As far as I understand, the basic reason why we are not able to solve QCD at low energies is because it is a strongly coupled quantum field theory, and a part from exceptional situations in simplified models (involving for instance supersymmetry), we have no tools to fully analyze such regimes. Correct me if I am wrong, but this is a rather technical obstruction.

Is there thus a simple, non-technical reason why proving confinement is such a difficult problem?

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    $\begingroup$ Aside from not having any tools to analyze such regimes? $\endgroup$ – user1504 Jun 18 '13 at 12:06
  • $\begingroup$ "Not having a tool" is a technical reason. I would expect that a problem such as confinement would require some new physical concepts to be solved, but that might be completely wrong of course. $\endgroup$ – Bru Jun 18 '13 at 12:19
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    $\begingroup$ @Bru - I am puzzled by what you are asking for. You have essentially already answered your own question in a single sentence right at the start - 'We are not able to solve QCD at low energies is because it is a strongly coupled quantum field theory' i.e. there are no small parameters to do perturbation theory in. Are you asking for someone to explain this sentence in more detail? $\endgroup$ – DJBunk Jun 18 '13 at 14:00
  • $\begingroup$ @DJBunk: Take for example quantum gravity. This is an outstanding problem in theoretical physics. Now: do you believe this is a pure technical problem to find the theory of quantum gravity (if it makes sense), or do you think one should also understand conceptually some issues like background independence, absence of local gauge invariant observables, etc ? In my opinion, the second possibility prevails, and this is an example of non-technical reason why quantum gravity is a difficult problem. I am wondering if there is a similar story for confinement. $\endgroup$ – Bru Jun 18 '13 at 14:17

The current understanding is indeed that a technical barrier (i.e. the failure of perturbation series due to asymptotic freedom) is complicating the analysis of the low energy regime.

However, one might look at the problem from a slightly conceptual point of view. The Lagrangian of QCD, which defines the theory, is defined in terms of fundamental fermionic degrees of freedom, which we associate with quarks. At high energies, there is a direct correspondence between those theoretical degrees of freedom and the observable free ones. At low energy, the case is not that simple. The observed degrees of freedom at low energies are hadrons, they are no longer fundamental but show inner structure. The question is now: in which way does the inner structure of hadrons relate directly to the quark degrees of freedom of the high energy theory? There is no clear answer to this question. One may view hadrons as consisting of so-called "constituent quarks", which models baryons as three-quark and mesons as two-quark bound states. As a result of the strong interaction involving gluons, there is no direct correspondence between the quarks in the QCD Lagrangian at high energy and the constituent quarks at low energy.

This might not be the expected answer, but I hope it could nevertheless give some insight into the problem.

  • $\begingroup$ I don't think that saying that "the failure of perturbation series [is] due to asymptotic freedom" makes this answer clear. Asymptotic freedom allows to use perturbation theory in the UV. Of course, using the same RG equations in the IR makes the theory strongly coupled, but maybe this should be made clearer. $\endgroup$ – Adam Sep 29 '13 at 18:08
  • $\begingroup$ This is the exactly the problem: QCD in the IR is strongly coupled and hence perturbation theory breaks down. This is also the regime where QCD is confined, hence the connection. But the question was not about a technical explanation, that's why I did not go into more detail. $\endgroup$ – Frederic Brünner Sep 29 '13 at 19:16

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