On discrete spectrum of one two-particle lattice Hamiltonian

Abstract

Linear self-adjoint operators in the Friedrichs models arise in various fields, for instance, in the perturbation theory of spectra of self-adjoint operators, in the quantum field theory, in theory of two- and three-particle discrete Schrödinger operators, in hydrodynamics, etc. An operator $H$ in the Friedrichs model is a sum of two operators in the Hilbert space $L_2(\Omega)$, that is, $H=H_0+\varepsilon K$, $\varepsilon>0$, where $H_0$ is the operator by multiplication by a function and $K$ is a compact integral operator. For the operators in the Friedrichs models we need to solve the following problems: 1) Under which conditions the discrete spectrum is an empty set? 2) Under which conditions the discrete spectrum is a non-empty set? 3) Find conditions ensuring that an operator in the Friedrichs model is a finite set; 4) Find sufficient conditions guaranteeing that an operator in the Friedrichs model is an infinite set. It is known that if a kernel of an integral operator in the model is degenerate, then the discrete spectum of the correponding operator in the Friedrichs model is a finite set. Therefore, a necessary condition for the operator in the Friedrichs model to possess an infinite discrete spectrum is the non-degeneracy of the integral operator in the model. In the paper we consider linear bounded self-adjoint operator in the Friedrichs model, for which the integral operator has a non-degenerate kernel. In this work, we study the first and the fourth questions. We obtain one sufficient condition guaranteeing that the operators in Friedrichs model possesses an infinite discrete spectrum. We study the spectrum of one two-particle discrete Schrödinger operator $Q(\varepsilon)$ on the lattice $\mathbb{Z}^{\nu}\times\mathbb{Z}^{\nu}$, in which the Fourier transform of the operator $Q(\varepsilon)$ is represented as $H=H_0+\varepsilon K$, $\varepsilon>0$. It is shown that the structure of the Schrödinger operator $Q(\varepsilon)$ highly depends on the dimension $\nu$ of the lattice. It is proven that in the case $\nu=1,2$, for all $\varepsilon>0$ the discrete spectrum of the Schrödinger operator $Q(\varepsilon)$ is infinite, while in the case $\nu \geq3$, for sufficiently small $\varepsilon>0$, the discrete spectrum of the Schrödinger operator $Q(\varepsilon)$ is an empty set.