标题： 超流体玻色气体的声子激发是否是戈德斯通玻色子？ 显示英文标题

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

摘要： 人们普遍认为，超流体玻色气体中的声子是戈德斯通玻色子。 这是由自发对称性破缺（SSB）所解释的，通常定义如下：系统的哈密顿量在$U(1)$变换$\hat{\Psi}(\mathbf{r},t)\rightarrow e^{i\alpha}% \hat{\Psi}(\mathbf{r},t)$下保持不变，而序参量$\Psi(\mathbf{r},t)$不保持不变。 然而，SSB 的严格定义有所不同：哈密顿量和边界条件在对称变换下保持不变，而\emph{基态}不保持不变。 基于后一种标准，我们使用三种方法研究了无自旋、弱相互作用的玻色子有限系统：标准的博戈留波夫方法、粒子数守恒的博戈留波夫方法以及基于精确基态波函数的方法。 我们的结果表明，标题中的问题的答案是\textquotedblleft no\textquotedblright 。 因此，现实世界（有限）的超流体玻色气体中的声子类似于经典气体中的声波：它们不是戈德斯通玻色子，而是由原子间相互作用产生的量子化集体振动模式。 然而，在无限玻色气体的情况下，情况变得矛盾：基态可以被视为无限简并或非简并，使得声子既类似于戈德斯通玻色子，又与之不同。 显示英文摘要

摘要： 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 $\hat{\Psi}(\mathbf{r},t)\rightarrow e^{i\alpha}% \hat{\Psi}(\mathbf{r},t)$, whereas the order parameter $\Psi(\mathbf{r},t)$ is not. However, the strict definition of SSB is different: the Hamiltonian and the boundary conditions are invariant under a symmetry transformation, while the \emph{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 \textquotedblleft no\textquotedblright. 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.