On the spectrum of a multidimensional periodic magnetic Shrödinger operator with a singular electric potential

Abstract

We prove absolute continuity of the spectrum of a periodic $n$-dimensional Schrödinger operator for $n\geqslant 4$. Certain conditions on the magnetic potential $A$ and the electric potential $V+\sum f_j\delta_{S_j}$ are supposed to be fulfilled. In particular, we can assume that the following conditions are satisfied. (1) The magnetic potential $A\colon{\mathbb{R}}^n\to{\mathbb{R}}^n$ either has an absolutely convergent Fourier series or belongs to the space $H^q_{\mathrm{loc}}({\mathbb{R}}^n;{\mathbb{R}}^n)$, $2q>n-1$, or to the space $C({\mathbb{R}}^n;{\mathbb{R}}^n)\cap H^q_{\mathrm{loc}}({\mathbb{R}}^n;{\mathbb{R}}^n)$, $2q>n-2$. (2) The function $V\colon{\mathbb{R}}^n\to\mathbb{R}$ belongs to Morrey space ${\mathfrak{L}}^{2,p}$, $p\in \big(\frac{n-1}{2},\frac{n}{2}\big]$, of periodic functions (with a given period lattice), and $$\lim\limits_{\tau\to+0}\sup\limits_{0<r\leqslant\tau}\sup\limits_{x\in{\mathbb{R}}^n}r^2\bigg(\big(v(B^n_r)\big)^{-1}\int_{B^n_r(x)}|{\mathcal{V}}(y)|^pdy\bigg)^{1/p}\leqslant C,$$ where $B^n_r(x)$ is a closed ball of radius $r>0$ centered at a point $x\in{\mathbb{R}}^n$, $B^n_r=B^n_r(0)$, $v(B^n_r)$ is volume of the ball $B^n_r$, $C=C(n,p;A)>0$. (3) $\delta_{S_j}$ are $\delta$-functions concentrated on (piecewise) $C^1$-smooth periodic hypersurfaces $S_j$, $f_j\in L^p_{\mathrm{loc}}(S_j)$, $j=1,\ldots,m$. Some additional geometric conditions are imposed on the hypersurfaces $S_j$, and these conditions determine the choice of numbers $p\geqslant n-1$. In particular, let hypersurfaces $S_j$ be $C^2$-smooth, the unit vector $e$ be arbitrarily taken from some dense set of the unit sphere $S^{n-1}$ dependent on the magnetic potential $A$, and the normal curvature of the hypersurfaces $S_j$ in the direction of the unit vector $e$ be nonzero at all points of tangency of the hypersurfaces $S_j$ and the lines $\{x_0+te\colon t\in\mathbb{R}\}$, $x_0\in{\mathbb{R}}^n$. Then we can choose the number $p>\frac{3n}{2}-3$, $n\geqslant 4$.