Superfluidity is often explained in terms of spontaneous breaking of global $U(1)$ symmetry. However, we know that in real, finite-size quantum systems, this symmetry can never be broken. Quantum fluctuations will couple degenerate states, and the true ground state of the system will be a symmetric combination of broken symmetry states. In the thermodynamic limit, there can be spontaneous symmetry breaking, and for very large (but finite) systems the symmetry can be approximately broken, in that it takes an exponentially long time for the system to fluctuate between different symmetry-breaking states.
When a continuous global symmetry is spontaneously broken, as in a superfluid in the infinite-size limit, we know that there will be a gapless Nambu-Goldstone (NG) mode in the excitation spectrum. I am curious about the interpretation of the NG mode in the case of finite quantum systems without exact symmetry breaking (but with symmetry breaking in the thermodynamic limit). For a superfluid, we know that this mode is the gapless phonon with linear dispersion, and this mode is present in real systems, as observed, e.g., in inelastic neutron scattering experiments. Strictly speaking, however, there is no spontaneous symmetry breaking in real systems, so the gapless NG mode should not exist.
I understand that for all practical purposes we can treat the symmetry as being spontaneously broken since the time to fluctuate out of the symmetry breaking state is so long. But what if we set aside practicality and assume that we can observe a superfluid (say, with neutron scattering, or with whatever method you believe is relevant) for arbitrarily long times? What would would we conclude about the presence or absence of the NG mode, and about the shape of the excitation spectrum? Would a gap open, or would dispersion relation change? If the degenerate states are coupled by quantum fluctuations on long time scales, I think there should be some experimental signature of the quantum fluctuations; what is this signature?
