On the number of negative eigenvalues of a Schrödinger operator with point interactions
Let \(X=(x_1,\dots,x_n)\in \mathbb R^n\) and \(\alpha=(\alpha_1,\dots,\alpha_n)\in \mathbb R^n\). The one-dimensional Schrödinger operator \(L_{X,\alpha}\) is defined on \(\mathbb R\) by the expression \(-d^2/dx^2+\sum_{i=1}^n\alpha_i\delta(x-x_i)\), where \(\delta\) is the Dirac \(\delta\)-function. The state function \(\psi\) belongs to the Sobolev space \(W_2^2(R\setminus X)\), satisfying in the points of \(X\) (`points of interaction') conjugation conditions \(\psi(x_i+0)=\psi(x_i-0)\), \(\psi'(x_i+0)-\psi'(x_i-0)=\alpha_i\psi(x_i)\). It is known that the operator \(L_{X,\alpha}\) is selfadjoint and that its spectrum consists of the positive semiaxis and not more than \(n\) simple negative eigenvalues. \textit{S. Albeverio} and \textit{L. Nizhnik} proved in [Lett. Math. Phys. 65, No.~1, 27--35 (2003; Zbl 1056.34093)] that in the case \(n=1\) a negative eigenvalue exists only if \(\alpha_1<0\). The present authors consider the case when \(n>1\) and the number \(m\) of negative intensities \(\alpha_i\) is also greater than~1. In this situation, they obtain sufficient conditions for the operator \(L_{X,\alpha}\) to have at least \(m\) negative eigenvalues.
scientific article; zbMATH DE number 5544980