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arXiv:2501.00893 · Journal of Physics A Mathematical and Theoretical

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

Maksim Tomchenko

Published Jul 17, 2026• 0 citations
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Abstract

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 Ψ ^ ( r , t ) → e i α Ψ ^ ( r , t ) , whereas the order parameter Ψ ( 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 quantized 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.

DOI
10.1088/1751-8121/ae8c94

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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 Ψ ^ ( r , t ) → e i α Ψ ^ ( r , t ) , whereas the order parameter Ψ ( 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 quantized 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.
We study the superfluid properties of a system of interacting bosons on a lattice, which, moreover, are coupled to the vibrational modes of this lattice, treated here in terms of Einstein phonon modes. The ground state corresponds to two correlated condensates: that of the bosons and that of the phonons. Two competing effects determine the common collective sound-wave-like mode with sound velocity v,v,v, arising from gauge symmetry breaking. (i) The sound velocity v0{v}_{0}v0​ (corresponding to a weakly interacting Bose system on a rigid lattice) in the lowest-order approximation is reduced due to reduction of the repulsive boson-boson interaction, arising from the attractive part of the phonon-mediated interaction in the static limit. (ii) The second-order correction to the sound velocity is enhanced as compared to that of bosons on a rigid lattice when the boson-phonon interaction is switched on due to the retarded nature of the phonon-mediated interaction. The overall effect is that the sound velocity is essentially unaffected by the coupling with phonons, indicating the robustness of the superfluid state. The induction of a coherent state in the phonon system driven by the condensation of the bosons could be of experimental significance, permitting spectroscopic detection of superfluid properties of bosons. Our results are based on an extension of the Beliaev-Popov formalism for a weakly interacting Bose gas on a rigid lattice to one on a deformable lattice with which it interacts.
We discuss the mechanism of spontaneous symmetry breaking and the elementary excitations for a weakly-interacting Bose gas at a finite temperature. We consider both the non-relativistic case, described by the Gross-Pitaevskii equation, and the relativistic one, described by the cubic nonlinear Klein-Gordon equation. We analyze similarities and differences in the two equations and, in particular, in the phase and amplitude modes (i.e., Goldstone and Higgs modes) of the bosonic matter field. We show that the coupling between phase and amplitude modes gives rise to a single gapless Bogoliubov spectrum in the non-relativistic case. Instead, in the relativistic case the spectrum has two branches: one is gapless and the other is gapped. In the non-relativistic limit we find that the relativistic spectrum reduces to the Bogoliubov one. Finally, as an application of the above analysis, we consider the Bose-Hubbard model close to the superfluid-Mott quantum phase transition and we investigate the elementary excitations of its effective action, which contains both non-relativistic and relativistic terms.
It is usually assumed that the Bose crystal at T=0T=0T=0 corresponds to the genuine ground state of a Bose system, i.e., this state is non-degenerate and is described by the wave function without nodes. By means of symmetry analysis we show that the ground state of a Bose system of any density should correspond to a liquid or gas, but not to a crystal. The main point is that any anisotropic state of a system of spinless bosons is degenerate. We prove this for an infinite three-dimensional (3D) system and a finite ball-shaped 3D system. One can expect that it is true also for a finite system of any form. Therefore, the anisotropic state cannot be the genuine ground state. Hence, a zero-temperature natural 3D crystal should correspond to an excited state of a Bose system. The wave function Ψ0cΨ^{c}_{0}Ψ0c​ of a zero-temperature 3D Bose crystal is proposed for zero boundary conditions. Apparently, such Ψ0cΨ^{c}_{0}Ψ0c​ corresponds to a local minimum of energy (absolute minimum corresponds to a liquid). Those properties yield the possibility of existence of superfluid liquid H2_{2}2​, Ne, Ar, and other inert elements. We propose several possible experimental ways of obtaining them.
Bragg spectroscopy shows the evolution of gapless Goldstone modes and single-particle-like excitations in an atomic Fermi superfluid as it crosses from a Bardeen–Cooper–Schrieffer superfluid to the Bose–Einstein condensate regime. Spontaneous symmetry breaking is a central paradigm of elementary particle physics1, magnetism2, superfluidity3 and superconductivity4. According to Goldstone’s theorem, phase transitions that break continuous symmetries lead to the existence of gapless excitations in the long-wavelength limit5. These Goldstone modes can become the dominant low-energy excitation, showing that symmetry breaking has a profound impact on the physical properties of matter. Here, we present a comprehensive study of the elementary excitations in a homogeneous strongly interacting Fermi gas through the crossover from a Bardeen–Cooper–Schrieffer (BCS) superfluid to a Bose–Einstein condensate (BEC) of molecules using two-photon Bragg spectroscopy. The spectra exhibit a discrete Goldstone mode, associated with the broken-symmetry superfluid phase, as well as pair-breaking single-particle excitations. Our techniques yield a direct determination of the superfluid pairing gap and speed of sound in close agreement with strong-coupling theories.
Elementary excitation in the presence of a bosonic superfluid vortex line includes a kelvon and a Bogoliubov phonon as Nambu-Goldstone bosons, associated with spontaneous breaking of translational and U(1)\text{U}(1)U(1) symmetries, respectively. Using the Bogoliubov approximation, we construct an effective Hamiltonian to describe the interplay among kelvons and Bogoliubov phonons in a microscopic manner and apply it for the decay process of an excited kelvon with Bogoliubov-phonon emission. Such a decay process governs the late-time behavior of vortex-excitation dynamics observed in cold-atom experiments and also in follow-up numerical simulations using the time-dependent Gross-Pitaevskii equation. We show our numerical result of the decay process and compare it with these previous studies. Our formalism provides the interaction Hamiltonian in terms of wave functions and energy spectra of the elementary excitations, thus potentially being useful in the description of other low-energy phenomena caused by the kelvon--Bogoliubov-phonon interaction.