Operators with Singular Continuous Spectrum, Iv. Hausdorff Dimensions, Rank One Perturbations, and Localization

Although concrete operators with singular continuous spectrum have proliferated recently [7,11,13,17,34,35,37,39], we still don’t really understand much about singular continuous spectrum. In part, this is because it is normally defined by what it isn’t — neither pure point nor absolutely continuous. An important point of view, going back in part to Rodgers and Taylor [27,28], and studied recently within spectral theory by Last [22] (also see references therein), is the idea of using Hausdorff measures and dimensions to classify measures. Our main goal in this paper is to look at the singular spectrum produced by rank one perturbations (and discussed in [7,11,33]) from this point of view. A Borel measure μ is said to have exact dimension α ∈ [0, 1] if and only if μ(S) = 0 if S has dimension β < α and if μ is supported by a set of dimension α. If 0 < α < 1, such a measure is, of necessity, singular continuous. But, there are also singular continuous measures of exact dimension 0 and 1 which are “particularly close” to point and a.c. measures, respectively. Indeed, as we’ll explain, we know of “explicit” Schrödinger operators with exact dimension 0 and 1, but, while they presumably exist, we don’t know of any with dimension α ∈ (0, 1). While we’re interested in the abstract theory of rank one perturbations, we’re especially interested in those rank one perturbations obtained by taking a random Jacobi matrix and making a Baire generic perturbation of the potential at a single point. It is a disturbing fact that the strict localization (dense point spectrum with ‖xe−itHδ0‖2 = (e−itHδ0, x2e−itHδ0) bounded in t), that holds a.e. for the random case, can be destroyed by arbitrarily small local perturbations [7,11]. We’ll ameliorate this discovery in the present paper in three ways: First, we’ll see that, in this case, the spectrum is always of dimension zero, albeit sometimes pure point and sometimes singular continuous. Second, we’ll show that not