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

TL;DR

The paper rigorously assesses whether phonon excitations in a finite superfluid Bose gas are Goldstone bosons by analyzing three frameworks: the standard Bogoliubov theory, a particle-number-conserving Bogoliubov approach, and an exact ground-state wave function. It shows that in finite systems the ground state remains invariant under $U(1)$ and phonons are not Goldstone modes, with any apparent symmetry breaking arising from approximation artifacts. In the thermodynamic limit, ground-state degeneracy can appear due to particle-number indeterminacy, leading to a nuanced, paradoxical view where the system can be regarded as both degenerate and non-degenerate depending on interpretation. These results separate the concepts of condensation and superfluidity from spontaneous symmetry breaking and emphasize caution when extrapolating finite-system results to the infinite limit. The findings have implications for understanding the true nature of phonons and the role (or lack) of Goldstone physics in real, finite quantum gases.

Abstract

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 $\hatΨ(\mathbf{r},t)\rightarrow e^{iα}% \hatΨ(\mathbf{r},t)$, whereas the order parameter $Ψ(\mathbf{r},t)$ 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.