Using number-conserving and exact wavefunction arguments, the paper concludes that finite superfluid Bose gases do not spontaneously break U(1) symmetry, so their phonons are not Goldstone bosons.
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Is a phonon excitation of a superfluid Bose gas a Goldstone boson?
Using number-conserving and exact wavefunction arguments, the paper concludes that finite superfluid Bose gases do not spontaneously break U(1) symmetry, so their phonons are not Goldstone bosons.