I was thinking about hadrons in general Yang-Mills theories and I have some doubts that I'd like to discuss with you.

Suppose that we have a Yang-Mills theory that, like QCD, tend to bind quarks into color singlet states. So far nothing strange, even QED tend to bind electromagnetic charges to form neutral systems. The twist in Yang-Mills theories comes when we consider the running of the coupling constant: $$ \alpha(\mu)\simeq\frac{1}{ln(\mu/\Lambda)} $$ which present asymptotic freedom, i.e. it decrease when the energy scale $\mu$ increase. This formula also suggest that when the energy scale approach the energy $\Lambda$ the interactions between two colored objects becomes very strong, so we assume that we cannot observe colored objects on scales greater than $1/\Lambda$.

Here my first question: We cannot see free quark in our world because the scale $1/\Lambda$ happen to be smaller than hadronic typical dimension so we cannot pull quarks enough outside hadrons to see them in non hadronic environment? And, if yes, could exist Yang Mills theories where this not happen, i.e. where the hadron typical scale is smaller than $1/\Lambda$ and then we can see free quarks?

In QCD, being the hadrons system of strongly coupled quarks, we cannot study them by perturbative approach. However the presence of a spontaneously broken chiral symmetry allows us to study some of their properties.

Here my second question:We do not need this SSB to form hadrons, right? There could be some theory were we have hadrons but not this SSB?

If something is unclear tell me and I'll try to explain better.


2 Answers 2


1) Note that $\Lambda$ is the only scale in the problem, and there is no sense in which $\Lambda$ can be small or large. In particular, $1/\Lambda$ cannot be much larger than hadronic sizes, because $1/\Lambda$ is the size of a hadron.

The only thing you can do is insert another scale into the theory, such as the mass of a heavy quark, $M\gg\Lambda$. In that case you get a theory of Coulomb bound states (size $1/M$ up to logs). These states can be ionized into very long strings. Top quarks satisfy this condition, but they are too unstable to form bound states.

1b) Another example is electroweak theory, which is based on a confining gauge theory, $SU(2)_W$, but we talk about free electrons (see the answer to this question: Weak isospin confinement?). The reason we can do this is that the weak confinement scale $\Lambda_W$ is much smaller than the Higgs vev $v$. If the ordering was reversed, the physics would be mostly unchanged, but we would view the electron as a confined bound state of a bare electron and the Higgs.

2) In QCD chiral symmetry breaking is implied by confinement (this follows from the anomaly matching conditions), but there are QCD-like theories with confinment but no chiral symmetry breaking. An example is ${\cal N}=1$ SUSY QCD with $N_f=N_c+1$ flavors. This is a theory of confined massless hadrons.

Postcript: (Anomaly Matching) T'Hooft argued that the anomalies in the microscopic theory and the effective low energy theory must match. (He has a clever argument based on adding extra fermion fields to make the theory anomaly free, and then gauging the flavor symmetries, but this seems intuitively obvious). If chiral symmetry breaks we can always represent anomalies by WZ terms in the chiral lagrangian. However, if chiral symmetry does not break (and the theory is confining) then the situation is more complicated, because now the anomaly has to represented by the triangle anomalies of massless composite fermions. In general this does not work, because the quantum numbers of composite fermions are different from the quantum numbers of fundamental fermions. In three flavor QCD, for example, massless quarks are flavor triplets with fractional charges, and massless baryons are octets with integer charges.

  • $\begingroup$ 1)Yes, i was thinking exactly to that. If the mass of the quarks would have been greater then $\Lambda$ we could have seen free quark, right? 2)So if we have a theory with explicit mass terms for the quarks but no SSB we still have massive hadrons? $\endgroup$
    – Cervantes
    Jan 6, 2016 at 16:50
  • $\begingroup$ 1) It's a funny theory, because strictly speaking it is still confining, except that at any finite temperature, or in a finite volume you don't notice. 1b) Another example is electroweak theory, see my edit. 2) Yes, except that for finite m the notion of SSB becomes fuzzy (because now the symmetry is already explicitly broken). $\endgroup$
    – Thomas
    Jan 6, 2016 at 17:55
  • $\begingroup$ 1)But this is not the case of the top quark? It has a mass of $\simeq163$ GeV which is bigger than $\Lambda_{QCD}\simeq300$ MeV, so why we do not see free top quarks? $\endgroup$
    – Cervantes
    Jan 6, 2016 at 18:04
  • 1
    $\begingroup$ No. The size of the hydrogen atom is $1/m_e$ not $1/m_p$, or $1/m_H$. Similarly, the size of a B meson is $1/\Lambda$. This is formalized in heavy quark effective theory (HQET). $\endgroup$
    – Thomas
    Jan 6, 2016 at 20:13
  • 1
    $\begingroup$ Any text book on particle physics will have a short discussion on quarkonia (charmonium, most notably the J/psi, bottomonium, in particular the upsilon). A more technical discussion is in chapter 6 of Yndurain, or reviews like this arxiv.org/abs/1111.0165 . $\endgroup$
    – Thomas
    Jan 25, 2016 at 3:59

This is regarding your SECOND QUESTION

" Here my second question:We do not need this SSB to form hadrons, right? There could be some theory were we have hadrons but not this SSB? "

Such a solution has indeed been suggested recently in: https://www.researchgate.net/profile/Afsar_Abbas/publications and see article: Gravity, Anomaly Cancellation, Anomaly Matching, and the Nucleuse

What is called the BIZ solution therein is exactly a non-SSB solution based entirely on anomaly cancellations and where gravity plays a highly non-trival role at low energy physics.


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