In quantum field theory, as long as the interaction is weak, one can use perturbation theory, i.e. Feynman diagrams and the virtual particle formalism. The particle spectrum of the interacting theory is the same as that of the free theory, when the interaction is "turned off".

If however the interaction is strong, this no longer holds and the particle spectra might be different.

I can't see though how a strong interaction can alter the particle spectrum of the free fields. What happens?

The reason I ask is this part of an answer:

"In perturbation theory, when the interactions are weak, the formalism of virtual particles is used and is in fact very useful! This is essentially what Feynman diagrams describe.

You can do this when interactions are weak because you can assume that the particles that appear in the theory are the same particles which appear when the interactions are turned off. So the fields that you used to construct the QFT still correspond to the particles in the theory, and have a standard interpretation in terms of creation and annihilation operators.

When interactions become strong, there is no guarantee that the particle spectrum is the same as when the interactions are turned off. The fields you used to construct the QFT therefore no longer necessarily correspond to particles in the theory. A description of the interaction in terms of the 'free' virtual particles no longer makes much sense "

The quote is from the answer to this question: Why do operator valued distributions in QFT refer to free fields only?

  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Dec 5, 2022 at 20:07
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    $\begingroup$ Please add a link to the original post you are quoting $\endgroup$ Dec 5, 2022 at 21:05
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    $\begingroup$ You're asking the wrong question. If you take one hamiltonian (the free particle one) and replace it by another (the interacting one), why should their spectra be at all related? $\endgroup$ Dec 5, 2022 at 21:10

1 Answer 1


What happens is gluon-induced confinement. People have been thinking about it for half a century and have not fully seen it to their satisfaction, either—but they have learned an awful lot about it, since.

The strong interaction you mention increases in strength with growing distance to completely alter the degrees of freedom of the theory, as the original fields of the Lagrangian, quarks and gluons, have little to do with the excitations and the spectrum of the effective hadronic Lagrangian, and its free fields. At distances larger than a fermi or so, the effective Lagrangian hadron fields are colorless.

You merely have a very different-looking theory of massive hadrons and light (Goldstone) pions (and Kaons, etc...), whose mass scale is dramatically higher than the scale of quarks (as small as a few MeVs; the gluons are massless), a large fraction of a GeV, due to chiral symmetry breaking.

These large mass scales: the mass gap, the confinement scale, and the chiral symmetry breaking scale, are interdependent in a complex theory that has not been fully worked out—only empirically simulated on a computer, but do not crucially depend on the mass scales of the perturbative short distance theory you started from.


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