Considering the centrality of Goldstone quasiparticles in condensed matter theories, I was wondering if the converse of the theorem might also be true: Does the existence of a gapless excitation imply some broken symmetry?
For example, I would expect the some phonon-like excitation in a Bose gas even at temperatures above which it becomes disordered and does not break the usual $U(1)$ symmetry. Similarly, a Fermi gas has gapless bosonic excitations by exciting particles immediately across the Fermi surface, and no $U(1)$ symmetry is broken.
This led me to think about a few other follow-up questions:
This wiki lists the above modes as due to a broken Galilean symmetry. Recognizing that the Galilean boost is $$\hat{U}\psi(x,t)=e^{-i\frac{1}{2}mv^2+ipv }\psi(x-vt,t),$$ it is easy to see that it is broken if translation symmetry or $U(1)$ are broken. However, can it be broken independently of these? What is the appropriate order parameter? In particular, I can't easily think of a Hamiltonian eigenstate state that is Galilean-symmetric besides the vacuum - is that the case that this symmetry is broken almost by construction?
In the above example of a Fermi gas, all excitation are at positive frequency $\epsilon_{k>k_F} - \epsilon_{k'<k_F} > 0$ at $T=0$, as I would expect of collective excitations. With $T>0$, transitions can also take place from higher states to lower, and negative poles will enter the dynamical structure factor $\langle [n(r,t), n(0,0)]\rangle$. Yet the ground state density matrix at finite temperature $e^{-\beta H}$ is that which minimizes the free energy $F$, and therefore any transition will increase the free energy of the system, including such high-to-low negative frequency transitions. How should these two facts be taken in context - is energy added or removed from the system upon such excitations? Is there any notion of a free energy eigenmode expansion?
(I thought the issue of Goldstone mode redundancy might be at play, but reading through e.g. this article by Watanabe & Murayama, I didn't find an answer to my questions)
