Regularity of the Density of States of Random Schrödinger Operators

Abstract

In this paper we solve a long standing open problem for Random Schrödinger operators on $L^2({\mathbb {R}}^d)$ with i.i.d single site random potentials. We allow a large class of free operators, including magnetic potential, however our method of proof works only for the case when the random potentials satisfy a complete covering condition. We require that the supports of the random potentials cover ${\mathbb {R}}^d$ and the bump functions that appear in the random potentials form a partition of unity. For such models, we show that the Density of States (DOS) is m times differentiable in the part of the spectrum where exponential localization is valid, if the single site distribution has compact support and has Hölder continuous $m+1$ st derivative. The required Hölder continuity depends on the fractional moment bounds satisfied by appropriate operator kernels. Our proof of the Random Schrödinger operator case is an extensions of our proof for Anderson type models on $\ell ^2(\mathbb {G})$, $\mathbb {G}$ a countable set, with the property that the cardinality of the set of points at distance N from any fixed point grows at some rate in $N^\alpha , \alpha >0$. This condition rules out the Bethe lattice, where our method of proof works but the degree of smoothness also depends on the localization length, a result we do not present here. Even for these models the random potentials need to satisfy a complete covering condition. The Anderson model on the lattice for which regularity results were known earlier also satisfies the complete covering condition.

Appendix

We collect a few Lemmas in this appendix that are used in the main part of the paper. All these Theorems are well known and proved elsewhere in the literature, but we state them in the form we need and also give their proofs for the convenience of the reader.

Lemma A.1

Consider a positive function $\rho \in L^1({\mathbb {R}}, dx)$ and $J \subset {\mathbb {R}}$ an interval. Let $F(z) = \int \frac{1}{x - z} ~ \rho (x) dx$. Then, for any $m \in {\mathbb {N}}$,

$$\begin{aligned} \mathop {\hbox {ess sup}}\limits _{x \in J} \left| \frac{d^m}{dx^m}\rho \right| (x) < \infty \end{aligned}$$

whenever

$$\begin{aligned} \sup _{z \in {\mathbb {C}}^+, ~ \mathfrak {R}(z) \in J} \left| \frac{d^{m}}{dx^{m}} \mathfrak {I}F\right| (z) < \infty . \end{aligned}$$

Proof

Since $\rho (x) dx$ is a finite positive measure, F is analytic in ${\mathbb {C}}^+$, and the assumption on F implies that functions $\frac{d^\ell }{dz^\ell }\mathfrak {I}F$ are bounded harmonic functions in the strip $\{z \in {\mathbb {C}}^+ : \mathfrak {R}(z) \in J\}$, $0 \le \ell \le m$. Therefore the boundary values

$$\begin{aligned} h_\ell (E) = \lim _{\epsilon \rightarrow 0} \frac{d^\ell }{dz^\ell } \mathfrak {I}F (E+i\epsilon ) \end{aligned}$$

exist for Lebesgue almost every $E \in J$ and $h_\ell$ are essentially bounded in J, $0 \le \ell \le m$. For any $E_0 \in J$ for which $h_\ell (E_0)$ is defined for all $0 \le \ell \le m$ and we have for $0 \le \ell \le m-1$,

$$\begin{aligned} \frac{\partial ^\ell }{\partial x^\ell }(\mathfrak {I}F)(E+i\epsilon ) - \frac{\partial ^\ell }{\partial x^\ell }(\mathfrak {I}F) (E_0+i\epsilon ) = \int _{E_0}^E \frac{\partial ^{\ell +1}}{\partial x^{\ell +1}}(\mathfrak {I}F)(x+i\epsilon ) ~ dx, ~~ E \in J. \end{aligned}$$

(A.1)

Since the integrands above are Harmonic functions in the strip, their boundary values exist, they are uniformly bounded in the strip, so by the dominated convergence theorem the integral converges to

$$\begin{aligned} \int _{E_0}^E h_{\ell +1}(x) ~ dx, ~~ E \in J. \end{aligned}$$

On the other hand the left hand side of Eq. (A.1) converges to $h_{\ell }(E) - h_\ell (E_0)$, showing that $h_\ell (E)$ is differentiable in J. Since, $\rho (x) = \frac{1}{\pi }h_0(x), ~ x \in J$, a simple induction argument now gives the Lemma. $\quad \square$

Lemma A.2

On a separable Hilbert space $\mathscr {H}$, let A and B be two bounded operators generating strongly differentiable contraction semi-groups $e^{tA}, e^{tB}$ respectively, then for any $0< s < 1$,

$$\begin{aligned} \left\Vert e^{ t A}-e^{ t B}\right\Vert \le 2^{1-s}|t|^s \left\Vert A-B\right\Vert ^s. \end{aligned}$$

Proof

Since $e^{tA}, e^{tB}$ are strongly differentiable, the fundamental Theorem of calculus gives the bound,

$$\begin{aligned} \left\Vert e^{ t A}-e^{ t B}\right\Vert =\left\Vert \int _0^t e^{(t-s)A} (A - B)e^{sB} ~ ds \right\Vert \le |t| \left\Vert A - B\right\Vert . \end{aligned}$$

Since $e^{tA}, e^{tB}$ are contractions we have the trivial bound

$$\begin{aligned} \left\Vert e^{tA} - e^{tB}\right\Vert \le 2, \end{aligned}$$

so the Lemma follows by interpolation. $\quad \square$

Lemma A.3

Let g be a probability density with a $\tau$-Hölder continuous derivative. Suppose A is a bounded operator on a separable Hilbert space $\mathscr {H}$ with $\mathfrak {I}A > 0$ and satisfies

$$\begin{aligned} \left\Vert (A+\lambda I)^{-1}\right\Vert<C<\infty , ~~\lambda \in supp(g). \end{aligned}$$

Then

$$\begin{aligned} \int g(\lambda )(A+\lambda I)^{-1}d\lambda =- \int _0^\infty e^{i t A}\left( \int g(\lambda )e^{i t \lambda }d\lambda \right) dt. \end{aligned}$$

(A.2)

Proof

Since $(A+\lambda I)^{-1}$ is bounded we have, in the strong sense,

$$\begin{aligned} (A+\lambda I)^{-1}=\lim _{\epsilon \downarrow 0}(A+\epsilon +\lambda I)^{-1}. \end{aligned}$$

Since $\mathfrak {I}A >0$, the bounded operator $(A+i\epsilon )$ is the generator of a contraction semi-group, so using [59, Corollary 1, Section IX.4] we have

$$\begin{aligned} \int g(\lambda ) (A+i\epsilon +\lambda I)^{-1} d\lambda&=\int g(\lambda )\int e^{i t (A+i\epsilon +\lambda I)} dt d\lambda \nonumber \\&= \int \int g(\lambda ) e^{(-\epsilon + \lambda )t} e^{i t A} dt d\lambda . \end{aligned}$$

(A.3)

Since g has a $\tau$-Hölder continuous derivative, its Fourier transform is a bounded integrable function. Therefore by Fubini we can interchange the $\lambda$ and t integrals on the right hand side of the above equation to get the right hand side of Eq. (A.2). On the other hand using the fact that $\left\Vert (A+\epsilon +\lambda I)^{-1}\right\Vert <2C$ for $0<\epsilon <\frac{1}{2C}$ and g is a probability density, we have

$$\begin{aligned}&\lim _{\epsilon \downarrow 0} \int g(\lambda ) (A+i\epsilon +\lambda I)^{-1} d\lambda \qquad =\int g(\lambda ) \left[ \lim _{\epsilon \downarrow 0}(A+i\epsilon +\lambda I)^{-1}\right] d\lambda \\&\qquad =\int g(\lambda ) (A+\lambda I)^{-1} d\lambda . \end{aligned}$$

This set of equalities when applied to the left hand side of the Eq. (A.2) gives the Lemma after letting $\epsilon$ go to zero. $\quad \square$

We give the Lemma below which is a consequence of proofs of results in Stollmann [56] and Combes et al. [13]. These papers essentially prove the result, but we write it here since it does not occur in the form we need to use.

Lemma A.4

Suppose A is a self-adjoint operator on a separable Hilbert space $\mathscr {H}$ and suppose B is a non-negative bounded operator. Consider the operators $A(t) = A + t B, ~~ t \in {\mathbb {R}}$, $\phi \in Range(B)$ and $\nu ^\phi _{A(t)}$ the spectral measure of A(t) associated with the vector $\phi$. Suppose $\mu$ is a finite absolutely continuous measure with bounded density, then

$$\begin{aligned} \sup _{z \in {\mathbb {C}}^+} \int \mathfrak {I}( \langle \phi , (A(t) - z)^{-1}\phi \rangle ) d\mu (t) < \infty . \end{aligned}$$

(A.4)

In particular the measure $\int \nu ^\phi _{A(t)} ~ dt$ has bounded density.

Proof

We set $\tilde{\nu } = \int \nu ^\phi _{A(t)} ~ dt$, then $\tilde{\nu }$ is a positive finite measure. We recall that the modulus of continuity of a measure $\nu$ is defined as

$$\begin{aligned} s(\nu , \epsilon ) = \sup \{ \nu ([a, a+\epsilon ]) : a \in {\mathbb {R}}\}. \end{aligned}$$

This definition immediately implies that an absolutely continuous measure $\mu$ with bounded density $\rho$, satisfies $s(\mu , \epsilon ) \le \Vert \rho \Vert _\infty \epsilon$. Therefore the Theorem 3.3 of Stollman [56], implies that

$$\begin{aligned} s(\tilde{\nu }, \epsilon ) \le 6\Vert B\Vert \Vert \phi \Vert s(\mu , \epsilon ) \le C\Vert \rho \Vert _\infty \epsilon . \end{aligned}$$

This inequality implies that the density of $\tilde{\nu }$ is bounded. Since the function

$$\begin{aligned} F(z) = \int \mathfrak {I}( \langle \phi , (A(t) - z)^{-1}\phi \rangle ) ~ d\mu (t) = \int \mathfrak {I}(\frac{1}{x - z}) ~ d\tilde{\nu } \end{aligned}$$

is positive Harmonic in ${\mathbb {C}}^+$, by the maximum principle its supremum is attained on ${\mathbb {R}}$. The boundary values of F on ${\mathbb {R}}$ exist and equal the density of the measure $\tilde{\nu } = \int \nu ^\phi _{A(t)} ~ dt$ Lebesgue almost everywhere , by Theorem 1.4.16 of Demuth and Krishna [21], giving the result. $\quad \square$

Lemma A.5

Consider the operators $H^\omega$, $H_\Lambda ^\omega$ given in Eq. (4.1) and the discussion following it. Then for any finite $E \in {\mathbb {R}}$, the operators $u_0 E_{H^\omega _\Lambda }((-\infty , E))$, $u_0 E_{H^\omega }((-\infty , E))$ are trace class for all $\omega$. The traces of these operators are uniformly bounded in $\omega$ for fixed E.

Proof

We will give the proof for $H^\omega$, the proof for the others is similar. The hypotheses on $H^\omega$ imply that it is bounded below and the pair $H_0, H^\omega$ are relatively bounded with respect to each other, being bounded perturbations of each other, the operators $(H_0+a)^d E_{H^\omega }((-\infty , E))$ are bounded for any fixed $(E, a, \omega )$. So taking a in the resolvent set of $H_0$ and using the fact that $u_0(H_0+a)^{-d}$ is trace class we see that

$$\begin{aligned} u_0 E_{H^\omega }(-\infty , E)=u_0 (H_0+a)^{-d}(H_0+a)^{d} E_{H^\omega }(-\infty , E), \end{aligned}$$

is a product of a trace class operator and a bounded operator for each fixed $(\omega , a, E)$ with a positive and large. Therefore $u_0 E_{H^\omega }(-\infty , E)$ is also trace class for each $E, \omega$. The uniform boundedness statement is obvious from the assumptions on the random potential. $\quad \square$