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theorems:goldstones_theorem

# Goldstone's theorem

## Concrete

Long after phonons were understood, Jeffrey Goldstone started to think about broken symmetries and order parameters in the abstract. He found a rather general argument that, whenever a continuous symmetry (rotations, translations, SU(3), …) is broken, long–wavelength modulations in the symmetry direction should have low frequencies. The fact that the lowest energy state has a broken symmetry means that the system is stiff: modulating the order parameter will cost an energy rather like that in equation 2. In crystals, the broken translational order introduces a rigidity to shear deformations, and low frequency phonons (figure 8). In magnets, the broken rotational symmetry leads to a magnetic stiff- ness and spin waves (figure 9a). In nematic liquid crystals, the broken rotational symmetry introduces an orientational elastic stiffness (it pours, but resists bending!) and rotational waves (figure 9b). In superfluids, the broken gauge symmetry leads to a stiffness which results in the superfluidity. Superfluidity and superconductivity really aren’t any more amazing than the rigidity of solids. Isn’t it amazing that chairs are rigid? Push on a few atoms on one side, and 109 atoms away atoms will move in lock–step. In the same way, decreasing the flow in a superfluid must involve a cooperative change in a macroscopic number of atoms, and thus never happens spontaneously any more than two parts of the chair ever drift apart. The low–frequency Goldstone modes in superfluids are heat waves! (Don’t be jealous: liquid helium has rather cold heat waves.) This is often called second sound, but is really a periodic modulation of the temperature which passes through the material like sound does through a metal.

O.K., now we’re getting the idea. Just to round things out, what about superconductors? They’ve got a broken gauge symmetry, and have a stiffness to decays in the superconducting current. What is the low energy excitation? It doesn’t have one. But what about Goldstone’s theorem? Well, you know about physicists and theorems . . .

That’s actually quite unfair: Goldstone surely had conditions on his theorem which excluded superconductors. Actually, I believe Goldstone was studying superconductors when he came up with his theorem. It’s just that everybody forgot the extra conditions, and just remembered that you always got a low frequency mode when you broke a continuous symmetry. We of course understood all along why there isn’t a Goldstone mode for superconductors: it’s related to the Meissner effect. The high energy physicists forgot, though, and had to rediscover it for themselves. Now we all call the loophole in Goldstone’s theorem the Higgs mechanism, because (to be truthful) Higgs and his high–energy friends found a much simpler and more elegant explanation than we had.

(11In condensed-matter language, the Goldstone mode produces a chargedensity wave, whose electric fields are independent of wavelength. This gives it a finite frequency (the plasma frequency) even at long wavelength. In high-energy language the photon eats the Goldstone boson, and gains a mass. The Meissner effect is related to the gap in the order parameter fluctuations (~ times the plasma frequency), which the high-energy physicists call the mass of the Higgs boson.)

Goldstone’s theorem shows that the existence of an observable with a nonvanishing vacuum expectation value implies the existence of states whose energy goes to zero as the momentum does; that is, as E(p) r 0 p r 0. In relativistic field theory, this implies the existence of massless particles since . For an intuitive picture, imagine ap- 2 24 22 E p m c p c plying the operator corresponding to the broken symmetry to the vac- Qˆ uum state F0S. The result would be a distinct vacuum state, but with the same energy since commutes with the Hamiltonian. Now consider in- Qˆ stead the operator defined over some finite region V; the states Qˆ V should have the same energy as F0S except for boundary terms. But Qˆ VF0S since this operator implements a continuous symmetry, the region V can be smoothly deformed so that the boundary terms vanish as , which V r 0 implies that the energy of the state must go to zero for short wave- Qˆ VF0S lengths. To make this (slightly) more rigorous, consider an observable whose commutator with has a nonzero vacuum expectation value, A Q ˆ ˆ . 6 Rewriting as an integral of the charge ˆˆ ˆ limVr A0F[QV V , A]F0S p c ( 0 Q density, we have . Assuming that the cur- 3 0ˆ ˆ limVr ∫V d xA0F[ j (x), A]F0S p c rent is conserved, if the boundary terms vanish then this integral will be time invariant. Manipulation of this expression shows that massive particles would, however, lead to explicit time dependence. For the left-hand side to be nonzero, there must be states FnS such that , with ˆ0 A0Fj FnS ( 0 vanishing spatial momenta; these states are the massless Goldstone modes.7 http://www.jstor.org/stable/pdf/10.1086/518324.pdf

Examples

Landau phonons in Bose-Einstein condensates
"The Bose-Einstein condensation is characterized by the breaking of a global U(1) gauge group (acting on the Bose particle field as the U(1) group of Example 1), as very clearly displayed by the free Bose gas.5 The U(1) breaking leads to the existence of Goldstone modes, the so-called Landau phonons, and the existence of such excitations may in turn indicate the presence of a broken U(1) symmetry" Source

## Abstract

It was known from perturbative investigations of self-interacting scalar fields by Goldstone that the local current conservation may lead to a divergent global charge resulting from the contribution of a massless scalar (”Goldstone”) boson which impedes the large distance convergence and in this way causes a situation which was appropriately referred to as spontaneous symmetry breaking (SSB). Kastler, Swieca and Robinson showed that this cannot happen in the presence of a mass gap [12], and in a follow up paper (based on the use of the Jost-Lehmann-Dyson representation) Swieca together with Ezawa [13] succeeded to prove the Goldstone theorem in a model- and perturbation- independent way.

The Goldstone theorem states that a Noether symmetry in QFT is spontaneously broken precisely if a massless scalar ”Goldstone boson” prevents the convergence of some of the global charge $Q= \int j_0 = \infty.$

This quasiclassical prescription leads to a model-defining first order interaction density which maintains the conservation of the symmetry currents in all orders. There are symmetry-representing unitary operators for each finite spacetime region O but the global charges $Q= \int j_0$ of same symmetry generating currents diverge. This is the definition of SSB whereas the shift in field space procedure is a way to prepare such a situation whenever SSB is possible. For the later presentation of the Higgs model it is important to be aware of a fine point about SSB whose nonobservance led to a still lingering confusion. As soon as scalar self-interacting fields are coupled to s = 1 potentials the physical interpretation of the field shift manipulation on a Mexican hat potential as a SSB is incorrect; one obtains the Higgs model for the wrong physical reasons and misses the correct reasons why there can be no self-interacting massive vectormesons without the presence of a H-field. Although this can be described correctly in the gauge theoretic formulation, a better understanding is obtained in the positivity preserving string-local setting of LQP (see section 6)

## Why is it interesting?

Goldstone's theorem states that whenever a continuous global symmetry is spontaneously broken, there exists a massless excitation about the spontaneously broken vacuum. Decomposing $\Phi(x)=|\Phi(x) |e^{i\rho(x)}$, $\rho$ transforms as $\rho(x) \to \rho(x) + \theta$. Hence the Lagrangian can depend on $\rho$ only via the derivative = $\partial_\mu \rho$; there cannot be any mass term for $\rho$, and it is a massless field. $\rho$ — identified as the field which transforms inhomogeneously under the broken symmetry — is referred to as the Goldstone boson.

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