Optimal Dimensionality Reduced Quantum Walk and Noise Characterization

Abstract

In a recent work by Novo et al. (Sci. Rep. $\mathbf 5$, 13304, 2015), the invariant subspace method was applied to the study of continuous-time quantum walk (CTQW). In this work, we adopt the aforementioned method to investigate the optimality of a perturbed quantum walk search of a marked element in a noisy environment on various graphs. We formulate the necessary condition of the noise distribution in the system such that the invariant subspace method remains effective and efficient. Based on the noise, we further formulate how to set the appropriate coupling factor to preserve the optimality of the quantum walker. Thus, a quantum walker based on an N by N Hamiltonian can be efficiently implemented using the near-term quantum technology.

A Appendix A: Reduction Using Lanczos Algorithm

If an $N \times N$ matrix is reduced to a $3 \times 3$ matrix by Lanczos algorithm, the reduced matrix in the $\{v_1, v_2, v_3\}$ basis is

$$\begin{aligned} \begin{bmatrix} \alpha _1&\beta _2&0 \\ \beta _2&\alpha _2&\beta _3 \\ 0&\beta _3&\alpha _3\\ \end{bmatrix} \end{aligned}$$

(44)

By using the Lanczos algorithm on a UCPG configuration given in Sect. 2, we start with $v_1 =| \omega \rangle$ and $A = H_a$. Immediately we know that $\alpha _1 = 0$ and that leads to $\beta _2 = \sqrt{N-m_0}$. At iteration $j=2$, we can obtain

$$\begin{aligned} v_2= & {} \frac{1}{\sqrt{N-m_0}}\sum _{i\in V, i\notin V_0}| i \rangle \end{aligned}$$

(45)

$$\begin{aligned} w'_2= & {} \frac{1}{\sqrt{N-m_0}}((N-m_0)| \omega \rangle + (N-m_0)\sum _{i\in V_0, i\ne \omega }| i \rangle \nonumber \\&+ (N-m_0-m_1)\sum _{i\in V, i\notin V_0} | i \rangle )\end{aligned}$$

(46)

$$\begin{aligned} w_2= & {} \frac{1}{\sqrt{N-m_0}}(N-m_0)\sum _{i\in V_0, i\ne \omega }| i \rangle \end{aligned}$$

(47)

with $\alpha _2 = (N-m_0-m_1)$. At iteration $j=3$, we obtain $\beta _3 = \sqrt{(N-m_0)(m_0-1)}$ with

$$\begin{aligned} v_3= & {} \frac{1}{\sqrt{m_0-1}}\sum _{i\in V_0, i \ne \omega }| i \rangle \end{aligned}$$

(48)

$$\begin{aligned} w'_3= & {} {\sqrt{m_0-1}}\sum _{i\in V, i\notin V_0} | i \rangle \end{aligned}$$

(49)

and then we have $\alpha _3 =0, w_3=0$. Readers should be reminded that in 5 the matrix is written in the $\{v_1, v_3, v_2\}$ basis, instead of the $\{v_1, v_2, v_3\}$ basis.