The Bose-Hubbard model is well-defined description of a Bose solid which may be realistic for cold atoms in a periodic optical lattice. We show that contrary to accepted theories it can never have as a ground state a perfect Mott insulator solid and that it has a low-energy spectrum of vortex-like phase fluctuations. Whether the ground state is necessarily commensurate remains an open question.

arXiv:1102.4797Feb 23, 20112 citations

The ground state of the Hubbard model

E. O. Aiyohuyin, A. A. Oni-Ojo

Journal of the Nigerian Association of Mathematical PhysicsJun 13, 20080 citations

Bulletin of the American Physical Society

Ground-State of the Bose-Hubbard Model

Jay D. Mancini, Vassilios Fessatidis, S.P. Bowen, Robert K. Murawski, Jiri Maly

Published Mar 17, 2016• 0 citations

OpenAlex

Related papers

Ground state properties of one- and two-component Bose-Hubbard model

Barbara Capogrosso-Sansone, Şebnem Güneş Söyler, Nikolay V. Prokof'ev +1 more

Bulletin of the American Physical SocietyMar 12, 20080 citations

The ground state phases of the disordered Bose-Hubbard model with attractive interactions

Olli Mansikkamäki, Matti Silveri

Bulletin of the American Physical SocietyMar 17, 20210 citations

Lieb's Theorem for Bose Hubbard Models

Zhongchao Wei, Chong Ke Zhao

Using a cone-theoretical method, we prove the uniqueness of the ground state for two Bose Hubbard models. The first model is the usual Bose Hubbard model with real hopping coefficients and attractive interactions. The second model is a two-component Bose Hubbard model. Under certain conditions, we show that the ground state in the subspace with particle number $N=2n$ ( $n$ is a positive integer) is unique for both models. For the second model, we show that the ground state has spin along the z-axis $S^{z}=0$ . When the hopping coefficients are real, it has zero spin quantum number, i.e., it is a singlet. Our proofs work equally well for any arbitrary finite-size lattice.Using a cone-theoretical method, we prove the uniqueness of the ground state for two Bose Hubbard models. The first model is the usual Bose Hubbard model with real hopping coefficients and attractive interactions. The second model is a two-component Bose Hubbard model. Under certain conditions, we show that the ground state in the subspace with particle number $N=2n$($n$ is a positive integer) is unique for both models. For the second model, we show that the ground state has spin along the z-axis $S^{z}=0$. When the hopping coefficients are real, it has zero spin quantum number, i.e., it is a singlet. Our proofs work equally well for any arbitrary finite-size lattice.

arXiv:2506.10319Jun 12, 20250 citations

Ground States of the Hubbard Model

Assa Auerbach

Graduate texts in contemporary physicsJan 1, 19940 citations

The ground state of the bose-hubbard model is a supersolid

Philip Warren Anderson

arXiv:1102.4797Feb 23, 20112 citations

The ground state of the Hubbard model

E. O. Aiyohuyin, A. A. Oni-Ojo

Journal of the Nigerian Association of Mathematical PhysicsJun 13, 20080 citations

N=2n

(

n

is a positive integer) is unique for both models. For the second model, we show that the ground state has spin along the z-axis

S^{z}=0

. When the hopping coefficients are real, it has zero spin quantum number, i.e., it is a singlet. Our proofs work equally well for any arbitrary finite-size lattice.