# Why isn't the Goldstone boson on the Standard Model map?

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## Summary:

I cannot seem to find any information on these particles other than they are bosons and they may have something to do with the Higgs field or weak iso spin, but they don't appear on the Standard Model.

## Main Question or Discussion Point

They seem to have something to do with Z or W particles, maybe, and they have something to do with mediating weak iso spin, I think, and they may be related to the Higgs field. But they don't have a box dedicated to them on the Standard Model of particle physics. Can someone tell me why this is, and also what the hell are they. Are they just theoretical, or have they been superseded. There appears to be a dearth of information about them for the lay person. I mean if they are actual real bosons, shouldn't they be listed? I'm a bit confused about this and some enlightenment would be appreciated.

Thanks.

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Demystifier
Gold Member
Goldstone boson is not one particular kind of boson, that's why it's not on the list of particles. It is a large class of bosons defined by a certain type of theory that describes them.

atyy
Summary:: I cannot seem to find any information on these particles other than they are bosons and they may have something to do with the Higgs field or weak iso spin, but they don't appear on the Standard Model.
As @Demystifier says, Goldstone bosons are a general term and apply in other areas of physics. In the historical context of the standard model and the Higgs, Goldstone bosons were not wanted as they are not observed in nature as elementary particles, but it was not clear how to construct a theory in which some particles have mass without also having a Goldstone boson in the theory. The Higgs mechanism enables us to construct such a theory that matches observation.
http://www.scholarpedia.org/article/Englert-Brout-Higgs-Guralnik-Hagen-Kibble_mechanism

vanhees71
vanhees71
Gold Member
2019 Award
There are no elementary Goldstone bosons in nature. That's why they are not given in tables of elementary constituents which are leptons, quarks, gauge bosons, and the Higgs boson in the Standard Model. It's likely that this is however not the complete zoo as there's a lot of evidence that there is "dark matter" not yet identified as measured particles but only seen by its gravitational effects on stars in galaxies and as part of the cosmological standard model to fit the high-precision data on the cosmic microwave background.

A Goldstone boson always occurs if you have spontaneous symmetry breaking of a global symmetry, which implies that the ground state of the theory is degenerate and does not obey the symmetry that governs the Hamiltonian and thus the dynamics of the system.

One example realized in nature is only approximate but very important: The strong interaction on the fundamental level is described by Quantum Chromodynamics, which however can be treated with perturbative methods only for collisions at very high energies, where the "running coupling" is small (aysmptotic freedom). In the low-energy region, you cannot even observe the fundamental constituents, the quarks and gluons due to confinement, i.e., the asymptotic free states (which we see as particles) are color-charge neutral bound states, the socalled hadrons. Most of them are either mesons (a quark-antiquark boundstate) and baryons (three-quark bound states).

Fortunately in the light-quark sector (up and down quarks) QCD has an approximate chiral symmetry, because the quark masses of a few MeV are smaller than the typical hadronic masses like the proton mass, so that you can approximately treat these quarks as massless. In this limit QCD has a chiral ##\mathrm{SU}(2)_{\text{L}} \times \mathrm{SU}(2)_{\text{R}}## symmetry, which would imply (if realized in the usual "Wigner-Weyl mode") that for each hadron there should be another hadron of the same mass but opposite parity. That's however not observed in nature. The reason is that the strong interaction is so strong that the vacuum (i.e., the ground state of QCD) is not the naive "perturbative" vacuum state but a state, where a socalled quark condensate forms, i.e., ##\langle \bar{\psi} \psi \rangle \neq 0##, where ##\psi## are the quark fields. Now this quark condensate is not invariant under chiral transformations but only under the socalled isovector transformations. This means the chiral symmetry is broken to ##\mathrm{SU}(2)_{\text{V}}##. Now QFT implies that then there must as many massless Goldstone bosons (or better Nambu-Goldstone bosons) as the dimension of the symmetry groups. The chiral symmetry is a 6-dimensional symmetry group, the subgroup which is a symmetry of the ground state is 3 dimensional, which implies that there are 3 massless modes, and these are in our context identified with the pions.

Now the pions are not really massless but have a mass of about 140 MeV. That's due to the fact that the u- and d-quarks are not really massless. So in addition to the above described spontaneous breaking of chiral symmetry it's also slightly explicitly broken, but this you can consider as a correction you can imply by perturbation theory. So the pion acquires its mass due to this explicit breaking. In this sense it's approximately the Goldstone bosons of the approximate chiral symmetry of the light-quark sector of QCD. This is very important, because it tells us how to build effective quantum field theories for hadrons, i.e., the low-energy sector of QCD with the observable hadronic degrees of freedom described by quantum fields.

atyy, dextercioby and haushofer