Stationary Measures of Three-State Quantum Walks with Defect on the One-Dimension Lattice

Jinling Gao^*, Mingjun Zhang

School of Information Engineering, Lanzhou University of Finance and Economics, Lanzhou, China.

DOI: 10.4236/ojapps.2023.134038                

Abstract

In this paper, we focus on the space-inhomogeneous three-state on the one-dimension lattice, a one-phase model and a two-phase model include. By using the transfer matrices method by Endo et al., we calculate the stationary measure for initial state concrete eigenvalue. Finally we found the transfer matrices method is more effective for the three-state quantum walks than the method obtained by Kawai et al.

Keywords

Three-State Quantum Walks, Stationary Measure, One-Phase, Two-Phase, Transfer Matrices

1. Introduction

There is an abundance of research on discrete-time quantum walks since 1993 [1] [2] [3] . Then quantum walks as a quantum mechanical attract large number of scholars [4] [5] [6] . For instance, the two-phase quantum walks are related to the research of topological insulator [7] [8] , and one-defect quantum walks are applied to quantum search algorithms [9] [10] .

Recently, the asymptotic behaviors of the quantum walks have received much attention [11] [12] . Konno gave the uniform measure as a stationary measure of the one-dimensional discrete-time quantum walks [13] . Endo et al., solve the eigenvalue problem and present a stationary measure by using SGF method [14] . Then Wang et al., obtain the stationary measures of three-state Wojcik walk by adopting SGF method [15] . Shortly afterwards, Kawai et al. raised Reduced matrix method [16] . Lately, Endo et al. got the transfer matrices and solve the eigenvalue [17] . In this paper, we will use this method to further derive one-phase and two-phase model of space-inhomogeneous three-state quantum walks.

2. Three-State Discrete-Time Quantum Walks

In this section, we give the definition of three-state quantum walk on $\mathbb{Z}$ , where $\mathbb{Z}$ is the set of integers. The discrete-time quantum walk on $\mathbb{Z}$ defined by a unitary matrix;

$U_x=\left[\begin{array}{ccc}{a}_x& b_x& c_x\\ d_x& e_x& f_x\\ g_x& h_x& i_x\end{array}\right]$ (2.1)

We let $\mathbb{N}$ be the set of nonnegative integers, and ${\Psi }_n\left(x\right)={\left({\Psi }_n^L\left(x\right)\mathrm{,}{\Psi }_n^O\left(x\right)\mathrm{,}{\Psi }_n^R\left(x\right)\right)}^{\text{T}}$ be the amplitude of the wave function corresponding to the chiralities “L”, “O”, and “R” at position $x\in \mathbb{Z}$ and time $n\in \mathbb{N}$ . Obviously, for each position $x\in \mathbb{Z}$ , the matrix $U_x$ can be divided into three parts.

$U_x=U_x^L+U_x^O+U_x^R$ (2.2)

Through these matrix, we can define time evolution of a quantum walk in the following way:

${\Psi }_{n+1}\left(x\right)\equiv U_{x+1}^L{\Psi }_n\left(x+1\right)+U_x^O{\Psi }_n\left(x\right)+U_{x-1}^R{\Psi }_n\left(x-1\right)$ (2.3)

Then let

$\begin{array}{l}{\Psi }_n={\left(\cdots \mathrm{,}{\Psi }_n\left(-1\right)\mathrm{,}{\Psi }_n\left(0\right)\mathrm{,}{\Psi }_n\left(1\right)\right)}^{\text{T}}\mathrm{,}\\ U^{\left(s\right)}=\left[\begin{array}{ccccccc}\ddots & ⋮& ⋮& ⋮& ⋮& ⋮& \cdots \\ \cdots & U_{-2}^O& U_{-1}^L& O& O& O& \cdots \\ \cdots & U_{-2}^R& U_{-1}^O& U_0^L& O& O& \cdots \\ \cdots & O& U_{-1}^R& U_0^O& U_1^L& O& \cdots \\ \cdots & O& O& U_0^R& U_1^R& U_2^L& \cdots \\ \cdots & O& O& O& U_1^R& U_2^O& \cdots \\ \cdots & ⋮& ⋮& ⋮& ⋮& ⋮& \ddots \end{array}\right],O=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\end{array}$ (2.4)

Then the sate at time n can be expressed as

${\Psi }_n={\left(U^{\left(s\right)}\right)}^n{\Psi }_0,n\ge 0$ (2.5)

where ${\Psi }_0$ is the initial state.

Definition 2.1. The one-phase model of space-inhomogeneous three-state quantum walk is defined on the set $\mathbb{Z}$ of integers. which is characterized by a chirality-state space $\left\{|L\rangle \mathrm{,}|O\rangle \mathrm{,}|R\rangle \right\}$ and a position space $\left\{|x\rangle \mathrm{<mo>\; :\; </mo>}x\in \mathbb{Z}\right\}$ , and the chiralities “L”, “R” and “O” express the left, right and neutral state for the motion of the walker. Its time evolution is determined by the following $3\times 3$ unitary matrices

$U_x=\frac{{\text{e}}^{i{\theta }_x}}{3}\left[\begin{array}{ccc}-1& 2& 2\\ 2& -1& 2\\ 2& 2& -1\end{array}\right]\mathrm{,}x\in \mathbb{Z}$ (2.6)

where

$\begin{array}{l}{\theta }_x=\{\begin{array}{ll}0,\hfill & x=\pm 1,\pm 2,\cdots ,\hfill \\ 2\pi \tau ,\hfill & x=0,\hfill \end{array}\hfill \end{array}$

with $\tau \in \left(\mathrm{0,1}\right)$ , which ${\theta }_x$ shows the phase $2\pi \tau$ of the walk.

Then

$U_x^L=\frac{{\text{e}}^{i{\theta }_x}}{3}\left[\begin{array}{ccc}-1& 2& 2\\ 0& 0& 0\\ 0& 0& 0\end{array}\right],U_x^O=\frac{{\text{e}}^{i{\theta }_x}}{3}\left[\begin{array}{ccc}0& 0& 0\\ 2& -1& 2\\ 0& 0& 0\end{array}\right],U_x^R=\frac{{\text{e}}^{i{\theta }_x}}{3}\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 2& 2& -1\end{array}\right].$

Definition 2.2. The two-phase model of space-inhomogeneous three-state quantum walk is defined on the set $\mathbb{Z}$ of integers, which is characterized by a chirality-state space $\left\{|L\rangle \mathrm{,}|O\rangle \mathrm{,}|R\rangle \right\}$ and a position space $\left\{|x\rangle \mathrm{<mo>\; :\; </mo>}x\in \mathbb{Z}\right\}$ . Its time evolution is determined by the following unitary matrices

$\begin{array}{l}{U}_x=\{\begin{array}{ll}{U}_{+},\hfill & x\ge 1,\hfill \\ U_0,\hfill & x=0,\hfill \\ U_{-},\hfill & x\le -1,\hfill \end{array}\hfill \end{array}$ (2.7)

where

$U_{\pm }=\left[\begin{array}{ccc}-\frac{1+g_{\pm }}{2}& \frac{{\hslash }_{\pm }}{\sqrt{2}}& \frac{1-g_{\pm }}{2}\\ \frac{{\hslash }_{\pm }}{\sqrt{2}}& g_{\pm }& \frac{{\hslash }_{\pm }}{\sqrt{2}}\\ \frac{1-g_{\pm }}{2}& \frac{{\hslash }_{\pm }}{\sqrt{2}}& -\frac{1+g_{\pm }}{2}\end{array}\right],U_0=\frac{\xi }{3}\left[\begin{array}{ccc}-1& 2& 2\\ 2& -1& 2\\ 2& 2& -1\end{array}\right].$ (2.8)

where $g_{\pm }=\cos {\gamma }_{\pm },{\hslash }_{\pm }=\sin {\gamma }_{\pm },{\gamma }_{\pm }\in \left[0,2\pi \right),\xi ={\text{e}}^{2\pi i\tau },\tau \in \left(0,1\right)$ . Then

$\begin{array}{l}{U}_x^L=\{\begin{array}{ll}{U}_{x_{+}}^L,\hfill & x\ge 1,\hfill \\ U_x^L,\hfill & x=0,\hfill \\ U_{x_{-}}^L,\hfill & x\le -1,\hfill \end{array}{U}_x^O=\{\begin{array}{ll}{U}_{x_{+}}^O,\hfill & x\ge 1,\hfill \\ U_0^O,\hfill & x=0,\hfill \\ U_{x_{-}}^O,\hfill & x\le -1,\hfill \end{array}{U}_x^R=\{\begin{array}{ll}{U}_{x_{+}}^R,\hfill & x\ge 1,\hfill \\ U_0^R,\hfill & x=0,\hfill \\ U_{x_{-}}^R,\hfill & x\le -1,\hfill \end{array}\hfill \end{array}$ (2.9)

$U_{x_{+}}^L=\left[\begin{array}{ccc}-\frac{1+g_{+}}{2}& \frac{{\hslash }_{+}}{\sqrt{2}}& \frac{1-g_{+}}{2}\\ 0& 0& 0\\ 0& 0& 0\end{array}\right],U_{x_{+}}^O=\left[\begin{array}{ccc}0& 0& 0\\ \frac{{\hslash }_{+}}{\sqrt{2}}& g_{+}& \frac{{\hslash }_{+}}{\sqrt{2}}\\ 0& 0& 0\end{array}\right],U_{x_{+}}^R=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ \frac{1-g_{+}}{2}& \frac{{\hslash }_{+}}{\sqrt{2}}& -\frac{1+g_{+}}{2}\end{array}\right].$ (2.10)

$U_{x_{-}}^L=\left[\begin{array}{ccc}-\frac{1+g_{-}}{2}& \frac{{\hslash }_{-}}{\sqrt{2}}& \frac{1-g_{-}}{2}\\ 0& 0& 0\\ 0& 0& 0\end{array}\right],U_{x_{-}}^O=\left[\begin{array}{ccc}0& 0& 0\\ \frac{{\hslash }_{-}}{\sqrt{2}}& g_{-}& \frac{{\hslash }_{-}}{\sqrt{2}}\\ 0& 0& 0\end{array}\right],U_{x_{-}}^R=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ \frac{1-g_{-}}{2}& \frac{{\hslash }_{-}}{\sqrt{2}}& -\frac{1+g_{-}}{2}\end{array}\right].$ (2.11)

$U_0=\frac{\xi }{3}\left[\begin{array}{ccc}-1& 2& 2\\ 0& 0& 0\\ 0& 0& 0\end{array}\right],U_0=\frac{\xi }{3}\left[\begin{array}{ccc}0& 0& 0\\ 2& -1& 2\\ 0& 0& 0\end{array}\right],U_0=\frac{\xi }{3}\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 2& 2& -1\end{array}\right].$ (2.12)

3. Stationary Measure

In the present section, we first recall some fundamental notions and facts about stationary measure. Firstly, we introduced a mapping $\varphi \mathrm{<mo>\; :\; </mo>}{\left({ℂ}^3\right)}^{\mathbb{Z}}\to {ℝ}_{+}^{\mathbb{Z}}$

$\varphi \left(x\right)={\left(\cdots \mathrm{,}{\Vert \Psi \left(-1\right)\Vert }^2\mathrm{,}{\Vert \Psi \left(0\right)\Vert }^2\mathrm{,}{\Vert \Psi \left(1\right)\Vert }^2\mathrm{,}\cdots \right)}^{\text{T}}\in {ℝ}_{+}^{\mathbb{Z}}\mathrm{,}$

where $ℂ$ is the set of complex number and $\Vert \cdot \Vert$ is the norm in ${ℂ}^3$ and $\Psi ={\left(\cdots ,\Psi \left(-1\right),\Psi \left(0\right),\Psi \left(1\right),\cdots \right)}^{\text{T}}\in {\left({ℂ}^3\right)}^{\mathbb{Z}}$ . For every $x\in \mathbb{Z}$ , we note that

$\varphi \left(\Psi \right)\left(x\right)={\Vert \Psi \left(x\right)\Vert }^2\mathrm{,}\Psi \in {\left({ℂ}^3\right)}^{\mathbb{Z}}\mathrm{.}$ (3.1)

Then, the function $x\to \varphi \left(\Psi \right)\left(x\right)$ gives a measure $\mu$ on $\mathbb{Z}$ by $\mu (\cdot )=\varphi \left(\Psi \right)(\cdot )$ for $\Psi$ .

Definition 3.1. Let

$\begin{array}{l}{\mathcal{M}}_s=\{\varphi \left({\Psi }_0\right)\in {ℝ}_{+}^{\mathbb{Z}}\backslash \left\{0\right\}\mathrm{<mo>\; :\; </mo>}\text{there}\text{exists}{\Psi }_0\text{such}\text{that}\stackrel{}{}\\ \varphi \left({\left(U^{\left(s\right)}\right)}^n{\Psi }_0\right)=\varphi \left({\Psi }_0\right)\text{for}\text{any}n\ge 0\}\end{array}$ (3.2)

where $0$ is the zero vector. We call the element of ${\mathcal{M}}_s$ the stationary measure of quantum walk. If $\mu \in {\mathcal{M}}_s$ , then ${\mu }_n=\mu$ , where ${\mu }_n\left(x\right)=\varphi \left({\Psi }_n\left(x\right)\right)$ is the measure of the quantum walk at position $x\in \mathbb{Z}$ and at time $n\in \mathbb{N}$ .

Next we consider the eigenvalue problem:

$U^{\left(s\right)}\Psi =\lambda \Psi \mathrm{,}\left(\lambda \in ℂ\mathrm{,}\left|\lambda \right|=1\right)$ (3.3)

4. Main Results and Proofs

In this section, we obtain the stationary measure of the three-state quantum walk with one defect by following lemma.

Lemma 4.1. [17] Let ${\left\{U_y\right\}}_{y\in \mathbb{Z}}$ be the set of y-parameterized unitary matrices of the three-state inhomogeneous quantum walk, and $\Psi \left(x\right)={\left[{\Psi }^L\left(x\right)\mathrm{,}{\Psi }^O\left(x\right)\mathrm{,}{\Psi }^R\left(x\right)\right]}^{\text{T}}$ be the probability amplitude. Note that there is a restriction for the initial state $\Psi \left(0\right)$ [18] Then the solutions for $U^{\left(s\right)}\Psi =\lambda \Psi \left(\Psi \in Map\left(\mathbb{Z}\mathrm{,}{ℂ}^3\right)\mathrm{,}\lambda \in S^1\right)$ , where $S^1=\left\{\mathfrak{z}\in ℂ\mathrm{<mo>\; :\; </mo>}\left|\mathfrak{z}\right|=1\right\}$ , are

$\begin{array}{l}{\Psi }_x=\{\begin{array}{ll}{\displaystyle {\prod }_{y=1}^x{T}_y^{(+)}\Psi }\left(0\right),\hfill & x\ge 1,\hfill \\ \Psi \left(0\right),\hfill & x=0,\hfill \\ {\displaystyle {\prod }_{y=-1}^x{T}_y^{(-)}\Psi }\left(0\right),\hfill & x\le -1,\hfill \end{array}\hfill \end{array}$ (4.1)

where $T_y^{(\pm )}$ are the transfer matrices defined by

$T_y^{(+)}=\left[\begin{array}{ccc}{t}_{11}^{+}& t_{12}^{+}& t_{13}^{+}\\ t_{21}^{+}& t_{22}^{+}& t_{23}^{+}\\ t_{31}^{+}& t_{32}^{+}& t_{33}^{+}\end{array}\right],T_y^{(-)}=\left[\begin{array}{ccc}{t}_{11}^{-}& t_{12}^{-}& t_{13}^{-}\\ t_{21}^{-}& t_{22}^{-}& t_{23}^{-}\\ t_{31}^{-}& t_{32}^{-}& t_{33}^{-}\end{array}\right],$ (4.2)

with

$t_{11}^{+}=\frac{\left(\lambda -e_y\right)\left({\lambda }^2-g_{y-1}{c}_y\right)-g_{y-1}{b}_y{f}_y}{\lambda \left[a_y\left(\lambda -e_y\right)+b_y{d}_y\right]},t_{12}^{+}=-\frac{h_{y-1}\left[b_y{f}_y+c_y\left(\lambda -e_y\right)\right]}{\lambda \left[a_y\left(\lambda -e_y\right)+b_y{d}_y\right]}$

$t_{13}^{+}=-\frac{i_{y-1}\left[b_y{f}_y+c_y\left(\lambda -e_y\right)\right]}{\lambda \left[a_y\left(\lambda -e_y\right)+b_y{d}_y\right]},t_{21}^{+}=\frac{{\lambda }^2{d}_y+g_{y-1}\left(a_y{f}_y-c_y{d}_y\right)}{\lambda \left[a_y\left(\lambda -e_y\right)+b_y{d}_y\right]}$

$t_{22}^{+}=\frac{h_{y-1}\left(a_y{f}_y-c_y{d}_y\right)}{\lambda \left[a_y\left(\lambda -e_y\right)+b_y{d}_y\right]},t_{23}^{+}=\frac{i_{y-1}\left(a_y{f}_y-c_y{d}_y\right)}{\lambda \left[a_y\left(\lambda -e_y\right)+b_y{d}_y\right]}$

$t_{31}^{+}=\frac{g_{y-1}}{\lambda },t_{32}^{+}=\frac{h_{y-1}}{\lambda },t_{33}^{+}=\frac{i_{y-1}}{\lambda },$

and

$t_{11}^{-}=\frac{a_{y+1}}{\lambda },t_{12}^{-}=\frac{b_{y+1}}{\lambda },t_{13}^{-}=\frac{c_{y+1}}{\lambda },$

$t_{21}^{(-)}=-\frac{a_{y+1}\left(f_y-g_y-i_y{d}_y\right)}{\lambda \left[h_y{f}_y+i_y\left(\lambda -e_y\right)\right]},t_{22}^{(-)}=-\frac{b_{y+1}\left(f_y-g_y-i_y{d}_y\right)}{\lambda \left[h_y{f}_y+i_y\left(\lambda -e_y\right)\right]},$

$t_{23}^{(-)}=\frac{{\lambda }^2{f}_y-c_{y+1}\left(f_y-g_y-i_y{d}_y\right)}{\lambda \left[h_y{f}_y+i_y\left(\lambda -e_y\right)\right]},t_{31}^{(-)}=-\frac{a_{y+1}\left[h_y{d}_y+g_y\left(\lambda -e_y\right)\right]}{\lambda \left[h_y{f}_y+i_y\left(\lambda -e_y\right)\right]},$

$t_{32}^{(-)}=-\frac{b_{y+1}\left[h_y{d}_y+g_y\left(\lambda -e_y\right)\right]}{\lambda \left[h_y{f}_y+i_y\left(\lambda -e_y\right)\right]},t_{33}^{(-)}=-\frac{\left(\lambda -e_y\right)\left({\lambda }^2-g_y{c}_{y+1}\right)-h_y{c}_{y+1}{d}_y}{\lambda \left[h_y{f}_y+i_y\left(\lambda -e_y\right)\right]}.$

We now state the stationary measure of one-phase model with one defect.

Theorem 4.1. Let ${\Psi }_n\left(x\right)=\left({\Psi }_n{\left(x\right)}^L\mathrm{,}{\Psi }_n{\left(x\right)}^O\mathrm{,}{\Psi }_n{\left(x\right)}^R\right)$ be the wave function of probability amplitude, and ${\Psi }_0^L=\alpha ,{\Psi }_0^O=\beta ,{\Psi }_0^R=\gamma$ be the initial state. We take $\alpha =-\gamma ,\beta =0$ and $\lambda =-1$ . Then through the definition (2.1)

$U_x=\frac{{\text{e}}^{i{\theta }_x}}{3}\left[\begin{array}{ccc}-1& 2& 2\\ 2& -1& 2\\ 2& 2& -1\end{array}\right],x\in \mathbb{Z}$

where

${\theta }_x=\{\begin{array}{ll}0,\hfill & x=\pm 1,\pm 2,\cdots ,\hfill \\ 2\pi \tau ,\hfill & x=0.\hfill \end{array}$

We obtain the stationary measure

$\mu \left(x\right)=\{\begin{array}{ll}\left(4-2\Re \xi \right){\left|\alpha \right|}^2,\hfill & x=\pm 1,\hfill \\ 2{\left|\alpha \right|}^2,\hfill & \text{others}.\hfill \end{array}$ (4.3)

Proof. Put $\alpha ={\Psi }^L\left(0\right),\beta ={\Psi }^O\left(0\right)$ and $\gamma ={\Psi }^R\left(0\right)$ . Now we take $\alpha =-\gamma ,\beta =0$ and $\lambda =-1$ , then the solutions for $U^{\left(s\right)}\Psi =\lambda \Psi$ are

${\Psi }_x=\{\begin{array}{ll}{\displaystyle {\prod }_{y=1}^x{T}_y^{(+)}\Psi }\left(0\right),\hfill & x\ge 1,\hfill \\ \Psi \left(0\right),\hfill & x=0,\hfill \\ {\displaystyle {\prod }_{y=-1}^x{T}_y^{(-)}\Psi }\left(0\right),\hfill & x\le -1,\hfill \end{array}$

where $\begin{array}{l}{T}_y^{\pm }\hfill \end{array}$ are

$T_1^{(+)}=\left[\begin{array}{ccc}1& 0& 0\\ \frac{2\xi }{3}-1& \frac{2\xi }{3}& -\frac{\xi }{3}\\ -\frac{2\xi }{3}& -\frac{2\xi }{3}& \frac{\xi }{3}\end{array}\right],T_1^{(-)}=\left[\begin{array}{ccc}\frac{\xi }{3}& -\frac{2\xi }{3}& -\frac{2\xi }{3}\\ -\frac{\xi }{3}& \frac{2\xi }{3}& \frac{2\xi }{3}-1\\ 0& 0& 1\end{array}\right].$

$T_y^{(+)}=\left[\begin{array}{ccc}1& 0& 0\\ -\frac{1}{3}& \frac{2}{3}& -\frac{1}{3}\\ -\frac{2}{3}& -\frac{2}{3}& \frac{1}{3}\end{array}\right],T_y^{(-)}=\left[\begin{array}{ccc}\frac{1}{3}& -\frac{2}{3}& -\frac{2}{3}\\ -\frac{1}{3}& \frac{2}{3}& -\frac{1}{3}\\ 0& 0& 1\end{array}\right]\left(\left|y\right|\ge 2\right).$

Then through the formula (4.1), we can obtain

$\begin{array}{l}\Psi \left(x\right)=\{\begin{array}{ll}\left[\begin{array}{c}\alpha \\ 0\\ -\alpha \end{array}\right],\hfill & \left|x\right|\ne 1,\hfill \\ \left[\begin{array}{c}\alpha \\ \left(\xi -1\right)\alpha \\ -\xi \alpha \end{array}\right],\hfill & x=1,\hfill \\ \left[\begin{array}{c}\xi \alpha \\ \left(1-\xi \right)\alpha \\ -\alpha \end{array}\right],\hfill & x=-1.\hfill \end{array}\hfill \end{array}$ (4.4)

Therefore the corresponding stationary measure is given by

$\begin{array}{l}\mu \left(x\right)=\{\begin{array}{ll}\left(4-2\Re \xi \right){\left|\alpha \right|}^2,\hfill & x=\pm 1,\hfill \\ 2{\left|\alpha \right|}^2,\hfill & \text{others}.\hfill \end{array}\hfill \end{array}$

¨

Next we state the stationary measure of two-phase model by transfer matrices method.

Theorem 4.2. Let ${\Psi }_n\left(x\right)=\left({\Psi }_n{\left(x\right)}^L\mathrm{,}{\Psi }_n{\left(x\right)}^O\mathrm{,}{\Psi }_n{\left(x\right)}^R\right)$ be the wave function of probability amplitude, and ${\Psi }_0^L=\alpha ,{\Psi }_0^O=\beta ,{\Psi }_0^R=\gamma$ be the initial state. We take $\alpha =-\gamma ,\beta =0$ and $\lambda =-1$ . Then through the definition (2.2)

$\begin{array}{l}{U}_x=\{\begin{array}{ll}{U}_{+},\hfill & x\ge 1,\hfill \\ U_0,\hfill & x=0,\hfill \\ U_{-},\hfill & x\le -1,\hfill \end{array}\hfill \end{array}$

where

$U_{\pm }=\left[\begin{array}{ccc}-\frac{1+g_{\pm }}{2}& \frac{{\hslash }_{\pm }}{\sqrt{2}}& \frac{1-g_{\pm }}{2}\\ \frac{{\hslash }_{\pm }}{\sqrt{2}}& g_{\pm }& \frac{{\hslash }_{\pm }}{\sqrt{2}}\\ \frac{1-g_{\pm }}{2}& \frac{{\hslash }_{\pm }}{\sqrt{2}}& -\frac{1+g_{\pm }}{2}\end{array}\right],U_0=\frac{\xi }{3}\left[\begin{array}{ccc}-1& 2& 2\\ 2& -1& 2\\ 2& 2& -1\end{array}\right],$

where $g_{\pm }=\cos {\gamma }_{\pm },{\hslash }_{\pm }=\sin {\gamma }_{\pm },{\gamma }_{\pm }\in \left[0,2\pi \right),\xi ={\text{e}}^{2\pi i\tau },\tau \in \left(0,1\right)$ . Then the stationary measure is

$\mu \left(x\right)=\left(2+\Delta \right){\left|\alpha \right|}^2$ (4.5)

where

$\begin{array}{l}\Delta =\{\begin{array}{ll}\frac{{\hslash }_{-}^2}{2},\hfill & x<1,\hfill \\ \frac{{\hslash }_{-}^2\left(1-\Re \xi \right)}{{\left(1+g_{-}\right)}^2},\hfill & x=-1,\hfill \\ 0,\hfill & x=0,\hfill \\ \frac{{\hslash }_{+}^2\left(1-\Re \xi \right)}{{\left(1+g_{+}\right)}^2},\hfill & x=1,\hfill \\ \frac{{\hslash }_{+}^2}{2},\hfill & x<1\hfill \end{array}\hfill \end{array}$ (4.6)

Proof. Put $\alpha ={\Psi }^L\left(0\right),\beta ={\Psi }^O\left(0\right)$ and $\gamma ={\Psi }^R\left(0\right)$ . Now we take $\alpha =-\gamma ,\beta =0$ and $\lambda =-1$ , then the solutions for $U^{\left(s\right)}\Psi =\lambda \Psi$ are

$\begin{array}{l}{\Psi }_x=\{\begin{array}{ll}{\displaystyle {\prod }_{y=1}^x{T}_y^{(+)}\Psi }\left(0\right),\hfill & x\ge 1,\hfill \\ \Psi \left(0\right),\hfill & x=0,\hfill \\ {\displaystyle {\prod }_{y=-1}^x{T}_y^{(-)}\Psi }\left(0\right),\hfill & x\le -1,\hfill \end{array}\hfill \end{array}$

where $T_y^{\pm }$ are

$T_1^{(+)}=\left[\begin{array}{ccc}1& 0& 0\\ -\frac{{\hslash }_{+}\left(3-2\xi \right)}{3\sqrt{2}\left(1+g_{+}\right)}& \frac{2{\hslash }_{+}\xi }{3\sqrt{2}\left(1+g_{+}\right)}& -\frac{{\hslash }_{+}\xi }{3\sqrt{2}\left(1+g_{+}\right)}\\ -\frac{2\xi }{3}& -\frac{2\xi }{3}& \frac{\xi }{3}\end{array}\right]\mathrm{,}$

$T_1^{(-)}=\left[\begin{array}{ccc}\frac{\xi }{3}& -\frac{2\xi }{3}& -\frac{2\xi }{3}\\ -\frac{{\hslash }_{-}\xi }{3\sqrt{2}\left(1+g_{-}\right)}& \frac{2{\hslash }_{-}\xi }{3\sqrt{2}\left(1+g_{-}\right)}& -\frac{{\hslash }_{-}\left(3-2\xi \right)}{3\sqrt{2}\left(1+g_{-}\right)}\\ 0& 0& 1\end{array}\right]\mathrm{.}$

$\begin{array}{l}{T}_y^{(+)}=\left[\begin{array}{ccc}1& 0& 0\\ \frac{{\hslash }_{+}}{2\sqrt{2}}& \frac{1-g_{+}}{2}& -\frac{{\hslash }_{+}}{2\sqrt{2}}\\ -\frac{1-g_{+}}{2}& -\frac{{\hslash }_{+}}{\sqrt{2}}& \frac{1+g_{+}}{2}\end{array}\right],\\ T_y^{(-)}=\left[\begin{array}{ccc}\frac{1+g_{-}}{2}& -\frac{{\hslash }_{-}}{\sqrt{2}}& -\frac{1-g_{-}}{2}\\ \frac{{\hslash }_{-}}{2\sqrt{2}}& \frac{1-g_{-}}{2}& -\frac{{\hslash }_{-}}{2\sqrt{2}}\\ 0& 0& 1\end{array}\right]\left(\left|y\right|\ge 2\right).\end{array}$

Then through the formula (4.1), we can obtain

$\begin{array}{l}\Psi \left(x\right)=\{\begin{array}{ll}\left[\begin{array}{c}\alpha \\ \frac{{\hslash }_{-}}{\sqrt{2}}\alpha \\ -\alpha \end{array}\right],\hfill & x<-1,\hfill \\ \left[\begin{array}{c}\xi \alpha \\ \frac{{\hslash }_{-}\left(1-\xi \right)}{\sqrt{2}\left(1+g_{-}\right)}\alpha \\ -\alpha \end{array}\right],\hfill & x=-1,\hfill \\ \left[\begin{array}{c}\alpha \\ 0\\ -\alpha \end{array}\right],\hfill & x=0,\hfill \\ \left[\begin{array}{c}\alpha \\ \frac{{\hslash }_{+}}{\sqrt{2}\left(1+g_{+}\right)}\alpha \\ -\xi \alpha \end{array}\right],\hfill & x=1,\hfill \\ \left[\begin{array}{c}\alpha \\ \frac{{\hslash }_{+}}{\sqrt{2}}\alpha \\ -\alpha \end{array}\right],\hfill & x>1.\hfill \end{array}\hfill \end{array}$

Therefore we obtain the stationary measure

$\mu \left(x\right)=\left(2+\Delta \right){\left|\alpha \right|}^2$

where

$\begin{array}{l}\Delta =\{\begin{array}{ll}\frac{{\hslash }_{-}^2}{2},\hfill & x<1,\hfill \\ \frac{{\hslash }_{-}^2\left(1-\Re \xi \right)}{{\left(1+g_{-}\right)}^2},\hfill & x=-1,\hfill \\ 0,\hfill & x=0,\hfill \\ \frac{{\hslash }_{+}^2\left(1-\Re \xi \right)}{{\left(1+g_{+}\right)}^2},\hfill & x=1,\hfill \\ \frac{{\hslash }_{+}^2}{2},\hfill & x<1\hfill \end{array}\hfill \end{array}$

¨

5. Summary

In this paper, we derive the stationary measure of three-state walks with one dimension via transfer matrices. As a future work, we would investigate spectral theory and localization of three-state quantum walks.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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