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arXiv:2010.04070

Can a crystal be the ground state of a Bose system?

Maksim Tomchenko

Published Oct 8, 2020• 1 citation
arXivHTMLPDF

Abstract

It is usually assumed that the Bose crystal at $T=0$ 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 $Ψ^{c}_{0}$ of a zero-temperature 3D Bose crystal is proposed for zero boundary conditions. Apparently, such $Ψ^{c}_{0}$ corresponds to a local minimum of energy (absolute minimum corresponds to a liquid). Those properties yield the possibility of existence of superfluid liquid H$_{2}$, Ne, Ar, and other inert elements. We propose several possible experimental ways of obtaining them.

DOI
10.48550/arxiv.2010.04070

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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.
As is known, the ground state (GS) of a system of spinless bosons must be non-degenerate and must be described by a nodeless wave function. With the help of the general quantum mechanical analysis we show that any \textit{anisotropic} state of a system of spinless bosons is degenerate. We prove this for a two-dimensional (2D) system, infinite or finite circular, and for a three-dimensional (3D) system, infinite or finite ball-shaped. It is natural to expect that this is valid for finite 2D and 3D systems of any shape. Hence, GS of a Bose system of any density is isotropic and, therefore, corresponds to a liquid or gas. Therefore, the lowest state of a 2D or 3D natural crystal consisting of spinless bosons should be described by a wave function with nodes. This leads to nontrivial experimental predictions. We propose a possible ansatz for the wave function of the lowest state of a 3D Bose crystal and discuss possible experimental consequences.
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 perform the symmetry analysis of the properties of the ground state of a finite system of interacting spinless bosons for the three most symmetric boundary conditions (BCs): zero BCs with spherical and circular symmetries, as well as periodic BCs. The symmetry of the system can lead to interesting properties. For instance, the density of a periodic Bose system is an exact constant: ρ(r) = const. Moreover, in the case of perfect spherical symmetry of BCs, the crystalline state cannot produce the Bragg peaks. The main result of the article is that symmetry properties and general quantum-mechanical theorems admit equally both crystalline and liquid ground state for a Bose system of any density.
At the present time, it is considered that Bose crystals are formed at the cooling of a fluid, because the state of crystal is more favorable by energy. It is also believed [1,2] that no ordering factor forming a crystal is present, except for the interatomic interaction. However, the available solutions [1,2,3] for the wave functions (WFs) of the ground and excited states of a crystal are approximate and are obtained for cyclic boundary conditions, which are not realized in the Nature. Here, we present the exact solutions for the WFs of a Bose crystal with rectangular lattice under natural zero boundary conditions. The structure of WFs implies that 1) a crystal is formed by a standing wave in the probability field; 2) a crystal in the ground state contains a condensate of atoms with the wave vector \textbf{k}_l=(π/\bar{R}_x, π/\bar{R}_y, π/\bar{R}_z) (\bar{R}_x, \bar{R}_y, \bar{R}_z are the periods of the lattice) that is equal to a half of the vector of the reciprocal lattice. These solutions indicate that the ordering factor forming a crystal is an intense standing wave similar to a sound one. Thus, the periodicity of a lattice is caused by that of a sound wave, but not only by the energy minimum principle. Apparently, the crystals of other types and with different lattices have the wave nature as well. The condensate opens a possibility to explain the nonclassical inertia moment discovered by Kim and Chan [4,5] in solid He-4, which testifies, probably, to the presence of a superfluid subsystem in the crystal.
A calculation scheme suitable for studying the ground state of many-body Bose quantum systems with strong forces is developed. The calculation scheme is based on the nodeless property of the ground-state wave function and is capable of yielding in some cases upper and lower bounds on the ground-state energy and corresponding wave functions. From a treatment of liquid helium in the ground state, including a short-range interaction form, it is concluded that the calculation scheme is suitable for problems with involved interactions and is capable of reducing the problem to the mathematical one of calculation of the energy and lattice structure of a classical crystal at absolute zero of temperature.
In this paper we suggest a new ground-state wave function and low-temperature density matrix for a strongly interacting system of bosons. Our basic assumption is that the system can support long-wavelength phonons and that these can propagate independently of any other mode of motion. We therefore write the ground-state function as the product of two factors. One factor arises from the zero-point motion of the phonons, and we show that it has the form ${\ensuremath{\Pi}}_{i<j}f({r}_{\mathrm{ij}})$, where lnf(r)\mathrm{ln}f(r)lnf(r) has an infinite range. The other factor is assumed to have this same form but with lnf(r)\mathrm{ln}f(r)lnf(r) of finite range; it takes into account the short-range correlations arising from the strong repulsive part of the interparticle potential. The function we have chosen to represent the short-range correlations is not new; functions of this kind were first introduced by Bijl and later by Jastrow. At finite temperatures we use a density matrix for an ensemble of excited phonon states. We find that for small wave vectors kkk, the structure factor S(k)S(k)S(k) is equal to $\frac{\ensuremath{\hbar}k}{2mc}$, a result that was first derived by Feynman. At a finite temperature TTT, S(k)S(k)S(k) tends to the constant value kBTmc2\frac{{k}_{B}T}{m{c}^{2}}mc2kB​T​, where mmm is the mass of the particles and ccc the velocity of propagation of the phonons. The momentum distribution nk{n}_{k}nk​ has a ${k}^{\ensuremath{-}1}$ singularity at absolute zero and a stronger, ${k}^{\ensuremath{-}2}$ singularity at finite temperatures. The small-kkk behavior of both S(k)S(k)S(k) and nk{n}_{k}nk​ is completely controlled by the correlations introduced by the phonons. Both the ground-state function and the density matrix imply that there is a finite fraction of particles in the zero-momentum state in three dimensions; this fraction does not seem to be appreciably affected by the infinite-range correlations introduced by the phonons. We find, however, that these correlations imply that a one-dimensional Bose system does not exhibit Bose-Einstein condensation at any temperature, while a two-dimensional system exhibits Bose-Einstein condensation only at absolute zero.