Symmetry Unbreaking

1. Jan 28, 2016

jlcd

Symmetry breaking separates the electromagnetic and weak force from the electroweak force. Is there an opposite procedure or symmetry unbreaking in which the em and weak force can be made to combine?

What other forces or examples in physics where you can do reverse symmetry breaking or symmetry unbreaking?

2. Jan 28, 2016

Simon Bridge

The opposite of symmetry breaking is establishing symmetry.
An example of re-establishing a broken symmetry is when you stand a column back on it's end after it has fallen over or when you mould a clay blob back into a sphere after it has been deformed.

3. Jan 29, 2016

vanhees71

What you seem to be talking about is "spontaneous symmetry breaking". Strictly speaking a local gauge symmetry cannot be spontaneously broken, but usually the Higgs mechanism is presented as such a spontaneous symmetry breaking.

Of course, I cannot explain this in full detail in such a forum posting, and it's not B level, but some questions simply can't be answered on B level! I'll keep it qualitative.

First let's understand true spontaneous symmetry breaking. This occurs for global symmetries. An important example is the (approximate) chiral symmetry of the light-quark sector of QCD, which is spontaneously broken to the (as well only approximate) isospin symmetry (let's restrict ourselves to the two lightest quarks u and d). If chiral symmetry were realized in the usual way ("Wigner-Weyl mode"), the ground state of the theory (vacuum) were invariant under the full chiral symmetry group, and this would imply that for any hadron there's a chiral-partner hadron of opposite parity with the same mass (in the chiral limit, i.e., current-quark masses set to 0). This is not observed in nature. The reason is that the strong interaction is attractive in the quark-antiquark system, which leads to a non-vanishing chiral condensate, i.e., the vacuum expectation value $\langle \overline{q} q \rangle \neq 0$. This is invariant under isospin transformations but not under chiral transformations. Thus the symmetry group of the vacuum is reduced to a subgroup. Nevertheless the Hamiltonian of the theory is of course still symmetric under the full group, and this implies that the vacuum (ground state) is degenerate, i.e., there are different eigenstates to the same lowest energy eigenvalue. The observable consequence is that there are excitations of the vacuum that do not cost any energy. Quantum field theory leads to the conclusion that there are always massless scalar or pseudoscalar field excitations (i.e., massless scalar or pseudoscalar particles). Since in our case the chiral transformations are spontaneously broken, the excitations are pseudoscalars and in this specific case it must be three distinct pseudoscalar particles. These are the pions, which come indeed as the charged pions $\pi^{\pm}$ and the neutral pion $\pi^0$, i.e., as three different particles. This is known as the Nambu-Goldstone theorem and the corresponding particles are the Nambu-Goldstone bosons of the spontaneously broken symmetry. Since the up- and down-quark masses are not exactly 0, the pions have a small mass of around 140 MeV, but this is small compared to the typical hadronic scale of 1 GeV and thus can be treated as a perturbation. Chiral perturbation theory is one of the most important techniques to build low-energy effective theories compatible with Quantum Chromodynamics (QCD) to describe hadrons at not too high energies.

Of course, as your question implies, it's well thinkable that a spontaneously broken symmetry can get restored. This is indeed the case, and from the above explanation it's clear that this somehow means that you should try to "melt" the quark condensate. This is indeed a question of high interest in the heavy-ion community. The idea is that if you bring hadrons in a very hot and dense state, they start to overlap and vigorously collide with each other. Now QCD is an asymptotically free theory, i.e., the interaction should get weaker the more energetic the collisions become. That lead to the idea that the hadrons at a high enough temperature and/or density may dissolve into a soup of quasifreely moving quarks and gluons, the socalled quark-gluon plasma, and this implies that the chiral symmetry should become restored and the particle spectrum should become degenerate, i.e., chiral partner excitations should show the same mass spectrum if you only make the temperature and/or density of the soup of strongly interacting particles large enough. The only way we can try to do so in our laboratories is to smash large atomic nuclei together, and that's done at various labs, among them the Relativistic Heavy Ion Collider (RHIC) at the Brookhaven National Lab in Upton, NY and at the Large Hadron Collider at CERN in Geneva. The result of these heavy-ion programs so far is that very likely, one indeed produces such a Quark Gluon Plasma for a very short moment (the rapidly expanding fireballs live for about several 10 fm/c), and that chiral symmetry might be restored, although there is not a clear final empirical evidence for it, because we can only measure the spectral properties of vector mesons pretty accurately by studying the emission of dileptons (i.e., electron-positron and muon-antimuon pairs) in the heavy-ion collisions. All data indicate that the light vector mesons, $\rho$, $\omega$, and $\phi$, indeed become very broad in the dense and hot medium and the corresponding production rates for dileptons merge into the expectation from emission from a QGP (where the main production mechanism is the quark-annihilation process $q + \overline{q} \rightarrow \mathrm{e}^+ + \mathrm{e}^-$. Unfortunately one cannot easily measure the spectral properties of the axial-vector meson partners to directly prove empirically the degeneracy of their spectra at high temperatures and densities, which would be the clearest proof for chiral-symmetry restoration.

Now to the electroweak standard model: Here you have something very similar to spontaneous symmetry breaking, known as the Anderson-Higgs mechanism. Here, the symmetry is realized as a local symmetry, implying the existence of socalled gauge bosons. At first glance, also here the ground state seems to be degenerate, but that's not really the case, because to connect the putative different vacua you can use local gauge transformations, but the math of the theory tells you that anything connected by a local gauge transformation describes precisely the same physics as the original description. This is very welcome and was the breakthrough in electroweak theory, because it implies that in this case there are no massless Nambu-Goldstone modes in the particle spectrum, but these field-degrees of freedom get (choosing a particular gauge) part of the gauge bosons, which become massive. This is exactly what's needed for electroweak phenomenology. The "pseudo-spontaneous-breaking" in the Standard model is indeed such that you get three massive gauge bosons (the $W^{\pm}$ and $Z^0$ bosons) and a massless gauge boson (the photons).

Also here, the gauge symmetry should be restored at high temperatures/densities, but that's at much higher temperatures and densities than we can make in the lab. So the only time where this was realized in nature was shortly after the Big Bang.