An effective Hamiltonian approach to quantum random walk

Abstract

In this article we present an effective Hamiltonian approach for discrete time quantum random walk. A form of the Hamiltonian for one-dimensional quantum walk has been prescribed, utilizing the fact that Hamiltonians are generators of time translations. Then an attempt has been made to generalize the techniques to higher dimensions. We find that the Hamiltonian can be written as the sum of a Weyl Hamiltonian and a Dirac comb potential. The time evolution operator obtained from this prescribed Hamiltonian is in complete agreement with that of the standard approach. But in higher dimension we find that the time evolution operator is additive, instead of being multiplicative (see Chandrashekar, Sci. Rep. 3, 2829 (18)). We showed that in the case of two-step walk, the time evolution operator effectively can have multiplicative form. In the case of a square lattice, quantum walk has been studied computationally for different coins and the results for both the additive and the multiplicative approaches have been compared. Using the graphene Hamiltonian, the walk has been studied on a graphene lattice and we conclude the preference of additive approach over the multiplicative one.

Appendices

Appendix A: Derivation of Hamiltonian from evolution operators in 1D

1.1 A.1 Derivation of H _T

If the Hamiltonian corresponding to T is defined as $H_{T} \otimes \mathbb {1}$, then it can be derived as follows:

$$\begin{array}{@{}rcl@{}} &&{\kern-.6pc}\mathrm{e}^{-i H_{T} {\Delta} t \otimes \mathbb{1}} ={} \sum\limits_{x} \left( \begin{array}{cc} \vert x - {\Delta} x\rangle \langle x \vert & 0 \\ 0 & \vert x + {\Delta} x \rangle \langle x \vert \end{array}\right) \\ &&{\kern4pc}={} \sum\limits_{x} \left( \begin{array}{cc} \mathrm{e}^{i k {\Delta} x} & 0 \\ 0 & \mathrm{e}^{- i k {\Delta} x} \end{array}\right)\left( \mathbb{1}_{2} \otimes \vert x \rangle \langle x \vert \right) \\ &&{\kern4pc}={} \left( \!\begin{array}{cc} \mathrm{e}^{i k {\Delta} x} & 0 \\ 0 & \mathrm{e}^{- i k {\Delta} x} \end{array}\!\right)\left( \! \because \sum\limits_{x} \vert x\rangle \langle x\vert = \mathbb{1}\!\right) \\ &&{\kern-.9pc}\Rightarrow H_{T} {\Delta} t \otimes \mathbb{1} \,=\, i \ln \left( \begin{array}{cc} \mathrm{e}^{i k {\Delta} x} & 0 \\ 0 & \mathrm{e}^{- i k {\Delta} x} \end{array}\right) \\ &&{\kern4pc}=\! -k {\Delta} x \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right) \\ &&{\kern-.9pc}\Rightarrow H_{T} \otimes \mathbb{1} \,=\, - k v \sigma_{z} \,\,\,\,\,\,\,\,\,\,\,\,\, \left( \because v = \frac{\Delta x}{\Delta t} \right). \end{array} $$

Thus, we see that the derived Hamiltonian $H_{T} \otimes \mathbb {1}$ is Hermitian.

Let us find the expression for H _T now. Taking projection onto x-space we get

$$\begin{array}{@{}rcl@{}} &&{\kern-.8pc}\left( \mathbb{1}_{2} \otimes \langle x \vert \right) H_{T} \otimes \mathbb{1} \left( \mathbb{1}_{2} \otimes \vert \alpha \rangle \right)\\ &&\quad= - \left( \mathbb{1}_{2} \otimes \langle x \vert \right) k v \sigma_{z} \left( \mathbb{1}_{2} \otimes \vert \alpha \rangle \right)\\ &&\quad [ \vert \alpha \rangle \textrm{ is some ket in the position Hilbert space}] \\ &&{\kern-.9pc}\Rightarrow H_{T} \langle x \vert \alpha \rangle = - \left( \begin{array}{cc} \langle x \vert & 0 \\ 0 & \langle x \vert \end{array}\right) \left( \begin{array}{cc} k v & 0 \\ 0 & - k v \end{array}\right) \left( \begin{array}{cc} \vert \alpha \rangle & 0 \\ 0 & \vert \alpha \rangle \end{array}\right)\\ &&{\kern3.8pc}= - \left( \begin{array}{cc} \langle x \vert k v \vert \alpha \rangle & 0 \\ 0 & - \langle x \vert k v \vert \alpha \rangle \end{array}\right)\\ &&{\kern3.8pc}= i v \left( \begin{array}{cc} \frac{\partial \langle x \vert \alpha \rangle}{\partial x } & 0 \\ 0 & - \frac{\partial \langle x \vert \alpha \rangle}{\partial x } \\ \end{array}\right)\\ &&{\kern3.8pc}= i v \sigma_{z} \frac{\partial}{\partial x} \langle x \vert \alpha \rangle\\ &&{\kern-.9pc}\Rightarrow H_{T} = i v \sigma_{z} \frac{\partial}{\partial x}. \end{array} $$

1.2 A.2 Derivation of H _S

If we denote the Hamiltonian governing the chirality flip as ${\sum }_{m} H_{S} \otimes \vert m \rangle \langle m \vert$, we can write

$$\begin{array}{@{}rcl@{}} &&{} \mathrm{e}^{-i {\sum}_{m} H_{S} {\Delta} \tau \otimes \vert m \rangle \langle m \vert} = \sum\limits_{m} S \otimes \vert m \rangle \langle m \vert\\ &&{}\quad \Rightarrow \mathrm{e}^{-i {\sum}_{m} H_{S} {\Delta} \tau \otimes \vert m \rangle \langle m \vert}\left( \mathbb{1}_{2} \otimes \vert n \rangle \right)\\ &&{}\quad = \sum\limits_{m} S \otimes \vert m \rangle \langle m \vert \left( \mathbb{1}_{2} \otimes \vert n \rangle \right)\\ &&{}\quad \Rightarrow \mathrm{e}^{-i H_{S} {\Delta} \tau} \otimes \vert n \rangle = S \otimes \vert n \rangle\\ &&{}\quad \Rightarrow H_{S} = \frac{i}{\Delta \tau} \ln S. \end{array} $$

Now H _S is Hermitian by construction, because every unitary operator on a Hilbert space can be written as U=e^iA for some Hermitian A.

1.3 A.3 Derivation of H(x)

The total Hamiltonian H is given by

$$\begin{array}{@{}rcl@{}} H &=& {\sum}_{n} H_{S} \otimes \vert n\rangle \langle n \vert + H_{T} \otimes \mathbb{1}. \end{array} $$

Therefore, by taking projection over x-space we get

$$\begin{array}{@{}rcl@{}} &&{} \left( \mathbb{1}_{2} \otimes \langle x \vert \right) H \left( \mathbb{1}_{2} \otimes \vert \alpha \rangle \right) \\ &&{}(\vert \alpha \rangle \text{ is some ket in the position Hilbert space})\\ &&\quad = \sum\limits_{n} H_{S} \delta (x-n) \langle n \vert \alpha \rangle + H_{T} \langle x \vert \alpha \rangle\\ &&\quad = \left( \sum\limits_{n} H_{S} \delta (x-n) + H_{T} \right) \langle x \vert \alpha \rangle\\ &&\quad \Rightarrow H(x) = \sum\limits_{n} H_{S} \delta (x-n) + H_{T}. \end{array} $$

Here in the second step we have used a proper unit weighting factor.

Appendix B: Derivation of evolution operator from Hamiltonian in 2D square lattice

We have

$$\begin{array}{@{}rcl@{}} H(n,m) &=& H_{T} \otimes \mathbb{1} + H_{S}\otimes \vert n,m \rangle \langle n,m \vert. \end{array} $$

Now following the logic as described in §3.1.1, we can say that [H _S ,H _T ] Δt=0. Now,

$$\begin{array}{@{}rcl@{}} H_{S} &=& \frac{i}{\Delta \tau}\ln (S_{x} \otimes M_{1} + S_{y}\otimes M_{2} )\\ &\Rightarrow& \mathrm{e}^{-i H_{S} {\Delta} \tau \otimes \vert n,m \rangle \langle n,m \vert} \left( \mathbb{1}_{4} \otimes \vert n,m \rangle \langle n,m \vert \right)\\ &=& (S_{x} \otimes M_{1} + S_{y}\otimes M_{2}) \otimes \vert n,m \rangle \langle n,m \vert . \end{array} $$

Again,

$$\begin{array}{@{}rcl@{}} H_{T} &=& -\left( k_{x} v_{x} \left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right) + k_{y} v_{y} \left( \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right)\right) \otimes \sigma_{z} . \end{array} $$

Therefore,

$$\begin{array}{@{}rcl@{}} &&{}\mathrm{e}^{-i H_{T} {\Delta} t \otimes \mathbb{1}} \left( \mathbb{1}_{4} \otimes \vert x,y \rangle \langle x,y \vert \right) {}\\ &&\quad={} \left( {}\left( {}\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}{}\right) \otimes \left( {}\begin{array}{cc} \mathrm{e}^{i k_{x} {\Delta} x} & 0 \\ 0 & \mathrm{e}^{-i k_{x} {\Delta} x} \end{array}{}\right)\right. \\ &&\qquad\left.+ \left( {}\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}{}\right) \otimes \left( \begin{array}{cc} \mathrm{e}^{i k_{y} {\Delta} y} & 0 \\ 0 & \mathrm{e}^{-i k_{y} {\Delta} y} \end{array}\right) \right) \\&&\qquad\otimes \vert x,y \rangle \langle x,y \vert \\ &&\Rightarrow T(x,y){} = {}\left( \begin{array}{cccc} \mathrm{e}^{i k_{x} {\Delta} x} & 0 & 0 & 0 \\ 0 & \mathrm{e}^{- i k_{x} {\Delta} x} & 0 & 0 \\ 0 & 0 & \mathrm{e}^{i k_{y} {\Delta} y} & 0 \\ 0& 0 & 0 & \mathrm{e}^{- i k_{y} {\Delta} y} \end{array}\right)\\ &&\quad\otimes \vert x,y \rangle \langle x,y \vert . \end{array} $$

Therefore, we have

$$\begin{array}{@{}rcl@{}} W(n,m) &=& T(n,m) ((S_{x} \otimes M_{1}+ S_{y} \otimes M_{2})\\&&\otimes \vert n,m \rangle \langle n,m \vert ). \end{array} $$