Does there exist a classical limit of QCD? I mean in the sense of wave particle duality of eg photons. Is there any similar thing for gluons?

  • $\begingroup$ short answer: no, there are classical analogies to gluons and asymptotic freedom etc.. $\endgroup$
    – Nikos M.
    Aug 21, 2014 at 23:08

1 Answer 1


You can say that QCD is the opposite of QED. I don't know how much you know of Renormalization group, but QCD has an asymptotic freedom, that mean that when you go to higher energies, you can use perturbative theory to do cross section calculation. That because the coupling constant is running with energy, due to the Beta-function.

The goodness of your perturbative expansion is given by the coupling constant, for example in QED you have $$ \alpha_{QED} =\frac{1}{137} $$ Even $\alpha_{QED}$ is a running constant, but it changes very little.

So you have a better understanding of QCD at high energies-->small wavelenght.

At low energies the problem is that you have a non-perturbative theorem. Many people don't understand how much the problem is difficult, you can read the Maiani-Testa No Go theorem, that give a hard blow to Lattice-QCD years ago.

For the classical limit, I think that you can see something of HQET, or HQE. HQET you use the fact that heavy quark are "heavier" than $\Lambda_{QCD}$, so you find some funny things like a chromomagnetic field and chromoelectric field, if I remember correctly this was analogous to a gordon decomposition for a QCD current, and I was impressed at the time, but don't think too high of it, HQET has few limited uses.

Maybe even HQE (heavy quark expansion) can be useful for that, because you try to study heavy mesons, like $\Upsilon(4S)$ where heavy quarks can be treated like non-relativistic, but I haven't first experience of it.

How you can understand, a classical limit seem to arise at higher energies and so at short wavelength.

A few useful links Is there a strong force analog to magnetic fields?


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