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Page 15 of this note states,

If a continuous symmetry of the Lagrangian is spontaneously broken, and if there are no long-range forces, then exists a zero-frequency excitation at zero momentum.

i.e., in presence of long-range interactions symmetry breaking is not accompanied by the emergence of Goldstone bosons. It also says that

The absence of long-range forces, which may tend to couple spins at large distances, is necessary for the existence of a mode with $\omega\to 0$ as $k\to 0$.

It's worth pointing out that similar stuff is also mentioned on page 432 of Chaikin and Lubensky's Principles of Condensed Matter Physics:

In this case (i.e., in the absence of long-range forces), the Goldstone theorem implies that there is a mode whose frequency goes continuously to zero as the wavenumber goes to zero.

What does the author mean by the phrase "if there are no long-range forces"? Does it refer to the presence of a gauge field as in case of superconductivity?

Is this phrase trying to point out the familiar fact that in presence of a gauge field the Goldstone bosons are "eaten up" in the unitary gauge? But I doubt this because once the symmetry is broken, the gauge field acquires a mass and it's no longer a long-range force.

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  • $\begingroup$ By long range forces, I think they mean that the bare propagator decays fast enough (in real space). This is equivalent to some locality condition in QFT. (For long-range spin interactions, the theory is usually mean-field, and the discussion is somewhat different.) $\endgroup$
    – Adam
    Commented Jan 8, 2018 at 7:30

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In a theory -

  • One can have continuous global symmetries (such as flavor symmetry). These do not have any massless gauge bosonic fields associated with them.

  • One can have continuous gauge symmetries (such as the U(1) electromagnetic symmetry). These have massless gauge bosons associated to them. These mediate long range forces.

Goldstone's theorem then states -

  • When a continuous global symmetry is spontaneously broken, there exists a zero-frequency excitation at zero momentum. This is a massless d.o.f. which is called the Goldstone boson.

  • When a continuous gauge symmetry is spontaneously broken, the massless goldstone boson associated to the gauge symmetry becomes massive.

In essence what happens is that the Goldstone mode of the spontaneous symmetry breaking is "eaten up" by the gauge boson and it appears as the longitudinal mode which makes it massive. In terms of representation theory, $$2 (\text{massless spin-1 d.o.f}) + 1 (\text{Goldstone/longitudinal mode}) = 3 (\text{massive spin-1 d.o.f})$$

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  • $\begingroup$ Yes. I'm aware of this as I've mentioned in the last paragraph of my question. But I am suspicious because of two reasons. 1. Once the symmetry is broken, the gauge fields acquire mass, and they can no longer produce a long-range interaction. 2. Why should we say that a gauge field produces a long-range interaction even when it is massless? The interaction term in the $U(1)$ theory (such as scalar QED) is given by $\sim qA_\mu(\phi^*\partial_\mu\phi-\phi\partial_\mu\phi^*)+q^2A_\mu A^\mu|\phi|^2$. Can I call this interaction long-range just because the field $A_\mu$ exists everywhere? @Prahar $\endgroup$
    – SRS
    Commented Dec 30, 2017 at 13:48
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    $\begingroup$ 1. I believe that the author means to say "in the absence of long-range forces before spontaneous symmetry breaking". 2. If a boson is massless its interaction is long-range. This has to do with the effective potential which drops exponentially for massive states, but only polynomially for massless fields. $\endgroup$
    – Prahar
    Commented Dec 30, 2017 at 13:50
  • $\begingroup$ By long-range interaction, I understand, for example, an interaction term that goes like the $1/r$. @Prahar $\endgroup$
    – SRS
    Commented Dec 30, 2017 at 13:51
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    $\begingroup$ @SRS - Yes, every massless bosonic gauge field mediates a long-range interaction. $\endgroup$
    – Prahar
    Commented Dec 30, 2017 at 13:57
  • $\begingroup$ 1. Agreed. 2. Is there a way I can show that this "effective potential" (as I understand, you're probably talking about finding the Coulomb-like potential by Fourier transforming the scattering amplitude) goes like some $r^{-n}$ ? In Heisenberg Hamiltonian, the interaction is short-range already at the Hamiltonian level because the coefficient $J$ in the Hamiltonian $H=-J\textbf{s}_i\cdot\textbf{s}_j$ represents the strength of exchange interaction which is already short-range. In case of scalar QED, the short-range"ness" of the interaction is not obvious at the Lagrangian level. $\endgroup$
    – SRS
    Commented Dec 30, 2017 at 13:57

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