# What is spontaneous symmetry breaking in QUANTUM GAUGE systems?

Wen's question What is spontaneous symmetry breaking in QUANTUM systems? is cute, but here's an even cuter question. What is spontaneous symmetry breaking in QUANTUM GAUGE systems?

There are some gauge models where the Higgs phase and the confinement phase are continuously connected, you know. Why are the Higgs phase and the confinement phase identical in Yang-Mills-Higgs systems? You mention cluster decomposition? Sorry, because bilinear operators have to be connected by a Wilson line.

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Since gauge symmetry is not a symmetry in quantum theory, there is no spontaneous symmetry breaking of QUANTUM GAUGE symmetry.

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You can still break the global part. The gauge transformations on the boundary are physical, although if you only do a gauge transformation on the interior, it's unphysical, just a redundancy. – Ron Maimon Jun 23 '12 at 8:17
@Ron: The gauge symmetry is not a symmetry in quantum theory since both the local and the global gauge transformations are do-nothing transformations when act on quantum states. The quantum states simply do not change under those transformations, regardless if the system has boundary or not. So even the global part on the boundary is a do-nothing transformation and cannot break. – Xiao-Gang Wen Jun 23 '12 at 12:48
@Xiao-Gan-Wen: This is true of the local symmetry, but not the global symmetry. You can see this because the Noether prescription for a global phase rotation in a U(1) symmetric theory still gives a nontrivial conserved current. This is true of energy momentum in gravity, of supercurrents in supergravity, of every gauge current. Further, the global phase rotations are physically inequivalent--- you can translate a system relative to the boundaries, even if translations are gauged. – Ron Maimon Jun 23 '12 at 20:00
I absolutely agree with @RonMaimon. The true symmetry group is the global group that is equivalent to the gauge group $G$ modulo $G_*$, where $G_*$ is the redundancy group, that is, the gauge group whose elements go to the identity in the boundaries. – Diego Mazón Jul 26 '12 at 23:58
@drake: In my language for the situation that you described, $G_*$ is the gauge symmetry which cannot break. $G/G_*$ is the physical symmetry which can break. – Xiao-Gang Wen May 7 '13 at 11:53

It's all about RELATIVE symmetry breaking in QUANTUM GAUGE theories. RELATIVE to the Higgs field direction, symmetry is broken. This is a novel concept of RELATIVE symmetry breaking. A spin-0 helium atom quantum state might not break rotational symmetry, but RELATIVE to the direction of one of the electrons relative to the nucleus, the rotational symmetry for the other electron is RELATIVELY broken.

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It's only RELATIVE to the boundary value of the Higgs field. You can still arbitrarily phase rotate the Higgs direction in the interior. – Ron Maimon Jul 23 '12 at 7:50

Spontaneous breaking of a gauge symmetry is breaking the global symmetry. All gauge symmetries are not symmetries in the local part, if you do a local gauge transformation, you have the same exact state, But the global part of the gauge symmetry is physical, and leads to a Noether current. when you break the gauge symmetry, you are breaking the global part.

For example, in quantum electrodynamics, if you perform a phase rotation of the electron field by a constant factor $e^{i\theta}$, you do nothing to the vector potential. This transformation produces a nontrivial Noether current, it is the electric current. You can break this global symmetry with no problem, just using a charged condensate. This is the Higgs mechanism for U(1), or superconductivity.

A completely local version would be to do a local phase rotation by $\theta(x)$ where $\theta$ is a function which is nonzero only in a compact region. This transformation adds a gradient to A, and now if you do Noether's prescription, you find a conserved charge which is identically zero.

For gauge theories, the Noether currents derived from gauge tranformations are of the kind that they reduce to boundary integrals. In any theory, Noether currents are such that the charge can be evaluated at the initial time surface and the final surface, and you get the same answer. In a gauge theory, these charges can be evaluated on a large sphere on the initial surface and the final surface, using only the gauge field. In electrodynamics, this is Gauss's law. The Noether charge becomes a boundary object in a gauge theory, which is determined by the asymptotic values of the fields.

This is discussed to some extent on the Wikipedia page on infraparticle.

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Gauge symmetries are redundancies of the description, not a part of physics; gauge fields have surplus structures (e.g., non-physical polarizations) one brings in to describe the system more conveniently (for instance in a local and manifestly Lorentz covariant form). You can describe a gauge system in a language that does not have gauge symmetry at all. A famous example is AdS/CFT: you can describe an SU(N) gauge theory at large N by a theory of gravity; there is no SU(N) on the gravity side. Note that the global part of a gauge symmetry is as non-physical as its local part. It is however the case that ordinary gauge field theories (like QCD; not like GR) do have, aside from the non-physical gauge symmetry, a physical global symmetry which yields charge conservation. This physical global symmetry looks exactly like the global part of the (non-physical) gauge transformation, but one should differentiate between the two: gauge transformations are just like changing coordinates, but the physical global transformation involves changing the dynamical degrees of freedom. (In a similar situation in GR one should differentiate between the non-physical diff invariance and physical symmetries due to the existence of Killing vectors. The global part of the diff group is not only poorly defined, it is as non-physical as its local part. It is the isometries of the spacetime that are of physical significance. For example the existence of a conserved energy is due to the existence of a time-like Killing vector, not the meaningless "global part of the diff group".) (The case of non-abelian gauge theories with topological sectors has one further subtlety: your physical degrees of freedom are the equivalence classes of gauge field configurations that are continuously deformable to each other within each class.)

Gauge symmetry being non-physical, gauge symmetry breaking is also not a matter of physics, but a matter of changing the description. In the abelian Higgs model (which resembles superconductivity) for instance, if one changes their description from A) 2 components of scalar Higgs + 2 polarizations of the photon, to B) 1 component of Higgs + (1 component of Higgs + 2 components of the photon), one is breaking (or sacrificing) the symmetry in the initial description, to more clearly see the mass spectrum of the theory.

However, the "Higgs mechanism" is not non-physical as the above paragraph might initially suggest. It is physical in the following sense. Abelian gauge theories have three well-known phases: Landau phase, Coulomb phase, and the Higgs phase. Massless electrodynamics is an example that is always in the Landau phase (with logarithmic charge fall-off). Massive electrodynamics has a Landau regime for distances less than $m_{e}^{-1}$ and a Coulomb regime (with $1/r$ potential) for large distances. The abelian Higgs model has a Coulomb phase (with $m_{v}=0$) if the Higgs potential is minimum at the origin, and a Higgs phase (with $m_{v}\neq 0$) if the Higgs potential is minimum away from the origin. It is due to this dynamical phase transition that the change of description comes useful to understand the mass spectrum. (Look at John Preskill's note http://www.theory.caltech.edu/~preskill/ph230/notes2000/230Lectures27-29-Page347-402.pdf)

The important point to remember is: In the phenomena commonly referred to as spontaneous breaking of gauge symmetries, no physical (real) symmetry breaks. Physically, no global or local symmetry gets broken, only a phase transition happens that one can track by the mass of the vector boson as the order parameter. In the previous sentence, by local symmetry I mean some physical local symmetry (as the word "Physically" at the beginning of the sentence shows), like conformal symmetry in 2D; I'm not even thinking about non-physical symmetries like gauge symmetry, because they do not have anything to do with the nature.

I have limited the discussion to the abelian Higgs model, but the essence of the argument is the same for non-abelian gauge fields as well. If this answer has not been illuminating, I recommend reading the aforementioned notes by John Preskill.

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