In the case of QCD, the $\Lambda_{QCD}$ introduces a scale in the theory that can be also modified in presence of strongly interacting fermions. This mass-scale breaks the classical scale invariance by means of quantum corrections(even if one deals with massless fermions) It is interpreted as the mass scale of the bound states of the theory (protons in QCD or glueballs in pure Yang-Mills).

Suppose one applies the same to QED.

In QED (in d=3+1) one can't have bound states with massless fermions due to the lack of some invariant mass (QED can not give rise to flux tubes to form bound states from massless fermions).

Consequently to have a bound state in QED the scale invariance should be necessarily broken classically in advance.

The question is that if $\Lambda_{QCD}$ is a mass scale for the theory, then what about the Landau pole in QED (or $\Lambda_{QED}$)?

Of course, perturbation theory fails in both cases, but I'm searching for something beyond that, just like the mass scale of QCD presenting itself as a Landau pole.

  • 4
    $\begingroup$ Asymptotically free theories don't have Landau poles... $\endgroup$ Jan 29, 2022 at 21:14
  • $\begingroup$ I fixed the problem with terminology. @ConnorBehan $\endgroup$ Jan 30, 2022 at 12:18

1 Answer 1


In case of QCD the pole represents the scale of the mass of the lightest baryon(massive bound states of the theory).

Most probably, in case of QED the pole represents an upper limit on the mass of the bound states of the theory. Though physically nonsense, since it's way beyond the Planck scale where not only QED but presumably the whole framework of QFT fails.


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