Saved in:
| Main Author: | |
|---|---|
| Format: | Preprint |
| Published: |
2025
|
| Subjects: | |
| Online Access: | https://arxiv.org/abs/2501.00893 |
| Tags: |
Add Tag
No Tags, Be the first to tag this record!
|
| _version_ | 1866915811460907008 |
|---|---|
| author | Tomchenko, Maksim |
| author_facet | Tomchenko, Maksim |
| contents | 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. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2501_00893 |
| institution | arXiv |
| publishDate | 2025 |
| record_format | arxiv |
| spellingShingle | Is a phonon excitation of a superfluid Bose gas a Goldstone boson? Tomchenko, Maksim Quantum Gases Quantum Physics 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. |
| title | Is a phonon excitation of a superfluid Bose gas a Goldstone boson? |
| topic | Quantum Gases Quantum Physics |
| url | https://arxiv.org/abs/2501.00893 |