condensed matter - 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?

Answer

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 $\mathbb{R}$.