2025/01/01 by Maksim Tomchenko, Tomchenko, Maksim
Physics and Astronomy · #FOS: Physical sciences #Quantum Gases (cond-mat.quant-gas) #Quantum Physics (quant-ph) #cond-mat.quant-gas #quant-ph
paper · pdf · doi:10.48550/arxiv.2501.00893
published as J. Phys. A: Math. Theor. 59, 305202 (2026) · 14 pages, v2: significant changes in section 3; language amendments throughout
arxiv created 2026/02/23 · arxiv updated 2026/07/30
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)→ eiα% 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 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.