Generator of an abstract quantum walk

Abstract

We give an explicit formula of the generator of an abstract Szegedy evolution operator in terms of the discriminant operator of the evolution. We also characterize the asymptotic behavior of a quantum walker through the spectral property of the discriminant operator by using the discrete analog of the RAGE theorem.

Appendix

1.1 Proof of Proposition 2.3

We present a proof of Proposition 2.3. Let H be the generator of an evolution $(U, \{\mathcal {H}_v \}_{v \in V}) \in \mathscr {F}_\mathrm{QW}$. Throughout this subsection, we assume that $\mathrm{dim}\mathcal {H}_v < \infty$ ($v \in V$). Let $\mathcal {H}_1$ be the set of vectors $\Psi _0 \in \mathcal {H}$ satisfying

$$\begin{aligned} \lim _{N \rightarrow \infty } \frac{1}{N}\sum _{n=0}^{N-1} \nu _n^{\Psi _0}(R) = 0 \end{aligned}$$

for any finite subset R of V, and $\mathcal {H}_2$ the set of vectors $\Psi _0 \in \mathcal {H}$ satisfying

$$\begin{aligned} \lim _{m \rightarrow \infty } \sup _{n} \nu _n^{\Psi _0}(R_m^\mathrm{c}) = 0 \end{aligned}$$

for any sequence $\{R_m\}$ of finite subsets of V such that $R_m \subset R_{m+1}$ and $V=\cup _m R_m$. Because $\nu _n^{\alpha \Psi _0 + \beta \Phi _0}(R) \le 2 \left( |\alpha |^2 \nu _n^{\Psi _0}(R) + |\beta |^2 \nu _n^{\Phi _0}(R) \right)$, we know that $\mathcal {H}_1$ and $\mathcal {H}_2$ are subspaces of $\mathcal {H}$. Let $P_R = \sum _{x \in R} P_x$ ($R \subset V$). Then,

$$\begin{aligned} \nu _n^{\Psi _0}(R) = \left\| P_R \mathrm{e}^{in H} \Psi _0 \right\| ^2. \end{aligned}$$

Lemma 6.1

$\mathcal {H}_1 \perp \mathcal {H}_2$.

Proof

Let $\Psi _0 \in \mathcal {H}_1$ and $\Phi _0 \in \mathcal {H}_2$. Then, for all $R \subset V$,

$$\begin{aligned} |\langle \Psi _0, \Phi _0 \rangle |&= \frac{1}{N} \sum _{n=0}^{N-1} |\langle \Psi _n, \Phi _n \rangle | \\&\le \frac{1}{N} \sum _{n=0}^{N-1} |\langle P_R \Psi _n, P_R \Phi _n \rangle | + \frac{1}{N} \sum _{n=0}^{N-1} |\langle P_{R^\mathrm{c}} \Psi _n, P_{R^\mathrm{c}} \Phi _n \rangle | \\&\le \Vert \Phi _0\Vert \left( \frac{1}{N} \sum _{n=0}^{N-1} \Vert P_R \Psi _n\Vert \right) + \Vert \Psi _0\Vert \left( \frac{1}{N} \sum _{n=0}^{N-1} \Vert P_{R^\mathrm{c}} \Phi _n\Vert \right) . \end{aligned}$$

We first estimate the first term. By the Cauchy–Schwarz inequality,

$$\begin{aligned} \frac{1}{N} \sum _{n=0}^{N-1} \Vert P_R \Psi _n\Vert \le \left( \frac{1}{N} \sum _{n=0}^{N-1} \Vert P_R \Psi _n\Vert ^2 \right) ^{1/2} = \bar{\nu }_N^{\Psi _0}(R)^{1/2}. \end{aligned}$$

The second term is estimated as follows:

$$\begin{aligned} \frac{1}{N} \sum _{n=0}^{N-1} \Vert P_{R^\mathrm{c}} \Phi _n\Vert \le \sup _{n \ge 0} \Vert P_{R^\mathrm{c}} \Phi _n\Vert = \sup _{n \ge 0} \nu _n^{\Phi _0}(R^\mathrm{c})^{1/2}. \end{aligned}$$

Combining these inequalities yields the result that

$$\begin{aligned} |\langle \Psi _0, \Phi _0 \rangle | \le \Vert \Phi _0\Vert \bar{\nu }_N^{\Psi _0}(R)^{1/2} + \Vert \Psi _0\Vert \sup _{n \ge 0} \nu _n^{\Phi _0}(R^\mathrm{c})^{1/2}. \end{aligned}$$

(7.1)

Let $\epsilon >0$ and $\{R_m\}_{m\ge 1}$ be a family of finite subsets of V such that $R_m \subset R_{m+1}$ and $V = \cup _{m \ge 1} R_m$. Because $\Phi _0 \in \mathcal {H}_2$, there exists an $m_0 \in \mathbb {N}$ such that $\nu _n^{\Phi _0}(R_m^\mathrm{c}) < \epsilon ^2/\Vert \Psi _0\Vert ^2$ ($m \ge m_0$). Because $\Psi _0 \in \mathcal {H}_1$, it follows from (7.1) that

$$\begin{aligned} \lim _{N \rightarrow \infty } |\langle \Psi _0, \Phi _0 \rangle | \le \epsilon , \end{aligned}$$

which completes the proof. $\square$

Lemma 6.2

• (i) $\mathcal {H}_\mathrm{c}(H) \subset \mathcal {H}_1$;

• (ii) $\mathcal {H}_\mathrm{p}(H) \subset \mathcal {H}_2$.

Proof

Let $\Psi _0 \in \mathcal {H}_\mathrm{c}(H)$. For any finite set R,

$$\begin{aligned} \bar{\nu }_N^{\Psi _0}(R) = \sum _{x \in R} \sum _{j =1}^{\mathrm{dim}\mathcal {H}_x} \bar{\nu }_N (\phi _{x,j}), \end{aligned}$$

(6.2)

where $\{\phi _{x,j}\}$ is a complete orthonormal system of $\mathcal {H}_x$ and $\bar{\nu }_N (\phi )$ $:= \frac{1}{N} \sum _{n=0}^{N-1}$ $|\langle \phi , \mathrm{e}^{inH}\Psi _0 \rangle |^2$. Because, by assumption, the sum in (6.2) runs over a finite set, it suffices to show that $\lim _{N \rightarrow \infty } \bar{\nu }_N (\phi ) = 0$. Let $\omega (x) = \mathrm{e}^{in x}$ and $g_N(\omega ) = \frac{1}{N}\sum _{n=0}^{N-1} \omega ^n$. Then, $g_N(\omega ) = \frac{1-\omega ^N}{N(1-\omega )}$ if $\omega \not =1$ and $g_N(1) = 1$. By the Fubini theorem,

$$\begin{aligned} \bar{\nu }_N(\phi ) = \int _0^{2\pi } \int _0^{2\pi } g_N(\omega (\lambda -\mu )) \mathrm{d}\langle P_\mathrm{c}(H)\phi , E_H(\lambda ) \Psi _0 \rangle \mathrm{d}\langle \Psi _0, E_H(\mu ) P_\mathrm{c}(H) \phi \rangle , \end{aligned}$$

where $P_\mathrm{c}(H)$ is the projection onto $\mathcal {H}_\mathrm{c}(H)$. By the polarization identity, there exists $\{\psi _j\}_{j=1,2,3,4} \subset \mathcal {H}_\mathrm{c}(H)$ such that

$$\begin{aligned} \bar{\nu }_N(\phi ) \le \mathrm{const.} \sum _{j,k=1,2,3,4} \int _0^{2\pi } \int _0^{2\pi } |g_N(\omega (\lambda -\mu ))| \mathrm{d}\Vert E_H(\lambda ) \psi _j\Vert ^2 \mathrm{d}\Vert E_H(\mu ) \psi _k\Vert ^2. \end{aligned}$$

Because $F_j := \Vert E_H(\cdot ) \psi _j\Vert ^2$ is continuous,

$$\begin{aligned} \int \int _{\{(\lambda , \mu ) \mid \lambda =\mu \}} \mathrm{d}F_j(\lambda ) \mathrm{d}F_k(\mu )&\le \int _0^{2\pi } \mathrm{d}F_k(\mu ) \int _{\mu -\epsilon }^{\mu + \epsilon } \mathrm{d}F_j(\lambda ) \\&= \int _0^{2\pi } \mathrm{d}F_k(\mu ) (F_j(\mu + \epsilon ) - F_j(\mu - \epsilon )) \rightarrow 0, \end{aligned}$$

as $\epsilon \rightarrow 0$. Because $\sup _{|\omega |=1} |g_N(\omega )| \le 1$ and $\lim _{N \rightarrow \infty } g_N(\omega (\lambda -\mu )) = 0$ ($\lambda \not =\mu$), we obtain $\lim _{N\rightarrow 0} \bar{\nu }_N(\phi ) =0$ by the dominated convergence theorem. This completes the proof of (i).

Let $\Psi _0 \in \mathcal {H}_\mathrm{p}(H)$. For any $\epsilon > 0$, there exist eigenvectors $\{ \phi _j\}_{j=1}^M$ ($M \in \mathbb {N}$) of H such that $\Vert \Psi _0 - \sum _{j=1}^M \langle \phi _j, \Psi _0 \rangle \phi _j \Vert < \epsilon$. Let $\{R_m\}$ be a sequence of finite subsets of V such that $R_m \subset R_{m+1}$ and $\cup _m R_m = V$. It follows that

$$\begin{aligned} \nu _n^{\Psi _0}(R_m^\mathrm{c})^{1/2} \le \sum _{j=1}^M |\langle \phi _j, \Psi _0 \rangle | \Vert P_{R_m^\mathrm{c}} \phi _j \Vert + \epsilon , \end{aligned}$$

which proves $\lim _{m \rightarrow \infty } \sup _n \nu _n^{\Psi _0}(R_m^\mathrm{c}) = 0$. Hence, we have (ii). $\square$

Proof of Proposition 2.3

Combining Lemmas 6.1 and 6.2 yields the result that

$$\begin{aligned} \mathcal {H}_2 \subset \mathcal {H}_1^\perp \subset \mathcal {H}_\mathrm{p}(H) \subset \mathcal {H}_2, \quad \mathcal {H}_1 \subset \mathcal {H}_2^\perp \subset \mathcal {H}_\mathrm{c}(H) \subset \mathcal {H}_1, \end{aligned}$$

which proves the proposition. $\square$

1.2 Proof of Equation (2.4)

In this subsection, we prove the following:

Lemma 6.3

Let $(U, \{\mathcal {H}_v\}_{v \in V}) \in \mathscr {F}_{QW}$ and $\Psi _0 \in \mathcal {H}$ satisfy

$$\begin{aligned} \lim _{m \rightarrow \infty } \sup _n \nu _n^{\Psi _0}(R_m^\mathrm{c}) = 0 \end{aligned}$$

for an increasing sequence $\{R_m \}$ of finite subsets of V. Then, (2.4) holds. In particular, (2.4) holds for all $\Psi _0 \in \mathcal {H}_\mathrm{p}(H)$.

Proof

By assumption, we know that for any $\epsilon > 0$, there exists $m_0 \in \mathbb {N}$ such that $\sup _n \nu _n^{\Psi _0}(R_{m_0}^\mathrm{c}) < \epsilon$. Hence,

$$\begin{aligned} \limsup _{n \rightarrow \infty } \nu _n^{\Psi _0} (R_{m_0}) \ge 1- \epsilon . \end{aligned}$$

(6.3)

If $\limsup _n \nu _n^{\Psi _0}(x) = 0$ for any $x \in R_{m_0}$, then

$$\begin{aligned} \limsup _n \nu _n^{\Psi _0}(R_{m_0}) = \sum _{x \in R_{m_0}} \limsup _n \nu _n^{\Psi _0}(x) = 0, \end{aligned}$$

which contradicts (6.3). Therefore, (2.4) holds for some $x \in R_{m_0}$. $\square$