Complementarity in quantum walks

Complementary observables play an important role in quantum information science. Primarily, they are the cornerstone of quantum cryptography [1, 2]. In addition, the quantum Fourier transform [3], which changes between the eigenbases of two such observables, is the main ingredient of the Shor's factoring algorithm [4]. Formally, two observables, A and B, are complementary if they do not share a common set of eigenvectors. Furthermore, such observables are maximally complementary if their eigenvectors, $\{|a_i\rangle\}_{i = 1}^D$ and $\{|b_j\rangle\}_{j = 1}^D$, form mutually unbiased bases (MUB) [5], i.e.

$$\begin{equation} |\langle a_i |b_j \rangle| = \frac{1}{\sqrt{D}}. \end{equation} \tag{ 1 }$$

In realistic situations perfect MUB are hard to implement. This lead to the concept of approximate MUB (AMUB) [6, 7] in complex [8, 9] and in real spaces [10, 11]. In case of AMUB one relaxes the condition (1) and instead looks for bases whose vectors obey

$$\begin{equation} |\langle a_i |b_j \rangle| \leqslant \frac{c}{\sqrt{D}}, \end{equation} \tag{ 2 }$$

where c is a real positive constant independent of D [9].

Despite significant impact on our understanding of nature, majority of research on MUB and AMUB is purely mathematical. Note that only few strongly complementary observables have a clear physical meaning [12], e.g. position and momentum, or spin-$1/2$ projections onto three mutually orthogonal axes. It is therefore important to look for some physical models in which complementarity is exhibited by observables that provide an intuitive description of states or dynamics. The goal of this work is to show that there exists a family of dynamical models, known as quantum walks (QWs) [13, 14], whose evolution operators are strongly complementary and whose eigenbases form AMUB.

QWs are dynamical models that describe quantum particles moving on a lattice. In this work we focus on their discrete-time versions (DTQWs) [13, 14]. Such models can simulate dynamics of various physical systems, e.g. [15–24], and are known to be capable of universal quantum computation [25–28], and as a consequence, of universal quantum simulation [29]. Moreover, they were already implemented on many experimental platforms [30]. One of the most appealing features of DTQWs is the fact that relatively simple and finite models can be used to investigate highly-nontrivial and complex phenomena. In this sense DTQWs are considered to be quantum analogues of classical cellular automata [31].

We study a DTQW on a d-cycle in which a single particle acquires a phase-shift that depends on its position and on a state of its coin. The coin is an auxiliary degree of freedom that decides whether the particle moves right or left. It is described by a two-dimensional (2D) subsystem, therefore the DTQW's Hilbert space has dimension $D = 2d$. We note that DTQWs with position-dependent coins were already studied [32–52] in various contexts, but as far as we know their complementarity properties were never explored. We show that the eigenvectors of the evolution operator exhibit a peculiar dependence on the amplitude of the phase-shift $\varphi = 2\pi q/d$, where $q = 0,1,\ldots d-1$ is a parameter that governs the amplitude. In particular, we observe that for prime d the two different DTQW evolution operators, governed by $q \neq q^{^{\prime}}$, the corresponding eigenvectors form AMUB, i.e. $|\langle v_q|v^{^{\prime}}_{q^{^{\prime}}} \rangle| \leqslant \sqrt{2/D}$ for all $|v_q\rangle$ and $|v^{^{\prime}}_{q^{^{\prime}}}\rangle$.

In addition, in the second part of this work we focus on the continuous limit of our model, which is known to describe a Dirac particle [53, 54]—an elementary quantum relativistic system such as electron. We find that the observed complementarity relations still occur in the continuous version for which the factorability of d is not an issue anymore. More precisely, we show that eigenvectors of Hamiltonians of one-dimensional (1D) Dirac particles subjected to different gauge fields obey analogous AMUB relations as in DTQW case for prime d.

We consider a 1D DTQW on a d-cycle [55]. The system consists of a particle that can be located at one of d positions $x = 0,1,\ldots,d-1$ (we assume $x \equiv x ~\text{mod}~ d$) and of a coin, a two-level system that can be in one of two states $b = \pm 1$. We represent these states in the following way: $|+\rangle = (1~0)^T$ and $|-\rangle = (0~1)^T$. The coin can be either a particle's internal degree of freedom, akin to a spin, or an external system. However, this choice is of no importance here. The general pure state of the system at time t is given by

$$\begin{equation} |\psi_t\rangle = \sum_{x=0}^{d-1}\sum_{b=\pm 1} \alpha_{x,b}(t) |x\rangle \otimes |b\rangle. \end{equation} \tag{ 3 }$$

A single step of the evolution is generated by the unitary operator

$$\begin{equation} |\psi_{t+1}\rangle = U |\psi_{t}\rangle. \end{equation} \tag{ 4 }$$

It is of the form

$$\begin{equation} U = S (\unicode{x1D7D9} \otimes C) F, \end{equation} \tag{ 5 }$$

where

$$\begin{equation} S = \sum_x \left(|x+1\rangle\langle x| \otimes \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + |x-1\rangle\langle x| \otimes \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \right) \end{equation} \tag{ 6 }$$

is the conditional translation and C is an arbitrary quantum coin toss operator

$$\begin{equation} C = e^{i\delta} \begin{pmatrix}c & s\\-s^{*} & c^{*}\end{pmatrix}\in \mathcal{U}(2), \end{equation} \tag{ 7 }$$

where

$$\begin{equation} c=\cos\theta e^{i\gamma},~s=\sin\theta e^{i\sigma}, \end{equation} \tag{ 8 }$$

and δ, θ, γ, σ are arbitrary angles. Finally, F is a position and coin-dependent phase-shift operator

$$\begin{equation} F = \sum_x |x\rangle\langle x| \otimes \begin{pmatrix} e^{i\varphi x} & 0 \\ 0 & e^{-i\varphi x} \end{pmatrix},~~~~\left(\varphi = \frac{2\pi q}{d}= \varepsilon q \right). \end{equation} \tag{ 9 }$$

The parameter $q = 0,1,\dots,d-1$ determines the phase-shift's magnitude.

The above discrete spacetime model simulates a particle hopping on a ring that is subjected to an artificial gauge field. The gauge field is generated by F and its magnitude is given by $\varphi = \varepsilon q$. We will come back to this interpretation later when we will discuss the continuous spacetime limit.

The eigen-problem for the unitary evolution operator (5) is studied in the appendix A. Here we present its solution for d-prime and θ ≠ 0. The eigenvalues for a given q are

$$\begin{equation} \lambda_{m,\tau}^{(q)} = e^{i\left(\delta+q\frac{\varepsilon}{2} +\tau \xi_m^{(q)}\right)}, \end{equation} \tag{ 10 }$$

where $\tau = \pm1$, $m = 0,{\ldots},d-1$ and

$$\begin{equation} \cos \xi_{m}^{(q)}=(-1)^q\cos\theta\cos\left((m+q)\varepsilon-\gamma\right). \end{equation} \tag{ 11 }$$

The corresponding eigenvectors are

$$\begin{equation} |\psi_{m,\tau}^{(q\neq 0)}\rangle = N_{m,\tau}^{(q)}\frac{1}{\sqrt{d}}\sum_{j=0}^{d-1} e^{i\chi_{m,j}^{(q)}}|jq\rangle \otimes \begin{pmatrix} 1 \\ \beta_{m,\tau}^{(q)} e^{i2q\varepsilon j} \end{pmatrix}, \end{equation} \tag{ 12 }$$

$$\begin{equation} |\psi_{m,\tau}^{(q= 0)}\rangle = N_{m,\tau}^{(0)}~|m\rangle \otimes \begin{pmatrix} 1 \\ \beta_{m,\tau}^{(0)} \end{pmatrix}, \end{equation} \tag{ 13 }$$

where

$$\begin{align} \beta_{m\tau}^{(q\neq0)}=\frac{(-1)^qe^{i\tau \xi_{m}^{(q)}}e^{-i(m+q)\varepsilon}-\cos\theta e^{-2i(m+q)\varepsilon}e^{i\gamma}}{\sin\theta e^{i\sigma}} \end{align} \tag{ 14 }$$

$$\begin{align} \beta_{m,\tau}^{(q= 0)} = \frac{e^{i\tau \xi_{m}^{(0)}}-\cos\theta e^{i\sigma}}{\sin\theta e^{i\sigma}}, \end{align} \tag{ 15 }$$

$$\begin{align} \chi_{m,j}^{(q)}=-\frac{q\varepsilon}{2}j^2+(m\varepsilon+q\pi)j, \end{align} \tag{ 16 }$$

$$\begin{align} N_{m,\tau}^{(q)}=\frac{1}{\sqrt{1+|\beta_{m,\tau}^{(q)}|^2}}, \end{align} \tag{ 17 }$$

and the vectors $|jq\rangle$ and $|m\rangle$ in equations (107) and (113) are given by the following formula

$$\begin{equation} |k\rangle=\sum_{x=0}^{d-1}e^{ik\varepsilon x}|x\rangle, \end{equation} \tag{ 18 }$$

where $k = 0,1,\ldots,d-1$.

The above results allow us to formulate the following theorem

Theorem 1. If d is prime and $q\neq q^{^{\prime}}$, then for all $m,m^{^{\prime}},\tau,\tau^{^{\prime}}$ the following holds

$$\begin{equation} | \langle\psi_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})} |\psi_{m,\tau}^{(q)}\rangle| \leqslant \sqrt{\frac{2}{D}} = \frac{1}{\sqrt{d}}. \end{equation} \tag{ 19 }$$

This theorem is proven in the appendix B. In simple words, it states that for prime d the overlap between the eigenvectors corresponding to $q\neq q^{^{\prime}}$ is never greater than $\sqrt{2/D}$. This implies strong complementarity between the two eigenbases. Nevertheless, not all overlaps are the same, hence the eigenvectors corresponding to $q\neq q^{^{\prime}}$ are not MUB. However, the modulus square of their overlap is bounded by twice the inverse of the system's dimension, therefore in large Hlibert spaces the corresponding eigenbases are AMUB [6–11]. It is natural to ask what happens for non-prime d. We observed that for some choices of $q\neq q^{^{\prime}}$ the overlaps between the corresponding bases are bounded by $\sqrt{2/D}$ as well, however in general this overlap exceeds $\sqrt{2/D}$. Moreover, for non-prime d the overlap between the eigenvectors corresponding to $q\neq q^{^{\prime}}$ can be quite large. For example, for $\gamma = \delta = \sigma = 0$ and $\theta = \pi/4$ it can reach $\sqrt{1/3}$ (d = 18), or $\sqrt{1/2}$ (d = 16).

There is an interesting case corresponding to θ = 0. In this situation the coin operator $\unicode{x1D7D9} \otimes C$ commutes with both, the phase-shift operator F and the conditional translation operator S, and its action can be effectively ignored. The eigen-problem corresponding to this case is studied in the appendix C. In this case the eigenvectors of the evolution operator are of the form

$$\begin{equation} |\Psi_{m\tau}^{(q)}\rangle = \frac{1}{\sqrt{d}}\sum_{j=0}^{d-1}e^{i\chi_{m\tau j}^{(q)}}|\,jq\rangle\otimes |\tau\rangle, \end{equation} \tag{ 20 }$$

where

$$\begin{equation} \chi_{m\tau j}^{(q)}=-\tau\frac{q\varepsilon}{2}j^2+\left(m\varepsilon+q\pi\right)j, \end{equation} \tag{ 21 }$$

$m = 0,1,\ldots,d-1$ and $\tau = \pm 1$. The spatial part of these eigenvectors recovers perfect MUB relations, since for $q\neq q^{^{\prime}}$

$$\begin{equation} |\langle\Psi_{m\tau}^{(q)}|\Psi_{m^{^{\prime}}\tau^{^{\prime}}}^{(q^{^{\prime}})}\rangle|^2 = \delta_{\tau \tau^{^{\prime}}}\frac{1}{d}. \end{equation} \tag{ 22 }$$

The reason for the above stems from the following fact. Note that for θ = 0 we ignore the action of C, hence the evolution operator can be written as U = SF. In addition, one can represent

$$\begin{equation} S = X \otimes \frac{(\unicode{x1D7D9}+\sigma_z)}{2} + X^{\dagger} \otimes \frac{(\unicode{x1D7D9}-\sigma_z)}{2}, \end{equation} \tag{ 23 }$$

and

$$\begin{equation} F = Z^{q} \otimes \frac{(\unicode{x1D7D9}+\sigma_z)}{2} + (Z^q)^{\dagger} \otimes \frac{(\unicode{x1D7D9}-\sigma_z)}{2}, \end{equation} \tag{ 24 }$$

where σ_z is the Pauli z-matrix and

$$\begin{eqnarray} & &X|x\rangle = |x+1 ~\text{mod}~ d\rangle, \end{eqnarray} \tag{ 25 }$$

$$\begin{eqnarray} & &Z|x\rangle = e^{i\frac{2\pi x}{d}}|x\rangle, \end{eqnarray} \tag{ 26 }$$

are the Weyl–Heisenberg operators. It is known that for prime d the eigenbases of the following set of Weyl–Heisenberg operators form d + 1 MUB [5]

$$\begin{equation} \{Z,X,XZ,XZ^2,\ldots,XZ^q,\ldots,XZ^{d-1}\}. \end{equation} \tag{ 27 }$$

Finally, let us consider the case of imperfect phase-shifts that may occur in realistic implementations. We assume that the phase shift at each position x is affected by a random disturbance, i.e. $\varphi x \rightarrow \varphi x + \delta_{\epsilon}$, where δ_ε is a noise parameter sampled from a Gaussian distribution of variance ε. We numerically studied the modification of theorem 1

$$\begin{equation} | \langle\psi_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})} |\psi_{m,\tau}^{(q)}\rangle| \leqslant \frac{c(\epsilon)}{\sqrt{D}}, \end{equation} \tag{ 28 }$$

where in this case c is a function of ε. We found that for small ε the corresponding bases are still good AMUB—see figure 1. The value of $c(\epsilon)$ blows up as the noise amplitude becomes of the same order as the phase shift.

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The above DTQW exhibits the strong complementarity property if its dimension is a doubled prime. One may ask if this finding is just a peculiarity, resulting from a discrete spacetime formulation, that might disappear in the continuous limit. To answer this question we considered the continuous limit for a particular coin operator, corresponding to $\gamma = \delta = \sigma = 0$, which was shown to describe a Dirac particle [53, 54] . In this case the elements of the evolution operator can be rewritten with the help of position, momentum and Pauli operators as

$$\begin{equation} S = e^{i p \sigma_z},~~C=-ie^{i \frac{\pi}{2}(s \sigma_x + c \sigma_z)},~~F=e^{i \varphi x \sigma_z}, \end{equation} \tag{ 29 }$$

and that the above DTQW evolution operator can emerge as a result of a Trotterization of

$$\begin{equation} U = e^{-i H \Delta t} = e^{-i \left((p-e A) \sigma_z + m \sigma_x\right) \Delta t}, \end{equation} \tag{ 30 }$$

where $-\Delta t = 1$, $eA\Delta t = \varphi x + \frac{\pi}{2}c$, $-m \Delta t = \frac{\pi}{2}s$, and we omitted the global phase of −i. The above Hamiltonian H is an analogue of the 1D Dirac Hamiltonian, for which e is the particle's charge, m the mass, and A is the x-component of the vector potential.

We introduce $A = \mu x$, where µ is a continuous parameter that is an analog of the discrete ϕ. The evolution operator (30) is associated to the following Dirac equation

$$\begin{equation} \begin{pmatrix} -i\frac{\partial}{\partial x} - \mu x & m \\ m & i\frac{\partial}{\partial x} + \mu x \end{pmatrix} \begin{pmatrix} \alpha(x) \\ \beta(x) \end{pmatrix} = E \begin{pmatrix} \alpha(x) \\ \beta(x) \end{pmatrix}, \end{equation} \tag{ 31 }$$

where $\alpha(x)$ and $\beta(x)$ are the two spinor components of the eigenfunction and E is the system's energy. We adopted the standard convention in which $\hbar = 1$ and the velocity of light c = 1. We also assumed that the particle's charge is e = 1.

The solution of (31) is provided in the appendix D. The energies of the particle are given by

$$\begin{equation} E_{k,\pm} = \pm \sqrt{k^2 + m^2}, \end{equation} \tag{ 32 }$$

where $-\infty \lt k\lt \infty$ is the particle's momentum. On the other hand, the corresponding eigenfunctions are

$$\begin{equation} \psi_{k,\pm}^{\mu}(x) = {\mathcal N}_{k,\pm} e^{i(\frac{\mu}{2} x^2 + kx)}\begin{pmatrix} m \\ \pm \sqrt{k^2+m^2}-k \end{pmatrix}, \end{equation} \tag{ 33 }$$

where

$$\begin{equation} {\mathcal N}_{k,\pm} = 2(k^2+m^2 \mp k\sqrt{k^2+m^2}). \end{equation} \tag{ 34 }$$

Interestingly, only the eigenfunctions depend on µ. Since the particle is unbounded ($-\infty \lt x\lt \infty$), the above eigenfunctions are unnormalized and we have $|\psi^{\mu}_{k,\pm}(x)|^2 = 1$, but $\int_{-\infty}^{\infty} dx|\psi_{k,\pm}^{\mu}(x)|^2 = \infty$.

Now we are able to formulate the next theorem

Theorem 2. For all $\mu\neq \mu^{^{\prime}},k,k^{^{\prime}}$ and $z,z^{^{\prime}} = \pm 1$ the following holds

$$\begin{equation} \left| \int_{-\infty}^{\infty} dx \psi^{\mu}_{k,z}(x)\left(\psi^{\mu^{^{\prime}}}_{k^{^{\prime}},z^{^{\prime}}}(x)\right)^{\ast} \right| \leqslant \sqrt{\frac{2\pi}{|\mu - \mu^{^{\prime}}|}}. \end{equation} \tag{ 35 }$$

This theorem states that the overlap between any two eigenvectors corresponding to two different Dirac Hamiltonians (with $\mu \neq \mu^{^{\prime}}$) never exceeds a certain finite value. It is a continuous analog of theorem 1. Its proof is given in appendix E.

At this point it is worth to relate the above result to a study of MUB in continuous-variable systems [56]. It is known that the eigenfunctions of the position and momentum operators are complementary and that in fact the position and momentum eigenbases are MUB since

$$\begin{equation} |\langle x|p\rangle| = \frac{1}{\sqrt{2\pi}}. \end{equation} \tag{ 36 }$$

Once again we assume that $\hbar = 1$. The above equation is a continuous version of equation (1). We can also define a continuous analog of equation (2) as

$$\begin{equation} |\langle x^{^{\prime}}|p^{^{\prime}}\rangle| \leqslant \frac{c}{\sqrt{2\pi}} \end{equation} \tag{ 37 }$$

where $\{|x^{^{\prime}}\rangle\}$ and $\{|p^{^{\prime}}\rangle\}$ are some new bases and $c\geqslant 1$ is a constant that depends on the relations between these bases. It was shown in [56] that if one defines the operator

$$\begin{equation} x_{\theta} = x \cos{\theta} + p \sin{\theta} , \end{equation} \tag{ 38 }$$

then all eigenvectors of x_θ and $x_{\theta^{^{\prime}}}$ obey

$$\begin{equation} \left| \langle x_{\theta}|x_{\theta^{^{\prime}}}\rangle \right| = \frac{1}{\sqrt{2\pi |\sin(\theta - \theta^{^{\prime}})|}} = \frac{c(\theta - \theta^{^{\prime}})}{\sqrt{2\pi}}, \end{equation} \tag{ 39 }$$

where

$$\begin{equation} c(\theta - \theta^{^{\prime}}) = \frac{1}{\sqrt{|\sin(\theta - \theta^{^{\prime}})|}} \geqslant 1. \end{equation} \tag{ 40 }$$

It was concluded that the above bases are not exactly MUB, since the overlap depends on the difference $\theta-\theta^{^{\prime}}$, i.e. it depends on the choice of the operators x_θ and $x_{\theta^{^{\prime}}}$, not on a particular Hilbert space. Nevertheless, these bases exhibit many properties of MUB in a sense that the overlap between any two vectors from two different bases is the same and are finite, despite the fact that the vectors are unnormalized.

In the Dirac particle case the situation is similar, although the overlaps are not the same. In fact, the relations between the eigenvectors are analogous to AMUB. In addition, note that if we substitute $\gamma \sin\theta = 1$ and $\gamma \cos\theta = -\mu$, so that $\gamma = 1/\sin\theta$ and $\theta = \cot^{-1}(-\mu)$, then our Dirac Hamiltonian can be rewritten as

$$\begin{equation} H = \gamma x_{\theta} \sigma_z + m \sigma_x. \end{equation} \tag{ 41 }$$

The above clearly shows that in our case the complementarity properties of x_θ and $x_{\theta^{^{\prime}}}$ are affected by the addition of a 2D spinor space, which results in the imperfect AMUB relation stated in theorem 2. However, if we consider massless particles (m = 0) the Dirac Hamiltonian becomes

$$\begin{equation} H = \gamma x_{\theta} \sigma_z \end{equation} \tag{ 42 }$$

and the relations (39) are recovered. This is in perfect analogy to the DTQW scenario with θ = 0 studied in appendix C.

Majority of research on MUB and AMUB focuses on their mathematical properties and on their applications in quantum information science. The main contribution of this work is a proof that AMUB can naturally emerge as eigenstates of physically relevant observables, i.e. evolution operators and Hamiltonians.

Our finding can lead to some new physically meaningful uncertainty relations. In addition, it may lead to some new dynamical DTQW effects. In particular, it would be interesting to study quenches [57] in DTQW model, i.e. evolutions of energy eigenstates after a sudden change of some parameter in the Hamiltonian. Note, that in our DTQW model a sudden change of phase implies a complete change of the evolution's operator eigenbasis, which can lead to unexpected evolutions.

This research is supported by the Polish National Science Centre (NCN) under the Maestro Grant No. DEC-2019/34/A/ST2/00081. J W acknowledges support from IDUB BestStudentGRANT (No. 010/39/UAM/0010). Part of numerical studies in this work have been carried out using resources provided by Wroclaw Centre for Networking and Supercomputing (wcss.pl), Grant No. 551 (A.S.S.).

All data that support the findings of this study are included within the article (and any supplementary files).

A.1. Model

Here we derive the eigenvalues and the eigenvectors of the unitary evolution operator that governs the dynamics of our model. Recall that we consider the evolution operator of the form

$$\begin{equation} U = S (I\otimes C) F, \end{equation} \tag{ 43 }$$

with the phase operator

$$\begin{equation} F = e^{i\varphi x \sigma_z} = \sum_x |x\rangle\langle x| \otimes \begin{pmatrix} e^{i\varphi x} & 0 \\ 0 & e^{-i\varphi x} \end{pmatrix}, \end{equation} \tag{ 44 }$$

where

$$\begin{equation} \varphi = 2\pi\frac{q}{p}=nq\varepsilon,~~~~\varepsilon = \frac{2\pi}{d},~~~~np=d. \end{equation} \tag{ 45 }$$

Step operator is given by

$$\begin{equation} S |x,\rightarrow \rangle = |x+1,\rightarrow \rangle, \end{equation} \tag{ 46 }$$

$$\begin{equation} S |x,\leftarrow \rangle = |x-1,\leftarrow \rangle. \end{equation} \tag{ 47 }$$

We use the following notation

$$\begin{equation} a_x |x,\rightarrow \rangle + b_x |x,\leftarrow \rangle = |x\rangle \otimes \begin{pmatrix} a_x\\b_x\end{pmatrix}. \end{equation} \tag{ 48 }$$

The coin operator is

$$\begin{equation} C = e^{i\delta} \begin{pmatrix}c & s\\-s^{*} & c^{*}\end{pmatrix}\in \mathcal{U}(2), \end{equation} \tag{ 49 }$$

where

$$\begin{equation} c=\cos\theta e^{i\gamma},~~~~s=\sin\theta e^{i\sigma},~~~~\det C = e^{2i\delta}, \end{equation} \tag{ 50 }$$

and we chose

$$\begin{equation} 0<\theta\leqslant\frac{\pi}{2}. \end{equation} \tag{ 51 }$$

(The case of θ = 0 is considered in appendix C). In the following we will consider only coins from the $\mathcal{SU}(2)$

$$\begin{equation} C =\begin{pmatrix}c & s\\-s^{*} & c^{*}\end{pmatrix}, \end{equation} \tag{ 52 }$$

because parameter δ only shifts energies and can be included after diagonalization.

A.2. Ansatz

Our calculations are based on the following ansatz on the form of eigenvectors

$$\begin{equation} |\Psi_r\rangle = \sum_{j=0}^{p-1}e^{i\chi_j}|r+jqn\rangle_p \otimes \begin{pmatrix}1\\ \beta_r e^{i\eta_j}\end{pmatrix}, \end{equation} \tag{ 53 }$$

with $r = 0,1,{\ldots},n-1$ and where χ_j and η_j are independent on r whereas β_r is independent on j. Moreover we will assume that

$$\begin{equation} \chi_j=Aj^2+Bj, \end{equation} \tag{ 54 }$$

$$\begin{equation} \eta_j=Cj, \end{equation} \tag{ 55 }$$

with $A, B, C$ j-independent. In the above ansatz we used the momentum states

$$\begin{equation} |k\rangle _p=\frac{1}{\sqrt{d}}\sum_{x=0}^{d-1}e^{ik\varepsilon x}|x\rangle, \end{equation} \tag{ 56 }$$

$k = 0,1,{\ldots},d-1$.

A.3. Derivation

Note how previously defined operators S and F act on momentum states

$$\begin{align} S|k\rangle_p|\rightarrow\rangle = e^{-ik\varepsilon}|k\rangle_p|\rightarrow\rangle, \end{align} \tag{ 57 }$$

$$\begin{align} S|k\rangle_p|\leftarrow\rangle = e^{ik\varepsilon}|k\rangle_p|\leftarrow\rangle, \end{align} \tag{ 58 }$$

$$\begin{align} F|k\rangle_p|\rightarrow\rangle = |k+nq\rangle_p|\rightarrow\rangle, \end{align} \tag{ 59 }$$

$$\begin{align} F|k\rangle_p|\leftarrow\rangle = |k-nq\rangle_p|\leftarrow\rangle. \end{align} \tag{ 60 }$$

So

$$\begin{align} F|\Psi_r\rangle & = \sum_{j=0}^{p-1}e^{i\chi_{j}}|r+(\,j+1)qn\rangle_p|\rightarrow\rangle +\sum_{j=0}^{p-1} e^{i\chi_j}\beta_r e^{i\eta_j}|r+(\,j-1)qn\rangle_p |\leftarrow\rangle, \end{align} \tag{ 61 }$$

which can be written as

$$\begin{equation} F|\Psi_r\rangle = \sum_{j=0}^{p-1}e^{i\chi_{j-1}}|r+jqn\rangle_p \otimes \begin{pmatrix}1\\ \beta_r e^{i(-\chi_{j-1}+\chi_{j+1}+\eta_{j+1})}\end{pmatrix}. \end{equation} \tag{ 62 }$$

It follows that

$$\begin{eqnarray} & & (\unicode{x1D7D9} \otimes C) F|\Phi_r\rangle=\sum_{j=0}^{p-1}e^{i\chi_{j-1}}|r+jqn\rangle_p \begin{pmatrix}c & s\\-s^{*} & c^{*}\end{pmatrix}\begin{pmatrix}1\\ &&\beta_r e^{i(-\chi_{j-1}+\chi_{j+1}+\eta_{j+1})}\end{pmatrix} \end{eqnarray} \tag{ 63 }$$

and finally

$$\begin{align} U|\Phi_r\rangle=\sum_{j=0}^{p-1}e^{i\chi_{j-1}}|r+jqn\rangle_p \otimes \begin{pmatrix}e^{-i(r+jqn)\varepsilon}&0\\0&e^{i(r+jqn)\varepsilon}\end{pmatrix} \begin{pmatrix}c & s\\-s^{*} & c^{*}\end{pmatrix}\begin{pmatrix}1\\ \beta_r e^{i(-\chi_{j-1}+\chi_{j+1}+\eta_{j+1})}\end{pmatrix},\nonumber\\ \end{align} \tag{ 64 }$$

This should be equal to

$$\begin{equation} \lambda |\Phi_r\rangle=\lambda \sum_{j=0}^{p-1}e^{i\chi_j}|r+jqn\rangle_p \otimes \begin{pmatrix}1\\ \beta_r e^{i\eta_j}\end{pmatrix}. \end{equation} \tag{ 65 }$$

Therefore, our goal is to solve the equation

$$\begin{eqnarray} & & e^{i\chi_{j-1}} \begin{pmatrix}e^{-i(r+jqn)\varepsilon}&0\\0&e^{i(r+jqn)\varepsilon}\end{pmatrix} \begin{pmatrix}c & s\\-s^{*} & c^{*}\end{pmatrix} \begin{pmatrix}1\\ &&\beta_r e^{i(-\chi_{j-1}+\chi_{j+1}+\eta_{j+1})}\end{pmatrix}=\lambda e^{i\chi_j} \begin{pmatrix}1\\ &&\beta_r e^{i\eta_j}\end{pmatrix}, \nonumber\\ \end{eqnarray} \tag{ 66 }$$

which can be rearranged in the form

$$\begin{eqnarray} & & \begin{pmatrix}c & s\\-s^{*} & c^{*}\end{pmatrix}\begin{pmatrix}1\\ \beta_r e^{i(-\chi_{j-1}+\chi_{j+1}+\eta_{j+1})}\end{pmatrix}= \lambda e^{i\chi_j}e^{-i\chi_{j-1}}\begin{pmatrix}e^{i(r+jqn)\varepsilon}&0\\0&e^{-i(r+jqn)\varepsilon}\end{pmatrix}\begin{pmatrix}1\\ &&\beta_r e^{i\eta_j}\end{pmatrix}.\nonumber\\ \end{eqnarray} \tag{ 67 }$$

The above is the system of two equations

$$\begin{align} c+s\beta_r e^{i(-\chi_{j-1}+\chi_{j+1}+\eta_{j+1})} = \lambda e^{i\chi_j}e^{-i\chi_{j-1}} e^{ir\varepsilon}e^{ijqn\varepsilon}, \end{align} \tag{ 68 }$$

and

$$\begin{align} -s^{*}+c^{*}\beta_r e^{i(-\chi_{j-1}+\chi_{j+1}+\eta_{j+1})} = \lambda e^{i\chi_j}e^{-i\chi_{j-1}} e^{i\eta_j} e^{-ir\varepsilon}e^{-ijqn\varepsilon}\beta_r. \end{align} \tag{ 69 }$$

We insist that exponents should be j independent. This gives us three equations with constants $\mathcal{A}, \mathcal{B}, \mathcal{C}$

$$\begin{align} \chi_{j-1}-\chi_j-jnq\varepsilon=\mathcal{A}, \end{align} \tag{ 70 }$$

$$\begin{align} \chi_{j+1}+\eta_{j+1}-\chi_{j-1}=\mathcal{B}, \end{align} \tag{ 71 }$$

$$\begin{align} \chi_{j-1}-\chi_j-\eta_{j}+jnq\varepsilon=\mathcal{C}. \end{align} \tag{ 72 }$$

Now our system of equations reads

$$\begin{align} c+s\beta_r e^{i\mathcal{B}} = \lambda e^{-i\mathcal{A}} e^{ir\varepsilon}, \end{align} \tag{ 73 }$$

$$\begin{align} -s^{*}+c^{*}\beta_r e^{i\mathcal{B}} = \lambda e^{-i\mathcal{C}} e^{-ir\varepsilon}\beta_r. \end{align} \tag{ 74 }$$

Therefore

$$\begin{equation} \beta_r=\frac{\lambda e^{-i\mathcal{A}}e^{-ir\varepsilon}-c}{s}e^{-i\mathcal{B}}, \end{equation} \tag{ 75 }$$

and we obtain quadratic formula for eigenvalues λ

$$\begin{align} \lambda^2-\lambda(ce^{i\mathcal{A}}e^{-ir\varepsilon}+c^{*}e^{i(\mathcal{B}+\mathcal{C})}e^{ir\varepsilon})+e^{i(\mathcal{A}+\mathcal{B}+\mathcal{C})}=0. \end{align} \tag{ 76 }$$

Plugging equations (54) and (55) to equation (70) one obtains

$$\begin{equation} -(2A+nq\varepsilon)j+(A-B)=\mathcal{A}, \end{equation} \tag{ 77 }$$

hence

$$\begin{equation} A=-\frac{nq\varepsilon}{2}, \end{equation} \tag{ 78 }$$

$$\begin{align} A-B=\mathcal{A}. \end{align} \tag{ 79 }$$

On the other hand plugging equations (54) and (55) to equation (71) leads to

$$\begin{align} (4A+C)j+(2B+C)=\mathcal{B}, \end{align} \tag{ 80 }$$

hence

$$\begin{equation} C=2nq\varepsilon. \end{equation} \tag{ 81 }$$

Again plugging equations (54) and (55) to equation (72) gives us following formula

$$\begin{equation} -(2A+C-nq\varepsilon)j+(A-B)=\mathcal{C}, \end{equation} \tag{ 82 }$$

therefore

$$\begin{equation} \mathcal{C}=A-B=\mathcal{A}. \end{equation} \tag{ 83 }$$

We demand that periodic condition should be fulfilled i.e. $\chi_{j+p}-\chi_j$ must be integer multiple of 2π. We have

$$\begin{align} \chi_{j+p}-\chi_j= Ap^2+Bp+2 Apj \end{align} \tag{ 84 }$$

and by taking into account that $2 A p j = -jq2\pi$ we get condition

$$\begin{equation} B=\frac{2\pi m}{p}-Ap=mn\varepsilon+q\pi, \end{equation} \tag{ 85 }$$

with

$$\begin{equation} m=0,1,{\ldots},p-1. \end{equation} \tag{ 86 }$$

Let us summarize what we obtained so far

$$\begin{align} A=-\frac{1}{2}nq\varepsilon,~~B=mn\varepsilon+q\pi,~~C=2nq\varepsilon, \end{align} \tag{ 87 }$$

$$\begin{align} \mathcal{C}=\mathcal{A}=\frac{qn\varepsilon}{2}-((m+q)n\varepsilon+q\pi), \end{align} \tag{ 88 }$$

$$\begin{align} \mathcal{B}=2((m+q)n\varepsilon+q\pi). \end{align} \tag{ 89 }$$

In our quadratic formula equation (76) linear coefficient can be now expressed as

$$\begin{equation} ce^{i\mathcal{A}}e^{-ir\varepsilon}+c^{*}e^{i(\mathcal{B}+\mathcal{C})}e^{ir\varepsilon}=2e^{iqn\varepsilon/2}\cos \xi, \end{equation} \tag{ 90 }$$

where

$$\begin{align} \cos \xi=\cos\theta\cos\Omega, \end{align} \tag{ 91 }$$

$$\begin{align} \Omega=\gamma-\Xi-r\varepsilon, \end{align} \tag{ 92 }$$

$$\begin{align} \Xi=(m+q)n\varepsilon+q\pi. \end{align} \tag{ 93 }$$

Now the formula for eigenvalues take the form

$$\begin{equation} \lambda^2-2\lambda e^{iqn\varepsilon/2}\cos\xi+e^{iqn\varepsilon}, \end{equation} \tag{ 94 }$$

which can be easly solved

$$\begin{equation} \lambda_\pm=e^{iqn\varepsilon/2}e^{\pm i \xi}. \end{equation} \tag{ 95 }$$

We also have

$$\begin{equation} \beta_r=\frac{\lambda e^{-i\left(\frac{qn\varepsilon}{2}+\Xi\right)}e^{ir\varepsilon}-e^{-2i\Xi}c}{s}. \end{equation} \tag{ 96 }$$

A.4. Summary—case q ≠ 0

Now we can summarize above formulae. The eigenvalues for a given q and coin given by equation (49) are

$$\begin{equation} \lambda_{m,r,\tau}^{(q)} = e^{i\left(\delta+q n\frac{\varepsilon}{2} +\tau \xi_{m,r}^{(q)}\right)}, \end{equation} \tag{ 97 }$$

where $m = 0,{\ldots},p-1$, $r = 0,{\ldots},n-1$, $\tau = \pm1$ and

$$\begin{align} \cos \xi_{m,r}^{(q)}=(-1)^q\cos\theta\cos\left((m+q)n\varepsilon +r\varepsilon -\gamma\right). \end{align} \tag{ 98 }$$

The corresponding eigenvectors are

$$\begin{align} |\psi_{m,r,\tau}^{(q\neq 0)}\rangle = N_{m,\tau}^{(q)}\frac{1}{\sqrt{d}}\sum_{j=0}^{d-1} e^{i\chi_{m,j}^{(q)}}|r+jqn\rangle_p \otimes \begin{pmatrix} 1 \\ \beta_{m,r,\tau}^{(q)} e^{i2nq\varepsilon j} \end{pmatrix}, \end{align} \tag{ 99 }$$

$$\begin{align} |\psi_{m,\tau}^{(q= 0)}\rangle = N_{m,\tau}^{(0)}~|m\rangle_p \otimes \begin{pmatrix} 1 \\ \beta_{m,\tau}^{(0)} \end{pmatrix}, \end{align} \tag{ 100 }$$

where

$$\begin{equation} \beta_{m,r,\tau}^{(q\neq0)}=\frac{(-1)^qe^{i\tau \xi_{m,r}^{(q)}}e^{-i((m+q)n\varepsilon-r\varepsilon}-\cos\theta e^{-2i(m+q)n\varepsilon}e^{i\gamma}}{\sin\theta e^{i\sigma}} \end{equation} \tag{ 101 }$$

$$\begin{align} \beta_{m,\tau}^{(q= 0)} = \frac{e^{i\tau \xi_{m}^{(0)}}-\cos\theta e^{i\sigma}}{\sin\theta e^{i\sigma}}, \end{align} \tag{ 102 }$$

$$\begin{align} \chi_{m,j}^{(q)}=-\frac{qn\varepsilon}{2}j^2+(mn\varepsilon+q\pi)j, \end{align} \tag{ 103 }$$

and

$$\begin{align} N_{m,r, \tau}^{(q)}=\frac{1}{\sqrt{1+|\beta_{m,r, \tau}^{(q)}|^2}}. \end{align} \tag{ 104 }$$

A.5. Summary—case q ≠ 0 and d-prime

Here we summarize solution for d-prime. In this case it must be that p = d, n = 1 and r = 0.

The eigenvalues for a given q are

$$\begin{align} \lambda_{m,\tau}^{(q)} = e^{i\left(\delta+q\frac{\varepsilon}{2} +\tau \xi_m^{(q)}\right)}, \end{align} \tag{ 105 }$$

where $\tau = \pm1$, $m = 0,{\ldots},d-1$ and

$$\begin{align} \cos \xi_{m}^{(q)}=(-1)^q\cos\theta\cos\left((m+q)\varepsilon-\gamma\right). \end{align} \tag{ 106 }$$

The corresponding eigenvectors are

$$\begin{equation} |\psi_{m,\tau}^{(q\neq 0)}\rangle = N_{m,\tau}^{(q)}\frac{1}{\sqrt{d}}\sum_{j=0}^{d-1} e^{i\chi_{m,j}^{(q)}}|jq\rangle_p \otimes \begin{pmatrix} 1 \\ \beta_{m,\tau}^{(q)} e^{i2q\varepsilon j} \end{pmatrix}, \end{equation} \tag{ 107 }$$

where

$$\begin{equation} \beta_{m\tau}^{(q\neq0)}=\frac{(-1)^qe^{i\tau \xi_{m}^{(q)}}e^{-i(m+q)\varepsilon}-\cos\theta e^{-2i(m+q)\varepsilon}e^{i\gamma}}{\sin\theta e^{i\sigma}} \end{equation} \tag{ 108 }$$

$$\begin{equation} \chi_{m,j}^{(q)}=-\frac{q\varepsilon}{2}j^2+(m\varepsilon+q\pi)j, \end{equation} \tag{ 109 }$$

and

$$\begin{equation} N_{m,\tau}^{(q)}=\frac{1}{\sqrt{1+|\beta_{m,\tau}^{(q)}|^2}}. \end{equation} \tag{ 110 }$$

A.6. Summary—case q = 0

In this case eigenvalues are given by

$$\begin{align} \lambda_{m,\tau}^{(q=0)} = e^{i\left(\delta+\tau \xi_m^{(q=0)}\right)}, \end{align} \tag{ 111 }$$

where

$$\begin{align} \cos \xi_{m}^{(q=0)}=\cos\theta\cos\left(m\varepsilon-\gamma\right) \end{align} \tag{ 112 }$$

and $\tau = \pm1$, $m = 0,{\ldots},d-1$. The eigenvectors are given by

$$\begin{align} |\psi_{m,\tau}^{(q= 0)}\rangle = N_{m,\tau}^{(0)}~|m\rangle_p \otimes \begin{pmatrix} 1 \\ \beta_{m,\tau}^{(0)} \end{pmatrix}, \end{align} \tag{ 113 }$$

with

$$\begin{equation} \beta_{m,\tau}^{(q= 0)} = \frac{e^{i\tau \xi_{m}^{(0)}}-\cos\theta e^{i\sigma}}{\sin\theta e^{i\sigma}}. \end{equation} \tag{ 114 }$$

In this appendix we provide a proof of the theorem 1.

B.1. Case $q^{^{\prime}}\neq0$, q ≠ 0

We first observe that

$$\begin{eqnarray} & \langle\psi_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})} |\psi_{m,\tau}^{(q)}\rangle =N_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})}N_{m,\tau}^{(q)}\frac{1}{d}\sum_{j,j^{^{\prime}}=0}^{d-1} e^{i(\chi_{m,j}^{(q)}-\chi_{m^{^{\prime}},j^{^{\prime}}}^{(q^{^{\prime}})})} \langle \, j^{^{\prime}}q^{^{\prime}}|jq\rangle_p \begin{pmatrix} 1 \\ &&\beta_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})} e^{i2q^{^{\prime}}\varepsilon j^{^{\prime}}} \end{pmatrix}^{\dagger} \begin{pmatrix} 1 \\ &&\beta_{m,\tau}^{(q)} e^{i2q\varepsilon j} \end{pmatrix}.\nonumber\\ \end{eqnarray} \tag{ 115 }$$

For a given pair $q,q^{^{\prime}}$ ($q^{^{\prime}},q\lt d$, $~~q^{^{\prime}}\neq q$) and j let us define unique $0\leqslant \tilde{j}\leqslant d$ such that $\tilde{j}q^{^{\prime}}$ is congruent with jq modulo d

$$\begin{align} \tilde{j}q^{^{\prime}}\equiv jq~ (\mathrm{mod}~d). \end{align} \tag{ 116 }$$

The condition that d is prime guarantees the existence of $\tilde{j}$. Note that

$$\begin{align} \langle \, j^{^{\prime}}q^{^{\prime}}|jq\rangle_p=\delta_{j^{^{\prime}}\tilde{j}}, \end{align} \tag{ 117 }$$

where we use Kronecker delta $\delta_{j^{^{\prime}}\tilde{j}}$. Now

$$\begin{align} \langle\psi_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})} |\psi_{m,\tau}^{(q)}\rangle =N_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})}N_{m,\tau}^{(q)}\frac{1}{d}\sum_{j=0}^{d-1} e^{i(\chi_{m,j}^{(q)}-\chi_{m^{^{\prime}},\tilde{j}}^{(q^{^{\prime}})})} \begin{pmatrix} 1 \\ \beta_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})} e^{i2q^{^{\prime}}\varepsilon \tilde{j}} \end{pmatrix}^{\dagger} \begin{pmatrix} 1 \\ \beta_{m,\tau}^{(q)} e^{i2q\varepsilon j} \end{pmatrix}. \end{align} \tag{ 118 }$$

Equation (116) implies

$$\begin{equation} e^{i2\varepsilon (qj-q^{^{\prime}}\tilde{j})}=1, \end{equation} \tag{ 119 }$$

therefore

$$\begin{equation} \begin{pmatrix} 1 \\ \beta_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})} e^{i2q^{^{\prime}}\varepsilon \tilde{j}} \end{pmatrix}^{\dagger} \begin{pmatrix} 1 \\ \beta_{m,\tau}^{(q)} e^{i2q\varepsilon j} \end{pmatrix}= 1+(\beta_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})})^*\beta_{m,\tau}^{(q)}. \end{equation} \tag{ 120 }$$

We arrive at

$$\begin{align} \langle\psi_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})} |\psi_{m,\tau}^{(q)}\rangle = N_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})}N_{m,\tau}^{(q)}\left(1+(\beta_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})})^*\beta_{m,\tau}^{(q)}\right)\frac{1}{d}\sum_{j=0}^{d-1} e^{i(\chi_{m,j}^{(q)}-\chi_{m^{^{\prime}},\tilde{j}}^{(q^{^{\prime}})})} \end{align} \tag{ 121 }$$

and

$$\begin{align} | \langle\psi_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})} |\psi_{m,\tau}^{(q)}\rangle|^2=\frac{1}{d^2}~\left(N_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})}N_{m,\tau}^{(q)}\right)^2\left|1+(\beta_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})})^*\beta_{m,\tau}^{(q)}\right|^2\left|\sum_{j=0}^{d-1} e^{i(\chi_{m,j}^{(q)}-\chi_{m^{^{\prime}},\tilde{j}}^{(q^{^{\prime}})})}\right|^2. \end{align} \tag{ 122 }$$

Next, let us observe that (the proof is given latter)

$$\begin{equation} \left|\sum_{j=0}^{d-1} e^{i(\chi_{m,j}^{(q)}-\chi_{m^{^{\prime}},\tilde{j}}^{(q^{^{\prime}})})}\right|^2 = d. \end{equation} \tag{ 123 }$$

Using the above we get

$$\begin{equation} | \langle\psi_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})} |\psi_{m,\tau}^{(q)}\rangle|^2=\frac{1}{d}~\left(N_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})}N_{m,\tau}^{(q)}\right)^2~\left|1+(\beta_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})})^*\beta_{m,\tau}^{(q)}\right|^2. \end{equation} \tag{ 124 }$$

Obviously

$$\begin{equation} \left|1+(\beta_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})})^*\beta_{m,\tau}^{(q)}\right|^2\leqslant \left(1+|\beta_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})}||\beta_{m,\tau}^{(q)}|\right)^2, \end{equation} \tag{ 125 }$$

therefore (see the proof at the end of this appendix)

$$\begin{eqnarray} & & \left(N_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})}N_{m,\tau}^{(q)}\right)^2~\left|1+(\beta_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})})^*\beta_{m,\tau}^{(q)}\right|^2 \leqslant \frac{\left(1+|\beta_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})}||\beta_{m,\tau}^{(q)}|\right)^2}{\left(1+|\beta_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})}|^2\right)\left(1+|\beta_{m,\tau}^{(q)}|^2\right)}\leqslant 1. \end{eqnarray} \tag{ 126 }$$

It follows that

$$\begin{equation} | \langle\psi_{m^{^{\prime}},\tau^{^{\prime}}}^{(q^{^{\prime}})} |\psi_{m,\tau}^{(q)}\rangle|^2\leqslant \frac{1}{d}=\frac{2}{D}. \end{equation} \tag{ 127 }$$

B.2. Case $q^{^{\prime}} = 0,q\neq 0$

We have

$$\begin{align} \langle\psi_{m^{^{\prime}},\tau^{^{\prime}}}^{(0)} |\psi_{m,\tau}^{(q)}\rangle=N_{m^{^{\prime}},\tau^{^{\prime}}}^{(0)}N_{m,\tau}^{(q)}\frac{1}{\sqrt{d}}\sum_{j=0}^{d-1} e^{i\chi_{m,j}^{(q)}}\langle m^{^{\prime}}|jq\rangle \begin{pmatrix} 1 \\ \beta_{m^{^{\prime}},\tau^{^{\prime}}}^{(0)}\end{pmatrix}^{\dagger} \begin{pmatrix} 1 \\ \beta_{m,\tau}^{(q)} e^{i2q\varepsilon j} \end{pmatrix}. \end{align} \tag{ 128 }$$

Since $\langle m^{^{\prime}}|jq\rangle\neq 0$ for $j = \tilde{j}$, where $m^{^{\prime}}\equiv \tilde{j}q~(\mathrm{mod}~d)$

$$\begin{align} \langle\psi_{m^{^{\prime}},\tau^{^{\prime}}}^{(0)} |\psi_{m,\tau}^{(q)}\rangle=N_{m^{^{\prime}},\tau^{^{\prime}}}^{(0)}N_{m,\tau}^{(q)}\frac{1}{\sqrt{d}}~e^{i\chi_{m,\tilde{j}}^{(q)}}\left(1+(\beta_{m^{^{\prime}},\tau^{^{\prime}}}^{(0)})^*\beta_{m,\tau}^{(q)}e^{i2q\varepsilon \tilde{j}}\right) \end{align} \tag{ 129 }$$

and

$$\begin{align} \left|\langle\psi_{m^{^{\prime}},\tau^{^{\prime}}}^{(0)} |\psi_{m,\tau}^{(q)}\rangle\right|^2=\left(N_{m^{^{\prime}},\tau^{^{\prime}}}^{(0)}\right)^2\left(N_{m,\tau}^{(q)}\right)^2\frac{1}{d}~\left|1+(\beta_{m^{^{\prime}},\tau^{^{\prime}}}^{(0)})^*\beta_{m,\tau}^{(q)}e^{i2q\varepsilon \tilde{j}}\right|^2. \end{align} \tag{ 130 }$$

Since

$$\begin{equation} \left|1+(\beta_{m^{^{\prime}},\tau^{^{\prime}}}^{(0)})^*\beta_{m,\tau}^{(q)}e^{i2q\varepsilon \tilde{j}}\right|^2\leqslant \left(1+|\beta_{m^{^{\prime}},\tau^{^{\prime}}}^{(0)}||\beta_{m,\tau}^{(q)}|\right)^2 \end{equation} \tag{ 131 }$$

one has

$$\begin{equation} \left|\langle\psi_{m^{^{\prime}},\tau^{^{\prime}}}^{(0)} |\psi_{m,\tau}^{(q)}\rangle\right|^2\leqslant\frac{1}{d}~\frac{\left(1+|\beta_{m^{^{\prime}},\tau^{^{\prime}}}^{(0)}||\beta_{m,\tau}^{(q)}|\right)^2}{\left(1+|\beta_{m^{^{\prime}},\tau^{^{\prime}}}^{(0)}|^2\right)\left(1+|\beta_{m^{^{\prime}},\tau^{^{\prime}}}^{(q)}|^2\right)}\leqslant \frac{1}{d}. \end{equation} \tag{ 132 }$$

B.3. Proof of equation (123)

For given $q,q^{^{\prime}}\neq q$

$$\begin{equation} S= \sum_{j=0}^{d-1} e^{i(\chi_{m,j}^{(q)}-\chi_{m^{^{\prime}},\tilde{j}}^{(q^{^{\prime}})})}= \sum_{j=0}^{d-1} e^{i\varepsilon(w_j+v_{m,j})}, \end{equation} \tag{ 133 }$$

where

$$\begin{equation} w_j=\left(\frac{qj(d-j)-q^{^{\prime}}\tilde{j}(d-\tilde{j})}{2}\right)_{\mathrm{mod}~d}, \end{equation} \tag{ 134 }$$

$$\begin{equation} v_{m,j}=(mj-m^{^{\prime}}\tilde{j})_{\mathrm{mod}~d}. \end{equation} \tag{ 135 }$$

Note that $j,\tilde{j},q,q^{^{\prime}},m,m^{^{\prime}}$ can be considered as elements of finite field ${\mathbb{F}}_d = {\mathbb{Z}}/ d{\mathbb{Z}}$. w_j and $v_{m,j}$ are also elements of ${\mathbb{F}}_d$ (note that $\left[qj(d-j)-q^{^{\prime}}\tilde{j}(d-\tilde{j})\right]$ is even). Let us define another element Q of this field with the equation $Q\circ q^{^{\prime}} = q$, where $\circ$ stands for the field multiplication. Due to this definition

$$\begin{equation} \tilde{j}=Q\circ j. \end{equation} \tag{ 136 }$$

Let us also define one more element of the field ${\mathbb{F}}_d$, h by $h\circ 2 = 1$. Equation (133) can be re-framed in the form

$$\begin{equation} w_j=h\circ\left[(q\circ j\circ(-j))~-_{\mathbb{F}}~(q^{^{\prime}}\circ\tilde{j}\circ(-\tilde{j}))\right]. \end{equation} \tag{ 137 }$$

It follows from equation (136) that

$$\begin{align} w_j & = h\circ\left[ (q^{^{\prime}}\circ Q \circ Q \circ Q \circ j \circ j{-}_{\mathbb{F}}~q \circ j \circ j) \right] = h\circ(Q{-}_{\mathbb{F}}~1)\circ q \circ j \circ j=x\circ j \circ j, \end{align} \tag{ 138 }$$

where $x = h\circ(Q-_{\mathbb{F}}~1)\circ q\in \mathbb{F}_d$. To put it another way $w_j = (xj^2)_{\mathrm{mod}~d}$. Analogously $v_{m,j} = (yj)_{\mathrm{mod}~d}$, where $y = m-_{\mathbb{F}}~m^{^{\prime}}\circ Q \in \mathbb{F}_d$

$$\begin{equation} S=\sum_{j=0}^{d-1} e^{i\varepsilon(xj^2+yj)}. \end{equation} \tag{ 139 }$$

For $q\neq q^{^{\prime}}$, Q ≠ 1 and x ≠ 0. It follows from theory of generalized quadratic Gauss sums that

$$\begin{equation} |S|^2=d. \end{equation} \tag{ 140 }$$

B.4. Proof of the second inequality in equation (126)

It is clear that

$$\begin{equation} (|a|-|b|)^2\geqslant0. \end{equation} \tag{ 141 }$$

This leads to

$$\begin{equation} |a|^2+|b|^2\geqslant 2|a||b|, \end{equation} \tag{ 142 }$$

and

$$\begin{equation} |a|^2+|b|^2+1+|a|^2|b|^2\geqslant2|a||b|+1+|a|^2|b|^2, \end{equation} \tag{ 143 }$$

which gives

$$\begin{equation} (1+|a|^2)(1+|b|^2) \geqslant(1+|a||b|))^2. \end{equation} \tag{ 144 }$$

Now let us solve the eigen-problem for the DTQW with θ = 0.

C.1. Ansatz

We make an ansatz on the form of eigenvectors

$$\begin{equation} |\Psi_{r,\tau}\rangle = \frac{1}{\sqrt{p}}\left(\sum_{j=0}^{p-1}e^{i\chi_j}|r+jqn\rangle_p \right)\otimes |\tau\rangle, \end{equation} \tag{ 145 }$$

where

$$\begin{equation} r=0,1,{\ldots},n-1,~~~\tau = \pm1. \end{equation} \tag{ 146 }$$

In above we use the convention

$$\begin{equation} |\rightarrow\rangle =|+1\rangle , \end{equation} \tag{ 147 }$$

$$\begin{equation} |\leftarrow\rangle =|-1\rangle. \end{equation} \tag{ 148 }$$

Moreover, we assume the following form of χ_j

$$\begin{equation} \chi_j=Aj^2+Bj, \end{equation} \tag{ 149 }$$

with A, B constant.

C.2. Derivation

Note that

$$\begin{equation} F|\Psi_{r,\tau}\rangle = \frac{1}{\sqrt{p}}\left(\sum_{j=0}^{p-1}e^{i\chi_{j-\tau}}|r+jqn\rangle_p \right)\otimes |\tau\rangle. \end{equation} \tag{ 150 }$$

It follows that

$$\begin{eqnarray} U|\Psi_{r,\tau}\rangle = \frac{1}{\sqrt{p}}\left(\sum_{j=0}^{p-1}e^{i\chi_{j-\tau}}|r+jqn\rangle_p \right) \otimes \begin{pmatrix}e^{-i(r+jqn)\varepsilon}&0\\0&e^{i(r+jqn)\varepsilon}\end{pmatrix} |\tau\rangle \end{eqnarray} \tag{ 151 }$$

which should be equal to $\lambda |\Psi_{r,\tau}\rangle$. Therefore, our goal is to solve the equation

$$\begin{equation} e^{i\chi_{j-\tau}} e^{-i\tau (r+jqn)\varepsilon}=\lambda e^{i\chi_j}. \end{equation} \tag{ 152 }$$

We insist that exponents are j-independent. This gives us an equation with a constant $\mathcal{A}_\tau$

$$\begin{equation} \chi_{j-\tau}-\chi_j-\tau jnq\varepsilon=\mathcal{A}_\tau, \end{equation} \tag{ 153 }$$

and a formula for the eigenvalue λ

$$\begin{equation} \lambda = e^{i\mathcal{A}_\tau}e^{-ir\varepsilon\tau}. \end{equation} \tag{ 154 }$$

Plugging equation (149) to equation (153) one obtains

$$\begin{equation} -(2A+\tau nq\varepsilon)j+(A-B)=\mathcal{A}_\tau, \end{equation} \tag{ 155 }$$

therefore

$$\begin{equation} A=-\frac{\tau nq\varepsilon}{2}, \end{equation} \tag{ 156 }$$

$$\begin{equation} A-B=\mathcal{A}_\tau. \end{equation} \tag{ 157 }$$

We demand that the periodicity condition should be fulfilled, i.e. $\chi_{j+p}-\chi_j$ must be an integer multiple of 2π. We have

$$\begin{equation} \chi_{j+p}-\chi_j= Ap^2+Bp+2 Apj \end{equation} \tag{ 158 }$$

and since $2 A p j = -\tau jq2\pi$ we get

$$\begin{equation} B=\frac{2\pi m}{p}-Ap=mn\varepsilon+\tau q\pi, \end{equation} \tag{ 159 }$$

where

$$\begin{equation} m=0,1,{\ldots},p-1. \end{equation} \tag{ 160 }$$

Let us summarize what we obtained so far:

$$\begin{equation} A=-\tau\frac{nq\varepsilon}{2},~~B=mn\varepsilon+\tau q\pi, \end{equation} \tag{ 161 }$$

$$\begin{equation} \mathcal{A}_\tau = A-B = -\tau \frac{nq\varepsilon}{2}-mn\varepsilon-\tau q\pi. \end{equation} \tag{ 162 }$$

The eigenvalue formula takes the form

$$\begin{equation} \lambda_{r,m,\tau} = e^{-i(\tau\frac{nq\varepsilon}{2}+mn\varepsilon+q\pi+r\varepsilon)}. \end{equation} \tag{ 163 }$$

In addition

$$\begin{equation} \chi_j=-\tau\frac{nq\varepsilon}{2}j^2+\left(mn\varepsilon+q\pi\right)j. \end{equation} \tag{ 164 }$$

C.3. Summary θ = 0, d prime

We get

$$\begin{align} \lambda_{m,\tau}^{(q)} = e^{-i(\tau\frac{q\varepsilon}{2}+m\varepsilon+q\pi)}, \end{align} \tag{ 165 }$$

$$\begin{align} m=0,{\ldots},d-1, \end{align} \tag{ 166 }$$

$$\begin{align} |\Psi_{m\tau}^{(q)}\rangle = \frac{1}{\sqrt{d}}\sum_{j=0}^{d-1}e^{i\chi_{m\tau j}^{(q)}}|jq\rangle_p \otimes |\tau\rangle, \end{align} \tag{ 167 }$$

$$\begin{align} \chi_{m\tau j}^{(q)}=-\tau\frac{q\varepsilon}{2}j^2+\left(m\varepsilon+q\pi\right)j. \end{align} \tag{ 168 }$$

The spatial part of these eigenvectors obeys perfect MUB relations for $q \neq q^{^{\prime}}$

$$\begin{align} |\langle\Psi_{m\tau}^{(q)}|\Psi_{m^{^{\prime}}\tau^{^{\prime}}}^{(q^{^{\prime}})}\rangle|^2 = \delta_{\tau \tau^{^{\prime}}}\frac{1}{d}. \end{align} \tag{ 169 }$$

C.4. Proof of equation (169)

Let us write

$$\begin{equation} |\Psi_{m\tau}^{(q)}\rangle = |\tilde{\Psi}_{m\tau}^{(q)}\rangle \otimes |\tau\rangle, \end{equation} \tag{ 170 }$$

where

$$\begin{equation} |\tilde{\Psi}_{m\tau}^{(q)}\rangle = \frac{1}{\sqrt{d}}\sum_{j=0}^{d-1}e^{i\chi_{m\tau j}^{(q)}}|jq\rangle_p. \end{equation} \tag{ 171 }$$

We need to show that

$$\begin{equation} |\langle \tilde{\Psi}_{m\tau}^{(q)}|\tilde{\Psi}_{m^{^{\prime}}\tau}^{(q^{^{\prime}})}\rangle|^2 =\frac{1}{d^2}\left|\sum_{j=0}^{d-1}e^{i(\chi_{m\tau j}^{(q)}-\chi_{m^{^{\prime}}\tau \tilde{j}}^{(q^{^{\prime}})})}\right|^2=\frac{1}{d}, \end{equation} \tag{ 172 }$$

where $\tilde{j}$ is defined by the equation

$$\begin{equation} \tilde{j}q^{^{\prime}} \equiv jq (mod~d). \end{equation} \tag{ 173 }$$

However, the above is equivalent to equation (123), that was already proven in appendix B.

As shown in the manuscript, the continuous limit of our DTQW model leads to the following Dirac equation

$$\begin{equation} \begin{pmatrix} -i\frac{\partial}{\partial x} - \mu x & m \\ m & i\frac{\partial}{\partial x} + \mu x \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = E \begin{pmatrix} \alpha \\ \beta \end{pmatrix}, \end{equation} \tag{ 174 }$$

where $\alpha = \alpha(x)$, $\beta = \beta(x)$, m is the particle's mass and µ describes the amplitude of the potential $A = \mu x$. It is assumed that $\hbar = c = 1$ and that the particle's charge is e = 1. The above leads to

$$\begin{equation} \beta = \frac{i}{m} \frac{\partial \alpha}{\partial x} + \frac{(\mu x + E)}{m} \alpha, \end{equation} \tag{ 175 }$$

and

$$\begin{equation} \frac{\partial^2 \alpha}{\partial x^2} - 2 i\mu x \frac{\partial \alpha}{\partial x} + (E^2-m^2-\mu^2 x^2 - i\mu)\alpha = 0. \end{equation} \tag{ 176 }$$

We make the following ansatz

$$\begin{equation} \alpha = {\mathcal N} e^{if(x)}, \end{equation} \tag{ 177 }$$

where f(x) is some real function and $\mathcal{N}$ is a normalization constant. Since we consider at most a second derivative over x, it is enough to take $f(x) = ax^2 + bx +c$ with a, b and c being real parameters. We take

$$\begin{equation} f(x) = \frac{\mu}{2}x^2 + kx, \end{equation} \tag{ 178 }$$

where k is a real parameter, and plug it into equation (176). We obtain

$$\begin{equation} E_{k,\pm}= \pm \sqrt{k^2 + m^2}. \end{equation} \tag{ 179 }$$

Moreover, the solution is

$$\begin{equation} \psi_{k,\pm}^{\mu} = \begin{pmatrix} \alpha_{k,\pm}^{\mu}(x) \\ \beta_{k,\pm}^{\mu}(x) \end{pmatrix}, \end{equation} \tag{ 180 }$$

where

$$\begin{equation} \alpha_{k,\pm}^{\mu}(x) = {\mathcal N} e^{i\left(\frac{\mu}{2}x^2 + kx \right)} \end{equation} \tag{ 181 }$$

and

$$\begin{equation} \beta_{k,\pm}^{\mu}(x) = \alpha_{k,\pm}^{\mu}(x) \left(\frac{E-k}{m}\right). \end{equation} \tag{ 182 }$$

The equation describes the particle moving in the unbounded space, therefore the solution is not normalized, i.e.

$$\begin{equation} \int_{-\infty}^{+\infty} dx \begin{pmatrix} \alpha^{\ast}(x) & \beta^{\ast}(x) \end{pmatrix} \begin{pmatrix} \alpha(x) \\ \beta(x) \end{pmatrix} = \infty. \end{equation} \tag{ 183 }$$

However, we choose $\mathcal N$ such that $|\alpha |^2 + |\beta|^2 = 1$. This leads to

$$\begin{equation} {\mathcal N}_{k,\pm} = 2(k^2+m^2 \mp k\sqrt{k^2+m^2}). \end{equation} \tag{ 184 }$$

Now we prove Theorem 2. First note that

$$\begin{equation} (\psi_{k^{^{\prime}},c^{^{\prime}}}^{\mu^{^{\prime}}})^{\ast} \psi_{k,c}^{\mu} = \Gamma_{k,k^{^{\prime}},c,c^{^{\prime}}} e^{i\left(\frac{(\mu-\mu^{^{\prime}})}{2}x^2 + (k-k^{^{\prime}})x \right)} \end{equation} \tag{ 185 }$$

where $c,c^{^{\prime}} = \pm 1$ and

$$\begin{equation} \Gamma_{k,k^{^{\prime}},c,c^{^{\prime}}} = {\mathcal N}_{k^{^{\prime}},c^{^{\prime}}}{\mathcal N}_{k,c} \left(m^2 + (E_{k,c}-k)(E_{k^{^{\prime}},c^{^{\prime}}}-k^{^{\prime}})\right). \end{equation} \tag{ 186 }$$

Note that equation (184) implies

$$\begin{equation} |\Gamma_{k,k^{^{\prime}},c,c^{^{\prime}}}| \leqslant 1. \end{equation} \tag{ 187 }$$

Next, we evaluate the overlap

$$\begin{align} \left| \int_{-\infty}^{+\infty} dx (\psi_{k^{^{\prime}},c^{^{\prime}}}^{\mu^{^{\prime}}})^{\ast} \psi_{k,c}^{\mu} \right| & = \left| \Gamma_{k,k^{^{\prime}},c,c^{^{\prime}}} \int_{-\infty}^{+\infty} dx e^{i\left(\frac{(\mu-\mu^{^{\prime}})}{2}x^2 + (k-k^{^{\prime}})x \right)} \right| \nonumber \\ & \leqslant \left| \int_{-\infty}^{+\infty} dx e^{i\left(\frac{(\mu-\mu^{^{\prime}})}{2}x^2 + (k-k^{^{\prime}})x \right)} \right|. \end{align} \tag{ 188 }$$

We introduce

$$\begin{equation} \Delta = \frac{\mu-\mu^{^{\prime}}}{2},~~~~\delta = k-k^{^{\prime}}, \end{equation} \tag{ 189 }$$

and evaluate

$$\begin{eqnarray} &&\int_{-\infty}^{+\infty} dx e^{i\left(\Delta x^2 + \delta x \right)} = e^{-i\frac{\delta^2}{4\Delta}} \int_{-\infty}^{+\infty} dx e^{i\left(\Delta x + \frac{\delta}{2\sqrt{\Delta}} \right)^2} = e^{-i\frac{\delta^2}{4\Delta}} \frac{(1+i)\sqrt{\pi}}{\sqrt{2\Delta}}. \end{eqnarray} \tag{ 190 }$$

Therefore

$$\begin{equation} \left| \int_{-\infty}^{+\infty} dx (\psi_{k^{^{\prime}},c^{^{\prime}}}^{\mu^{^{\prime}}})^{\ast} \psi_{k,c}^{\mu} \right| \leqslant \sqrt{\frac{2\pi}{|\mu -\mu^{^{\prime}}|}}. \end{equation} \tag{ 191 }$$