 Solution

A neutral superfluid can select a phase of its complex order parameter, breaking a global $U(1)$ symmetry. A spatially uniform change of that phase costs no energy. Slowly varying changes therefore produce a gapless Goldstone mode. For a dilute weakly repulsive Bose gas, the Bogoliubov quasiparticle dispersion isso $E_k\simeq \hslash c_{s}k$ at long wavelengths, with $c_s=\sqrt{g_B{n}_0/m_B}$. Density and phase fluctuations are canonically coupled components of this sound branch, rather than two independent positive-frequency modes. Superfluid helium and dilute atomic condensates supply neutral examples, although the dilute-gas formula is not quantitatively valid for strongly interacting liquid helium.

In a superconductor, the pair field is charged and its phase couples to the electromagnetic gauge field. Only combinations such as $\nabla \theta -q\mathbf{A}/\hslash$ are physically meaningful. The Anderson-Higgs mechanism in a superconductor produces electromagnetic screening and removes the neutral sound mode as an independent gapless charged-density oscillation. In a homogeneous three-dimensional continuum, long-range Coulomb forces give a longitudinal plasma gap, approximately ${\omega }^2={\omega }_p^2+c_s^2{k}^2$ at small $k$. The static transverse response gives the Meissner effect andwhen $n_s$ counts superconducting electrons of effective mass $m$. Counting pairs instead uses density $n_s/2$, charge $-2e$ and mass $2m$, giving the same stiffness. The condensate amplitude may also have a gapped Higgs mode, but that amplitude oscillation is distinct from the electromagnetic Higgs mechanism. Local gauge transformations remain a redundancy; this is not spontaneous breaking of an observable local symmetry in the same sense as the neutral global phase symmetry.