Is a phason a Goldstone mode? Suppose we have a lattice system whose ground state is an incommensurate charge-density wave (CDW).  Strictly speaking, this ground state does not have Goldstone modes because the only symmetry that is spontaneously broken is the discrete translational symmetry of the lattice.  But the possible symmetry-broken states vary continuously (they can be continuously parameterized by the arbitrary phase shift by which the CDW is translated relative to some reference configuration), which is why phasons (distortions of the CDW with arbitrarily long wavelengths and low energies) are gapless.  This seems like pretty much exactly what we'd expect from a Goldstone mode.  Should I think of a phason as a Goldstone mode?  Is the existence of a continuous degenerate ground-state manifold more important for "Goldstone-like" behavior than the existence of a continuous spontaneously broken symmetry?
 A: This question nagged me for most of Friday. It seems obvious that it is a Goldstone mode. You can translate the ICDW and the energy does not chage. However it is not clear what continuous symmetry remains since the lattice has already broken translation symmetry. To get to the bottom of the issue we should focus on the relevant Hamiltonian which is electron+phonon
$$
H=\sum_k\epsilon_k c_k^\dagger c_k+\sum_q\hbar\omega_q b_q^\dagger b_q+\sum_{k,q}g(k)c_{k+q}^\dagger c_kb_q+h.c.
$$
where $c_k^\dagger$ and $b_q^\dagger$ are electron and phonon creation operators respectively and $h.c.$ denotes Hermitian conjugate of the interaction term. This Hamiltonian is invariant under the continuous transformation
$$
c_k\to c_k e^{ika\varphi} \\
b_q\to b_q e^{iqa\varphi}
$$
for any choice of $\varphi$, and $a$ is the lattice constant. To see this is the relevant phase for CDW consider a simple 1D system with Peierls transition. There the CDW causes phonons to condense and the complex order parameter $\Delta$ is
$$
\Delta=|\Delta| e^{i\varphi}=g(2k_f)\langle b_{2k_f}+b_{-2k_f}^\dagger\rangle.
$$
The phase of $\Delta$ is chosen by spontaneous symmetry breaking with $\varphi$ parameterizing the continuous symmetry so there is a goldstone mode. For completeness the charge density is
$$
\rho_0+\delta\rho\cos(2k_f x+\varphi).
$$
The CDW order parameter I got from this reference.
Upon initial inspection the continuous symmetry here appears to be ordinary translation invariance ($\psi_k\to \psi_k e^{ikr}$). This cannot be correct as translation symmetry was already broken in forming the lattice and phonons. The continuous symmetry that $H$ posses is a $U(1)$ symmetry that is a remnant of the full translation symmetry $\mathbb{R}=\mathbb{Z}\times U(1)$. The $U(1)$ component of $\mathbb{R}$ is a translation symmetry with translation only defined within a unit cell of the lattice. Translation by multiple unit cells comes from the $\mathbb{Z}$ factor of $\mathbb{R}$.
A: References
I have found multiple references in scientific literature referring to phasons as a Goldstone mode. Below are some examples with links:


*

*"...the soft, amplitudon and phason (Goldstone) modes..." from Phase Transitions in Liquid Crystals edited by Arthur N. Chester (google book format open to the correct page here)

*" The Goldstone modes evolving from the magnetic satellites consist of transverse spin-wave modes and longitudinal phason modes..." from the abstract of Goldstone Modes and Low-Frequency Dynamics of Incommensurate Chromium Alloys by R.S. Fishman and S.H. Liu (the abstract can be found here)

*"...the strongest mode is the phason (Goldstone) mode..." from Probability Measures on Semigroups: Convolution Products, Random Walks and Random Matrices by Goran Hognas and Arunava Mukherjea (google book format open to the correct page here)

*"...the phason (Goldstone) mode..." from Relaxation Phenomena: Liquid Crystals, Magnetic Systems, Polymers, High-Tc Superconductors, Metallic Glasses edited by Wolfgang Haase (google book format open to the correct page here)


This google search contains more references. This makes it seem like (as tparker stated in the comments) it is "morally" though not strictly a Goldstone mode. This now begs the question, what is the actual relationship between phason/Goldstone modes.
Relationship
After looking around for a more in-depth description of the relationship between phasons and Goldstone modes, I found this physics.SE question. Below the one answer, the second comment says

In my understanding, the Goldstone mode always corresponds to the fluctuation of the "phase" while the fluctuation of "amplitude" may be called Higgs mode. So when we talk about the gapless Goldstone mode in SDW, it should relates to the fluctuation of spin directions rather than spin length. Thus I think there is only "Phason" excitations in SDW, but conventionally we call the excitations "magnons" or spin-wave (classical partner). I hope this comment may be helpful to you.

This paper also has some relevant sections on phasons and Goldstone modes. This book has some information that might be helpful, but unfortunately I cannot find a free copy online and the google book sample I've linked too doesn't include some of the relevant pages.
I also found a quote from the book Liquid Crystals in the Nineties and Beyond (edited by S. Kumar; google book format open to the correct page here) - "The Goldstone mode, which is a phason mode with the wave vector in the center of the dispersion..." which makes it seem like a Goldstone mode is a phason mode, not the other way around.
I found a paper that might also help: Phason dynamics in nonlinear photonic
quasicrystals by Barak Freedman, Ron Lifshitz, Jason Fleischer, and Mordechai Segev (the pdf can be downloaded here). The section that seems to be relevant to your Hamiltonian question is on the last page, in the last paragraph...". As such, the observed phason behaviour is representative of a more general hamiltonian dynamics commonly found in non-equilibrium pattern-forming systems". There are, of course, other relevant sections.
Hope this helps! I will continue to update this as I find more information.
A: 
Is the existence of a continuous degenerate ground-state manifold more important for "Goldstone-like" behavior than the existence of a continuous spontaneously broken symmetry?

A couple of recent papers by Takahaski and Nitta address this question: https://arxiv.org/pdf/1404.7696v3.pdf
https://arxiv.org/pdf/1410.2391v2.pdf
The authors use Bogoliubov theory to show that when the ground-state exhibits emergent symmetries--i.e. ones which have generators which do not commute with the Hamiltonian--we still anticipate gapless modes arising from the generators of the continuous ground-state degeneracy, called quasi-Goldstone-modes. Essentially, one may identify zero-modes (eigenvectors of $H_{k=0}$ with zero eigenvalue) associated with symmetry generators of the ground-state. However I think this result relies on Bogoliubov theory and is therefore most relevant to Bose-Einstein condensates.
A: I'm thinking of exactly the same question two weeks ago. And I was thinking this symmetry seemed to be a U(1) symmetry or something, because the system is invariant under continuous variation of that phase theta, just as you said.
Later I find this is actually the chiral symmetry of the left-moving and right-moving fields, and if you write the order parameters as a two-component vector, varying $\theta$ is same as rotating this "order-parameter-vector". On the other hand, varying $\theta$ is same as the translation of the charge distribution. So a continuous chiral symmetry is same as a continuous translation symmetry of the charge distribution. That's why phason mode is gapless. An infinitesimal deviation d$\theta$ is like a very long wavelength acoustic phonon mode.
