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Cambridge University Press eBooks

Boson Systems: Bose–Einstein Condensation and Superfluidity

Michael El-Batanouny

Published Mar 6, 2020• 0 citations
Publisher

Abstract

Treats the Bose-Einstein condensation, and explains superfluidity from the Bogoliubov and Ginzburg-Landau perspectives. It also describes the concept of spontaneous symmetry breaking and Goldstone modes.
DOI
10.1017/9781108691291.018

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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.

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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.
Significant experimental progress has been made recently for observing long-sought supersolidlike states in Bose-Einstein condensates, where spatial translational symmetry is spontaneously broken by anisotropic interactions to form a stripe order. Meanwhile, the superfluid stripe ground state was also observed by applying a weak optical lattice that forces the symmetry breaking. Despite the similarity of the ground states, here we show that these two symmetry breaking mechanisms can be distinguished by their collective excitation spectra. In contrast to gapless Goldstone modes of the spontaneous stripe state, we propose that the excitation spectra of the forced stripe phase can provide direct experimental evidence for the gapped pseudo-Goldstone modes. We characterize the pseudo-Goldstone mode of such lattice-induced stripe phase through its excitation spectrum and static structure factor. Our work may pave the way for exploring spontaneous and forced or approximate symmetry breaking mechanisms in different physical systems.
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.