Studying Higgs mechanism in EW theory and QCD I have a couple of questions that I would like to clarify:

1) The quark mass term in QCD Lagrangian should come from Higgs mechanism in EW sector of SM. I mean that you don't have a contribution to the mass coming from Higgs and other from a Dirac term that only satisfies $SU(3)_c$ symmetry but not $SU(2)_L$. Is this correct?

2) Let's imagine a world with a Higgs such that its vacuum expected value is zero. Then, Higgs mechanism does not break $SU(2)_L$ symmetry and, hence, leptons and quarks keep massless. In this world, with quarks but massless we could have hadrons but their masses would come out from QCD gluonic interactions among themselves and the gluonic and EW interactions of the virtual sea quarks, that of course are massless per se. We'd lose contribution from the mass given by Higgs mechanism (which is massless too) but not the resulting from electromagnetic interactions between valence and virtual quarks.

But if quarks and leptons are massless, due to Special Relativity, they move at the speed of light so, is this a problem for bounding to create hadrons? Moreover, the massless property of quarks would allow proton to decay into neutron, so this world would be lifeless.

Despite all these, there is no way to get a mass for gauge bosons or charged leptons since we keep unchanged the SM symmetry, or am I ignoring some fancy way?

Maybe, since quarks are now energetically equivalent in this world CKM matrix is matrix of ones up to a complex phase to keep the CP violation.

What else do you think could be different from our real world?

  • $\begingroup$ Frank Wilczek discusses hadron mass ignoring the Higgs mechanism in one or two of his Core articles, but I don't remember how he deals with the problem of massless particles moving at $c$. $\endgroup$
    – PM 2Ring
    May 2, 2019 at 8:10
  • 1
    $\begingroup$ The quark masses (along with other fermion masses) don't directly come from the Higgs mechanism; they come from a technically-separate Yukawa coupling to the Higgs boson. The Higgs mechanism only directly generates the mass of the electroweak bosons and the Higgs boson itself. I'm also not sure that you can say that the difference in quark masses is responsible for the stability of the proton relative to the neutron. The mass of the valence quarks contributes only a tiny fraction of the hadron mass; the vast majority is binding energy that comes from non-perturbative QCD effects. $\endgroup$ May 2, 2019 at 12:08
  • $\begingroup$ @probably_someone My reasoning is this: QCD effects are flavour independent so the part coming from QCD is the same in proton and neutron. Therefore, the difference of mass between them comes from the proper mass of quarks and electromagnetics effects. Don't you think this is correct? $\endgroup$
    – Vicky
    May 3, 2019 at 2:53
  • $\begingroup$ But of course Higgs mechanism generates the mass of quarks and leptons, those couplings that you refers are among Higgs bosons and these particles so since Higgs' VEV is non-zero, you have mass for quarks and leptons. Otherwise, the inclusion of this masses violates $SU(2)_L$ $\endgroup$
    – Vicky
    May 3, 2019 at 2:55

2 Answers 2


In the absence of the Higgs field, QCD strong interaction can generate masses for quarks as well via quark-antiquark condensation that breaks the chiral symmetry, $$ \langle \bar{q}q\rangle \sim \int \frac{1}{\not p - m} = \int \frac{m}{p^2 - m^2}, $$ where $\langle \bar{q}q\rangle \neq 0$ only if the dynamically generated effective mass $m \neq 0$. A side note: The above bears a resemblance to the gap equation in the BCS superconductivity theory, where the condensation is the indication of forming cooper pairs.

Mesons are the resultant Nambu-Goldstone bosons. Of course, if non-zero Higgs VEV is present, the chiral symmetry is NOT exact, which renders the mesons Pseudo-Nambu-Goldstone bosons.

Historically, Technicolor (and some extended variations) is an ambitious and failed attempt to get rid of the Higgs mechanism and replace it with a QCD-on-steroids scheme.

  • $\begingroup$ There is something that I don't understand about that condensate. It's said that if $\langle \bar{q}q\rangle \neq 0$ then you have mass generation. How is that? And other thing: those mesons that are pseudo-Goldstone bosons have actually mass due to the symmetry that arises from considering quarks as massless is not exact due to they have masses. But if we are in a universe where quarks are indeed massless, not as an appoximeted fact, but as an exact one, mesons would be massless for being real Goldstone bosons. $\endgroup$
    – Vicky
    May 3, 2019 at 3:00
  • $\begingroup$ @Vicky, please see the updates. $\endgroup$
    – MadMax
    May 3, 2019 at 14:40
  • $\begingroup$ But what I'm proposing is a universe where we have an actual Higgs such as its VEV is equal to zero, exacty, Then, breaking chiral symmetry you produce mesons as real Goldstone bosons, so they are massless. Then you can't even produce mass by chiral symmetry breaking because chiral symmetry is not approximate in the universe I'm proposing $\endgroup$
    – Vicky
    May 3, 2019 at 20:35
  • $\begingroup$ @Vicky, you are apparently confused. Zero mass applies to the meson as Goldstone boson, while dynamical chiral symmetry breaking produces NON-ZERO mass for fermions such as quarks in your hypothetical no-Higgs universe. $\endgroup$
    – MadMax
    May 6, 2019 at 15:36
  • $\begingroup$ Ok, I'm going to write another post focusing on this issue becuase I don't even know what you mean by 'dynamical' chiral symmetry breaking. And thanks $\endgroup$
    – Vicky
    May 6, 2019 at 20:53

These questions have been addressed in the Big Bang model, for the time before symmetry breaking and the Higgs field has zero vev.


Note that quark confinement into hadrons comes after weak symmetry breaking. Everything is different from our present world, before symmetry breaking time, of $10^{-12}$ sec from the Big Bang.


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