4 like - 0 dislike

On page 8 in http://arxiv.org/pdf/hep-th/9704139v1.pdf David Gross makes the following comment:

"This theory [QCD] has no ultraviolet divergences at all. The local (bare) coupling vanishes, and the only infinities that appear are due to the fact that one sometimes expresses observables measured at finite distances in terms of those measured at infinitely small distances. "

1) First of all, is this statement even correct?

2) Now my main question is: certainly, a naive application of the Feynman rules and regularization leads to non zero counterterms. (See chap 16 of Peskin). So what scheme could one work in such that the counterterms vanish?

This post imported from StackExchange Physics at 2014-04-01 13:24 (UCT), posted by SE-user DJBunk
asked Jun 14, 2012 in Theoretical Physics by DJBunk (80 points) [ no revision ]
retagged Apr 1, 2014
I don't undersand this too!! I learned all this divergence/regularization/renormalization stuff from calculating the QCD $\beta$-function which amounts to calculating a number of ultraviolet divergent integrals!!

This post imported from StackExchange Physics at 2014-04-01 13:24 (UCT), posted by SE-user stupidity
Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files, e.g., arxiv.org/abs/hep-th/9704139

This post imported from StackExchange Physics at 2014-04-01 13:24 (UCT), posted by SE-user Qmechanic

2 Answers

+ 3 like - 0 dislike

What Gross means is that QCD is well defined in the ultraviolet, so that if you take a lattice version and send the lattice spacing to zero, there is no divergence in the coupling as you take the lattice spacing small. Instead, the coupling goes to zero as the inverse logarithm of the lattice spacing, so very slowly.

This doesn't mean that QCD perturbation theory doesn't have ultraviolet divergences, it has those like any other nonsupersymmetric unitary interacting field theory in 4d. These ultraviolet divergences though are not a sign of a problem with the theory, since the lattice definition works fine. This is in contrast to, say, QED, where the short lattice spacing limit requires the bare coupling to blow up, and it is likely that the theory blows up to infinite coupling at some small but finite distance. This is certainly what happens in the simplest interacting field theory, the quartically self-interacting scalar.

There is no proof that the limit of small lattice spacing gives a proper continuum limit for QCD, but the difficulties are of a stupid technical nature--- there is absolutely no doubt that it is true. The full proof will require a better handle on the best way to define continuum limits for statistical fluctuating fields within mathematics.

answered Jun 15, 2012 by Ron Maimon (7,720 points) [ revision history ]
+ 1 like - 0 dislike

After renormalization (with infinite counterterms), QCD is utraviolet finite, which means that it has a sensible renormalized perturbation theory at sufficiently high energies = small distances.

The vanishing bare coupling referred to by David Gross means that there are divergences at finite coupling, but the true coupling is infinitesimal (i.e., vanishes in the limit of no regularization) and the renormalization technology extracts finite contributions in the limit of inifinitesimal coupling.

However, QCD still has severe infrared divergences (at small energies = large distances, which are not cured by renormalization. Because of confinement of quarks, these IR divergences are much more severe than in QED (where they can be resolved using coherent states), and techniques for handling these are not well developed.

In addition, QCD has - like all field theories - only an asymptotic perturbation series, which means that the series itself will also diverge if all terms are summed.

This post imported from StackExchange Physics at 2014-04-01 13:24 (UCT), posted by SE-user Arnold Neumaier
answered Jun 15, 2012 by Arnold Neumaier (15,757 points) [ no revision ]
Most voted comments show all comments

Yes, but all these questions are moot, since the theory is well defined on a lattice, where you don't need perturbations.

@RonMaimon: The lattice itself is a regularization, not the full theory. And on an infinite lattice, the IR problems are fully present. Working in the approximation but claiming its results for the exact theory without qualification is poor scientific style.

This post imported from StackExchange Physics at 2014-04-01 13:24 (UCT), posted by SE-user Arnold Neumaier
@ArnoldNeumaier: I think Ron's point is that the QFT we call QCD is defined to be a continuum limit of lattice regularizations. This does not involve any UV divergences; the only UV divergences which appear result from the degenerate nature of continuum perturbation theory.

This post imported from StackExchange Physics at 2014-04-01 13:24 (UCT), posted by SE-user user1504

@user1504: QCD is defined by its action, not by a questionable continuum limit of lattice regularizations. (If the limiting process were understood, the confinement problem would have been solved.) Indeed, different workers on QCD use (and need) quite different regularizations, depending on what they want to compute.

This post imported from StackExchange Physics at 2014-04-01 13:24 (UCT), posted by SE-user Arnold Neumaier

@ArnoldNeumaier: I take issue with the assertion that the short distance perturbation theory defines QCD. First, to make a definition with perturbation theory, you have to choose a renormalization scheme; the action alone isn't sufficient. So you're defining the theory via its effective approximations in any case. And the lattice definition generalizes the perturbation theory for the short distance degrees of freedom; this is why perturbation-improved lattice gauge theory works.

This post imported from StackExchange Physics at 2014-04-01 13:24 (UCT), posted by SE-user user1504
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The limiting process for lattice QCD is mathematically sound, it just hasn't been rigorously proven. The lack of rigorous proof is not because the limiting process is hard to understand, it's very simple.

In mathematics, one may claim soundness only if a rigorous proof has been established. Please try to give the very simple proof in a rigorous fashion, and you'll see that it isn't simple at all. Excellent mathematicians have tried and got stuck.

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