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It is well known how and why the running coupling of QCD grows as we reduce energy or equivalently increase distance. Hence, if the effective color charge increases with distance then why doesn't this infrared slavery fully explain quark confinement?

Diracology
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  • My understanding is that infrared slavery is just a phrase that describes quark confinement and not a theoretical principle. I first heard the phrase in the mid 1970s and that is what it meant then. – Lewis Miller Apr 17 '18 at 14:14
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    We only know about that running of the coupling in perturbation theory. Theoretically, we do not know if the coupling will keep getting stronger beyond the perturbative domain and actually cause confinement (though we expect something like that because of empirical facts). So to prove confinement requires a non-perturbative analysis. Lattice calculations seem to support confinement, as does the Seiberg-Witten model which realizes the monopole condensation idea of confinement in QCD. – rparwani Apr 17 '18 at 15:39
  • @rparwani So the confinement problem may be equivalent to finding the non perturbative running coupling? Isn't that already done in super Yang-Mills? That would mean that there is no confinement problem in SYM. – Diracology Apr 17 '18 at 16:09
  • @LewisMiller In fact infrared slavery is not a theoretical principle, but I understand it as a result that comes from first principles. We postulate a lagrangian, a gauge group and the representations of the particles and then we can calculate how the running coupling changes with energy (at least perturbatively). – Diracology Apr 17 '18 at 16:13
  • This question's been asked and answered in a slightly different form here: https://physics.stackexchange.com/questions/102480/are-confinement-and-asymptotic-freedom-two-sides-of-the-same-coin/102494#102494 Asymptotic freedom doesn't imply confinement. – user1504 Apr 17 '18 at 17:26
  • @Diracology No. Its not equivalent. The running of the coupling that we know of now is a perturbative effect. If the effective coupling keeps getting stronger, it might lead to a dramatic change (phase transition) in the state of the system, eg monopole condensation. So an increasing coupling strength at long distances is a hint that something new might happen, but its not enough to explain what actually happens. Eg In low dimensional models, the potential between charges is linear and so confining, and a similar linear confining potential is believed to occur in the confining phase of QCD. – rparwani Apr 18 '18 at 02:30
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    @rparwani Why don't you elaborate your comments as an answer? – Diracology Apr 18 '18 at 12:11
  • @Diracology Done as requested – rparwani Apr 18 '18 at 14:30

2 Answers2

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Confinement is not about the running coupling. By confinement we mean a qualitative change - a phase transition - in how the physics between two quarks works compared to e.g. two electrons.

In particular, confinement is the phenomenon that, in the confining phase, the force between two quark rises linearly with the distance between them, or at least that the energy required to separate two quarks is infinite. No amount of running coupling can explain this, since this is not related to the value of a coupling at all, but to the the functional form of the force law/free energy functional.

To see that a running coupling cannot explain this, consider the traditional force law in QED (i.e. Coulomb's law), then crank up the coupling. No matter how high the coupling goes, the energy required to separate two oppositely charged particles is finite.

The traditional lattice treatment (see also this answer of mine) identifies the Polyakov loop's expectation value as the order parameter of the confining/deconfining phase transition, and the Polyakov loop is essentially (and a bit handwavingly) the exponential of $-E$, where $E$ is the energy required to separate a two-quark system. As long as the Polyakov loop is zero, $E$ is infinite, i.e. the quarks are confined.

ACuriousMind
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  • Actually, the force between two quarks is believed to be constant for large distances (the potential grows with distance). In any case, I do not see why the running of the coupling constant cannot be the responsible for the behavior of the potential. In QED, the coupling goes to zero for large distances therefore there is no chance of confinement. On the other hand, imagine a theory with a Coulomb-like potential but with a running coupling that grows with distance square. Wouldn't that be confining? Wouldn't this confinement be explained by the running coupling? – Diracology Apr 17 '18 at 17:36
  • @Diracology I'm not sure I follow you. Two things: 1. The running coupling is just a trick to make perturbation theory work, see e.g. this answer of mine, it does not reflect some non-perturbative property of the theory. Confinement is definitely a non-perturbative property, since it occurs at scales where the running coupling is large. 2. What do you mean by the coupling going "with distance"? The distance the running coupling depends on is the distance scale resolvable by a process, not the distance between two particles involved. – ACuriousMind Apr 17 '18 at 17:49
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    For example: up to second order, the running coupling of $\phi^4$ theory is $\lambda_P(\mu)=\lambda-3C\lambda^2\log(\Lambda^2/\mu)$, where $\lambda$ is the bare coupling, $\Lambda$ is the cutoff and $\mu$ is the energy scale of some scattering process. Isn't changing the energy sale of the scattering the same as changing the distance between particles? In this sense I always believed that the coupling depends on the actual distance between particles. The explanation of the running in terms of screening/anti-screening also suggests that the coupling depends on the actual distance. – Diracology Apr 17 '18 at 18:03
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    For some value of the fine-structure constant, though, if you try to seperate two oppositively-charged particles enough, the field in between them will have sufficient energy to support pair-production, which will then cause two new dipoles to form, though. At that point, I'd say that you functionally would have "lepton confinement" – Zo the Relativist Apr 17 '18 at 18:04
  • @JerrySchirmer That sounds as confinement. – Diracology Apr 17 '18 at 18:05
  • I know it's not exactly the same thing as confinement, but the leading term in positronium decay is $\frac{1}{\alpha^6}$, which shows that stronger QED means "more stable positronium", and I wouldn't be surprised to see some sort of change to stability as QED becomes non-perturbative. It'd be an interesting case study to figure out how this goes. – Zo the Relativist Apr 17 '18 at 18:10
  • https://arxiv.org/pdf/hep-ph/0310099.pdf – Zo the Relativist Apr 17 '18 at 18:11
  • @ACuriousMind I disagree with your remark: "The running coupling is just a trick to make perturbation theory work". Its actually a physically observable effect. In fact getting a negative beta function to explain experimental data was the driving force for developing QCD. Recent updates on a running coupling here: https://arxiv.org/pdf/hep-ex/0606035.

    I also disagree with the comment "....not the distance between two particles involved'. It is. Simply by the Fourier transform.

     – rparwani Apr 18 '18 at 02:48
  • @rparwani 1. fair enough, my comment about the running coupling is a bit too terse to get across what I really mean. 2. I do not budge on the distance, though - the distance scale for a scattering is usually explained as the distance resolvable by a process, why would that be the same as the distance between the particles involved? – ACuriousMind Apr 18 '18 at 06:32
  • @ACuriousMind I admit that I have my own doubts about my understanding of QFT. But I believe you might be doing some confusion between the scale of energy involved in the scattering (and this is associated to distance between particles ) and the scale given by the cutoff (wouldn't this scale be related to the distance resolvable by a process?). In fact the latter scale is not physical and the physical coupling cannot depend on it (the bare coupling on the other hand does). – Diracology Apr 18 '18 at 12:09
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(I am reproducing and elaborating on my comments above here as an Answer at the request of @Diracology, and take the opportunity to clarify some other comments.)

We only know about that running of the QCD coupling in perturbation theory. It confirms the idea of asymptotic freedom which early experiments had indicated, and for which Noble prizes were awarded. That such a running is a physical phenomenon is summarised by data in this paper.

However, we do not know if the coupling will keep getting stronger at low energies (long distances) beyond the perturbative domain in which we can do reliable calculations. If indeed the effective coupling keeps getting stronger, it is possible that strong interaction dynamics might lead to a dramatic change (phase transition) to a novel state of the system.

So to study the strong coupling domain we need non-perturbative tools, such as lattice gauge theory, or some toy models.

Indeed, lattice gauge theory indicates that QCD has a confining phase, characterised by a linear quark potential similar to what one finds in lower dimensional analytic models.

One physical picture for quark confinement is the t'Hooft-Mandelstam "monopole condesation" leading to a "dual superconductor" whereby colour electric flux forms strings between charges, leading to a linear confining potential (Recall that in the usual Meissner effect in superconductors the condensation of electric charges (Cooper pairs) leads to magnetic flux tubes).

Seiberg and Witten illustrated the monopole codensation idea for confinement in a SUSY model of QCD, adding support to a similar idea in usual QCD.

In summary, confinement in QCD seems to be associated with a different phase from that which we associate with the asymptotic freedom perturbative calculations. But it is unclear (to me at least) how much of a role the strong coupling dynamics plays in the transition to the confinement phase.

**As an aside, there are many ways of interpreting and calculating the beta function beyond what is done in textbooks which usually discuss it in the context of renormalisation (usual or Wilsonian). The running of the beta function with energy scale can be physically interpreted in position space (and also calculated, if you like. I am sure someone must have done it) as the effective interaction between a source charge and a probe charge, just like in Coulomb's law. The vacuum is polarised by virtual particles in QFT and so the effective interaction changes. Indeed in this picture its easy to see that the effective charge will depend on the distance between the source and probe. So the beta function can also be interpreted as telling us how the effective charges vary with distance between the charges.

If you want to amuse yourself, you can also calculate the same beta function by looking at the interaction of the charge with an external magnetic field, which provides the clearest physical picture of why spin one is different from spin 1/2 or spin 0. See this paper by Nielsen.

The basic reason why all the different methods give the same beta function is that deep down it physically corresponds to the effective interaction between charges. It does not matter whether you calculate it in momentum space or position space or between a charge and a current carrying wire, or between a charge and a external field etc.

rparwani
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