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It is widely accepted that phonons in a superfluid Bose gas are Goldstone bosons, and that the spontaneous breaking of $U(1)$ symmetry is the primary cause of superfluidity. The first statement 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 $\hat{\Psi}(\mathbf{r},t)\rightarrow e^{i\alpha}% \hat{\Psi}(\mathbf{r},t)$, but the order parameter $\Psi(\mathbf{r},t)$ is not. However, the strict definition of SSB is different: the Hamiltonian and the boundary conditions are invariant under a symmetry transformation, while the $\it{ground~ state}$ 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 SSB does not occur in a real-world (finite) superfluid Bose gas. Therefore, the phonons in such a gas are not Goldstone bosons and are similar to sound in a classical gas, and the superfluidity of such a system is not related to the spontaneous breaking of the $U(1)$ symmetry. In the case of an infinite Bose gas, however, the situation becomes paradoxical: the ground state can be regarded as either infinitely degenerate or non-degenerate; this means that the phonon is both similar to a Goldstone boson and different from it. [1]
[1] M. Tomchenko, Is a phonon excitation of a superfluid
Bose gas a Goldstone boson? J. Phys. A: Math. Theor. 59, 305202 (2026). https://doi.org/10.1088/1751-8121/ae8c94
