Stationary measures for the three-state Grover walk with one defect in one dimension (Mathematical Aspects of Quantum Fields and Related Topics)
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(2) 46. Definition of Our Model. 2. This section. gives. the definition of. three‐state. our. QW. with. one. defect at the. origin. on. \mathb {Z} The discrete‐time .. quantum version of the classical random walk with additional degree of freedom caned chirality. The chirality takes values left, stay, or right, and it means the motion of the walker. At each time step, if the walker has the left chirality, it moves one step to the left, and if it has the right chirality, it moves one step to the right. If it has the stay chirality, it stays at the same position.. QW. is. a. Let. where. us. define. L, O and. |L\rangle=\left\{ begin{ar y}{l 1\ 0\ 0 \end{ar y}\right\},|O)=\left\{ begin{ar y}{l 0\ 1\ 0 \end{ar y}\right\}|R\rangle=\left\{ begin{ar y}{l 0\ 0\ 1 \end{ar y}\right\}. R refer to the. left, stay. and. right chirality states, respectively. by a sequence of 3\times 3 unitary matrices \{U_{x} : x\in \mathbb{Z}\} where. The time evolution of the walk is determined. with. u_{x,jk}\in \mathbb{C}(x\in \mathbb{Z}jk=123). model,. we. ,. U_{x}=\left{\begin{ar y}{l u_{x,\mathrm{l}\ athrm{l}&u_{x,\mathrm{l}2&u_{x,\mathrm{l}3\ u_{x,21|}&u_{x,2}&u_{x,23}\ u_{x,31}&u_{x,32}&u_{x,3} \end{ar y}\right\}. and \mathb {C} is the set of complex numbers. To define the. dynamics of. U_{x}^L_{\overline{\neg} \left\{ begin{ar y}{l u_{x,\mathrm{l}\mathrm{l} &u_{x,12}&\primeu_{x,13}\ 0& 0\ 0& 0 \end{ar y}\right\}, U_{x}^o=\left\{ begin{ar y}{l 0&0&0\ \mathrm{u}_x,2\mathrm{l} &u_{x,2 }&u_{x,23}\ 0&0&0 \end{ar y}\right\} U_{x}^{R}=\left\{ begin{ar y}{l 0&0&0\ 0&0&0\ u_{x,3\mathrm{l} &u_{x,32}&u_{x,3 } \end{ar y}\right\}. with. U_{x}=U_{x}^{L}+U_{x}^{O}+U_{x}^{R}. the left. (resp. right). at. .. The. important point. position. x. at each time. U_{x}^{L} (resp. U_{x}^{R} ) represents that the walker moves U_{x}^{O} represents that the walker stays at position x.. is that. step.. The model considered here is. where. $\omega$=e^{i $\theta$}( $\theta$\in[02 $\pi$. If $\theta$=0 , that. is,. Here. $\omega$=1 then. U_{G}. given by. U_{G}=\displaystyle \frac{1}{3}[_{2}^{-1}2 -12 -212].. on. $\Psi$_{n} denote the amplitude. to. U_{x}=\left\{ begin{ar ay}{l} $\omega$U_{G}&(x=0)\ U_{G}&(x=\pm1\pm2. \end{ar ay}\right.. is the Grover matrix. U_{x}=U_{G} for. usual three‐state Grover walk Let. our. divide U_{x} into three matrices:. any x\in \mathbb{Z}. .. So this. space‐homogeneous. model is. equivalent. to the. \mathbb{Z}. at time. n. of the. QW. as. follows.. $\Psi$_{n}= $\eta$\ldots$\Psi$_{n}^{L}(-1)$\Psi$_{n}^{O}(-1)$\Psi$_{n}^{R}(-1)$\Psi$_{n}^{L}(0)$\Psi$_{n}^{O}(0) , $\Psi$_{r $\iota$}^{R}(0)$\Psi$_{n}^{L}(1)$\Psi$_{n}^{O}(1)$\Psi$_{n}^{R}(1)\cdots]. where T. =^{T}[\cdots[_{$\Psi$_{n}^{R}(-1)}^{$\Psi$_{n}^{L}(-1)}$\Psi$_{n}^{o}(-1)][_{$\Psi$_{n}^{R}(0)}^{$\Psi$_{n}^{L}(0)}$\Psi$_{n}^{O}(0)][$\Psi$_{n}0_{(1)]\cdots]}$\Psi$_{n}^{L}(1). means. the. transposed operation. Then. the time evolution of the walk is defined. $\Psi$_{n+1}(x)=U_{x+1}^{L}$\Psi$_{n}(x+1)+U_{x}^{O}$\Psi$_{n}(x)+U_{x-1}^{R}$\Psi$_{n}(x-1). .. Now let. U^{(s)}=. U_{\frac}2 {o:OR}2 U_{\frac}O1 {oR}U_\frac:L1O U_{0}^R o {L}O^:. U_{1}^R OU_{1}^Lo: U_{2}^O Lo^{:}O :. :. :. with. 0=\left{\begin{ar y}{l 0& 0\ 0& 0\ 0& 0 \end{ar y}\right\}. by. (2.1).
(3) 47. Then the state of the. QW. at time. n. is. given by. $\Psi$_{n}=(U^{(s)})^{n}$\Psi$_{0}, for any n\geq 0. \mathbb{R}+=[0\infty ).. Let. .. Here. introduce. we. a. map. $\phi$. :. (\mathb {C}^{3})^{\mathrm{z} \rightar ow \mathb {R}_{+}^{\mathrm{Z}. such that if. $\Psi$=^{T}[\cdots[_{$\Psi$^{R}(-1)}^{$\Psi$^{L}(-1)}$\Psi$^{O}(-1)][_{$\Psi$^{R}(0)}^{$\Psi$^{L}(0)}$\Psi$^{O}(0)][_{$\Psi$^{R}(1)}^{$\Psi$^{L}(1)}$\Psi$^{O}(1)]\cdots]\in(\mathb {C}^{3})^{\mathrm{z}. then. $\phi$( $\Psi$)=^{T}[\ldots |$\Psi$^{L}(-1)|^{2}+|$\Psi$^{O}(-1)|^{2}+|$\Psi$^{R}(-1)|^{2}|$\Psi$^{L}(0)|^{2}+|$\Psi$^{O}(0)|^{2}+|$\Psi$^{R}(0)|^{2} |$\Psi$^{L}(1)|^{2}+|$\Psi$^{O}(1)|^{2}+|$\Psi$^{R}(1)|^{2} .]\in \mathb {R}_{+}^{\mathb {Z} . ... ,. That. is, for. any x\in \mathbb{Z}. $\phi$( $\Psi$)(x)= $\phi$( $\Psi$(x))=|$\Psi$^{L}(x)|^{2}+|$\Psi$^{O}(x)|^{2}+|$\Psi$^{R}(x)|^{2}. Moreover. we. define the. measure. of the. QW. at. position. x. by. $\mu$(x)= $\phi$( $\Psi$(x)) (x\in \mathrm{Z}) Now. we are. ready. to introduce the set of. \mathcal{M}_{8}=\mathcal{M}_{s}(U)=\{ $\phi$($\Psi$_{0})\in \mathb {R}_{+}^{\mathrm{Z} \backslash \{0\} where 0 is the Next. we. zero. :. stationary. measure:. there exists $\Psi$_{0} such that. vector. We call the element of. consider the. \mathcal{M}_{s} eigenvalue problem of the QW:. the. Remark that Let. | $\lambda$|=1. on. $\mu$_{n}(x). since. eigenvalue. be the. $\lambda$. .. measure. U^{( $\epsilon$)} Then. is. unitary.. we see. of the. QW. that. at. $\phi$( U^{(s)})^{n^{\triangleright} $\Psi$_{0})= $\phi$($\Psi$_{0}). stationary. U^{(s)} $\Psi$ 1= $\lambda \Psi$ ( $\lambda$\in \mathbb{C}) dependence. .. \displaystyle \lim_{n\rightar ow\infty}$\mu$_{n}(x). exists for any x\in \mathbb{Z} then. we. and at time. define the limit. n. Splitted Generating. In this. method. in order to. emphasize. the. i.e.,. .. measure. $\mu$_{\infty}(x)=\displaystyle \lim_{n\rightar ow\infty}$\mu$_{n}(x) (x\in \mathbb{Z}) 3. $\Psi$=$\Psi$^{( $\lambda$)}. We sometime write x. n\geq 0\}, (2.2). $\phi$($\Psi$^{( $\lambda$)})\in \mathcal{M}_{ $\epsilon$}.. position. any. QW.. .. $\mu$_{n}(x)= $\phi$($\Psi$_{n}(x)) (x\in \mathbb{Z}) If. measure. of the. for. $\mu$_{\infty}(x) by. .. Function Method. we give solutions of eigenvalue problem, U^{(s)} $\Psi$= $\lambda \Psi$ by the splitted generating function developed in the previous stndies [11, 3]. First we see that U^{(s)} $\Psi$= $\lambda \Psi$ is equivalent to the following. section,. relations:. $\lambda$[_{$\Psi$^{1\mathrm{i}(1)}^{$\Psi$^{L}(1)}$\Psi$^{O}(1)]=\displaystyle\frac{1}{3}[2$\Psi$^{L}2(1)-$\Psi$^{O}(1)+2$\Psi$^{R}(1)] $\lambda$\displayst le\left\{ begin{ar y}{l $\alpha$\ $\beta$\ $\gam a$ \end{ar y}\right\}= frac{1}3[2$\omega\ lpha$- \omega$\sqrt{}+2$\omega\gam a$] $\lambda$[_{ \Psi$^{R}(-1)}^{$\Psi$^{L}(-1)}$\Psi$^{O}(-1)]=\displaystle\frac{1}3[_{2} _{$\Psi$^{L}(-2)+ $\Psi$^{O}(-2)$\Psi$^{R}(-2)}^{-$\omega\lpha$+2$\omega\beta$+2$\omega\gam a$} \Psi$^{L}(-1)$\Psi$^{O}(-1)+2$\Psi$^{R}(-1)].
(4) 48. and for. x\neq-101. $\lambda$[_{$\Psi$^{R}(x)^{$\Psi$^{L}(x)$\Psi$^{O}(x)]=\displayst le\frac{1}3[_{2$\Psi$^{L}(x-1)+2$\Psi$^{O}(x-1)$\Psi$^{R}(x-1)}^{-$\Psi$^{L}(x+1) 2$\Psi$^{O}(x+1) 2$\Psi$^{R}(x+1)}2$\Psi$^{L}(x)-$\Psi$^{O}(x)+2$\Psi$^{R}(x)] where. $\Psi$^{L}(0)= $\alpha \Psi$^{O}(0)= $\beta$, $\Psi$^{R}(0)= $\gamma$. Here. we. introduce six. with. generating functions. | $\alpha$|^{2}+| $\beta$|^{2}+| $\gamma$|^{2}>0. as. follows:. f_{+}^{j}\displaystyle \langle z)=\sum_{x=1}^{\infty}$\Psi$^{j}(x)z^{x}, f_{-}^{j}(z)=\sum_{x=-1}^{-\infty}$\Psi$^{j}(x)z^{x} (j=L. OR) Then the. following lemma. LEMMA3.1. was. given Uy Wang. et al.. [17].. We put. A=[-\displayte\frac{2}3.$\lambd$+\frac{1}3-\frac{2z}3-\frac{2}3z where. .. -\displayst le\frac{2}3z}. f_{\pm}(z)=\left{bginary}{l f_\pm}^{L(z)\ f_{pm}^O(z)\ f_{pm}^R(z) \end{ary}\ight,. -\displaystle\frac{2}3 $\lambda$+\displaystyle\frac{z}3. a_{+}(z)=\left{bginary}{l -$\ambd\alph$\el 0\frac{$omega$z(2\alph$+2\beta$-\gma$)}{3 \end{ary}\ight a_{-}(z)=[\displaystle\frac{$\omega$(- \alph$+2\sqrt{}+2$\gam a$)}{3z0]. | $\alpha$|^{2}+| $\beta$|^{2}+| $\gamma$|^{2}>0. .. Then. we. have. Af_{\pm}(z)=a\pm(z). .. We should remark that. \displaystyle \det A=\frac{ $\lambda$( $\lambda$-1)}{3z}\{z^{2}+3( $\lambda$+\frac{4}{3}+\frac{1}{ $\lambda$}.)z+1\}. Then. $\theta$_{s}. and. $\theta$_{l}(\in \mathbb{C}). are. defined. by. \displaystyle \det A=\frac{ $\lambda$( $\lambda$-1)}{3z}(z+$\theta$_{s})(z+$\theta$_{l}) where. Wang. |$\theta$_{s}|\leq 1\leq|$\theta$_{l}| [17].. et al.. .. Note that. $\theta$_{s}$\theta$_{l}=1 Lemma .. 3.1. gives the following lemma which. was. also shown. by.
(5) 49. LEMMA 3.2. Here. $\Psi$^{L}(x)=\left\{ begin{ar y}{l $\alpha$(- \thea$_{s}^L(+)^{x}&(x\geq1)\ -\frac{(3$\lambda$+1)$\Delta$(-) \omega$-6($\lambda$+1)$\gam a$}{3($\lambda$-1)}($\thea$_{s}^L(-)^{x}&(x\leq-1), \end{ar y}\right. $\Psi^{O}(x)=\left{\begin{ar y}{l -\frac{2($\Delta$(+) \omega$-3 \alph$)}{3($\lambd$-1)}($\thea$_{s}^O(+)^{x}&(\geq1)\ -frac{2($\Delta$\ngle-)$\omega$-3 \gam $)}{3($\lambd$-1\rangle}(-$\thea$_{s}^O(-)^{x}&(\leq-1) \end{ar y}\right. $\Psi$^{R}(x)=\left\{begin{ar y}{l -\frac{(3$\lambda$+1)\triangle(+)$\omega$-6($\lambda$+1)$\alph$}{3($\lambda$-1)}($\thea$_{s}^R(+)^{x}&(x\geq1)\ $\gam a$(-\thea$_{s}^R(-)^{x}&(x\leq-1). \end{ar y}\right.. \triangle(+)=2 $\alpha$+2\sqrt{}- $\gamma \Delta$. =- $\alpha$+2\sqrt{}+2 $\gamma$ and. $\theta$_{s}^{L}(+)=-\displaystyle \frac{2( $\lambda$+1)\triangle(+) $\omega$-3$\lambda$^{2}(3 $\lambda$+1) $\alpha$}{3 $\lambda$( $\lambda$-1) $\alpha$} $\theta$_{s}^{L} =\frac{( $\lambda$-1) $\Delta$(-) $\omega$}{ $\lambda$\{(3 $\lambda$+1) $\Delta$(-) $\omega$-6( $\lambda$+1) $\gam a$\} ) $\theta$_{s}^{O}(+)=\displaystyle\frac{$\Delta$(+)$\omega$-3$\lambda$^{2}$\alpha$}{$\lambda$(\triangle(+)$\omega$-3$\alpha$)}$\theta$_{8}^{O}=\frac{$\Delta$(-)$\omega$-3$\lambda$^{2}$\gam a$}{$\lambda$( \Delta$(-)$\omega$-3$\gam a$)} $\theta$_{s}^{R}(+)=\displaystyle \frac{( $\lambda$-1) $\Delta$(+) $\omega$}{ $\lambda$\{(3 $\lambda$+1) $\Delta$(+) $\omega$-6( $\lambda$+1) $\alpha$\} $\theta$_{s}^{R} =-\frac{2( $\lambda$+1) $\Delta$(-) $\omega$-3$\lambda$^{2}(3 $\lambda$+1) $\gam a$}{3 $\lambda$( $\lambda$-1) $\gam a$}. From. now. on,. find out. we. a. necessary and sufficient condition for. $\theta$_{s}^{L}(+).=$\theta$_{s}^{o}(+)=$\theta$_{s}^{R}(+)=$\theta$_{8}^{L} =$\theta$_{8}^{o} =$\theta$_{s}^{R} First. we see. that. $\theta$_{S}^{L}(+)=$\theta$_{ $\epsilon$}^{R}. and. $\theta$_{s}^{R}(+)=$\theta$_{s}^{L}. give. ( $\alpha$- $\gamma$)( $\alpha$+ $\gamma$-2 $\beta$)=0. In. a. similar. Moreover. \mathrm{f}\mathrm{a}s\mathrm{h}\mathrm{i}\mathrm{o}\mathrm{n}$\theta$_{s}^{O}(+)=$\theta$_{s}^{O}. $\theta$_{8}^{L}(+)=$\theta$_{8}^{O}(+). implies. ( $\alpha$- $\gamma$)( $\alpha$+ $\gamma$-2 $\beta$)( $\lambda$+1)( $\lambda$-1), =0.. and. $\theta$_{ $\epsilon$}^{O}(+)=$\theta$_{\dot{s} ^{R}(+). give. ( $\lambda$+1)\{9 $\alpha$( $\omega \Delta$(+)-2 $\alpha$)$\lambda$^{2}-6 $\alpha \omega \Delta$(+) $\lambda$-w $\Delta$(+)(2 $\omega \Delta$(+)-9 $\alpha$)\}=0. Similarly combining. $\theta$_{S}^{L}. =$\theta$_{ $\delta$}^{o}. with. $\theta$_{$\delta$}^{o}. =$\theta$_{ $\epsilon$}^{R}. implies. ( $\lambda$+1)\{9 $\gamma$( $\omega \Delta$(-) 2 $\gamma$)$\lambda$^{2}-6 $\gamma \omega \Delta$(-) $\lambda$- $\omega \Delta$(-)(2 $\omega$\triangle(-) 9 $\gamma$)\}=0. From. $\theta$_{s}^{L}(+)=$\theta$_{S}^{R}(+). we. get. (3 $\lambda$+1)( $\lambda$+.1)\{9 $\alpha$( $\omega \Delta$(+)-2 $\alpha$)$\lambda$^{2}-6 $\alpha \omega \Delta$(+) $\lambda$- $\omega \Delta$(+)(2 $\omega \Delta$(+)-9 $\alpha$)\}=0. \mathrm{F}\mathrm{u}\mathrm{r}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{o}\mathrm{r}\mathrm{e}$\theta$_{s}^{L}. =$\theta$_{s}^{R}. gives. (3 $\lambda$+1)( $\lambda$+1)\{9 $\gamma$( $\omega$\triangle(-) 2 $\gamma$)$\lambda$^{2}-6 $\gamma \omega \Delta$(-) $\lambda$- $\omega \Delta$(-)(2 $\omega \Delta$(-) 9 $\gamma$)\}=0. Therefore. we. have.
(6) 50. LEMMA 3.3 A necessary and. sufficient. condition. for. $\theta$_{s}^{L}(+)=$\theta$_{s}^{O}(+)=$\theta$_{s}^{R}(+)=$\theta$_{8}^{L} =$\theta$_{s}^{O} =$\theta$_{8}^{R} is that. $\alpha \beta \gamma$. and. $\lambda$(\in \mathbb{C}). with. |\mathrm{a}|^{2}+| $\beta$|^{2}+| $\gamma$|^{2}>0. and. | $\lambda$|=1 satisfy. $\beta$=\displayst le\frac{2$\omega$( \alpha$+ \gam a$)}{3$\lambda$+ \omega$}. (3.3) (3.4). ( $\alpha$- $\gamma$)( $\alpha$+ $\gamma$-2 $\beta$)=0. ( $\lambda$+1)\{9 $\alpha$( $\omega \Delta$(+)-2 $\alpha$)$\lambda$^{2}-6 $\alpha \omega \Delta$(+) $\lambda$- $\omega$\triangle(+)(2 $\omega \Delta$(+)-9 $\alpha$)\}=0 ( $\lambda$+1)\{9 $\gamma$( $\omega \Delta$(-\rangle-2 $\gamma$)$\lambda$^{2}-6 $\gamma \omega \Delta$(-) $\lambda$- $\omega \Delta$(-)(2 $\omega \Delta$(-) 9 $\gamma$)\}=0. (3.5). (3.6). .. is missing in Wang et al. [17]. eigenvalue $\lambda$ for U^{(8)} $\Psi$= $\lambda \Psi$ as follows. spaceinhomogeneous.. We should note that. By is,. our. Lemma. QW. )(\mathrm{i})a= $\gamma$. is. 3.3,. We. case.. (a) $\alpha$\neq $\beta$. relation. a. we. that. see. case.. (3.3). obtain the. Eq. (3.5). If $\alpha$=0 then. is. equivalent. to. Here. we assume. Eq. (3.6), since $\Delta$(+)= $\Delta$. Eq. (3.3) gives $\beta$=0 So .. we assume. that. $\omega$\neq 1. that. = $\alpha$+2 $\beta$.. $\alpha \beta \gamma$\neq 0. .. Then. Eq. (3.5) implies. \displaystyle \frac{27$\alpha$^{2} {(3 $\lambda$+ $\omega$)^{2} ( $\lambda$+1)\{3( $\omega$-2)$\lambda$^{4}+2(5 $\omega$-3) $\omega \lambda$^{3} +(3$\omega$^{2}-8 $\omega$+3) $\omega \lambda$^{2}+2(5-3 $\omega$)$\omega$^{2} $\lambda$+3$\omega$^{3}(1-2 $\omega$)\}=0. One solution of this. explicitly. So. (b) $\alpha$= $\beta$. $\lambda$= $\omega$. case. .. we. equation is $\lambda$_{1}=-1 The rest of ggt stationary measures. .. $\lambda$_{2}$\lambda$_{3}$\lambda$_{4}$\lambda$_{5}. solutions. are. not obtained. do not. If $\alpha$=0 then. Eq. (3.5) gives. Eq. (3.3) gives $\beta$=0 So .. we assume. $\alpha$\sqrt{} $\gamma$\neq 0. .. Then. Eq. (3.3) implies. 27 $\lambda \alpha$^{2}( $\lambda$+1)( $\lambda$-1)^{2}=0. Then. (ii). $\beta$=\displaystyle\frac{$\alpha$+$\gam a$}{2}. we. have $\lambda$=-1 , since. $\omega$\neq 1.. case.. (a) $\beta$=0. case.. $\alpha$=- $\gamma$. we. Combining Eq. (3.3) with $\beta$=( $\alpha$+ $\gamma$)/2 gives. $\alpha$=- $\gamma$. .. Eq. (3.5) and. Then from. have. 9$\alpha$^{2}($\lambda$+1)($\lambda$-\displaystyle\frac{$\omega$+\sqrt{6$\omega$( \omega$-1)^{2} {3$\omega$-2})($\lambda$-\frac{$\omega$-\sqrt{6$\omega$( \omega$-1)^{2} {3$\omega$-2})=0.. Remark that Eq.. (3.5). is. equivalent. to. Eq. (3.6). since. \triangle(+)=- $\Delta$(-)=3 $\alpha$ Thus, $\alpha$\neq 0 implies .. $\lambda$=-1,\displaystyle\frac{$\omega$\pm\sqrt{6$\omega$($\omega$-1)^{2} {3$\omega$-2} (b) $\beta$\neq 0. $\lambda$= $\omega$. case. we. Combining Eq. (3.3). with. (3.7). .. $\beta$=( $\alpha$+ $\gamma$)/2 gives. $\lambda$=.. $\omega$. have. 27$\alpha$^{2} $\lambda$( $\lambda$+1)( $\lambda$-1)^{2}=0. Similarly, combining Eq. (3.6). with $\lambda$= $\omega$. gives. 27$\gamma$^{2} $\lambda$( $\lambda$+1)( $\lambda$-1)^{2}=0. Then $\lambda$=-1 follows from above two. We note that $\alpha$= $\gamma$ and. equations.. $\beta$=( $\alpha$+ $\gamma$)/2 gives \mathrm{a}= $\beta$= $\gamma$. .. This. case. is. (i‐b).. .. Then from. Eq. (3.5) and.
(7) 51. 4. Stationary Measures. First. we. Then. we see. obtain stationary. measures. for. (ii‐a). case. with respect to the. following $\lambda$ :. $\lambda$(\displaystyle\pm)=\frac{$\omega$\pm\sqrt{6$\omega$( \omega$-1)^{2} {3$\omega$-2}. that for. OR. j=L,. $\theta$_{s}^{j}(\displaystyle\pm)=\frac{-(3$\omega$+2)+2\sqrt{6}e^{\frac{$\theta$}{2}i {(2-3$\omega$)(1+2\sqrt{3(1-\cos$\theta$)}i \frac{-(3$\omega$+2)+2\sqrt{6}e^{\frac{$\theta$}{2}i {(2-3$\omega$)(1-2\sqrt{3(1-\cos$\theta$)}i \displaystyle\frac{-(3$\omega$+2)-2\sqrt{6}e\Si}{(2-3$\omega$)(1+2\sqrt{3(1-\cos$\theta$)}i)}\frac{-(3$\omega$+2)-2\sqrt{6}e^{\frac{$\theta$}{2}i {(2-3$\omega$)(1-2\sqrt{3(1-\cos$\theta$)}i)}. Note that. $\theta$_{s}^{j}(\pm)(j=L, O, R). do not. depend. on. j\mathrm{a}\mathrm{n}\mathrm{d}\pm. ,. so we. put. $\theta$_{s}=$\theta$_{s}^{j}(\pm). .. Then. we. get. |$\theta$_{s}|^{2}=\displaystyle \frac{37+12\cos $\theta$\pm 20\sqrt{6}\cos( $\theta$/2)}{(13_{:}-12\cos $\theta$)^{2} . We should remark that if. 0\leq $\theta$\leq 4 $\pi$ with. axccos. then. (1/3)=1.2309\ldots\leq $\theta$\leq 4 $\pi$-\arccos(1/3)=11.3354\ldots,. |$\theta$_{B}|^{2}=\displaystyle \frac{37+12\cos $\theta$+20\sqrt{6}\cos( $\theta$/2)}{(13-12\cos $\theta$)^{2} \leq 1. \mathrm{S}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{l}\mathrm{a}x1\mathrm{y}. if. 0\leq $\theta$\leq 4 $\pi$ with. 4 $\pi$ then. 0\leq $\theta$\leq 2 $\pi$-\arccos(1/3)=5.0522\ldots. and. 2 $\pi$+\arccos(1/3)=7.5141\ldots\leq $\theta$\leq. |$\theta$_{s}|^{2}=\displaystyle \frac{37+12\cos $\theta$-20\sqrt{6}\cos( $\theta$/2)}{(13-12\cos $\theta$)^{2} \leq 1. From Lemma 3.2 for this case,. we. have the. eigenvalues for $\lambda$(\pm). as. follows.. $\Psi$^{L}(x)=$\alpha$\times\left\{ begin{ar y}{l (-$\theta$_{s})^{x}&(x\geq1)\ \frac{(3$\lambda$+1)$\omega$-2($\lambda$+1)}{$\lambda$-1}(-$\theta$_{s})^{-x}&(x\leq-1) \end{ar y}\right. \backslah$\Psi$^{O}(x)=$\alpha$\times\left\{ begin{ar y}{l -2\frac{$\omega$-1}{$\lambda$-1}($\thea$_{8})^{x}&(x\geq1)\ 2\frac{$\omega$-1}{$\lambda$-1}($\thea$_{8})^{-x}&(x\leq-1), \end{ar y}\right.. $\Psi$^{R}(x)=$\alpha$\times\left\{ begin{ar y}{l -\frac{(3$\lambda$+1)$\omega$-2($\lambda$+1)}{$\lambda$-1}(-$\theta$_{s})^{x}&(x\geq1)\ -($\theta$_{s})^{-x}&(x\leq-1). \end{ar y}\right. Here. we. note that. \displaystyle\frac{(3$\lambda$(\pm)+1)$\omega$-2($\lambda$(\pm)+1)}{$\lambda$-1}.=\frac{(3e^{i$\theta$}-2)(\sqrt{6}e^{i\frac{$\theta$}{2} +2)}{\sqrt{6}e^{i\frac{$\theta$}{2} -2}\frac{(3e^{i$\theta$}-2)(\sqrt{6}e^{i\frac{$\theta$}{2} +2)}{-\sqrt{6}e^{i\frac{$\theta$}{2} -2} \displaystyle\frac{(3e^{i$\theta$}-2)(-\sqrt{6'}e^{i\frac{$\theta$}{2}+2)}{\sqrt{6}e^{i\mathrm{s}_{-2} \frac{(3e^{i$\theta$}-2)(-\sqrt{6}e^{i\frac{$\theta$}{2}+2)}{-\sqrt{6}e^{i\frac{$\theta$}{2}-2}, and. \displaystyle\frac{$\omega$-1}{$\lambda$(\pm)-1}.=\frac{3e^{i$\theta$}-2}{-2+\sqrt{6}e^{i_{E}^{$\theta$} -2\sqrt{6}e^{i_{E}^{$\theta$} 3e^{i$\theta$}-2 Therefore. we. obtain the. following. main result:.
(8) 52. THEOREM 4.1. (02 $\pi$)). We consider the three‐state Grover walk with. $\alpha$=- $\gamma \beta$=0. and. .. Then. a. solution. of U^{( $\epsilon$)} $\Psi$= $\lambda \Psi$. one. defect. at the. origin. on. \mathb {Z}. .. Here. $\omega$=e^{i $\theta$}( $\theta$\in. with. $\lambda$=\displayst le\frac{$\omega$\pm\sqrt{6$\omega$( \omega$-1)^{2} {3$\omega$-2} is. given by. Moreover the. $\Psi(x)=alph-$\te_{8})^|xims\leftbgn{ary} [Mtix]&(\geq1) {[}Matrix]&(=0)\ {[}^frac3$lmbd+1)\oega$-2(lmbd+1)}{\frac2($omeg-1)\labd$}{m-1]&(x\leq). nd{ary}\ight. stationary. measure. of the. walk is. $\mu$(x)\cdot=|$\alpha$|^{2}\times\left\{ begin{ar y}{l |$\thea$_{8}|^{2x}\{1+( 3-12\cos$\thea$)(\frac{2}m_{1}+\frac{m_{2} m_{3})\}&/(x\neq0)\ 2&(x=0). \end{ar y}\right. Here. |$\theta$_{s}|^{2}=\displaystyle \frac{37+12\cos $\theta$\pm 20\sqrt{6}\cos( $\theta$/2)}{(13-12\cos $\theta$)^{2} m_{k}=5+(-1)^{n_{k} 2\sqrt{6}\cos( $\theta$/2) (k=123) with. ;. n_{k}\in\{01\}.. relation between stationary and limit measures of the walk. If $\omega$=1, ,then our space‐homogeneous three‐state Grover walk on \mathb {Z} As for the hmit measure for an initial state; $\Psi$_{0}(0)= $\tau$[\overline{ $\alpha$}\sqrt{}^{\sim}\overline{ $\gamma$}] and $\Psi$_{0}(x)= $\tau$[0, 00](x\neq 0) the following result is known (see Konno As. a. corollary,. we. give. a. model becomes the usual. [10],. for. .. example).. $\mu_{infty}(x)=le\bgin{ary}l \(3+sqt{6)|2ilde$\aph}+t{bea$|^2}+(3-\sqrt{6)ilde$ba}+2\t{gma$}|^2&\ -overlin{$aph}+\tdeba$il{\gma$}|^2ties(49-0\qr{6})^x&(ge1\ frac{6-2sqt}(|\overlin{$aph}+\overlin{$bta}|^2+\overlin{$bta}+2\degma$}|^{2)&(x=0,\ 3-sqrt{6})|2\ilde$aph}+ovrline{$\bta}|^2+(3sqr{6)\tilde$ba}+2{\gma$}|^2& -\tilde{$aph}+ \beta$ild{gma$}|^2\ties(49-0qr{6})^x&(\le-1. nd{ary}\ight..
(9) 53. Combining. this with the. On the other. corresponding (ii‐a). hand, Theorem. case. (\tilde{ $\alpha$}=-\tilde{ $\gamma$}, \sqrt{}^{\sim}=0). gives. $\mu$_{\infty}(x)=\left\{ begin{ar y}{l 24|\tilde{$\alpha$}|^{2}(49-20\sqrt{6})^{X}&(x\neq0),\ 4(5-2\sqrt{6})|\tilde{$\alpha$}|^{2}.&(x=0). \end{ar y}\right.. 4.1 with. (4.8). $\theta$=0\mathrm{a}\mathrm{n}\mathrm{d}-\mathrm{p}\mathrm{a}\mathrm{l}\cdot \mathrm{t}(n_{1}=1n_{2}=0n_{3}=1) implies. $\mu$(x)=\left\{ begin{ar y}{l 12|$\alpha$|^{2}(5+2\sqrt{6})(49-20\sqrt{6})^{|x}&(x\neq0)\ 2|$\alpha$|^{2}&(x=\ve 0). \end{ar y}\right. If. we. put. $\alpha$=\pm(2-\sqrt{6})\overline{ $\alpha$}. ,. then. given by Eq. (4.8). Finally, we give stationary. (i). $\alpha$= $\gamma$. a. stationary. measures. measure. for $\lambda$=-1. .. given by Eq. (4.9). Remark that. we see. is. (4.9). equivalent. $\theta$_{8}=-1 for. case.. Therefore the. $\Psi(x)=left{bgnary} [\c$ph{omega-3}2^\l$ +)frac{ph\omeg$(3 }a-1\omeg$)&(xq {[}frac4\omeglph${a-3,\}]&(x=0) {[frc$alph\omeg-3}2_{$ a+)\frclphomeg$}{a-3,(\1omeg$)&xlq-. \nd{ary}iht. corresponding stationary. measure. is. given by. $\mu$(x)=\displayst le\frac{6|$\alpha$|^{2} 5-3\cos$\thea$}\times\left\{ begin{ar y}{l 3\cos^{2}$\thea$-3\cos$\thea$+2&(x\neq0)\ 3-2\cos$\thea$&(x=0). \end{ar y}\right. (ii). $\beta$=\displaystyle \frac{ $\alpha$+ $\gam a$}{2}. case.. (a) $\beta$=0. case.. \dot{$Psi}(x)=\left{bginary}{l [Mtrix]&(\geq1) {[}Matrix]&(=0)\ {[}Matrix]&(\leq-1). nd{ary}\ight.. to. this. a. limit. case.. measure.
(10) 54. Therefore the. corresponding stationary. measure. is. given by. $\mu$(x)=2|$\alpha$|^{2}\times\left\{ begin{ar y}{l (2-\cos$\thea$)&(x\neq0)\ 1&(x=0). \end{ar y}\right. (b) $\beta$\neq 0. case.. Therefore the. 5. $\Psi(x)=left{bgnary} [Mtix]&(\geq1) {[}frac$\lph+gma $}{\m2]&(x=0)\ .[Matri]&(xleq-1). \nd{ary}ight. corresponding stationary. measure. is. given by. $\mu(x)=\left{bginary}{l 6|$\alph|^{2}&(x\geq1) \frac{5}4(|$\alph|^{2}+$\gam $|^{2})+\frac{1}4($\alph overlin{$\gam $}+\overlin{$\aph}$\gam $)&(x=0,\ 6|$gam $|^{2}&(x\leq-1)_{:} \endary}\ight.. Summary. We obtained. stationary measures for the three‐state Grover walk with one defect at the origin on \mathb {Z} by solving the corresponding eigenvalue problem. Moreover, we found out a relation between stationary and limit measures of the walk. As a future work, it would‐ be interesting to investigate the relation between stationary measure, (time‐averaged) limit measure, and rescaled weak limit measure [78] for QWs in the more general setting. Acknowledgment. This work is partially supported by the Grant‐in‐Aid for Scientific Research (Challeng‐ ing Exploratory Research) of Japan Society for the Promotion of Science (Grant \mathrm{N}\mathrm{o}.15\mathrm{K}13443 ).. References [1] Cantero, Inf.. M. J., Grünbaum, F. A., Moral, L., Velazquez, Process., 11, 1149‐1192 (2012). L.: The CGMV method for quantum walks.. [2] Endo,. S Endo, T., Konno, N., Segawa, E., Takei, M.: Limit theorems of one‐defect, Quantum Inf. Comput., 15, 1373−13g6 (2015). [3] Endo, T., Konno, Mathematical. [4J Endo; T.,. N.: The stationary measure of Journal, 6^{\{}0 33‐47 (2014). two‐phase quantum walk with. space‐inhomogeneous quantum. walk. on. the. linc, Yokohama. ,. \mathrm{K}\mathrm{a}\grave{\mathrm{w} ai,. arXiv: 1603.08948. a. a. Quantum. H., Konno, N.:. (2016). The stationary. measure. for. diagonal quantum walk with. ope defect.
(11) 55. [5] Endo, T., Konno, N., Obuse, arXiv: 1511.04230 (2015). H.: Relation between. [6] Endo, T., Konno, N., Segawa, E., Takei, Mathematical Journal, 60, 49‐90 (2014) [7] Konno,. N.:. [8] Konno,. N.: A. [9] Konno,. N.:. Quantum random. walks in. two‐phase quantum walks and. the. M.: A one‐dimensional Hadamard walk with. one. dimension.. topological invariant, one. defect, Yokohama. Quantum \mathrm{I}\mathrm{n}\mathrm{f} Process., 1, 345‐354 (2002) .. type of limit theorems for the one‐dimensional quantum random walk. J. Math. Soc. Jpn.,. new. 57, 1179‐1195 (2005). Quantum Walks.. Quantum Potential Theory, Franz, U., and Schürmann, M., Eds., Lecture 309‐452, Springer‐Verlag, HeidelUerg (2008). In:. Notes in Mathematics: Vol. 1954, pp.. [10] Konno,. N.: The uniform. measure. for discrete‐time quantum walks in. one. dimension.. Quantum \mathrm{I}\mathrm{n}\mathrm{f} Process., 13, .. 1103-1125' (2014). [11] Konno,. N., Luczak, T., Segawa, dimension.. .. [12] Konno, N., Takei, Quantum. \mathrm{I}\mathrm{n}\mathrm{f}. R.:. Quantum. [15] Wójcik, A., Luczak, quantum walk. on. J.:. 1060‐1075. measures. 33‐53. of. inhomogeneous discrete‐time quantum walks. stationary. measure. for discrete‐time quantum walks in. one. Quantum. \mathrm{I}\mathrm{n}\mathrm{f}. .. dimension,. Physical Implementation of Quantum Walks. Springer (2013). Algorithms. Springer (2013). T. S. E.:. one. (2015). Walks and Search. Quantum walks:. [17] Wang, C., Lu, X., Wang,. in. (2013). Kurzyński, P., Grudka, A., Gdala, T., Bednarska‐Uzdega, M.: Thapping the line. Phys. Rev. \mathrm{A}, 86 012329 (2012). [16] Venegas‐Andraca, the line.. Limit. M.: The non‐uniform. Comput., 15,. .. [13] Manouchehri, K., Wang,. [14] Portugal,. E.:. Quantum \mathrm{I}\mathrm{n}\mathrm{f} Process., 12,. W.: The. a. a. particle of. a. comprehensive review. Quantum \mathrm{I}\mathrm{n}\mathrm{f} Process., 11, 1015‐1106 (2012). stationary. .. measure. Process., 14, 867‐880 (2015). of. a. space‐inhomogeneous three‐state quantum walk. on.
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