-3

The running coupling constant ("hold that constant!) is a well known phenomenon in quantum field theory. The constant varies with the energy of the interacting particles. I think this is rather strange. Why should that be the case? Does this follow naturally, is this imposed, or does it have to do with the dressed particle? In conformal field theory the coupling constant is, well, constant at every scale, but why not in "ordinary" QFT?

Isn't this running constant caused by the interaction with the virtual field at small distances which results in diverging values for scales approaching zero, which are absorbed by renormalization? Which actually makes the naked coupling or mass infinite? An indication maybe that the point-like character of particles is a chimera?

MatterGauge
  • 1
  • 3
    That's just how renormalization works, see https://physics.stackexchange.com/a/178756/50583, https://physics.stackexchange.com/a/696023/50583. The derivation of running couplings should be in every QFT textbook. Can you be more specific what you want to know about them? – ACuriousMind Jun 13 '22 at 22:06
  • 1
    Renormalization is not unique to QFT. In lattice models as well, you can integrate out short wavelength degrees of freedom and find that what remains is well described by the original Hamiltonian with shifted couplings. – Connor Behan Jun 13 '22 at 22:37
  • @ACuriousMind What I think is strange is that at the vertex it is "decided" what the coupling should be on basis of the energy and momentum. – MatterGauge Jun 13 '22 at 22:41

1 Answers1

1

(Perhaps this should be a mere comment but I don't have the reputation to comment)

Maybe it helps to remember that the parameters that appear in the Lagrangian (including the coupling and the masses) are not observables. Cross-sections and decay widths are observables, and have coupling constant and mass dependece, but the Lagrangian parameters cannot be directly observed. As such, they can be defined in multiple ways. Renormalization makes use of this freedom to make the parameters finite, but in doing so the price of introducing scale dependence is paid.

Néstor González Gracia
  • 111