Absolutely continuous spectrum for Schrödinger operators with random decaying matrix potentials on the strip

Abstract

We consider a family of random Schrödinger operators on the discrete strip with decaying random $\ell ^2$ matrix potential. We prove that the spectrum is almost surely pure absolutely continuous, apart from random, possibly embedded, eigenvalues which may accumulate at band edges.

Appendix A. Transfer matrices and spectral averaging formula on the strip

As above, we consider operators of the form

$$\begin{aligned} (H\Psi )_n=-\Psi _{n-1}-\Psi _{n+1}+B _n \Psi _n \end{aligned}$$

on $\ell ^2({{\mathbb {Z}}}_+)\otimes {{\mathbb {C}}}^l$. Solving the eigenvalue equation $H\Psi =z\Psi$ leads to the transfer matrices

$$\begin{aligned} T^z_n=\begin{pmatrix} B_n-zI &{}\quad -I \\ I &{}\quad \textbf{0} \end{pmatrix} \quad \text {and the equation}\quad \begin{pmatrix} \Psi _{n+1} \\ \Psi _n \end{pmatrix}=T^z_n \begin{pmatrix} \Psi _n \\ \Psi _{n-1} \end{pmatrix} \end{aligned}$$

Then, for $n>m$ we define the products

$$\begin{aligned} T^z_{m,n}=T^z_n T^z_{n-1} \cdots T^z_{m+1} T^z_m \quad \text {leading to}\quad \begin{pmatrix} \Psi _{n+1} \\ \Psi _n \end{pmatrix}=T^z_{m,n} \begin{pmatrix} \Psi _m \\ \Psi _{m-1} \end{pmatrix} \end{aligned}$$

for a formal solution of $H\Psi =z\Psi$.

1.1 A.1. Transfer matrices and resolvent boundary data

Let $n>m$, be non-negative integers. With $H_{m,n}$, we denote the restriction of H to $\ell ^2(\{m,m+1,\ldots ,n\}) \otimes {{\mathbb {C}}}^l$, that is

$$\begin{aligned} H_{m,n}=\begin{pmatrix} B_m &{}\quad -I &{}\quad \\ -I &{}\quad B_{m+1} &{}\quad -I \\ {} &{}\quad \ddots &{}\quad \ddots &{}\quad \ddots \\ {} &{} &{} \quad \ddots &{}\quad \ddots &{}\quad -I \\ {} &{} &{} &{}\quad -I &{}\quad B_n \end{pmatrix} \end{aligned}$$

Then, we define the m to n boundary resolvent data for $z \not \in \sigma (H_{m,n})$ by

$$\begin{aligned} \begin{pmatrix} \alpha ^z_{m,n} &{}\quad \beta ^z_{m,n} \\ \gamma ^z_{m,n} &{}\quad \delta ^z_{m,n} \end{pmatrix}= \begin{pmatrix} P_m^* \\ P_n^* \end{pmatrix} (H_{m,n}-z)^{-1} \begin{pmatrix} P_m&\quad P_n \end{pmatrix} \end{aligned}$$

(A.1)

where $P_k$ is the natural embedding of $\ell ^2(\{k\})\otimes {{\mathbb {C}}}^l$ into $\ell ^2(\{m,m+1,\ldots ,n\})\otimes {{\mathbb {C}}}^l$ for $m\le k \le n$. This means, e.g., $\alpha ^z_{m,n}=P_m^* (H_{m,n}-z)^{-1} P_m$, and in this setup

$$\begin{aligned} P_m=\begin{pmatrix} I\\ \textbf{0}\\ \vdots \\ \textbf{0} \end{pmatrix}\,,\quad P_n=\begin{pmatrix} \textbf{0}\\ \vdots \\ \textbf{0}\\ I \end{pmatrix}\quad \in \;\;{{\mathbb {C}}}^{(n-m+1)l \times l}. \end{aligned}$$

Note that $\alpha ^z_{m,n}, \beta ^z_{m,n}, \gamma ^z_{m,n}, \delta ^z_{m,n}$ are all $l\times l$ matrices.

Proposition A.1

Let be given $n\ge m\in {{\mathbb {Z}}}_+$ and let $z\not \in \sigma (H_{m,n})$ and let $\beta ^z_{m,n}$ be invertible. Then,

$$\begin{aligned} T^z_{m,n}\,=\, \begin{pmatrix} (\beta ^z_{m,n})^{-1} &{} - (\beta ^z_{m,n})^{-1}\alpha ^z_{m,n}\\ \delta ^z_{m,n} (\beta ^z_{m,n})^{-1} &{} \gamma ^z_{m,n} - \delta ^z_{m,n} (\beta ^z_{m,n})^{-1} \alpha ^z_{m,n} \end{pmatrix} \end{aligned}$$

Proof

For $\Psi =(\Psi _n)_n$ with $\Psi _n \in {{\mathbb {C}}}^l$, we define the notations:

$$\begin{aligned} {\hat{\Psi }}_k:= \Psi _k, \quad k<m. \quad \quad \quad {\hat{\Psi }}_m:= \left( \begin{array}{cc} \Psi _m \\ \Psi _{m+1} \\ \vdots \\ \Psi _n \end{array} \right) , \qquad {\hat{\Psi }}_{m+1}: = \Psi _{n+1}, \end{aligned}$$

and we use $P_m$ and $P_n$ as in (A.1), then we have

$$\begin{aligned} \Psi _{m}=P_m^* {\hat{\Psi }}_m, \quad \Psi _n=P_n^* {\hat{\Psi }}_m . \end{aligned}$$

Moreover, let $\Phi =H\Psi$ and introduce similar notations ${\hat{\Phi }}_k$, then we get

$$\begin{aligned} {\hat{\Phi }}_m=H_{m,n}{\hat{\Psi }}_m-P_m {\hat{\Psi }}_{m-1}-P_n {\hat{\Psi }}_{m+1}. \end{aligned}$$

With z being the spectral parameter, ${\hat{\Phi }}_m = z {\hat{\Psi }}_m$ leads to

$$\begin{aligned} P_n {\hat{\Psi }}_{m+1} = (H_{m,n}-z){\hat{\Psi }}_m-P_m {\hat{\Psi }}_{m-1} \end{aligned}$$

Multiplying with $P_m^*(H_{m,n}-z)^{-1}$ from the left, noting that ${\hat{\Psi }}_{m+1}=\Psi _{n+1}$ and using (A.1) gives

$$\begin{aligned} \beta ^z_{m,n} \Psi _{n+1} =\Psi _m-\alpha ^z_{m,n} \Psi _{m-1} \implies \Psi _{n+1} =(\beta ^z_{m,n})^{-1}\Psi _m-(\beta ^z_{m,n})^{-1}\alpha ^z_{m,n} \Psi _{m-1} \end{aligned}$$

Multiplying from the left with $P_n^*(H_{m,n}-z)^{-1}$ instead of $P_m^*(H_{m,n}-z)^{-1})$ leads to

$$\begin{aligned} \delta ^z_{m,n} \Psi _{n+1}=\Psi _n-\gamma ^z_{m,n} \Psi _{m-1} \end{aligned}$$

Replacing $\Psi _{n+1}$ with the formula above and resolving for $\Psi _n$ leads to

$$\begin{aligned} \Psi _n=\delta ^z_{m,n} (\beta ^z_{m,n})^{-1}\Psi _m+(\gamma ^z_{m,n} -\delta ^z_{m,n} (\beta ^z_{m,n})^{-1} \alpha ^z_{m,n})\Psi _{m-1} \end{aligned}$$

Finally, we have:

$$\begin{aligned} \begin{pmatrix} \Psi _{n+1} \\ \Psi _{n} \end{pmatrix}=\begin{pmatrix} (\beta ^z_{m,n})^{-1} &{}\quad - (\beta ^z_{m,n})^{-1}\alpha ^z_{m,n}\\ \delta ^z_{m,n} (\beta ^z_{m,n})^{-1} &{}\quad \gamma ^z_{m,n} - \delta ^z_{m,n} (\beta ^z_{m,n})^{-1} \alpha ^z_{m,n} \end{pmatrix} \begin{pmatrix} \Psi _m \\ \Psi _{m-1} \end{pmatrix} \end{aligned}$$

As $\Psi _m, \Psi _{m-1}$ determine the solution to $H\Psi =z\Psi$ uniquely, the matrix must be $T^z_{m,n}$. $\square$

1.2 A.2. Spectral averaging formula

Here, we state the strip-equivalent of the spectral average formula from Carmona-Lacroix [7, Theorem III.3.2 and III.3.6]. It is a special case of [32, Theorem 1]. First, we need to fix a vector in the root-slice. Thus, we choose some ${\vec {x}}\in {{\mathbb {C}}}^l$ which we identify with $\delta _0 \otimes {\vec {x}}\in \ell ^2\{{{\textbf{Z}}}_+\} \otimes {{\mathbb {C}}}^l$. Let us assume that $\Vert {\vec {x}}\Vert =1$, so that ${{\vec {x}}}^* {\vec {x}}= 1$. Furthermore, identifying ${\vec {x}}\,^*$ with a linear map from ${{\mathbb {C}}}^l$ to ${{\mathbb {C}}}$, we have a $l-1$-dimensional kernel consisting of the vectors orthogonal to ${\vec {x}}$,

$$\begin{aligned} {{\mathbb {K}}}:=\hbox {ker}({\vec {x}}\,^*) = \{{\vec v}\in {{\mathbb {C}}}^l:\,{\vec {x}}\,^* {\vec v}= 0 \} \,=\,\{{\vec v}\in {{\mathbb {C}}}^l:\, {\vec {x}}\cdot {\vec v}= 0\}. \end{aligned}$$

Then, in this special case, the work of [32] simply replaces $T^z_0$ by the set of $2l \times 2$ matrices

$$\begin{aligned} {{\mathbb {T}}}^z_0\,=\, \left\{ \begin{pmatrix} (B_n-zI) ({\vec {x}}+{\vec v}) &{}\quad - {\vec {x}}+(B_n-zI) {\vec w}\\ {\vec {x}}+{\vec v}&{}\quad {\vec w} \end{pmatrix}\,:\, {\vec v},{\vec w}\in {{\mathbb {K}}}\; \right\} \;\subset \; {{\mathbb {C}}}^{2l \times 2}\;. \end{aligned}$$

(A.2)

Note that

$$\begin{aligned} T^z_0 = \begin{pmatrix} B_0 -zI &{}\quad -I \\ I &{}\quad \textbf{0} \end{pmatrix} \quad \text {and}\quad {{\mathbb {T}}}^z_0 = T^z_0\,\left\{ \begin{pmatrix} {\vec {x}}+{\vec v}&{}\quad {\vec w}\\ \textbf{0}&{}\quad {\vec {x}} \end{pmatrix}\,:\, {\vec v},{\vec w}\in {{\mathbb {K}}}\right\} \end{aligned}$$

(A.3)

where we adopt the notation that $T{{\mathbb {A}}}=\{TA:\, A \in {{\mathbb {A}}}\}$ for sets of matrices ${{\mathbb {A}}}$.

Moreover, we consider the spectral measure $\mu _{{\vec {x}}}$ at the vector ${\vec {x}}\equiv \delta _0 \otimes {\vec {x}}$, that means

$$\begin{aligned} \int f d\mu _{{\vec {x}}}\,=\, \langle \delta _0 \otimes {\vec {x}}, \, f(H)\,(\delta _0 \otimes {\vec {x}})\, \rangle . \end{aligned}$$

Now, using that the operator H cannot have compactly (finitely) supported eigenfunctions, Theorem 1 in [32] implies the following:

Proposition A.2

[32] In the sense of a weak limit for finite measures, one finds that

$$\begin{aligned} \textrm{d}\mu _{{\vec {x}}}(E)\,&=\,\lim _{n\rightarrow \infty } \,\frac{1}{\pi }\, \frac{\textrm{d} E}{ \min \limits _{T^E \in {{\mathbb {T}}}^E_0}\left\| T^E_{1,n} \, T^E \left( \begin{array}{cc} 1 \\ 0 \end{array} \right) \right\| ^2}\,=\, \lim _{n\rightarrow \infty } \frac{1}{\pi }\, \frac{\textrm{d}E}{\min \limits _{\vec v \in {{\mathbb {K}}}} \left\| T^E_{0,n} \left( \begin{array}{cc} {\vec {x}}+\vec v \\ 0 \end{array} \right) \right\| ^2 } \end{aligned}$$

Using the symplectic structure of the transfer matrices and the Banach–Alaoglu theorem, one can obtain a criterion for absolute continuity (see [32]).

Proposition A.3

If one finds $\vec {u}_{E,n} \in {{\mathbb {C}}}^m$ for $E \in (a,b)$, $n \in {{\mathbb {N}}}$, such that

$$\begin{aligned} \liminf _{n \rightarrow \infty } \;\int _a^b\, \left\| \,T^E_{0,n} \left( \begin{array}{cc} \vec {u}_{E,n} \\ {\vec {x}} \end{array} \right) \,\right\| ^4\;\textrm{d}E\,<\,\infty \end{aligned}$$

then, the measure $\mu _{{\vec {x}}}$ is absolutely continuous in the interval (a, b).

Proof

First, in [32] it was shown that the minimum $\min \limits _{\vec v \in {{\mathbb {K}}}} \left\| T^E_{0,n} \left( \begin{array}{cc} {\vec {x}}+\vec v \\ 0 \end{array} \right) \right\|$ is achieved at a very specific vector which we call $\vec {v}_{E,n}\in {{\mathbb {K}}}$. Defining

$$\begin{aligned} f_n(E):= \pi ^{-1} \left\| T^E_{0,n} \left( \begin{array}{cc} {\vec {x}}+\vec v_{E,n} \\ 0 \end{array} \right) \right\| ^{-2} \end{aligned}$$

we see from Proposition A.2 that $\mu _{{\vec {x}}}$ is the weak limit of $f_n(E) \textrm{d}E$ in the interval (a, b). Note that

$$\begin{aligned}&\left( T^E_{0,n} \left( \begin{array}{cc} \vec {u}_{E,n} \\ {\vec {x}} \end{array} \right) \right) ^* \left( \begin{array}{cc} \textbf{0}&{} -I \\ I &{} \textbf{0} \end{array} \right) T^E_{0,n} \left( \begin{array}{cc} {\vec {x}}+ \vec v_{E,n} \\ 0 \end{array} \right) \\&\quad =\, \left( \begin{array}{cc} \vec {u}^*_{E,n}&{\vec {x}}\,^* \end{array} \right) \, \big (T^E_{0,n} \big )^* \left( \begin{array}{cc} \textbf{0}&{} -I \\ I &{} \textbf{0} \end{array} \right) T^E_{0,n} \left( \begin{array}{cc} {\vec {x}}+ \vec v_{E,n} \\ 0 \end{array} \right) \\&\quad =\, \left( \begin{array}{cc} \vec {u}^*_{E,n}&{\vec {x}}\,^* \end{array} \right) \left( \begin{array}{cc} \textbf{0}&{} -I \\ I &{} \textbf{0} \end{array} \right) \left( \begin{array}{cc} {\vec {x}}+ \vec v_{E,n} \\ 0 \end{array} \right) \, =\, \left( \begin{array}{cc} \vec {u}^*_{E,n}&{\vec {x}}\,^* \end{array} \right) \left( \begin{array}{cc} 0 \\ {\vec {x}}+ \vec v_{E,n} \end{array} \right) \,=\,1 \end{aligned}$$

where we use $\Vert {\vec {x}}\Vert =1$ and ${\vec {x}}\,^* \vec v_{E,n}=0$ as $\vec v_{E,n} \in {{\mathbb {K}}}$. Now, using the Cauchy–Schwartz inequality, this gives

$$\begin{aligned} 1\,\le \, \left\| \,T^E_{0,n} \left( \begin{array}{cc} \vec {u}_{E,n} \\ {\vec {x}} \end{array} \right) \,\right\| \; \cdot \; \left\| T^E_{0,n} \left( \begin{array}{cc} {\vec {x}}+\vec v_{E,n} \\ 0 \end{array} \right) \right\| \end{aligned}$$

and hence

$$\begin{aligned} \pi ^2 |f_n(E)|^2\,=\,\frac{1}{\left\| T^E_{0,n} \left( \begin{array}{cc} {\vec {x}}+\vec v_{E,n} \\ 0 \end{array} \right) \right\| ^4}\,\le \, \left\| \,T^E_{0,n} \left( \begin{array}{cc} \vec {u}_{E,n} \\ {\vec {x}} \end{array} \right) \,\right\| ^4. \end{aligned}$$

Thus, the estimate given implies that

$$\begin{aligned} \liminf _{n\rightarrow \infty } \int _a^b |f_n(E)|^2\; \textrm{d}E\,<\, \infty . \end{aligned}$$

This means, along a suitable sub-sequence, the norm of $f_n$ in $L^2(a,b)$ is bounded. By Banach–Alaoglu, there is a sub-sequence ( o better, a sub-sub-sequence of the suitable sub-sequence) $f_{n_k}$ which converges weakly in $L^2(a,b)$ to a limit $f \in L^2(a,b)$. Noting that bounded continuous functions $g \in C_b(a,b)$ are also in $L^2(a,b)$, one has

$$\begin{aligned} \lim _{k \rightarrow \infty } \int _a^b g(E)\; f_{n_k}(E)\,\textrm{d}E\,=\, \int _a^b g(E) \,f(E)\,\textrm{d}E. \end{aligned}$$

for all $g\in C_b(a,b)$. But since $f_{n_k}(E)\textrm{d}E$ converges weakly to the measure $\mu _{{\vec {x}}}$, this means that in the interval (a, b) we have

$$\begin{aligned} \textrm{d}\mu _{{\vec {x}}}(E)\,=\,f(E)\,\textrm{d}E\; \end{aligned}$$

which is an absolutely continuous measure in (a, b) with a density in $L^2(a,b)$. $\square$