Lifshitz Tails for the Fractional Anderson Model

Abstract

We consider the d-dimensional fractional Anderson model $(-\Delta )^\alpha + V_\omega$ on $\ell ^2({\mathbb {Z}}^d)$ where $0<\alpha \leqslant 1$. Here $-\Delta$ is the negative discrete Laplacian and $V_\omega$ is the random Anderson potential consisting of iid random variables. We prove that the model exhibits Lifshitz tails at the lower edge of the spectrum with exponent $d/ (2\alpha )$. To do so, we show among other things that the non-diagonal matrix elements of the negative discrete fractional Laplacian are negative and satisfy the two-sided bound $\frac{c_{\alpha ,d}}{|n-m|^{d+2\alpha }} \leqslant -(-\Delta )^\alpha (n,m)\leqslant \frac{C_{\alpha ,d}}{|n-m|^{d+2\alpha }}$ for positive constants $c_{\alpha ,d}$, $C_{\alpha ,d}$ and all $n\ne m\in {\mathbb {Z}}^d$.

Appendix A. A Limit of the Continuous Fractional Laplacian

A crucial ingredient to the proof of Theorem 2.2 is understanding the behaviour of the continuous fractional Laplacian $(-\Delta _c)^\alpha$ applied to Schwartz functions for large $x\in {\mathbb {R}}^d$.

Lemma A.1

Let ${\mathscr {S}}({\mathbb {R}}^d)$ be the set of Schwartz functions, $\varphi \in {\mathscr {S}}({\mathbb {R}}^d)$, $0<\alpha <1$ and $(-\Delta _c)^\alpha$ be the continuous fractional negative Laplacian. Then

$$\begin{aligned} \lim _{|x|\rightarrow \infty } |x|^{d+2\alpha } \big ((-\Delta _c)^\alpha \varphi \big )(x) = - K_{d,\alpha } \int _{{\mathbb {R}}^d} \mathrm {d}y\, \varphi (y) \end{aligned}$$

(A.1)

where

$$\begin{aligned} K_{d,\alpha }= \frac{4^\alpha \Gamma (d/2+\alpha )}{\pi ^{d/2} |\Gamma (-\alpha )|}>0. \end{aligned}$$

(A.2)

Proof

Let $\varphi \in {\mathscr {S}}({\mathbb {R}}^d)$, $0<\alpha <1$ and $x\in {\mathbb {R}}^d$. Then the continuous fractional negative Laplacian can be written as

$$\begin{aligned} \big ((-\Delta _c)^\alpha \varphi \big )(x) = K_{d,\alpha } \lim _{b\rightarrow 0} \int _{{\mathbb {R}}^d\setminus B_b(x)} \mathrm {d}y\, \frac{\big (\varphi (x)-\varphi (y)\big )}{|x-y|^{d+2\alpha }} \end{aligned}$$

(A.3)

with $K_{d,\alpha }>0$ given above, see [15].

Let $x\ne 0$ and $0<b<\frac{|x|}{2}$. We split the integral in (A.3) into two parts as follows:

$$\begin{aligned}&\int _{{\mathbb {R}}^d\setminus B_b(x)} \mathrm {d}y\, \frac{\big (\varphi (x)-\varphi (y)\big )}{|x-y|^{d+2\alpha }} \nonumber \\&= \int _{B_{\frac{|x|}{2}}(x)\setminus B_b(x)} \mathrm {d}y\, \frac{\big (\varphi (x)-\varphi (y)\big )}{|x-y|^{d+2\alpha }} + \int _{B^c_{\frac{|x|}{2}}(x)} \mathrm {d}y\, \frac{\big (\varphi (x)-\varphi (y)\big )}{|x-y|^{d+2\alpha }}. \end{aligned}$$

(A.4)

We first consider the second part

$$\begin{aligned} \int _{B^c_{\frac{|x|}{2}}(x)} \mathrm {d}y\, \frac{\big (\varphi (x)-\varphi (y)\big )}{|x-y|^{d+2\alpha }} =&\int _{B^c_{\frac{|x|}{2}}(0)} \mathrm {d}y\, \frac{\varphi (x)}{|y|^{d+2\alpha }} - \int _{B^c_{\frac{|x|}{2}}(x)} \mathrm {d}y \frac{\varphi (y)}{|x-y|^{d+2\alpha }}. \end{aligned}$$

(A.5)

Since $\varphi \in {\mathscr {S}}({\mathbb {R}}^d)$ decays faster than any polynomial the above implies

$$\begin{aligned} \lim _{|x|\rightarrow \infty } |x|^{d+2\alpha } \int _{B^c_{\frac{|x|}{2}}(x)} \mathrm {d}y\, \frac{\big (\varphi (x)-\varphi (y)\big )}{|x-y|^{d+2\alpha }} = -\lim _{|x|\rightarrow \infty } \int _{B^c_{\frac{|x|}{2}}(x)} \mathrm {d}y \frac{|x|^{d+2\alpha } }{|x-y|^{d+2\alpha }} \varphi (y). \end{aligned}$$

(A.6)

One directly sees that for all $y\in {\mathbb {R}}^d$

$$\begin{aligned} \lim _{|x|\rightarrow \infty } 1_{B^c_{\frac{|x|}{2}}(x)}(y) \frac{|x|^{d+2\alpha } }{|x-y|^{d+2\alpha }} = 1 \end{aligned}$$

(A.7)

where $1_S$ stands here for the indicator function of a set $S\in \text {Borel}({\mathbb {R}}^d)$ and for all $x,y\in {\mathbb {R}}^d$

$$\begin{aligned} \Big | 1_{B^c_{\frac{|x|}{2}}(x)}(y) \frac{|x|^{d+2\alpha } }{|x-y|^{d+2\alpha }} \Big | \leqslant 2^{d+2\alpha }. \end{aligned}$$

(A.8)

Hence the dominated convergence theorem implies

$$\begin{aligned} \lim _{|x|\rightarrow \infty } \int _{B^c_{\frac{|x|}{2}}(x)} \mathrm {d}y \frac{|x|^{d+2\alpha } }{|x-y|^{d+2\alpha }} \varphi (y) = \int _{{\mathbb {R}}^d} \mathrm {d}y\, \varphi (y). \end{aligned}$$

(A.9)

For the remaining term in (A.4) we compute for $0<b<\frac{|x|}{2}$,

$$\begin{aligned} \int _{B_{\frac{|x|}{2}}(x)\setminus B_b(x)} \mathrm {d}y \frac{\big (\varphi (x)-\varphi (y)\big )}{|x-y|^{d+2\alpha }}&=\int _b^{\frac{|x|}{2}} \mathrm {d}r r^{d-1}\int _{\partial B_1(0)} \mathrm {d}S(w) \frac{\big (\varphi (x)-\varphi (x+rw )\big )}{r^{d+2\alpha }}\nonumber \\&=\int _b^{\frac{|x|}{2}} \mathrm {d}r\, \frac{r^{d-1}}{r^{d+2\alpha }} \big (\phi (0)-\phi (r)\big ), \end{aligned}$$

(A.10)

where $\phi (r) := \int _{\partial B_1(0)} \mathrm {d}S(w)\, \varphi (x+rw )$ and we denote by $\mathrm {d}S$ integration with respect to surface measure. Then

$$\begin{aligned} \phi '(r)&= \int _{\partial B_1(0)} \mathrm {d}S(w)\, \big ((\nabla \varphi )(x+rw)\big )\cdot w\nonumber \\&=\frac{1}{r^{d-1}} \int _{\partial B_r(x)} \mathrm {d}S(v)\, \big ((\nabla \varphi )(v)\big )\cdot \frac{v-x}{r}\nonumber \\&=\frac{1}{r^{d-1}} \int _{\partial B_r(x)} \mathrm {d}S(v)\, \frac{\partial \varphi }{\partial \nu }(v), \end{aligned}$$

(A.11)

where $\nu$ is the outer normal vector. Green’s formula implies that

$$\begin{aligned} \int _{\partial B_r(x)} \mathrm {d}S(v)\, \frac{\partial \varphi }{\partial \nu }(v) = \int _{B_r(x)} \mathrm {d}y\, (\Delta \varphi ) (y). \end{aligned}$$

(A.12)

and therefore we obtain

$$\begin{aligned} \int _b^{\frac{|x|}{2}} \mathrm {d}r\, \frac{r^{d-1}}{r^{d+2\alpha }} \big (\phi (0)-\phi (r)\big ) = -\int _b^{\frac{|x|}{2}} \mathrm {d}r\, \frac{1}{r^{2\alpha +1}}\int _0^r \mathrm {d}s \frac{1}{s^{d-1}}\int _{B_s(x)} \mathrm {d}y\, (\Delta \varphi ) (y). \end{aligned}$$

(A.13)

This implies the bound

$$\begin{aligned}&\Big |\int _{B_{\frac{|x|}{2}}(x)\setminus B_b(x)} \mathrm {d}y\, \frac{\big (\varphi (x)-\varphi (y)\big )}{|x-y|^{d+2\alpha }}\Big |\nonumber \\&\leqslant C_{d}^{(3)} \sup _{y\in B_{\frac{|x|}{2}}(x)} |(\Delta \varphi )(y)| \int _b^{\frac{|x|}{2}} \mathrm {d}r\, \frac{1}{r^{2\alpha +1}}\int _0^r \mathrm {d}s\, s\nonumber \\&= \frac{C_{d}^{(3)}}{2} \sup _{y\in B_{\frac{|x|}{2}}(x)} |(\Delta \varphi )(y)| \int _b^{\frac{|x|}{2}} \mathrm {d}r\, r^{1-2\alpha } \end{aligned}$$

(A.14)

for some constant $C_{d}^{(3)}>0$ depending on the dimension only. Therefore, we end up for fixed $0\ne x\in {\mathbb {R}}^d$ with

$$\begin{aligned} \limsup _{b\rightarrow 0} \Big |\int _{B_{\frac{|x|}{2}}(x)\setminus B_b(x)} \mathrm {d}y\, \frac{\big (\varphi (x)-\varphi (y)\big )}{|x-y|^{d+2\alpha }}\Big | \leqslant C_{d}^{(4)} \sup _{y\in B_{\frac{|x|}{2}}(x)} |(\Delta \varphi )(y)| |x|^{2-2\alpha } \end{aligned}$$

(A.15)

for some constant $C_{d}^{(4)}>0$. Since $\varphi \in \mathscr {S}({\mathbb {R}}^d)$, we have

$$\begin{aligned} \limsup _{|x|\rightarrow \infty }\ C_{d}^{(4)} \sup _{y\in B_{\frac{|x|}{2}}(x)} |(\Delta \varphi )(y)| |x|^{2-2\alpha } = 0. \end{aligned}$$

(A.16)

Equations (A.4), (A.6), (A.9), (A.15) and (A.16) imply

$$\begin{aligned} \lim _{b\rightarrow 0}\int _{{\mathbb {R}}^d\setminus B_b(x)} \mathrm {d}y\, \frac{\big (\varphi (x)-\varphi (y)\big )}{|x-y|^{d+2\alpha }} = -\int _{{\mathbb {R}}^d} \mathrm {d}y\, \varphi (y) \end{aligned}$$

(A.17)

which inserted in (A.3) gives the result. $\square$