✎ Goldstone's theorem [The Physics Travel Guide]

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====== Goldstone's theorem ======

<blockquote>
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.

<cite>https://arxiv.org/pdf/1703.05448.pdf</cite>
</blockquote>

<note tip>
Explanations in this section should contain no formulas, but instead colloquial things like you would hear them during a coffee break or at a cocktail party.
</note>
  
<tabbox Student>

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.)

<cite>https://arxiv.org/pdf/cond-mat/9204009.pdf</cite>
</blockquote>

<blockquote>
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 
<cite>http://www.jstor.org/stable/pdf/10.1086/518324.pdf</cite>
</blockquote>

<note tip>
The motto in this section is: //the higher the level of abstraction, the better//.
</note>

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