Some sources state that when the mass of a quark goes to zero, it allows for Spontaneous Breaking of Chiral Symmetry and gets a constituent mass of about $200\, \mathrm{MeV}$.

Other sources state that when the masses of the light quarks go to zero, so does the pion mass. In this case, the explicit breaking of chiral symmetry disappears, but the spontaneous chiral symmetry breaking is triggered too.

So, in this limit, is the pion massless but their quarks carry a $200\, \mathrm{MeV}$ mass? What is wrong, or misunderstood, here?


2 Answers 2


You are understanding correctly. In the massless up/down quark limit, chiral symmetry is restored, and the pion becomes massless but quarks are still confined, and baryons have about the same mass as they do now. This is exactly why the idea that the pion is made of quarks is nonsense.

In the 1980s, many in the new generation sought to undo the progress of the 1960s, and willfully ignored the revolutionary work of Nambu, Sakurai, Skyrme, and others, dismissing it as pre-quark nonsense. They decided that a pion is made up of two nonrelativistic quark-objects, they called these objects "constituent quarks", and they made up force laws for these to reproduce the Hadron spectrum. Georgi and Glashow even went so far as to invent a quark-quark coupling force which was designed to lower the mass of the pion by interquark interactions!

This work is a little embarassing to read. The proper model of the pion was the much earlier one due to Nambu and Weinberg, and this is now verified thanks to numerical lattice QCD, where the mass of the quark can be tuned at will. When you tune the mass of the quarks to zero, the pion mass vanishes according to the laws of chiral peturbation theory.

The pion is a mode of oscillation of the quark chiral condensate, a material filling all of space. It is made out of quarks which are created by the independent fluctuations of the gluon field.

The gluon field completely randomizes on a Baryon scale, meaning that a quark going in a closed path larger than a proton circumference will get a completely random pick from SU(3) as its holonomy. A random gauge field will create large numbers of objects whose mass scale is much lower than this randomization scale, and in this case, the objects it creates are the light up and down quarks, and to a lesser extend strange quarks. These quarks condense in pairs in the vacuum, making a condensate whose order parameter is much like a mass term in the Dirac equation: $m \bar\psi \psi$. This condensate is not invariant under rotations of the left and right-handed quarks into each other, but the Lagrangian is (more or less, except for the negligible quark mass).

The Goldstone modes of the broken symmetry are waves in this condensate, and these are the pions. The goldstone mode is due to oscillations where the left and right part of the condensate slosh in phase in opposite directions, and these are collective excitations of quarks. The pion is made of quarks to the same extent that a sound wave is made of atoms.

That the pions are Goldstone bosons was not only theoretically predicted by Nambu, it explains their strange derivative couplings at low energy, and this was spectacularly extended to a full theory by Weinberg's soft-pion theorems, and chiral perturbation theory. The condensates were further used to give nonperturbative corrections to QCD particle propagation at intermediate distances in the Shifman-Vainshtein-Zakharov sum rules. So really, everyone should have known better than constituent quarks.

It is not clear that the notion of "constituent quark" actually has any form of real meaning, or whether it is just a figment of the imagination. The only partial evidence in it's favor that I think is not easy to explain in other way is that the total cross sections for pions are about 2/3 the total cross section for protons, as if the pomeron hits 2 quarks instead of three. I don't know if this approximate equality is not just a coincidence.

  • $\begingroup$ The order parameter of the condensate, is it null when the mass of the quark is zero? Has this "mass term" got a name? $\endgroup$
    – arivero
    Nov 19, 2011 at 11:10
  • 2
    $\begingroup$ @arivero: I see--- sorry for using a confusing notation--- no, the order parameter is not proportional to the quark mass, it is just an operator with the same form as the Dirac mass, which is clearly not chirally invariant (it fixes the phase between the two chiralities). The condensate order parameter asymptotes to a finite limit at zero quark mass, and the up and down quarks are light enough to be close to this limit. $\endgroup$
    – Ron Maimon
    Nov 19, 2011 at 20:32
  • $\begingroup$ Greetings: In your second paragraph you write: " they called these objects "constituent quarks", and they made up force laws for these to reproduce the Hadron spectrum. Georgi and Glashow even went so far as to invent a quark-quark coupling force which was designed to lower the mass of the pion by interquark interactions!" . Are there any references that criticize this approach? Thanks. $\endgroup$ Apr 9, 2017 at 16:14

Wow that theory stuff is all just amazing. Intuitively can we say a pion needs only one gluon to bind q anti-q pair whereas a baryon would have minimum of three gluons, one between each quark pair. Plus these 3 gluons have interactions amongst themselves bringing in another 3 gluons or so, whereas the pion gluon is alone so doesn’t generate more. This suggests pion has 1 sixth the mass of the nucleon from the basic gluon field not including sea nor constituent quarks which come in around 10 % uma each making only an extra bit for the 3q baryon v 2q pion.


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