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A few days ago I asked a question about QCD chiral symmetry breaking and hopefully got a very good answer. But the problem is I am not well-versed in QCD to fully appreciate the answer there. I know the QCD Lagrangian and how to derive it. I am familiar with the fact that it is different from QED because it is a nonabelian gauge theory. Lastly, I have read the basic idea of chiral symmetry breaking though not much about the mathematics. I have heard that QCD becomes nonperturbative at low energies. But I have no idea why does QCD behave this way and how low is low for QCD to become nonperturbative. It would be enormously helpful if someone can explain (with some equations) why QCD behaves this way or suggest me one or two textbook references.

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    $\begingroup$ The confinement scale where the strong coupling diverges ("infrared slavery") is of the order of 300 MeV... $\endgroup$ Nov 26, 2019 at 15:53
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    $\begingroup$ Look at the leftmost point of figure 9.3 in the PDG review of QCD. This is the limit of applicability of perturbation theory. Anything to the left of that is "strong coupling". $\endgroup$ Nov 26, 2019 at 16:01

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The beta function of QCD is negative if you take into account all the relevant fermions and bosons in the standard model. On the contrary, the beta function of QED is positive. Using the definition of the beta function

$$\beta (g) = \frac{\partial g}{\partial \log{\mu}}$$

where $\mu$ is the energy scale of the physical process at hand, one can see that the QCD coupling becomes very large for small values of the coupling. At around the QCD scale (around 330 MeV), the coupling constant becomes large enough that below this scale you cannot treat $g$ perturbatively anymore. As a result, QCD in the infrared is understood only via non-perturbative methods. As an example, you cannot understand features like color confinement using perturbation theory.

On the other hand, since the beta function of QED is positive, the coupling grows at larger and larger values of the energy scale.

A nice reference textbook: Chapter 16 and 17 of Peskin and Schroeder.

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The main reason is asymptotic freedom. When one calculates the beta function of the strong coupling in the perturbative regime, one sees that it grows towards the infrared. This is different from QED, where the coupling behaves the opposite.

The growing coupling of QCD towards smaller scales at some point leads to the formation of bound states of the quarks, e.g. hadrons, which appear as effective degrees of freedom. They finally form a condensate that breaks the chiral symmetry.

But it is very hard to get a precise quantitative grip on all of this. One usually says that the strong coupling becomes non-perturbative (that means it will be so large that perturbation theory fails) at scales of around 1-1.5 GeV...

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  • $\begingroup$ "When one calculates the beta function of the strong coupling in the perturbative regime, one sees that it grows towards the infrared. This is different from QED, where the coupling behaves the opposite." Can the reason be traced back to the nonabelian nature of QCD? $\endgroup$ Mar 1, 2020 at 20:07
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    $\begingroup$ Yes, but nonabelian theories do not necessarily behave that way. It is possible to have a nonabelian theory that does not show asymptotic freedom. It depends on the numbers of flavors and colors involved. But in the QCD case it works out to have asymptotic freedom. $\endgroup$
    – dan-ros
    Mar 2, 2020 at 21:13

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