1 Introduction

The weak limit theorems for discrete time quantum walks have been studied in various models (for reviews, see [7, 12]). In his papers [5, 6], Konno first proved the weak limit theorem for a position-independent quantum walk on \(\mathbb {Z}\). Grimmett et al. [4] simplified the proof and extended the result to higher dimensions. For position-dependent qunatum walks on \(\mathbb {Z}\), the weak limit theorems were obtained by Konno et al. [9], Endo and Konno [2], and Endo et al. [3].

We consider a position-dependent quantum walk on \(\mathbb {Z}\) given by a unitary evolution operator U:

$$\begin{aligned} (U\Psi )(x) = P(x+1)\Psi (x+1) + Q(x-1)\Psi (x-1), \quad x \in \mathbb {Z}, \end{aligned}$$

where \(\Psi \) is a state vector in the Hilbert space \(\mathcal {H} = \ell ^2(\mathbb {Z};\mathbb {C}^2)\) of states and

$$\begin{aligned} P(x) = \begin{pmatrix} a(x) &{} b(x) \\ 0 &{} 0 \end{pmatrix}, \quad Q(x) = \begin{pmatrix} 0 &{} 0 \\ c(x) &{} d(x) \end{pmatrix} . \end{aligned}$$

Let \(C(x) = P(x) + Q(x) \in U(2)\) and S be a shift operator such that \(U = SC\). Suppose that there exists a unitary matrix \(C_0 = P_0 + Q_0 \in U(2)\) such that

$$\begin{aligned} \Vert C(x) - C_0\Vert \le c_1 |x|^{-1-\epsilon }, \quad x \in \mathbb {Z} \setminus \{0\} \end{aligned}$$
(1.1)

with positive \(c_1\) and \(\epsilon \) independent of x. Here \(\Vert M\Vert \) stands for the operator norm of a matrix \(M \in M_2(\mathbb {C})\). A typical example is the quantum walks with one defect [1, 8, 9, 13], which clearly satisfies (1.1). We note that the condition (1.1) allows not only finite but also infinite defects, whereas the models introduced in [2, 3] do not satisfy (1.1). The unitary operator \(U_0 = SC_0\) also defines an evolution of a position-independent quantum walk on \(\mathbb {Z}\) and satisfies

$$\begin{aligned} (U_0\Psi )(x) = P_0 \Psi (x+1) + Q_0\Psi (x-1), \quad x \in \mathbb {Z} \end{aligned}$$

with \(C_0 = P_0 + Q_0\). Let \(\hat{x}\) be the position operator defined by \((\hat{x} \Psi )(x) = x \Psi (x), \quad x \in \mathbb {Z}\). and \(\hat{x}_0(t) = U_0^{-t} \hat{x} U_0^t\) the Heisenberg operator of \(\hat{x}\) at time \(t \in \mathbb {N}\) with the evolution \(U_0\). In Grimmett et al. [4] essentially proved that the operator \(\hat{x}_0(t)/t\) weakly converges to the asymptotic velocity operator \(\hat{v}_0\) so that

$$\begin{aligned} \text{ w- }\lim _{t \rightarrow \infty } \mathrm{exp} \left( i \xi \frac{\hat{x}_0(t)}{t} \right) = \mathrm{exp} \left( {i \xi \hat{v}_0} \right) , \quad \xi \in \mathbb {R}. \end{aligned}$$
(1.2)

Let \(X^{(0)}_t\) be the random variable denoting the position of a quantum walker at time \(t \in \mathbb {N}\) with the evolution operator \(U_0\). Then, the characteristic function of \(X^{(0)}_t/t\) is given by

$$\begin{aligned} \mathbb {E}\left( e^{i \xi X^{(0)}_t/t}\right) = \left\langle \Psi _0, e^{i\xi \hat{x}_0(t)/t} \Psi _0 \right\rangle , \quad \xi \in \mathbb {R}, \end{aligned}$$

where \(\Psi _0\) is the initial state of the quantum walker. Hence, (1.2) means that the random variable \(X^{(0)}_t/t\) converges in law to a random variable \(V_0\), which represents the linear spreading of the quantum walk: \(X^{(0)}_t \sim t V_0\).

In this paper, we derive the asymptotic velocity \(\hat{v}_+\) for the Heisenberg operator \(\hat{x}(t) = U^{-t} \hat{x} U^t\) with the evolution U of the position-dependent quantum walk. The decaying condition (1.1) implies that \(U-U_0\) is a trace class operator and allows us to prove the existence and completeness of the wave operator

$$\begin{aligned} W_+ = \text{ s- }\lim _{t \rightarrow \infty } U^{-t}U_0^t\Pi _\mathrm{ac}(U_0) \end{aligned}$$

using a discrete analogue of the Kato–Rosenblum Theorem (see [11] for details), where \(\Pi _\mathrm{ac}(U_0)\) is the orthogonal projection onto the subspace of absolute continuity of \(U_0\). We also prove that

$$\begin{aligned} \text{ s- }\lim _{t \rightarrow \infty } \mathrm{exp} \left( i \xi \frac{\hat{x}_0(t)}{t} \right) = \mathrm{exp} \left( {i \xi \hat{v}_0} \right) , \quad \xi \in \mathbb {R} \end{aligned}$$

under a reasonable condition, which is essentially the same as that of [4]. Furthermore, we assume that U has no singular continuous spectrum. Then, we prove that

$$\begin{aligned} \text{ s- }\lim _{t \rightarrow \infty } \mathrm{exp}\left( i \xi \frac{\hat{x}(t)}{t} \right) = \Pi _\mathrm{p}(U) + \mathrm{exp}(i \xi \hat{v}_+) \Pi _\mathrm{ac}(U), \end{aligned}$$
(1.3)

where \(\Pi _\mathrm{p}(U)\) is the orthogonal projection onto the direct sum of all eigenspaces of U and \(\hat{v}_+ = W_+ \hat{v}_0 W_+^*\). We believe that the absence of a singular continuous spectrum can be checked with a concrete example such as the one-defect model. As a consequence of (1.3), we have the following weak limit theorem. Let \(X_t\) be the random variable denoting the position of a quantum walker at time \(t \in \mathbb {N}\) with the evolution operator U and the initial state \(\Psi _0\). We prove that \(X_t/t\) converges in law to a random variable V with a probability distribution

$$\begin{aligned} \mu _V = \Vert \Pi _\mathrm{p}(U)\Psi _0\Vert ^2\delta _0 + \Vert E_{\hat{v}_+}(\cdot ) \Pi _\mathrm{ac}(U)\Psi _0\Vert ^2, \end{aligned}$$

where \(\delta _0\) is the Dirac measure at zero and \(E_{\hat{v}_+}\) the spectral measure of \(\hat{v}_+\).

The remainder of this paper is organized as follows. In Sect. 2, we present the precise definition of the model and our results. Section 3 is devoted to the proof of the existence and completeness of the wave operator. In Sect. 4, we construct the asymptotic velocity.

2 Definition of the model

Let \(\mathcal {H} = \ell ^2(\mathbb {Z};\mathbb {C}^2)\) be the Hilbert space of the square summable functions \(\Psi :\mathbb {Z} \rightarrow \mathbb {C}^2\). We define a shift operator S and a coin operator C on \(\mathcal {H}\) as follows. For a vector \(\Psi = \begin{pmatrix} \Psi ^{(0)} \\ \Psi ^{(1)} \end{pmatrix} \in \mathcal {H}\), \(S\Psi \) is given by

$$\begin{aligned} (S\Psi )(x) = \begin{pmatrix} \Psi ^{(0)}(x+1) \\ \Psi ^{(1)}(x-1) \end{pmatrix}, \quad x \in \mathbb {Z}. \end{aligned}$$

Let \(\{C(x)\}_{x \in \mathbb {Z}} \subset U(2)\) be a family of unitary matrices with

$$\begin{aligned} C(x) =\begin{pmatrix} a(x) &{} b(x) \\ c(x) &{} d(x) \end{pmatrix}. \end{aligned}$$

\(C\Psi \) is given by

$$\begin{aligned} (C \Psi )(x) = C(x) \Psi (x), \quad x \in \mathbb {Z}. \end{aligned}$$

We define an evolution operator as \(U = SC\). U satisfies

$$\begin{aligned} (U\Psi )(x) = P(x+1)\Psi (x+1) + Q(x-1)\Psi (x-1), \quad x \in \mathbb {Z} \end{aligned}$$

with

$$\begin{aligned} P(x) = \begin{pmatrix} a(x) &{} b(x) \\ 0 &{} 0 \end{pmatrix}, \quad Q(x) = \begin{pmatrix} 0 &{} 0 \\ c(x) &{} d(x) \end{pmatrix}. \end{aligned}$$

For a matrix \(M \in M(2,\mathbb {C})\), we use \(\Vert M\Vert \) to denote the operator norm in \(\mathbb {C}^2\): \(\Vert M\Vert = \sup _{\Vert {\varvec{x}}\Vert _{\mathbb {C}^2}=1} \Vert M{\varvec{x}}\Vert _{\mathbb {C}^2}\). We suppose that:

  1. (A.1)

    There exists a unitary matrix \(C_0 = \begin{pmatrix} a_0 &{} b_0 \\ c_0 &{} d_0 \end{pmatrix}\in U(2)\) such that

    $$\begin{aligned} \Vert C(x) - C_0\Vert \le c_1|x|^{-1-\epsilon }, \quad x \in \mathbb {Z} \setminus \{0\} \end{aligned}$$

    with some positive \(c_1\) and \(\epsilon \) independent of x.

We denote by \(\mathscr {T}_1\) the set of trace class operators.

Lemma 2.1

Let U satisfy (A.1) and set \(U_0 = SC_0\). Then, \(U - U_0 \in \mathscr {T}_1\).

Proof

Let \(T = U-U_0\) and \(T(x) = C(x) - C_0\). Then,

$$\begin{aligned} T^*T = (C-C_0)^*(C-C_0) \end{aligned}$$
(2.1)

is the multiplication operator by the matrix-valued function \(T(x)^*T(x)\). Let \(t_i(x)\) (\(i=1,2\)) be the eigenvalues of the Hermitian matrix \(T(x)^*T(x) \in M(2,\mathbb {C})\) and take an orthonormal basis (ONB) \(\{ \tau _i(x)\}_{i=1,2}\) of corresponding eigenvectors for all \(x \in \mathbb {Z}\). We use \(|\xi \rangle \langle \eta |\) to denote the operator on \(\mathcal {H}\) defined by \(|\xi \rangle \langle \eta |\Psi = \langle \eta , \Psi \rangle \xi \). Then, we have

$$\begin{aligned} T^* T = \sum _{i=1,2} \sum _{x \in \mathbb {Z}} t_i(x) |\tau _{i,x} \rangle \langle \tau _{i,x}|, \end{aligned}$$
(2.2)

where \(\{\tau _{i,x}\}\) is the ONB given by

$$\begin{aligned} \tau _{i,x}(y) = \delta _{xy} \tau _i(x), \quad y \in \mathbb {Z}. \end{aligned}$$

Since \(T^*(x)T(x) \ge 0\), we have \(t_i(x) \ge 0\). By (A.1), we know that

$$\begin{aligned} \max _{i=1,2} t_i(x) \le c_1^2 |x|^{-2-2\epsilon }. \end{aligned}$$

Hence, we have

$$\begin{aligned} \mathrm{Tr}|T| = \sum _{x \in \mathbb {Z}} \sum _{i=1,2} t_i(x)^{1/2} \le 2 c_1 \sum _{x \in \mathbb {Z}} |x|^{-1-\epsilon } < \infty , \end{aligned}$$

which means that \(T \in \mathscr {T}_1\). Since \(\mathscr {T}_1\) is an ideal, \(U-U_0 = ST \in \mathscr {T}_1\). \(\square \)

Example 2.1

(one-defect model) Let \(C_0, C_0^\prime \in U(2)\) be unitary matrices with \(C_0 \not = C_0^\prime \) and set

$$\begin{aligned} C(x) = {\left\{ \begin{array}{ll} C_0^\prime , &{} x = 0 \\ C_0, &{} x \not =0. \end{array}\right. } \end{aligned}$$

\(U = SC\) satisfies (A.1), because \(C(x) - C_0 = 0\) if \(x \not = 0\).

Example 2.2

Let \(C_0 \in U(2)\) be a unitary matrix and \(\{C(x)\} \subset U(2)\) a family of unitary matrices. Assume that

$$\begin{aligned} \max _{i,j} |(C(x) - C_0)_{ij}| \le c_1|x|^{-1-\epsilon }, \quad x \in \mathbb {Z} \setminus \{0\}, \end{aligned}$$

where \(M_{ij}\) denotes the ij-component of a matrix M. Then, \(U = SC\) satisfies (A.1), because all norms on a finite-dimensional vector space are equivalent.

We prove the following theorem in Sect. 3 using a discrete analogue of the Kato–Rosenblum theorem.

Theorem 2.1

Let U and \(U_0\) be as above and assume that (A.1) holds. Then,

$$\begin{aligned} W_+ = \text{ s- }\lim _{t \rightarrow \infty } U^{-t} U_0^t \Pi _\mathrm{ac}(U_0) \end{aligned}$$

exists and is complete.

In what follows, we introduce the asymptotic velocity \(\hat{v}_0\), obtained first in [4], of the quantum walk with the evolution \(U_0\) as follows. Let

$$\begin{aligned} \hat{U}_0(k) = \begin{pmatrix} e^{ik} &{} 0 \\ 0 &{} e^{-ik} \end{pmatrix} C_0, \quad k \in [0,2\pi ). \end{aligned}$$

Since \(\hat{U}_0(k) \in U(2)\), \(\hat{U}_0(k)\) is represented as

$$\begin{aligned} \hat{U}_0(k) = \sum _{i=1,2} \lambda _i(k) |u_j(k) \rangle \langle u_j(k)|, \end{aligned}$$

where \(\lambda _j(k)\) is an eigenvalue of \(\hat{U}_0(k)\) and \(u_j(k)\) is the corresponding eigenvector with \(\Vert u_j(k)\Vert =1\). The function \(k \mapsto e^{ik}\) is analytic, and so is \(\lambda _j(k)\). We need the following assumption on \(u_j(k)\):

  1. (A.2)

    The functions \(k \mapsto u_j(k)\) are continuously differentiable in k with

    $$\begin{aligned} \sup _{k \in [0,2\pi )} \left\| \frac{\hbox {d}}{\hbox {d}k} u_j(k) \right\| _{\mathbb {C}^2} < \infty . \end{aligned}$$

Let \(\mathcal {K}\) be the Hilbert space of square integrable functions \(f{:} [0,2\pi ) \rightarrow \mathbb {C}^2\) with norm

$$\begin{aligned} \Vert f\Vert _{\mathcal {K}} = \left( \int _0^{2\pi } \frac{\hbox {d}k}{2\pi } \Vert f(k)\Vert _{\mathbb {C}^2}^2 \right) ^{1/2}. \end{aligned}$$

Let \(\mathcal {F}_0{:} \mathcal {H} \rightarrow \mathcal {K}\) be the discrete Fourier transform given by

$$\begin{aligned} (\mathscr {F}\Psi )(k) = \sum _{x \in \mathbb {Z}} e^{-ik\cdot x} \Psi (x), \quad \Psi \in \mathcal {H}. \end{aligned}$$

We also use \(\hat{\Psi }(k) = \begin{pmatrix} \hat{\Psi }^{(0)}(k) \\ \hat{\Psi }^{(1)}(k) \end{pmatrix}\) to denote the Fourier transform of \(\Psi \). The asymptotic velocity \(\hat{v}_0\) is the self-adjoint operator defined by

$$\begin{aligned} \hat{v}_0 = \mathscr {F}^{-1} \left( \int _{[0,2\pi )}^\oplus \frac{\hbox {d}k}{2\pi } \sum _{j=1,2} \left( \frac{ i \lambda _j^\prime (k) }{\lambda _j(k)} \right) |u_j(k) \rangle \langle u_j(k)| \right) \mathscr {F} \end{aligned}$$

The position operator \(\hat{x}\) is a self-adjoint operator defined by

$$\begin{aligned} (\hat{x} \Psi )(x) = x \Psi (x), \quad x \in \mathbb {Z} \end{aligned}$$

with domain

$$\begin{aligned} D(\hat{x}) = \left\{ \Psi \in \mathcal {H} ~\Big |~ \sum _{x \in \mathbb {Z}} |x|^2 \Vert \Psi (x)\Vert _{\mathbb {C}^2}^2 < \infty \right\} . \end{aligned}$$

Let \(\hat{x}_0(t) = U_0^{-t} \hat{x} U_0^t\) be the Heisenberg operator of \(\hat{x}\) for the evolution \(U_0\).

Theorem 2.2

Let \(\hat{v}_0\) and \(\hat{x}_0\) be as above. Suppose that (A.2) holds. Then,

$$\begin{aligned} \text{ s- }\lim _{t \rightarrow \infty } \mathrm{exp}\left( i \xi \frac{\hat{x}_0(t)}{t} \right) = \mathrm{exp}(i \xi \hat{v}_0 ), \quad \xi \in \mathbb {R}. \end{aligned}$$
(2.3)

Proof

By [10, TheoremVIII.21], (2.3) holds if and only if

$$\begin{aligned} \text{ s- }\lim _{t \rightarrow \infty } \left( \frac{\hat{x}_0(t)}{t} - z \right) ^{-1} = (\hat{v}_0 - z)^{-1}, \quad z \in \mathbb {C} \setminus \mathbb {R}, \end{aligned}$$

which is proved in Sect. 4.1. \(\square \)

Example 2.3

  1. (i)

    Let \(C_0 = \begin{pmatrix} 0 &{} 1 \\ 1 &{} 0 \end{pmatrix}\). Then, \(\hat{U}_0(k)\) has eigenvalues 1 and \(-1\), which are independent of k. By definition, \(\hat{v}_0\) = 0. Hence, the random variable \(X^{(0)}_t/t\) converges in law to a random variable \(V_0\) with a probability distribution \(\delta _0\).

  2. (ii)

    Let \(C_0 = \begin{pmatrix} 1 &{} 0 \\ 0 &{} -1 \end{pmatrix}\). \(\hat{U}_0(k)\) has eigenvalues \(e^{ik}\) and \(-e^{-ik}\). Hence, \(\hat{v}_0\) has eigenvalues \(-1\) and 1. The random variable \(X^{(0)}_t/t\) converges in law to a random variable \(V_0\) with a probability distribution \(\Vert \Psi ^{(0)}\Vert ^2 \delta _{-1} + \Vert \Psi ^{(1)}\Vert ^2 \delta _1\).

  3. (iii)

    Let \(C_0\) be the Hadamard matrix. The eigenvalues of \(\hat{U}_0(k)\) are given by \(\lambda _j(k) = ((-1)^j w(k) + i \sin k)/\sqrt{2}\) (\(j=1,2\)), where \(w(k) = \sqrt{1+\cos ^2 k}\). Hence, \(\hat{v}_0\) has no eigenvalue. The corresponding eigenvectors

    $$\begin{aligned} u_j(k) = \sqrt{\frac{w(k)+(-1)^j \cos k}{2w(k)}} \begin{pmatrix} e^{ik} \\ (-1)^j w(k) - \cos k \end{pmatrix} \end{aligned}$$

    form an ONB of \(\mathbb {C}^2\) and satisfy (A.2). The random variable \(X^{(0)}_t/t\) converges in law to a random variable \(V_0\) with a probability distribution \(\Vert E_{\hat{v}_0}(\cdot )\Psi _0\Vert ^2\), where \(E_{\hat{v}_0}\) is the spectral measure of \(\hat{v}_0\). Let us consider the Hadmard walk starting from the origin. Let the initial state \(\Psi _0\) satisfy \(\Psi _0(0)= \begin{pmatrix} \alpha \\ \beta \end{pmatrix}\) (\(|\alpha |^2 + |\beta |^2=1\)) and \(\Psi (x) = 0\) if \(x\not =0\). Then,

    $$\begin{aligned} d \Vert E_{\hat{v}_0}(v)\Psi _0 \Vert ^2 = (1 - c_{\alpha ,\beta } v) f_K\left( v;\frac{1}{\sqrt{2}}\right) \hbox {d}v, \end{aligned}$$

    where \(c_{\alpha , \beta } = |\alpha |^2-|\beta |^2+\alpha \bar{\beta } + \bar{\alpha } \beta \),

    $$\begin{aligned} f_K(v;r) = \frac{\sqrt{1-r^2}}{\pi (1-v^2)\sqrt{r^2-v^2}}I_{(-r,r)}(v) \end{aligned}$$

    is the Konno function, and \(I_A\) is the indicator function of a set A. For more details, the reader can consult [4, 7].

Let \(\hat{x}(t) = U^{-t} \hat{x} U\) be the Heisenberg operator of \(\hat{x}\) and define the asymptotic velocity \(\hat{v}_+\) for the evolution U by

$$\begin{aligned} \hat{v}_+ = W_+ \hat{v}_0 W_+^*. \end{aligned}$$

We need the following assumption:

  1. (A.3)

    The singular continuous spectrum of U is empty.

We are now in a psition to state our main result, which is proved in Sect. 4.2.

Theorem 2.3

Let \(\hat{x}(t)\) and \(\hat{v}_+\) be as above. Suppose that (A.1)–(A.3) hold. Then,

$$\begin{aligned} \text{ s- }\lim _{t \rightarrow \infty } \mathrm{exp} \left( i \xi \frac{\hat{x}(t)}{t} \right) = \Pi _\mathrm{p}(U) +\mathrm{exp} \left( i \xi \hat{v}_+ \right) \Pi _\mathrm{ac}(U), \quad \xi \in \mathbb {R}. \end{aligned}$$

Let \(X_t\) be the random variable denoting the position of the walker at time \(t \in \mathbb {N}\) with the initial state \(\Psi _0\). We use \(\Pi _\mathrm{p}(U)\) to denote the orthogonal projection onto the direct sum of all eigenspaces of U and \(E_{A}\) to denote the spectral projection of a self-adjoint operator A.

Corollary 2.4

Let \(X_t\) be as above. Suppose that (A.1)–(A.3) hold. Then, \(X_t/t\) converges in law to a random variable V with a probability distribution

$$\begin{aligned} \mu _V = \Vert \Pi _\mathrm{p}(U)\Psi _0\Vert ^2 \delta _0 + \Vert E_{\hat{v}_+}(\cdot )\Pi _\mathrm{ac}(U)\Psi _0\Vert ^2, \end{aligned}$$

where \(\delta _0\) is the Dirac measure at zero.

Proof

From Theorem 2.1, \(\text{ s- }\lim _{t \rightarrow \infty } U_0^{-t} U^t ~\Pi _\mathrm{ac}(U)\) exists and is equal to \(W_+^*\). Then, \(W_+\) is unitary from \(\mathrm{Ran}W_+^* = \mathrm{Ran}\Pi _\mathrm{ac}(U_0)\) to \(\mathrm{Ran}W_+ = \mathrm{Ran}\Pi _\mathrm{ac}(U)\). Since, by Lemma 4.1, \(U_0\) is strongly commuting with \(\hat{v}_0\), we know, from the intertwining property \(UW_+ = W_+ U_0\), that U is also strongly commuting with \(\hat{v}_+\). Hence, \(\hat{v}_+\) is strongly commuting with \(\Pi _\mathrm{ac}(U)\) and \(e^{i \xi \hat{v}_+} \Pi _\mathrm{ac}(U) = \Pi _\mathrm{ac}(U) e^{i \xi \hat{v}_+}\). Hence, by Theorem 2.3, \(\mathrm{exp}(i\xi \hat{x}(t)/t) \Psi _0\) converges strongly to \(\Pi _\mathrm{p}(U)\Psi _0 + e^{i \xi \hat{v}_+} \Pi _\mathrm{ac}(U)\Psi _0\) and

$$\begin{aligned} \lim _{t \rightarrow \infty } \mathbb {E}(e^{i \xi X_t/t})= & {} \langle \Psi _0, \Pi _\mathrm{p}(U)\Psi _0 + e^{i \xi \hat{v}_+} \Pi _\mathrm{ac}(U)\Psi _0 \rangle = \Vert \Pi _\mathrm{p}(U)\Psi _0\Vert ^2 \\&\quad + \int _{-\infty }^\infty e^{i \xi v} d\Vert E_{\hat{v}_+}(v) \Pi _\mathrm{ac}(U)\Psi _0\Vert ^2 = \int _{-\infty }^\infty e^{i\xi v} d\mu _V(v), \end{aligned}$$

which proves the corollary. \(\square \)

Example 2.4

Let \(C_0\) be the Hadmard matrix and C(x) satisfy (A.1). As seen in Example 2.3 (iii), (A.2) is satisfied and the spectrum of \(U_0\) is purely absolutely continuous. Let \(\Psi _+ \in \mathcal {H}\) satisfy \(\Psi _+(0) = \begin{pmatrix} \alpha \\ \beta \end{pmatrix}\) (\(|\alpha |^2 + |\beta |^2 = 1\)) and \(\Psi _+(x) = 0\) if \(x\not =0\). By Example 2.3,

$$\begin{aligned} d\Vert E_{\hat{v}_+}(v) \Pi _\mathrm{ac}(U) W_+ \Psi _+\Vert ^2 = d\Vert E_{\hat{v}_0}(v) \Psi _+ \Vert ^2 =(1- c_{\alpha , \beta } v) f_K\left( v;\frac{1}{\sqrt{2}}\right) \hbox {d}v. \end{aligned}$$

Let \(\Psi _\mathrm{p} \in \mathrm{Ran} \Pi _\mathrm{p}(U_0)\) be a unit vector and take the initial state \(\Psi _0\) as \(\Psi _0 = C_1 \Psi _\mathrm{p} + C_2 W_+ \Psi _+\) (\(|C_1|^2 + |C_2|^2 = 1\)). Suppose that \(U = SC\) satisfies (A.3). By Corollary 2.4, \(X_t/t\) converges in law to V with a probability distribution \(\mu _V\) and

$$\begin{aligned} \mu _V(\hbox {d}v) = |C_1|^2 \delta _0(\hbox {d}v) + |C_2|^2 (1- c_{\alpha , \beta } v) f_K\left( v;\frac{1}{\sqrt{2}}\right) \hbox {d}v. \end{aligned}$$

3 Wave operator

To prove Theorem 2.1, we use the following general proposition:

Proposition 3.1

Let U and \(U_0\) be unitary operators on a Hilbert space \(\mathcal {H}\) and suppose that \(U-U_0 \in \mathscr {T}_1\). The following limit exists:

$$\begin{aligned} W_+ = \text{ s- }\lim _{t \rightarrow \infty } U^{-t} U_0^t \Pi _\mathrm{ac}(U_0) \end{aligned}$$

Proof of Theorem 2.1

Since, by Lemma 2.1, \(U-U_0 \in \mathscr {T}_1\), the wave operator \(W_+\) exists. If we interchange the roles of U and \(U_0\), then the proposition says that the limit \(\text{ s- }\lim _{t \rightarrow \infty } U_0^{-t} U^t \Pi _\mathrm{ac}(U)\) also exists, which implies that \(W_+\) is complete. This completes the proof. \(\square \)

In the remainder of this section, we suppose that \(U-U_0 \in \mathscr {T}_1\) and prove Proposition 3.1. This is done by a discrete analogue of [11, Theorem 6.2]. We use \(\mathcal {H}_\mathrm{ac}\) and \(\mathcal {H}_\mathrm{p}\) to denote the subspaces of absolute continuity and the direct sum of all eigenspaces of \(U_0\). Let \(E_0\) be the spectral measure of \(U_0\) with \(E_0([0,2\pi )) = I\). Let

$$\begin{aligned} \mathcal {H}_\mathrm{ac,0} = \{ \psi \in \mathcal {H}_\mathrm{ac} \mid d \Vert E_0(\lambda ) \psi \Vert ^2 = G_\psi (\lambda )^2 d\lambda \text{ and } G_\psi \in L^2 \cap L^\infty \}, \end{aligned}$$

where \(L^2 = L^2([0,2\pi ))\) and \(L^\infty = L^\infty ([0,2\pi ))\). Although the following lemma may be well known, we give proofs for completeness.

Lemma 3.1

\(\mathcal {H}_\mathrm{ac,0}\) is dense in \(\mathcal {H}_\mathrm{ac}\).

Proof

For all \(\psi \in \mathcal {H}_\mathrm{ac}\), there exists a positive function \(F \in L^1\) such that \(d\Vert E_0(\lambda )\psi \Vert ^2 = F (\lambda ) d\lambda \). Let \(B_n = F^{-1}([0,n])\), and let \(\chi _{B_n}\) be the characteristic function of \(B_n\). We set \(G_n = \sqrt{F} \chi _{B_n}\) and \(\psi _n = E_0(B_n)\Psi \). Then, \(G_n \in L^2 \cap L^\infty \) and \(\Vert E_0(B) \psi _n\Vert ^2 = \int _B G_n(\lambda )^2 d\lambda \). Hence, \(\psi _n \in \mathscr {H}_\mathrm{ac,0}\) and \(\psi = \lim _n \psi _n\). This completes the proof. \(\square \)

Lemma 3.2

Let \(\phi \in \mathcal {H}\) and \(\psi \in \mathcal {H}_\mathrm{ac,0}\). Then,

$$\begin{aligned} \sum _{t \in \mathbb {Z}} \left| \left\langle \phi , U_0^t \psi \right\rangle \right| ^2 \le 2\pi \Vert \phi \Vert ^2 \sup _{\lambda } G_\psi (\lambda )^2. \end{aligned}$$

Proof

Let \(\psi \in \mathcal {H}_\mathrm{ac,0}\) and \(\mathcal {L} = L^2([0,2\pi ), G^2_\psi (\lambda ) d\lambda )\). Let \(H_0\) be the self-adjoint operator defined by \(\langle \xi , H_0 \eta \rangle = \int _0^{2\pi } \lambda d\langle \xi , E_0(\lambda ) \eta \rangle \) (\(\xi , \eta \in \mathcal {H}\)). Let \(\mathscr {U}{:} \mathcal {L} \rightarrow \mathcal {H}\) be an injection defined by \(\mathscr {U}f = f(H_0)\psi \) (\(f \in \mathcal {L}\)). Then \(\mathscr {U} 1 = \psi \) and \(\mathscr {U} e^{it\lambda } = U_0^t\psi \) (\(t \in \mathbb {N}\)). We use \(\Pi \) to denote the orthogonal projection onto \(U\mathcal {L}\). Let \(\phi \in \mathcal {H}\) and \(F = \mathscr {U}^{-1}\Pi \phi \in \mathcal {L}\). Then, we have

$$\begin{aligned} \langle \phi , U_0^t\psi \rangle = \int _0^{2\pi } e^{i t\lambda } \bar{F}(\lambda ) G_\psi (\lambda )^2 d\lambda = 2\pi \widehat{\bar{F} G_\psi ^2}(t). \end{aligned}$$

Hence, by Parseval’s identity, we obtain

$$\begin{aligned} \sum _{t \in \mathbb {Z}} |\langle \phi , U_0^t \psi \rangle |^2= & {} 2\pi \int _0^{2\pi } |\bar{F}(\lambda ) G_\psi (\lambda )^2|^2 d\lambda \\\le & {} 2\pi \sup _{\lambda } G_\psi (\lambda )^2 \int _0^{2\pi } |\bar{F}(\lambda )|^2 G_\psi (\lambda )^2 d\lambda \\\le & {} 2\pi \sup _{\lambda } G_\psi (\lambda )^2 \Vert \Pi \phi \Vert ^2. \end{aligned}$$

This completes the proof. \(\square \)

Let \(W_t = U^{-t} U_0^t \).

Lemma 3.3

Let \(t, s \in \mathbb {N}\) (\(s\not =t\)). Then, \(\text{ s- }\lim _{r \rightarrow \infty } (W_t - W_s)U_0^r\Pi _\mathrm{ac}(U_0) = 0\).

Proof

For \(t, s \in \mathbb {N}\) (\(t>s\)), we have \(W_t = \sum _{k=s+1}^t (W_k - W_{k-1}) + W_s\) and \(W_k - W_{k-1} = U^{-k}(-T)U_0^{k-1}\), where \(T = U-U_0 \in \mathscr {T}_1\). Since \(\mathscr {T}_1\) is an ideal, we know that

$$\begin{aligned} W_t -W_s = \sum _{k=s+1}^t U^{-k} (-T) U_0^{k-1} \in \mathscr {T}_1. \end{aligned}$$

In particular, \(W_t-W_s\) is compact. Let \(H_0\) be the self-adjoint operator defined in the proof of Lemma 3.2. Since \(\text{ w- }\lim _{r \rightarrow \infty } e^{irH_0}\Pi _\mathrm{ac}(H_0) = 0\), we have

$$\begin{aligned} \text{ s- }\lim _{r \rightarrow \infty } (W_t-W_s) U_0^r \Pi _\mathrm{ac}(U_0) = \text{ s- }\lim _{r \rightarrow \infty } (W_t-W_s) e^{irH_0}\Pi _\mathrm{ac}(H_0) = 0. \end{aligned}$$

This completes the proof. \(\square \)

Proof of Proposition 3.1

By Lemma 3.1, it suffices to prove that, for \(\psi \in \mathcal {H}_\mathrm{ac,0}\),

$$\begin{aligned} \Vert (W_t - W_s)\psi \Vert \rightarrow 0, \quad t,s \rightarrow \infty . \end{aligned}$$

Because

$$\begin{aligned} \Vert (W_t - W_s)\psi \Vert ^2 = \langle \psi , W_t^*(W_t - W_s)\psi \rangle - \langle \psi , W_s^*(W_t - W_s)\psi \rangle , \end{aligned}$$

we need only to prove that

$$\begin{aligned} \langle \psi , W_t^*(W_t - W_s)\psi \rangle \rightarrow 0, \quad t,s \rightarrow \infty . \end{aligned}$$

By direct calculation, we have, for \(r > 1\),

$$\begin{aligned}&W_t^*(W_t - W_s) - U_0^{-r} W_t^* (W_t - W_s) U_0^r \\&\quad =U_0^{-r} W_t^* W_s U_0^r - W_t^* W_s \\&\quad = \sum _{k=0}^{r-1} \left( U_0^{-k-1} W_t^* W_s U_0^{k+1} - U_0^{-k} W_t^* W_sU_0^k \right) . \end{aligned}$$

Since

$$\begin{aligned} U_0^{-k-1} W_t^* W_s U_0^{k+1} - U_0^{-k} W_t^* W_sU_0^k = U_0^{-k-t-1} \left( TU^{t-s} - U^{t-s} T\right) U_0^{s+k}, \end{aligned}$$

we obtain

$$\begin{aligned}&W_t^*(W_t - W_s) - U_0^{-r} W_t^* (W_t - W_s) U_0^r \\&\quad = \sum _{k=0}^{r-1} U_0^{-k-t-1} \left( TU^{t-s} - U^{t-s} T\right) U_0^{s+k}. \end{aligned}$$

Since, by Lemma 3.3, \(\text{ s- }\lim _{r \rightarrow \infty }U_0^{-r} W_t^* (W_t - W_s) U_0^r \psi = 0\), we have

$$\begin{aligned}&W_t^*(W_t - W_s)\psi = \sum _{k=0}^{\infty } U_0^{-k-t-1} \left( TU^{t-s} - U^{t-s} T\right) U_0^{s+k}\psi \\&\quad = Z_{t,s} ((U_0 T) U^{t-s} - (U_0 U^{t-s}) T) \psi , \end{aligned}$$

where

$$\begin{aligned} Z_{t,s}(A) = \sum _{k=0}^\infty U_0^{-k-t} A U_0^{k+s}. \end{aligned}$$

By Lemma 3.4 below, we know that

$$\begin{aligned} |\langle \psi , W_t^*(W_t - W_s)\psi \rangle |&\le |\langle \psi , Z_{t,s}((U_0 T) U^{t-s}) \psi \rangle | \\&\quad + |\langle \psi , Z_{t,s}(U_0 U^{t-s}) T) \psi \rangle | \rightarrow 0, \quad t,s \rightarrow \infty . \end{aligned}$$

This completes the proof. \(\square \)

Lemma 3.4

Let \(Y \in \mathscr {T}_1\) and \(\{Q(t,s)\}\) be a family of bounded operators with \(\sup _{t,s}\Vert Q(t,s)\Vert < \infty \). Then, for all \(\psi \in \mathcal {H}_\mathrm{ac,0}\),

  1. (1)

    \(\lim _{t, s \rightarrow \infty } \left\langle \psi , Z_{t,s}(Y Q(t,s) )\psi \right\rangle = 0\);

  2. (2)

    \(\lim _{t, s \rightarrow \infty } \left\langle \psi , Z_{t,s}(Q(t,s) Y)\psi \right\rangle = 0\).

Proof

Let \(Y = \sum _{n=1}^\infty \lambda _n |\psi _n \rangle \langle \phi _n |\) be the canonical expansion of the compact operator Y. Since \(Y \in \mathscr {T}_1\), \(\sum _n \lambda _n < \infty \). Then, by the Cauchy–Schwartz inequality, we have

$$\begin{aligned} \left| \left\langle \psi , Z_{t,s}(Y Q(t,s) )\psi \right\rangle \right|&\le \sum _{n=1}^{\infty } \sum _{k=0}^{\infty } \lambda _n \left| \left\langle U_0^{k+t} \psi , \psi _n \right\rangle \left\langle \phi _n, Q(t,s) U_0^{k+s}\psi \right\rangle \right| \\&\le I_1(t,s)^{1/2} \times I_2(t,s)^{1/2}, \end{aligned}$$

where

$$\begin{aligned} I_1(t)&= \sum _{n=1}^{\infty } \sum _{k=0}^{\infty } \lambda _n \left| \left\langle \psi _n, U_0^{k+t} \psi \right\rangle \right| ^2, \\ I_2(t,s)&= \sum _{n=1}^{\infty } \sum _{k=0}^{\infty } \lambda _n \left| \left\langle Q(t,s)^*\phi _n, U_0^{k+s}\psi \right\rangle \right| ^2. \end{aligned}$$

By Lemma 3.2, we have

$$\begin{aligned}&I_2(t,s) \le 2\pi \sup _\lambda G_\psi (\lambda )^2 \sup _{t,s}\Vert Q(t,s)\Vert \sum _n \lambda _n < \infty , \end{aligned}$$

where we have used the fact that \(\phi _n\) is a normalized vector. Let \(u_k= \sum _{n=1}^{\infty } \lambda _n |\left\langle \psi _n,\right. \left. U_0^k \psi \right\rangle |^2\). Then, similar to the above, we observe that \(\{u_k\} \in \ell ^1(\mathbb {Z})\). Hence, we have

$$\begin{aligned} \lim _{t \rightarrow \infty } I_1(t) = \lim _{t \rightarrow \infty } \sum _{k=t}^\infty u_k = 0. \end{aligned}$$

This proves (i). The same proof works for (ii). \(\square \)

4 Asymptotic velocity

4.1 Proof of Theorem 2.2

Let

$$\begin{aligned} \mathcal {H}_0 = \bigcup _{m=0}^\infty \left\{ \Psi \in \mathcal {H} \mid \Psi (x) = 0,~ |x| \ge m \right\} . \end{aligned}$$

We use \(\mathcal {D}\) to denote a subspace of vectors \(\Psi \in \mathcal {H}\) whose Fourier transform \(\hat{\Psi }\) is differentiable in k with

$$\begin{aligned} \sup _{k \in [0,2\pi )} \left\| \frac{\hbox {d}}{\hbox {d}k} \hat{\Psi }(k) \right\| < \infty . \end{aligned}$$

Note that \(\mathcal {H}_0\) is a core for \(\hat{x}\), and so is \(\mathcal {D}\). Let \(D = \mathscr {F} \hat{x} \mathscr {F}^{-1}\). Then, by direct calculation, we know that \((D \hat{\Psi })(k) = i\frac{\hbox {d}}{\hbox {d}k} \hat{\Psi }(k)\) for \(\Psi \in \mathcal {D}\). We prove the following theorem:

Theorem 4.1

Suppose that (A.2) holds. Then,

$$\begin{aligned} \text{ s- }\lim _{t \rightarrow \infty } \left( \frac{\hat{x}_0(t)}{t} - z \right) ^{-1} = (\hat{v}_0 - z)^{-1}, \quad z \in \mathbb {C} \setminus \mathbb {R}. \end{aligned}$$
(4.1)

Proof

For all \(\Psi \in \mathcal {H}\) and \(\epsilon > 0\), there exists a vector \( \Psi _\epsilon \in \mathcal {D}\) such that \(\Vert \Psi - \Psi _\epsilon \Vert \le \epsilon \). Because, by the second resolvent identity,

$$\begin{aligned}&\left\| \left( \frac{\hat{x}_0(t)}{t} - z \right) ^{-1}\Psi - (\hat{v}_0 - z)^{-1} \Psi \right\| \\&\quad \le \frac{2\epsilon }{|\mathrm{Im}z|} + \left\| \left( \frac{\hat{x}_0(t)}{t} - z \right) ^{-1} \Psi _\epsilon - (\hat{v}_0 - z)^{-1} \Psi _\epsilon \right\| \\&\quad \le \frac{2\epsilon }{|\mathrm{Im}z|} + \frac{1}{|\mathrm{Im}z|} \left\| \left( \hat{v}_0 - \frac{\hat{x}_0(t)}{t} \right) (\hat{v}_0 - z)^{-1} \Psi _\epsilon \right\| , \end{aligned}$$

it suffices to prove that

$$\begin{aligned} \lim _{t \rightarrow \infty } \left\| \left( \hat{v}_0 - \frac{\hat{x}_0(t)}{t} \right) (\hat{v}_0 - z)^{-1} \Psi \right\| = 0, \quad \Psi \in \mathcal {D}. \end{aligned}$$

Note that

$$\begin{aligned} (\hat{v}_0-z)^{-1} = \mathscr {F}^{-1} \left( \int _{[0,2\pi )}^\oplus \hbox {d}k \sum _{j=1,2} \left( \frac{ i \lambda _j^\prime (k) }{\lambda _j(k)} -z\right) ^{-1} |u_j(k) \rangle \langle u_j(k)| \right) \mathscr {F}. \end{aligned}$$

Since \(\lambda _j(k)\) is analytic and \(|\lambda _j(k)|=1\), we observe from (A.2) that \((\hat{v}_0 - z)^{-1}\) leaves \(\mathcal {D}\) invariant. Hence, we only need to prove that

$$\begin{aligned} \lim _{t \rightarrow \infty } \left\| \left( \hat{v}_0 - \frac{\hat{x}_0(t)}{t} \right) \Psi \right\| = 0, \quad \Psi \in \mathcal {D}. \end{aligned}$$

By direct calculation, we have

$$\begin{aligned}&\left\| \left( \hat{v}_0 - \frac{\hat{x}_0(t)}{t} \right) \Psi \right\| ^2 \\&\quad = \int _0^{2\pi } \hbox {d}k \left\| \sum _{j=1,2} \left( \frac{i \lambda _j^\prime (k)}{\lambda _j(k)} \right) \langle u_j(k), \hat{\Psi }(k) \rangle u_j(k) - \hat{U}(k)^{-t} \frac{D}{t} \hat{U}(k)^t \hat{\Psi }(k) \right\| ^2 \\&\quad = \int _0^{2\pi } \frac{\hbox {d}k}{t^2} \left\| \sum _{j=1,2} \lambda _j(k)^t \hat{U}(k)^{-t} \left( i \frac{\hbox {d}}{\hbox {d}k} \langle u_j(k), \hat{\Psi }(k) \rangle u_j(k) \right) \right\| ^2. \end{aligned}$$

By the definition of \(\mathcal {D}\) and (A.2), we know that

$$\begin{aligned} \sup _{k \in [0,2\pi )} \left\| \left( i \frac{{\hbox {d}}}{\hbox {d}k} \langle u_j(k), \hat{\Psi }(k) \rangle u_j(k) \right) \right\| < \infty . \end{aligned}$$

Hence, we have

$$\begin{aligned} \left\| \left( \hat{v}_0 - \frac{\hat{x}_0(t)}{t} \right) \Psi \right\| = O(t^{-1}), \end{aligned}$$

which completes the proof. \(\square \)

4.2 Proof of Theorem 2.3

The proof falls naturally into two parts:

Theorem 4.2

Let U be a unitary operator on \(\mathcal {H}\). \(\hat{x}(t) = U^{-t} \hat{x} U^t\) satisfies

$$\begin{aligned} \text{ s- }\lim _{t \rightarrow \infty } \mathrm{exp}\left( i \xi \frac{\hat{x}(t)}{t} \right) \Pi _\mathrm{p}(U) = \Pi _\mathrm{p}(U), \quad \xi \in \mathbb {R}. \end{aligned}$$

Theorem 4.3

Let \(U=SC\) and \(U_0 = SC_0\) satisfy (A.1) and (A.2). Then,

$$\begin{aligned} \text{ s- }\lim _{t \rightarrow \infty } \mathrm{exp}\left( i\xi \frac{\hat{x}(t)}{t} \right) \Pi _\mathrm{ac}(U) = \mathrm{exp}(i \xi \hat{v}_+)\Pi _\mathrm{ac}(U), \quad \xi \in \mathbb {R}. \end{aligned}$$

Proof of Theorem 2.3

By (A.3), we have

$$\begin{aligned} \text{ s- }\lim _{t \rightarrow \infty } \mathrm{exp}\left( i\xi \frac{\hat{x}(t)}{t} \right)&= \text{ s- }\lim _{t \rightarrow \infty } \mathrm{exp}\left( i\xi \frac{\hat{x}(t)}{t} \right) (\Pi _\mathrm{p}(U) + \Pi _\mathrm{p}(U)) \\&= \Pi _\mathrm{p}(U) + \mathrm{exp}(i \xi \hat{v}_+)\Pi _\mathrm{ac}(U). \end{aligned}$$

This prove the theorem. \(\square \)

It remains to prove Theorems 4.2 and 4.3.

Proof of Theorem 4.2

Let \(\mathcal {H}_\mathrm{p}(U)\) be the direct sum of all eigenspaces of U. It suffices to prove that, for \(\Psi \in \mathcal {H}_\mathrm{p}(U)\),

$$\begin{aligned} \text{ s- }\lim _{t \rightarrow \infty } \mathrm{exp}\left( i \xi \frac{\hat{x}(t)}{t} \right) \Psi =\Psi . \end{aligned}$$

Let \(\lambda _n\) be the eigenvalues of U and take an ONB \(\{\eta _n\}_{n=1}^\infty \) of \(\mathcal {H}_\mathrm{p}\) such that \(U\eta _n = \lambda _n \eta _n\). We have \(\Pi _\mathrm{p}(U) = \sum _n |\eta _n \rangle \langle \eta _n|\). Let \(\epsilon >0\). For sufficiently large N, \(\Psi _N = \sum _{n=1}^N \langle \eta _n, \Psi \rangle \eta _n\) satisfies \(\Vert \Psi - \Psi _N\Vert \le \epsilon \). Then,

$$\begin{aligned} \left\| \mathrm{exp}\left( i \xi \frac{\hat{x}(t)}{t} \right) \Psi - \Psi \right\| \le 2 \epsilon + \left\| \mathrm{exp}\left( i \xi \frac{\hat{x}(t)}{t} \right) \Psi _N - \Psi _N \right\| . \end{aligned}$$

By direct calculation, we have

$$\begin{aligned} \left\| \mathrm{exp}\left( i \xi \frac{\hat{x}(t)}{t} \right) \Psi _N - \Psi _N \right\|= & {} \left\| \left( \mathrm{exp}\left( i \xi \frac{\hat{x}}{t} \right) -1 \right) U^t \Psi _N\right\| \nonumber \\= & {} \left\| \sum _{n=1}^N \lambda _n^t \langle \eta _n, \Psi \rangle \left( \mathrm{exp}\left( i \xi \frac{\hat{x}}{t} \right) -1 \right) \eta _n \right\| \nonumber \\\le & {} \sum _{n=1}^N |\langle \eta _n, \Psi \rangle | \left\| \left( \mathrm{exp}\left( i \xi \frac{\hat{x}}{t} \right) -1\right) \eta _n \right\| . \end{aligned}$$
(4.2)

Since \(\lim _{t \rightarrow \infty } |1-e^{i \xi x/t}|=0\), \(|1-e^{i \xi x/t}| \le 2\) and \(\sum _{x} \Vert \eta _n(x)\Vert _{\mathbb {C}^2}^2 = \Vert \eta _n\Vert ^2 < \infty \), we have

$$\begin{aligned} \lim _{t \rightarrow \infty } \left\| \left( \mathrm{exp}\left( i \xi \frac{\hat{x}}{t} \right) -1\right) \eta _n \right\| ^2 = \lim _{t \rightarrow \infty } \sum _{x \in \mathbb {Z}} |e^{i \xi x/t}-1|^2 \Vert \eta _n(x)\Vert _{\mathbb {C}^2}^2 = 0, \end{aligned}$$

which, combined with (4.2), completes the proof. \(\square \)

Lemma 4.1

\([U_0, \mathrm{exp}(i \xi \hat{v}_0)] = 0\).

Proof

By direct calculation, we have

$$\begin{aligned}{}[ U_0, \mathrm{exp}(i \xi \hat{v}_0)]&= \text{ s- }\lim _{t \rightarrow \infty } \left[ U_0, \mathrm{exp}\left( i \xi \frac{\hat{x}_0(t)}{t}\right) \right] \\&= \text{ s- }\lim _{t \rightarrow \infty } U_0 \left\{ \mathrm{exp}\left( i \xi \frac{\hat{x}_0(t)}{t}\right) - \mathrm{exp}\left( i \xi \frac{\hat{x}_0(t+1)}{t}\right) \right\} = 0. \end{aligned}$$

Proof of Theorem 4.3

By (A.1) and (A.2), Theorems 2.1 and 2.2 hold. Then, \(W_+\) is a unitary operator from \(\mathcal {H}_\mathrm{ac}(U_0)\) to \(\mathcal {H}_\mathrm{ac}(U)\). Hence, we have

$$\begin{aligned} \mathrm{exp}(i \xi \hat{v}_+) \Pi _\mathrm{ac}(U) = W_+ \mathrm{exp}(i \xi \hat{v}_0) W_+^*\Pi _\mathrm{ac}(U). \end{aligned}$$

By direct calculation, we observe that

$$\begin{aligned} I(t)&:= \mathrm{exp}\left( i\xi \frac{\hat{x}(t)}{t} \right) \Pi _\mathrm{ac}(U)- \mathrm{exp}(i \xi \hat{v}_+)\Pi _\mathrm{ac}(U) \\&= W_t \mathrm{exp}\left( i\xi \frac{\hat{x}_0(t)}{t} \right) W_t^* \Pi _\mathrm{ac}(U)- W_+ \mathrm{exp}(i \xi \hat{v}_0) W_+^*\Pi _\mathrm{ac}(U) \\&=: \sum _{j=1}^3 I_j(t), \end{aligned}$$

where

$$\begin{aligned} I_1(t)&= W_t \mathrm{exp}\left( i\xi \frac{\hat{x}_0(t)}{t} \right) \left( W_t^* - W_+^* \right) \Pi _\mathrm{ac}(U), \\ I_2(t)&= W_t\left( \mathrm{exp}\left( i\xi \frac{\hat{x}_0(t)}{t} \right) - \mathrm{exp}(i \xi \hat{v}_0) \right) W_+^* \Pi _\mathrm{ac}(U), \\ I_3(t)&= \left( W_t - W_+ \right) \mathrm{exp}(i \xi \hat{v}_0) W_+^* \Pi _\mathrm{ac}(U). \end{aligned}$$

Because \(W_t\) and \(\mathrm{exp}(i \xi \hat{x}_0(t)/t)\) are uniformly bounded, we know from Theorems 2.1 and 2.2 that \(\text{ s- }\lim _{t \rightarrow \infty }I_1(t) = \text{ s- }\lim _{t \rightarrow \infty }I_2(t) = 0\). Hence, we have

$$\begin{aligned} I(t)&= \left( W_t - W_+ \right) \mathrm{exp}(i \xi \hat{v}_0) W_+^* \Pi _\mathrm{ac}(U) + o(1) \\&= \left( W_t - W_+ \right) \Pi _\mathrm{ac}(U_0) \mathrm{exp}(i \xi \hat{v}_0) W_+^* \Pi _\mathrm{ac}(U) \\&\quad + \left( W_t - W_+ \right) [ \mathrm{exp}(i \xi \hat{v}_0), \Pi _\mathrm{ac}(U_0)] W_+^* \Pi _\mathrm{ac}(U) + o(1), \end{aligned}$$

where we have used the fact that \(\mathrm{Ran} W_+^* =\mathcal {H}_\mathrm{ac}(U_0)\). Since, by Lemma 4.1, \([ \mathrm{exp}(i \xi \hat{v}_0), \Pi _\mathrm{ac}(U_0)]=0\), we obtain from Theorem 2.1, that \(\text{ s- }\lim _{t \rightarrow \infty } I(t)=0\). This completes the proof. \(\square \)