Spectral Rigidity of Random Schrödinger Operators via Feynman–Kac Formulas

Abstract

We develop a technique for proving number rigidity (in the sense of Ghosh and Peres in Duke Math J 166(10):1789–1858, 2017) of the spectrum of general random Schrödinger operators (RSOs). Our method makes use of Feynman–Kac formulas to estimate the variance of exponential linear statistics of the spectrum in terms of self-intersection local times. Inspired by recent results concerning Feynman–Kac formulas for RSOs with multiplicative noise (Gaudreau Lamarre in Semigroups for one-dimensional Schrödinger operators with multiplicative Gaussian noise, Preprint arXiv:1902.05047v3, 2019; Gaudreau Lamarre and Shkolnikov in Ann Inst Henri Poincaré Probab Stat 55(3):1402–1438, 2019; Gorin and Shkolnikov in Ann Probab 46(4):2287–2344, 2018) by Gorin, Shkolnikov, and the first-named author, we use this method to prove number rigidity for a class of one-dimensional continuous RSOs of the form $-\frac{1}{2}\Delta +V+\xi$, where V is a deterministic potential and $\xi$ is a stationary Gaussian noise. Our results require only very mild assumptions on the domain on which the operator is defined, the boundary conditions on that domain, the regularity of the potential V, and the singularity of the noise $\xi$.

Appendix A: Transition Density Bounds

Proposition A.1

There exist constants $0<c<C$ such that for every $t\in (0,1]$,

$$\begin{aligned} ct^{-1/2}\le \inf _{x\in I}\Pi _Z(t;x,x) \qquad \text {and}\qquad \sup _{(x,y)\in I^2}\Pi _Z(t;x,y)\le Ct^{-1/2}. \end{aligned}$$

(A.1)

Proof

In Case 1, the result follows directly from the fact that $\Pi _B(t;x,y)\le 1/\sqrt{2\pi t}$ and $\Pi _B(t;x,x)=1/\sqrt{2\pi t}$ for all x, y, and t. A similar argument holds for Case 2. Consider now Case 3. We recall that, by definition,

$$\begin{aligned} \Pi _Y(t;x,y):=\sum _{z\in 2b\mathbb Z\pm y}\mathscr {G}_t(x-z)=\frac{1}{\sqrt{2\pi t}}\left( \sum _{k\in \mathbb Z}\mathrm e^{-(x-2bk+y)^2/2t}+\mathrm e^{-(x-2bk-y)^2/2t}\right) . \end{aligned}$$

On the one hand, note that $t\mapsto \mathrm e^{-z/t}$ is increasing in $t>0$ for every $z\ge 0$; hence for every $t\in (0,1]$, one has

$$\begin{aligned}&\sup _{(x,y)\in (0,b)^2}\left( \sum _{k\in \mathbb Z}\mathrm e^{-(x-2bk+y)^2/2t}+\mathrm e^{-(x-2bk-y)^2/2t}\right) \\&\quad \le \sup _{(x,y)\in (0,b)^2}\left( \sum _{k\in \mathbb Z}\mathrm e^{-(x-2bk+y)^2/2}+\mathrm e^{-(x-2bk-y)^2/2}\right) <\infty . \end{aligned}$$

On the other hand, by isolating the $k=0$ term in $\sum _{k\in \mathbb Z}\mathrm e^{-(2bk)^2/2t}$,

$$\begin{aligned} \inf _{x\in (0,b)}\left( \sum _{k\in \mathbb Z}\mathrm e^{-(2x-2bk)^2/2t}+\mathrm e^{-(2bk)^2/2t}\right)\ge & {} \left( \inf _{x\in (0,b)}\sum _{k\in \mathbb Z}\mathrm e^{-(2x-2bk)^2/2t}\right) +1\\\ge & {} 1, \end{aligned}$$

concluding the proof. $\square$