**3. Computation of Stationary State**

*3.1. Preliminary*

Recall that |*L* and |*R* represent the standard basis of C2; that is, |*L* = [1, 0] & and *R* = [0, 1] &. Let *χ* : <sup>2</sup>(<sup>Z</sup>; C<sup>2</sup>) → <sup>2</sup>(<sup>Γ</sup>*M*; C<sup>2</sup>) be a boundary operator such that (*χψ*)(*a*) = *ψ*(*a*) for any *a* ∈ {(*x*; *<sup>R</sup>*),(*<sup>x</sup>*; *L*) | *x* ∈ <sup>Γ</sup>*M*}. Here, the adjoint *χ*<sup>∗</sup> : <sup>2</sup>(<sup>Γ</sup>*M*; C<sup>2</sup>) → <sup>2</sup>(<sup>Z</sup>; C<sup>2</sup>) is described by 

$$(\chi^\* \varphi)(a) = \begin{cases} \varphi(a) & : a \in \{ (\mathfrak{x}; \mathbb{R}), (\mathfrak{x}; L) \mid \mathfrak{x} \in \Gamma\_M \}, \\ 0 & : \text{otherwise}. \end{cases}$$

We put the principal submatrix of *U* with respect to the impurities by *EM* := *χ U χ*<sup>∗</sup>. The matrix form of *EM* with the computational basis *<sup>χ</sup>δ*0|*L*, *<sup>χ</sup>δ*0|*R*, ... , *χδ <sup>M</sup>*−<sup>1</sup>|*L*, *χδ <sup>M</sup>*−<sup>1</sup>|*R* is expressed by the following 2 *M* × 2*M* matrix:

$$E\_M = \begin{bmatrix} 0 & P & & & & \\ Q & 0 & P & & & \\ & Q & 0 & \ddots & \\ & & \ddots & \ddots & P \\ & & & Q & 0 \end{bmatrix} \tag{5}$$

We express the ((*x*; *J*),(*x* ; *J*)) element of *EM* by

$$((E\_M)\_{(\mathbf{x};l),(\mathbf{x'};l')} := \left\langle \chi \delta\_{\mathbf{x}} | l \rangle , E\_M \chi \delta\_{\mathbf{x'}} | l' \rangle \right\rangle\_{\mathbb{C}^{2M}}.$$

Putting *ψn* := *χ*Ψ*<sup>n</sup>*, we have

$$\begin{aligned} \psi\_{n+1} &= \chi \mathcal{U}(\chi^\* \chi + (1 - \chi^\* \chi)) \Psi\_n \\ &= E\_M \psi\_n + \chi \mathcal{U}(1 - \chi^\* \chi) \Psi\_n \\ &= E\_M \psi\_n + \mathfrak{e}^{-i(n+1)\mathfrak{f}} \chi \delta\_0 |R\rangle. \end{aligned}$$

Then, putting *φn* := *<sup>e</sup><sup>i</sup>*(*n*+<sup>1</sup>)*ξψ<sup>n</sup>*, we have

$$
\epsilon^{-i\ddagger}\phi\_{n+1} = E\_M \phi\_n + \chi \delta\_0 |R\rangle. \tag{6}
$$

From [14], *ϕ* := ∃ lim*n*→∞ *φ<sup>n</sup>*. Then, the stationary state restricted to Γ*M* satisfies

$$(e^{-i\tilde{\xi}} - E\_M)\phi\_{\infty} = \chi \delta\_0 |R\rangle. \tag{7}$$

About the uniqueness of this solution is ensured by the following Lemma since it includes the existence of the inverse of (*e*<sup>−</sup>*i<sup>ξ</sup>* − *EM*).

**Lemma 1.** *Let EM be the above with a* = 0*.*† *Then σ*(*EM*) ⊂ {*λ* ∈ C | |*λ*| < <sup>1</sup>}*.*

**Proof.** Let *ψ* ∈ <sup>2</sup>(<sup>Γ</sup>*M*, C<sup>2</sup>) be an eigenvector of eigenvalue *λ* ∈ *<sup>σ</sup>*(*EM*). Then

$$|\lambda|^2 ||\psi||^2 = ||E\_M \psi||^2 = \langle \mathcal{U} \chi^\* \psi, \chi^\* \chi \mathcal{U} \chi^\* \psi \rangle \le \langle \mathcal{U} \chi^\* \psi, \mathcal{U} \chi^\* \psi \rangle = ||\chi^\* \psi||^2 = ||\psi||^2. \tag{8}$$

Here, for the inequality, we used the fact that *χ*<sup>∗</sup>*χ* is the projection operator onto

$$\text{span}\{\delta\_{\mathfrak{X}}|L\rangle, \delta\_{\mathfrak{X}}|R\rangle \mid \mathfrak{x} \in \Gamma\_{M}\} \subset \ell^{2}(\mathbb{Z}; \mathbb{C}^{2})$$

while for the final equality, we used the fact that *χχ*<sup>∗</sup> is the identity operator on <sup>2</sup>(<sup>Γ</sup>*M*; <sup>C</sup><sup>2</sup>). If the equality in (8) holds, then *χ*<sup>∗</sup>*χUχ*<sup>∗</sup>*ψ* = *Uχ*<sup>∗</sup>*ψ* holds. Then, we have the eigenequation *Uχ*<sup>∗</sup>*ψ* = *λχ*∗*ψ* by taking *χ*<sup>∗</sup> to both sides of the original eigenequation *χUχ*<sup>∗</sup>*ψ* = *λψ*. However, there are no eigenvectors having finite supports in a position independent quantum walk on Z with *a* = 0 since its spectrum is described by only a continuous spectrum in general. Thus, |*λ*|<sup>2</sup> < 1.

Now, let us solve this Equation (7). The matrix representation of *EM* with the permutation of the labeling such that (*x*; *R*) ↔ (*x*; *L*) for any *x* ∈ Γ*M* to (5) is


Then, the Equation (7) is expressed by


Here, we changed the way of blockwise of *EM* and we put *z* = *e*<sup>−</sup>*iξ* . Putting

$$A\_z := \begin{bmatrix} 0 & z \\ -d & -c \end{bmatrix}, \; B\_z := \begin{bmatrix} -b & -a \\ z & 0 \end{bmatrix}'$$

we have

$$\begin{aligned} \begin{bmatrix} z & 0 \end{bmatrix} \vec{\varphi}(0) &= 1, & A\_z \vec{\varphi}(0) + B\_{\vec{z}} \vec{\varphi}(1) &= 0, & A\_{\vec{z}} \vec{\varphi}(1) + B\_{\vec{z}} \vec{\varphi}(2) &= 0, & \dots \\ & \dots, A\_{\vec{z}} \vec{\varphi}(M-2) + B\_{\vec{z}} \vec{\varphi}(M-1) &= 0, & \begin{bmatrix} 0 & z \end{bmatrix} \vec{\varphi}(M-1) &= 0, & \end{aligned} \tag{9}$$

where *ϕ*(*x*)=[*ϕ*(*x*; *<sup>R</sup>*), *ϕ*(*x*; *L*)]& for any *x* ∈ Γ*M*. The inverse matrix of *Bz* exists since *z* = 0. Then, we have

$$\vec{\varphi}(1) = T\vec{\varphi}(0), \ \vec{\varphi}(2) = T^2\vec{\varphi}(0), \dots, \vec{\varphi}(M-1) = T^{M-1}\vec{\varphi}(0),\tag{10}$$

where

$$T = -B\_z^{-1} A\_z = \frac{1}{az} \begin{bmatrix} \Delta|a|^2 & -\Delta a \bar{b} \\ -\Delta db & z^2 + \Delta|b|^2 \end{bmatrix}.$$

$$\text{Here } \Delta = \det(P + Q) = \det\begin{bmatrix} a & b \\ c & d \end{bmatrix}. \text{ For the boundaries, there exists } \kappa \text{ such that}$$

$$
\vec{\varphi}(0) = \begin{bmatrix} z^{-1} & \kappa \end{bmatrix}, \begin{bmatrix} 0 & z \end{bmatrix} \\
\vec{\varphi}(M-1) = 0. \tag{11}
$$

By (10) and (11), *κ* satisfies

$$\left\langle \begin{bmatrix} 0\\1 \end{bmatrix}, T^{M-1} \begin{bmatrix} z^{-1} \\ \kappa \end{bmatrix} \right\rangle = 0 \tag{12}$$

which is equivalent to

$$\kappa = -\frac{z^{-1}(T^{M-1})\_{2,1}}{(T^{M-1})\_{2,2}}.$$

Now, the problem is reduced to considering the *n*-th power of *T* because the eigenvector is expressed by *ϕ*(*n*) = *Tnϕ*(0). Since *T* is a just 2 × 2 matrix, we can prepare the following lemma.

**Lemma 2.** *Let A be a* 2*-dimensional matrix denoted by*

$$A = \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}.$$

*1.* (*α* − *δ*)<sup>2</sup> + 4*βγ* = 0 *and A* = *I for some case. Let λ* = (*α* + *<sup>δ</sup>*)/2*. Then*

$$A^n = \begin{bmatrix} \lambda^n + \frac{\kappa - \delta}{2} n \lambda^{n-1} & \beta n \lambda^{n-1} \\ \gamma n \lambda^{n-1} & \lambda^n - \frac{\kappa - \delta}{2} n \lambda^{n-1} \end{bmatrix}$$

$$\text{2. } \quad \text{Otherwise. Let } \mathbb{J}\_n := (\det(A)^{1/2})^{n-1} \mathcal{U}\_{n-1} (\frac{\text{tr}(A)}{2 \det(A)^{1/2}}) \text{ for } n \ge 1. \text{ Then}$$

$$A^n = \begin{bmatrix} \zeta\_{n+1} - \delta \zeta\_n & \beta \zeta\_n\\ \gamma \zeta\_n & \zeta\_{n+1} - a \zeta\_n \end{bmatrix} '$$

*where Un*(·) *is the n-th Chebyshev polynomial of the second kind.*

**Remark 2.** *The condition "*(*α* − *δ*)<sup>2</sup> +4*βγ* = 0 *and A* = *I" is equivalent to the non-diagonalizability of A.*

**Remark 3.** *For A* = *T case, the condition of 1. is reduced to*

$$\omega := \Delta^{-1/2} z \in \{\epsilon\_1 |a| + \epsilon\_2 i |b| \mid \epsilon\_1, \epsilon\_2 \in \{\pm 1\}\} =: \partial B.$$

**Remark 4.** *For A* = *T case, the variable of the Chebyshev polynomial in 2. is reduced to*

$$(\operatorname{tr}(T)/(2\det(T)^{1/2}) = (\omega + \omega^{-1})/(2|a|).$$

*Moreover, if ω* = *eik, the Chebyshev polynomial is described by <sup>U</sup>*−<sup>1</sup>(·) = 0*,*

$$dI\_n(\cos k/|a|) = \frac{\lambda\_+^{n+1} - \lambda\_-^{n+1}}{\lambda\_+ - \lambda\_-} \ (n \ge 0).$$

*Here, λ*± *in RHS are the roots of the quadratic equation*

$$
\lambda^2 - \frac{2\cos k}{|a|}\lambda + 1 = 0
$$

*with* |*λ*−|≤|*<sup>λ</sup>*+|*.*

*3.2. Transmitting and Reflecting Rates*

> Let us divide the unit circle in the complex plain as follows:

$$B\_{in} = \{ \varepsilon^{ik} \mid |\cos k| < |a| \}, \ \partial B = \{ \varepsilon^{ik} \mid |\cos k| = |a| \}, \ B\_{out} = \{ \varepsilon^{ik} \mid |\cos k| > |a| \}. \tag{13}$$

By the unitarity of 1*a b c d*2 and using the Chebyshev recursion; *Un*+<sup>1</sup>(*x*) = <sup>2</sup>*xUn*(*x*) − *Un*−<sup>1</sup>(*x*), we insert (1) and (2) in Lemma 2 into (10), and we have an explicit expression for the stationary state as follows.

**Theorem 1.** *Let the stationary state restricted to* Γ*M* = {0, 1, ... , *M* − 1} *be φ*∞ *and ϕ*(*n*) := [*φ*∞(*<sup>n</sup>*; *R*) *φ*∞(*<sup>n</sup>*; *<sup>L</sup>*)]&*. Then we have*

$$\vec{\varphi}(n) = \begin{cases} \frac{z^{-1} (a\Lambda^{-1/2})^{-n}}{\omega \zeta'\_M - |a\zeta'\_M|} \begin{bmatrix} a\zeta'\_{M-n} - |a|\zeta'\_{M-n-1} \\ a\partial \zeta'\_{M-n-1} \end{bmatrix} & : \omega \notin \partial B \\\\ \frac{\Lambda^{-1/2}\lambda^n}{\epsilon\_R|a| + i\epsilon\_I M|b|} \begin{bmatrix} \epsilon\_R a (\epsilon\_R |a| + i\epsilon\_I |b| (M-n)) \\ b(M-n-1) \end{bmatrix} & : \omega \in \partial B \end{cases} \tag{14}$$

*for n* = 0, 1, ... , *M* − 1*, where α* = *a*/|*a*| *and ζm* = *Um*−<sup>1</sup>( *ω*+*ω*<sup>−</sup><sup>1</sup> <sup>2</sup>|*a*| ) (*m* ≥ <sup>0</sup>)*, λ* = sgn(*R*)*α*<sup>−</sup><sup>1</sup> Δ1/2*. Here R* = sgn(*Re*(*ω*)) *and I* = sgn(*Im*(*ω*))

Since the transmitting and reflecting rates are computed by

$$\begin{aligned} T(\omega) &= \left| \left\langle \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \vec{\varphi}(M-1) \right\rangle \times d \right|^2, \\ R(\omega) &= \left| \left\langle \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \vec{\varphi}(0) \right\rangle \times a + \left\langle \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \vec{\varphi}(0) \right\rangle \times b \right|^2. \end{aligned}$$

we obtain explicit expressions for them as follows.

**Corollary 1.** *Assume abcd* = 0*. For any ω* ∈ <sup>R</sup>/(<sup>2</sup>*π*<sup>Z</sup>)*, we have*

$$T(\omega) = \frac{|a|^2}{|a|^2 + |b|^2 \zeta'^2\_M} \tag{15}$$

$$R(\omega) = \frac{|b|^2 \zeta'^2\_M}{|a|^2 + |b|^2 \zeta'^2\_M} \tag{16}$$

Note that the unitarity of the time evolution can be confirmed by *T* + *R* = 1. By Corollary 1, we can find a necessary and sufficient conditions for the perfect transmitting; that is , *T* = 1.

**Corollary 2.** *Assume abcd* = 0*. Let ω* = *eik with some real value k. Then the perfect transmitting happens if and only if*

$$\arccos\left(\frac{\cos k}{|a|}\right) \in \left\{\frac{\ell}{M}\pi \mid \ell \in \{0, \pm 1, \dots, \pm (M-1)\}\right\}.$$

*On the other hand, the perfect reflection never occurs.*

Remark that if *ω* ∈/ *Bin*, then the perfect transmitting never happens.

### *3.3. Energy in the Perturbed Region*

Taking the square modulus to *ϕ*(*n*) in Theorem 1, the relative probability at position *n* ∈ Γ*M* = {0, . . . , *M* − 1} can be computed as follows.

**Proposition 2.** *Assume abcd* = 0*. Then, the relative probability is described by*

$$||\vec{\varphi}(n)||^2 = \begin{cases} \frac{1}{|a|^2 + |b|^2 \zeta'^2\_M} \left( |a|^2 + |b|^2 \zeta'^2\_{M-n-1} + |b|^2 \zeta'^2\_{M-n} \right) & : \omega \notin \partial B\\\\ \frac{1}{|a|^2 + M^2 |b|^2} \left\{ |a|^2 + |b|^2 (M-n)^2 + |b|^2 (M-n-1)^2 \right\} & : \omega \in \partial B \end{cases} \tag{17}$$

**Proof.** Let us consider the case for *ω* ∈/ *∂B*. Using the property of the Chebyshev polynomial, we have *ζm*+<sup>1</sup>*ζm*−<sup>1</sup> = *ζ*2*m* − 1 and (*ω* + *ω*<sup>−</sup><sup>1</sup>)/|*a*| · *ζm* = *ζm*+<sup>1</sup> + *ζm*−1. It holds that

$$\begin{aligned} (\omega + \omega^{-1})\zeta\_m'\zeta\_{m-1}' &= |a|(\zeta\_{m+1}' + \zeta\_{m-1}')\zeta\_{m-1}'|\\ &= |a|(\zeta\_{m}'^2 + \zeta\_{m-1}'^2 - 1). \end{aligned}$$

Since *ζ m* ∈ R, we have

$$q(m) := |\omega \zeta\_m' - |a| \zeta\_{m-1}'|^2 = \zeta\_m'^2 + |a|^2 \zeta\_{m-1}'^2 - |a|^2 (\omega + \omega^{-1}) \zeta\_m' \zeta\_{m-1}'$$

$$= |b|^2 \zeta\_m'^2 + |a|^2 \omega$$

Then, we have

$$\begin{aligned} ||\vec{\varphi}(n)||^2 &= \frac{1}{q(M)} (q(M-n) + |b|^2 \zeta'^2\_{M-n-1}) \\ &= \frac{|b|^2 \zeta'^2\_{M-n} + |a|^2 + |b|^2 \zeta'^2\_{M-n-1}}{|b|^2 \zeta'^2\_M + |a|^2}. \end{aligned}$$

Then, we can see how much quantum walkers accumulate in the perturbed region Γ*M* = {0, . . . , *M* − 1} by

$$\mathcal{E}\_M(\omega) = \colon \sum\_{n=0}^{M-1} \left| \left| \vec{q}(n) \right| \right|^2.$$

We call it the energy of quantum walk. The dependency of the energy on *ω* is symmetric on the unit circle in the complex plain.

**Corollary 3.** *Let* E *M*(*ω*) *be the above and assume abcd* = 0*. Then we have*

$$\mathcal{E}\_M(\omega) = \frac{1}{|a|^2 + |b|^2 \zeta'^2\_M} \left\{ M|a|^2 + \frac{|b|^2}{(\lambda\_+ - \lambda\_-)^2} \left( \zeta'^2\_{M+1} - \zeta'^2\_{M-1} - 4M \right) \right\} \tag{18}$$

*In particular,* E *<sup>M</sup>*(·) *is continuous at every ω*∗ ∈ *∂B and*

$$\mathcal{E}\_M(\omega\_\*) = \frac{1}{3} \frac{M}{|a|^2 + |b|^2 M^2} \left( 3|a|^2 + |b|^2 + 2|b|^2 M^2 \right).$$

**Proof.** Using the properties of the Chebyshev polynomial for example, *U*<sup>2</sup> *n* − *Un*+<sup>1</sup>*Un*−<sup>1</sup> = 1, *Tn* = ( *Un* − *Un*−<sup>2</sup>)/2, we have

$$(\lambda\_+^{m-1} + \lambda\_-^{m-1})\mathbb{j}'\_M = 2T\_{m-1}\mathbb{U}\_{m-1} = \mathbb{j}'^2\_m - \mathbb{j}'^2\_{m-1} + 1.$$

Then, we have

$$\begin{split} \sum\_{n=0}^{m-1} \zeta\_{n}'^{'2} &= \sum\_{n=0}^{m-1} \left( \frac{\lambda\_{+}^{m} - \lambda\_{-}^{m}}{\lambda\_{+} - \lambda\_{-}} \right)^{2} \\ &= \frac{1}{(\lambda\_{+} - \lambda\_{-})^{2}} \left\{ (\lambda\_{+}^{m-1} + \lambda\_{-}^{m-1}) \zeta\_{m}' - 2m \right\} \\ &= \frac{1}{(\lambda\_{+} - \lambda\_{-})^{2}} (\zeta\_{-}^{'2} - \zeta\_{-}^{'2}{}\_{m-1} - 2m + 1) \end{split} \tag{19}$$

Then, we have

$$\begin{aligned} \sum\_{n=0}^{M-1} ||\vec{\varphi}(n)||^2 &= \frac{1}{|a|^2 + |b|^2 \xi'^2\_M} \left( M|a|^2 + |b|^2 \sum\_{n=0}^{M-1} \xi'^2\_{M-n-1} + \xi'^2\_{M-n} \right) \\ &= \frac{1}{|a|^2 + |b|^2 \xi'^2\_M} \left\{ M|a|^2 + \frac{|b|^2}{(\lambda\_+ - \lambda\_-)^2} \left( \xi'^2\_{M+1} - \xi'^2\_{M-1} - 4M \right) \right\} \end{aligned}$$

Here, we used (19) in the last equality.

If *ω* ∈ *∂B*, then by directly computation taking summation of (17) over *n* ∈ Γ*M* = {0, 1, ... , *M* − <sup>1</sup>}, we obtain the conclusion. Let us see E*M*(·) is continuous at *∂B*. We put *x* := (1/|*a*|) cos *k* and *ζm*(*x*) := *ζm*. Remark that *ω* → *ω*∗ implies |*x*| → 1. In the following, we consider *x* → 1 case. The Taylor expansion of *ζm*(*x*) around *x* = 1 is

$$
\zeta'\_m(1-\epsilon) = m - \frac{m}{3}(m^2 - 1)\epsilon + O(\epsilon^2).
$$

The reason for obtaining the expansion until 1 order is

$$
\zeta'^2\_{M+1} - \zeta'^2\_{M-1} - 4M = O(\epsilon^2).
$$

around *x* = 1. Note that (*<sup>λ</sup>*+ − *<sup>λ</sup>*−)<sup>2</sup> = <sup>4</sup>(*x*<sup>2</sup> − <sup>1</sup>). Then

$$(\lambda\_+ - \lambda\_-)^2 = -8\epsilon + O(\epsilon)$$

around *x* = 1. Then inserting all of them into (18), we obtain

$$\lim\_{\omega \to \omega\_\*} \mathcal{E}\_M(\omega) = \frac{M}{|a|^2 + |b|^2 M^2} \left( |a|^2 + \frac{|b|^2}{3} + \frac{2|b|^2}{3} M^2 \right).$$
