mass of the proton with massless quarks?

arivero

A usual lore from chiral perturbation theory is that the mass of the pion is proportional to the sum of the up and down masses, and then it is going to be zero when such masses are zero.

Now, for the proton, I notice the following remark from Chris Quigg
Why is it different of the pion?

Einj

I am not really sure, but I think the point is that the pion is a quasi-Goldstone boson for the chiral symmetry. It means that in the limit of exactly broken chiral symmetry (i.e. the mass quark vanishes) it should be completely massless. On the other hand, in this situation the nucleons (and so the protons) acquire a mass.
Again, I am not really sure.

arivero

Perhaps the question is, why the chiral expansion of the nucleon has a constant term while the pion hasn't?

It is a little bit as the expansions of cos(x) and sin(x), but in this later case we know that one of the expansions must be even and the other must be odd, so it is crystal-clear.

The_Duck

There's no reason for the nucleon to be massless at zero quark mass. In general, we should expect hadrons to have masses of order the characteristic scale of QCD; call it ~1 GeV.

The thing that needs explaining is why the pion is massless at zero quark mass. That happens because of chiral symmetry and Goldstone's theorem, as Einj said.

arivero

This also puzzles me... What happens here in chiral symmetry breaking is that chiral SU(2)RxSU(2)L breaks down spontaneusly via quark condensates, a condensation which should happen for any quark mass smaller than one hundred MeV, and then the pion mass should be zero if the surviving SU(2) vector part, aka Isospin, were exact. But this is true always that the mass of up is equal to the mass of down, so the mass of pion should depend of the mass difference between up and down, not of the mass average.

Of course chiral symmetry is also explicitly broken because of the quark masses, but I fail to see how this mechanism compete with the condensation.

The_Duck

Why do you say this? You get exactly massless particles when you spontaneously break an exact symmetry--that is, the pions should be massless only if the original *axial* symmetry was exact.

arivero

Well, but condensation always happen, so there is always an spontaneus breaking; it is only that we are spontaneusly breaking an approximate symmetry, and I wonder how much of this approximation is hidden under the carpet of the breaking scale. What I was thinking is, there are two sources of failure in the masslessness of the pion:

- First, the up and down are not massless. But they are light respect to the QCD chiral scale, which is about 100 MeV.

- Second, the up and down have not the same mass. So the SU(2)_V symmetri is approximate too.

I was thinking which could be the relative contribution of each source to the mass of the pion, and wondering if the second one could be relevant too, or even more relevant.

For instance, imagine the up is massless. Then, should we have an exact chiral U(1)L x U(1)R and a massless neutral pion, with massive charged pions due to the breaking of SU(2)_V?

tom.stoer

As said the mass scale of the nucleon is rather natural (~ 1GeV) whereas the nearly massless pions are explained via the Goldstone mechanism. It is interesting to see what happens w/o spontaneous chiral symmetry breaking. So let's look at the eta prime meson (η') meson which is the flavor-singulet of the SU(3) generated by Isospin and Strangeness.

The eta meson is a Goldstone boson with mass 548 MeV (rather large compared to pions Due to the mass of the strange quark) whereas the eta prime is NOT a Goldstone boson b/c the singulet U(1) symmetry is not broken via the Goldstone mechanism but via the axial anomaly. Therefore the eta prime has a mass of 958 MeV which is rather close to the nucleon mass.

arivero

Ok, it seems that I was mixing the pseudoscalars such as the pion with the scalar from Vafa-Witten theorem. See eg 9.4 of hep-ph/9911532v2

[tex]m_\chi^2 f_\chi^2= (m_d-m_u) (\langle \bar \psi_u \psi_u \rangle -\langle \bar \psi_d \psi_d \rangle )[/tex]