Is a phonon excitation of a superfluid Bose gas a Goldstone boson?

It is generally accepted that phonons in a superfluid Bose gas are Goldstone bosons. This is justified by spontaneous symmetry breaking (SSB), which is usually defined as follows: the Hamiltonian of the system is invariant under the $U(1)$ transformation $\widehat{\mathrm{\Psi }}(\mathbf{r},t)\to e^{i\alpha }$, whereas the order parameter $\mathrm{\Psi }(\mathbf{r},t)$ is not. However, the strict definition of SSB is different: the Hamiltonian and the boundary conditions are invariant under a symmetry transformation, while the ground state is not. Based on the latter criterion, we study a finite system of spinless, weakly interacting bosons using three approaches: the standard Bogoliubov method, the particle-number-conserving Bogoliubov method, and the approach based on the exact ground-state wave function. Our results show that the answer to the question in the title is ``no’'. Thus, phonons in a real-world (finite) superfluid Bose gas are similar to sound in a classical gas: they are not Goldstone bosons, but quantised collective vibrational modes arising from the interaction between atoms. In the case of an infinite Bose gas, however, the picture becomes paradoxical: the ground state can be regarded as either infinitely degenerate or non-degenerate, making the phonon both similar to a Goldstone boson and different from it.