Pion mass

1. Sep 14, 2015

arivero

Thinking aloud.

Most descriptions of chiral symmetry breaking nowadays present it as something happening in QCD. But it was defined well before of the quark theory, and then it was something related to isospin symmetry.

It is a bit puzzling because it seems as if pion mass were originated from two different forces. On one hand, nuclear weak force breaks the symmetry between protons and neutrons and induces the mass of the pion. On other hand, nonzero quark masses break the symmetry between flavours and again the pion gets mass.

In both scenarios the idea is that the pion should be massless because of the spontaneous symmetry breaking but it becomes massive because of the explicit symmetry breaking. Disturbing thing is that a pair of massive quarks up, down always break explicitly chiral symmetry but do not need to break, if they are equal mass, nuclear isospin symmetry.

Last edited: Sep 14, 2015
2. Sep 14, 2015

fzero

I think you are a bit confused about the difference between chiral symmetry and isospin. Chiral symmetry is $SU(2)_R\times SU(2)_L$ where the factors act independently on massless right and left-handed quarks. If we give the quarks equal but nonzero mass, then the chiral symmetry is broken to its diagonal subgroup which can be identified with the $SU(2)_I$ isospin. If we give the quarks different masses then we only preserve the $U(1)_B$ baryon number. I don't know why you think that any of this is disturbing.

On energy scales alone we don't expect to explain the pion masses from the between the u and d Lagrangian masses (factor of ~30) alone, but they are within a factor of 2 or so of the QCD scale. However the splitting between the charged and neutral pions is small, so it can be related to the u-d mass splitting.

Maybe you mean to say that, that chiral symmetry is already broken along with the isospin and flavor symmetries by the Higgs mechanism. However, at lower scales the QCD vacuum generates further chiral symmetry breaking effects and even larger effective contributions to hadron masses. Then we have an effective regime where we can expand in $\Delta m_\pi / m_\pi$, and the lowest order of this expansion is chiral EFT.

3. Sep 14, 2015

arivero

Forget the quark model. Before the quark model, people was already telling that the pion was a Goldstone boson and that it had a small mass relative to proton because it was not an exact symmetry. Which was the argument, then? Was still chiral symmetry, or just isospin?

4. Sep 14, 2015

fzero

The splitting between the pion and the proton is explained by chiral symmetry breaking. It has (almost) nothing to do with the splitting between the u and d masses, just the fact that they are nonzero. However, there are improved chiral Lagrangians that include u-d splitting at higher-orders. So these models would presumably estimate the splitting between the charged and neutral pions.

The pattern of symmetries is also the same for the chiral Lagrangians, though there the chiral symmetries $G$ are represented on a matrix $U \rightarrow L U R^\dagger$ and the preserved symmetry$H$ is still the diagonal subgroup.

5. Sep 14, 2015

arivero

I was not going so far, at the moment. To start with, I wonder how do I argue about chiral symmetry without mentioning "nonzero quark mass". Without mentioning quarks, even. I guess that it should be done by looking at weak nuclear force.

Last edited: Sep 14, 2015
6. Sep 14, 2015

fzero

You don't really talk about chiral symmetry without mentioning quarks. You might refresh your memory about the linear $\sigma$-model (see sect 5.4 here for instance) where the chiral symmetry on the matrix valued field $U$ above is tied to the Yukawa-coupling between some quarks $\psi$ and the pions $\vec{\pi}$: $g \bar{\psi} ( \sigma + i \vec{\pi} \cdot \vec{\tau} \gamma_5 ) \psi$. The second term is invariant under the chiral symmetry on $U$ if the quark field also transforms in a chiral way under the same symmetry group. However, the first term is not chiral invariant, since $\sigma$ is a singlet ($\langle \sigma \rangle$ is the quark mass). The theory proceeds by studying the internal dynamics of the meson effective theory, so it seems like we are "forgetting" the quarks perhaps, but we are really not.

There is no weak interaction in any of this.

7. Sep 14, 2015

arivero

Well, I mean, the $g_A , g_P$ of eq 5.49.

8. Sep 14, 2015

Avodyne

In the material you quote, $\psi$ represents the proton and neutron, not the up and down quarks. See the text immediately after eq.(5.56).

And this is the answer to the OP's question. Before quarks, chiral symmetry was represented in terms of left- and right-handed nucleon (proton and neutron) fields. The (equal) proton and neutron masses were due to the expectation value of the $\sigma$ field.

9. Sep 14, 2015

fzero

Yes, I spaced out for a bit there, even after having read that explicitly. Indeed I wanted to clarify the connection between the action on scalars with the action on chiral fermions.

10. Sep 14, 2015

arivero

Great! that is the idea I was missing... chiral nucleons! I still need do some rumination of this thing before eating it. I still need to see which is the explicit symmetry breaking -isospin breaking is obvious answer, but I do not feel it completely satisfactory yet- and which is the spontaneous symmetry breaking... We can not make a "nucleon condensate", do we? Or is that the magic the sigma-model does?

11. Sep 14, 2015

fzero

A vev for the $\sigma$ field will break chiral symmetry to isospin. Something like a pion condensate $\langle \pi^0\rangle \neq 0$ would break isospin to baryon number. I'm not sure that that's exactly how it works but there's a short discussion of how to break isospin in the sigma model at higher orders around eq (24) of http://www.scholarpedia.org/article/Chiral_perturbation_theory#Higher_orders

12. Sep 15, 2015

Avodyne

Isospin is explicitly broken by the different up and down quark masses.

Before quarks, it was understood that isospin was explicitly broken, and that the effect was small, but the source of the breaking was not known.

Last edited: Sep 15, 2015
13. Sep 15, 2015

arivero

He claims a lead role in the process to shift away from weak interactions:

And he points out some difference between pionic and quark versions of the chiral breaking

14. Sep 15, 2015

Avodyne

This is not a "difference between pionic and quark versions of the chiral breaking". Weinberg is explaining how the up-down quark mass difference is expressed in the chiral lagrangian for pions.

15. Sep 15, 2015

arivero

Yep, bad expression on my side. Still it is interesting because from the pion point of view one can get the impression that if p and n keep having the same mass (as should happen if mass of up and down are equal, up to electromagnetic corrections) then isospin is not explicitly broken and pion should still be massless.

16. Sep 15, 2015

Avodyne

This is wrong. You need to read Weinberg more carefully.

17. Sep 17, 2015

vanhees71

There's nothing in chiral symmetry suggesting the proton and pion mass should be degenerate. Pions are pseudo scalar particles. So the chiral partner should be a scalar particle, and indeed there's the $\sigma$ meson that has a higher mass of about 400-500 MeV. This mass difference is mostly due to the spontaneous breaking. The chiral partner of the proton is some $N^*$ resonance, and these are also not mass-degenerate with the proton, which is also mostly due to the spontaneous breaking.