Lifshitz Tails for Random Diagonal Perturbations of Laurent Matrices

Abstract

We study the integrated density of states of one-dimensional random operators acting on $\ell ^2({\mathbb {Z}})$ of the form $T + V_\omega$ where T is a Laurent (also called bi-infinite Toeplitz) matrix and $V_\omega$ is an Anderson potential generated by i.i.d. random variables. We assume that the operator T is associated with a bounded, Hölder-continuous symbol f, that attains its minimum at a finite number of points. We allow for f to attain its minima algebraically. The resulting operator T is long-range with weak (algebraic) off-diagonal decay. We prove that this operator exhibits Lifshitz tails at the lower edge of the spectrum with an exponent given by the integrated density of states of T at the lower spectral edge. The proof relies on generalizations of Dirichlet–Neumann bracketing to the long-range setting and an adaption of Temple’s inequality.

Appendix A

1.1 A.1 Off-Diagonal Decay of Laurent Matrices

Lemma A.1

Let $T_f$ be Laurent operator associated with a symbol f satisfying Assumptions (A1)–(A2). The matrix entries of $T_f$ decay as

$$\begin{aligned} |{a_{n-m}}|\leqslant \frac{1}{|{n - m}|^{1+\nu }}, \quad m, n \in {\mathbb {Z}} \end{aligned}$$

and therefore, $T_f$ is an, at most, long-range operator with polynomially decaying off-diagonal terms.

Proof

To see this, note that $f\in C^{1}({\mathbb {T}})$ except at finitely many points $x_j\in {\mathbb {T}}$, $j=1,..,m$ for some $m\in {\mathbb {N}}$ where f remains $\nu$-Hölder continuous for some $\nu >0$ at these points. An adaption of the proof of [6, Thm. 2.2] implies for symbols $h\in C^1({\mathbb {T}}) {\setminus }\{y\}\cap C^{\alpha }({\mathbb {T}})$ that

$$\begin{aligned} |T_h(n,m)|\leqslant \frac{C}{|n-m|^{1+\nu }}, \quad n,m\in {\mathbb {Z}}\end{aligned}$$

(A.1)

for some $C>0$. We write $g = \sum _{j=1}^m h_j$ with $h_j \in C^1({\mathbb {T}}){\setminus }\{x_j\} \cap C^\alpha ({\mathbb {T}})$. Now, we apply inequality (A.1) to each function $h_j$ and since there is only a finite number of those we end up with the result. $\square$

Proposition A.2

Let $T_f$ be the Laurent operator associated with a symbol f satisfying Assumptions (A1)–(A2). The limit

$$\begin{aligned} \widetilde{I}_f(E):=\lim _{L\rightarrow \infty } \frac{{{\,\mathrm{Tr}\,}}\big ( 1_{[-L,L]} 1_{\leqslant E} (T_{f}) \big )}{2L+1} \end{aligned}$$

(A.2)

exists and equals the IDS of $T_f$, $I_f(E)$, for all $E\in {\mathbb {R}}$, where $I_f$ is defined in (2.8). Moreover,

$$\begin{aligned} I_f(E)= \frac{1}{2\pi }\big | \big \{ k\in [-\pi ,\pi ]: \ f(k) \leqslant E\big \}\big |, \end{aligned}$$

(A.3)

Proof

Using the off-diagonal decay of $T_f$ shown above, we can follow the arguments in [6, Proposition 2.1] to show that $\widetilde{I}_f$ and $I_f$ are identical for all $E\in {\mathbb {R}}$.

Next, let $\tau _x:\ell ^2({\mathbb {Z}})\rightarrow \ell ^2({\mathbb {Z}})$ be the translation by $x\in {\mathbb {Z}}$ acting on $\ell ^2({\mathbb {Z}})$ by $\tau _x \varphi (n)=\varphi (n-x)$, for $\varphi \in \ell ^2({\mathbb {Z}})$. The translation invariance of $T_f$ implies that

$$\begin{aligned} \lim _{L\rightarrow \infty } \frac{{{\,\mathrm{Tr}\,}}\big ( 1_{[-L,L]} 1_{\leqslant E} (T_{f}) \big )}{2L+1}\!=\!\frac{1}{2L+1} \sum _{x\in [-L,L]}\langle {\delta _x,1_{\leqslant E} (T_{f})\delta _x}\rangle \!=\! \langle \delta _0, 1_{\leqslant E} (T_{f})\delta _0 \rangle \end{aligned}$$

(A.4)

Using Fourier transform in the r.h.s. of the last line and the fact that $T_f$ is unitary equivalent to the operator $M_f$, multiplication by f, yields (A.3). $\square$

1.2 A.2 Properties of the Ground-State Space ${\mathcal {G}}$ of Neumann Restrictions of Toeplitz Matrices

Let T be a Laurent matrix with bandwidth N associated with a symbol g of the form (4.3). Consider its restriction to the cube $\Lambda _L=[-L,L]\cap {\mathbb {Z}}$, with $L\in {\mathbb {N}}$, $L>2N+1$ with modified Neumann boundary conditions (see Sect. 3), denoted by $T_L^{{\mathcal {N}}}$. We define its ground-state space by

$$\begin{aligned} {\mathcal {G}}:=\big \{\varphi \in \ell ^2(\Lambda _L): T_L^{{\mathcal {N}}}\varphi = 0\big \} \end{aligned}$$

(A.5)

We recall from [5, Section 5] (see Sect. 3) that the ground-state space ${\mathcal {G}}$ is spanned by

$$\begin{aligned} {\mathcal {G}} = \text {span} \big \{\varphi ^j_{k,L} \in \ell ^2(\Lambda _L):\ k=1,\ldots ,M,\,\,j=0,\ldots , \overline{\alpha }-1 \big \} \end{aligned}$$

(A.6)

with

$$\begin{aligned} \varphi ^j_{k,L} := \frac{1}{K_{j,L}^{1/2}} \big ( (-L)^j e^{i L E_k },\ldots ,0^j,1^j e^{-i E_k},\ldots , L^je^{-i L E_k}\big )^T\in {\mathbb {R}}^{2L+1} \end{aligned}$$

(A.7)

for $k=1,\ldots ,M$ and $j=0,\ldots ,\overline{\alpha }-1$, and

$$\begin{aligned} K_{j,L} = 2\sum _{m=-L}^L |m|^{2j}. \end{aligned}$$

(A.8)

Lemma A.3

Let $\varphi \in {\mathcal {G}}$. Then, there exists a constant $C>0$ such that for all $L\in {\mathbb {N}}$, $r\ne k$ and $j,s\in \{0,\ldots ,\overline{\alpha }\}$

$$\begin{aligned} |\langle \varphi _{k,L}^j,\varphi _{r,L}^s\rangle | \leqslant \frac{C}{|\Lambda _L|}. \end{aligned}$$

(A.9)

Proof

We compute

$$\begin{aligned} |\langle \varphi _{k,L}^j,\varphi _{r,L}^s\rangle | = \frac{1}{(K_{j,L}K_{s,L})^{1/2}}\Big | \sum _{m=-L}^L m^{j+s} e^{i m (E_k - E_r)}\Big | \end{aligned}$$

(A.10)

Using summation by parts $j+s$ times and $|\sum _{m=-L}^L e^{i m (E_k - E_r)}| = O(1)$, we obtain that

$$\begin{aligned} \Big |\sum _{m=-L}^L m^{j+s} e^{i m (E_k - E_r)}\Big | = O(L^{j+s}). \end{aligned}$$

(A.11)

Now, $K_{j,L}K_{s,L} = O(L^{2j+2s+2})$ implies

$$\begin{aligned} |\langle \varphi _{k,L}^j,\varphi _{r,L}^s\rangle | = O(L^{-1}) \end{aligned}$$

(A.12)

and therefore there exists a constant $C_{k,j,r,s}$ depending on k, j, r, s but independent of L such that

$$\begin{aligned} |\langle \varphi _{k,L}^j,\varphi _{r,L}^s\rangle | \leqslant \frac{C_{k,j,r,s}}{L}. \end{aligned}$$

(A.13)

Taking $C:=\displaystyle \max _{j,s,r\ne k} C_{k,j,r,s}$ gives the assertion. $\square$

Lemma A.4

There exists $L_0$ such that for all $L\geqslant L_0$, $l\in \Lambda _L$ and $\varphi \in {\mathcal {G}}$

$$\begin{aligned} |\varphi (l)| \leqslant \frac{\sqrt{2N}}{\sqrt{|\Lambda _L|}}. \end{aligned}$$

(A.14)

Proof

Let $\varphi \in {\mathcal {G}}$ with $\Vert \varphi \Vert =1$. Then, $\varphi =\displaystyle \sum _{k=0}^{n-1}\sum _{j=0}^{\overline{\alpha } - 1} a_{j,k} \varphi _{k,L}^j$ for some $a_{j,k}\in {\mathbb {C}}$ and $\varphi _{k,L}^j$ given in (A.7). We compute

$$\begin{aligned} \langle \varphi ,\widetilde{V}_\omega \varphi \rangle = \sum _{l\in \Lambda _L} \tilde{V}_\omega (l) |\varphi (l)|^2 \end{aligned}$$

(A.15)

where $\varphi (l)$ stands for the lth component of the vector $\varphi$. The last lemma implies that $|\langle \varphi _{k,L}^j,\varphi _{r,L}^s\rangle | \leqslant \frac{C}{|\Lambda _L|}$ for some constant $C>0$ and all $k\ne r$. Hence, we obtain

$$\begin{aligned} 1&= \sum _{l\in \Lambda _L}|\varphi (l)|^2 = \sum _{k,r=0}^{n-1}\sum _{j,s=0}^{\overline{\alpha } - 1} \overline{a}_{k,j} a_{r,s} \langle \varphi _{k,L}^j, \varphi _{r,L}^s\rangle \nonumber \\&\geqslant \sum _{k=0}^{n-1}\sum _{j=0}^{\overline{\alpha } - 1} |a_{k,j}|^2 - \sum _{(k,j)\ne (r,s)} |a_{k,j}||a_{r,s}| \frac{C}{|\Lambda _L|} \nonumber \\&\geqslant \sum _{k=0}^{n-1}\sum _{j=0}^{\overline{\alpha } - 1} |a_{k,j}|^2 - \frac{1}{2}\sum _{(k,j)\ne (r,s) } \big (|a_{k,j}|^2 + |a_{r,s}|^2 \big ) \frac{C}{|\Lambda _L|} \nonumber \\&\geqslant \sum _{k=0}^{n-1}\sum _{j=0}^{\overline{\alpha } - 1} |a_{k,j}|^2 \Big ( 1- \frac{C N}{|\Lambda _L|}\Big ) \end{aligned}$$

(A.16)

where we used the inequality $|xy|\leqslant \frac{1}{2}(|x|^2 + |y|^2)$ for $x,y\in {\mathbb {R}}$. We choose $L_0\in {\mathbb {N}}$ such that for all $L\geqslant L_0$ we obtain $\frac{C N}{|\Lambda _L|}\leqslant \frac{1}{2}$ and therefore

$$\begin{aligned} \sum _{k=0}^{n-1}\sum _{j=0}^{\overline{\alpha } - 1} |a_{k,j}|^2 \leqslant 2. \end{aligned}$$

(A.17)

This implies

$$\begin{aligned} |\varphi (l)| \leqslant \frac{1}{\sqrt{|\Lambda _L|}} \sum _{k=0}^{n-1}\sum _{j=0}^{\overline{\alpha } - 1} |a_{k,j}| \leqslant \frac{\sqrt{N}}{\sqrt{|\Lambda _L|}} \big (\sum _{k=0}^{n-1}\sum _{j=0}^{\overline{\alpha } - 1} |a_{k,j}|\big )^{1/2} \leqslant \frac{\sqrt{2N}}{\sqrt{|\Lambda _L|}}. \end{aligned}$$

(A.18)

$\square$