Overview: recent development and applications of reduction and lackadaisicalness techniques for spatial search quantum walk in the near term

Abstract

The adjacency matrices of graphs provide the foundation for constructing the Hamiltonians of Continuous-Time Quantum Walks (CTQWs). Various classes of graphs have been identified to be highly reducible and the reduced Hamiltonian preserves the dynamics of the original system. This makes the CTQW implementation feasible in the near term for search problems of large size. Highly reducible Hamiltonians are desirable because existing quantum devices are of limited size in terms of the number of qubits. In this work, we review the recent developments of dimensionality reduction and coupling factor value finding techniques. The CTQWs based on a reduced Hamiltonian can search optimally when the correctly calculated coupling factor is used. We list identified highly reducible graphs and include their optimality proofs when correct coupling factors are used. In addition, we discuss the recent developments on Lackadaisical Quantum Walkers (LQW) (a type of coin-based discrete-time quantum walk) for one- and two-dimensional spatial search. The optimal lower upper bound remains open in one- and two-dimensional Discrete-Time Quantum Walk.

Appendices

Appendix A: Basis change with perturbation

Theorem 2

[43] Given a reduced 3 $\times$ 3 Hamiltonian $H_{\mathrm{seek}} = H^{(0)}+ H^{(1)}$ in the $(|\omega \rangle , |b_1\rangle , |b_2\rangle )$ basis where

$$\begin{aligned} H^{(0)} = \begin{bmatrix} -1 &{} 0 &{} 0 \\ 0 &{} 0 &{} v_1 \\ 0 &{} v_1 &{}v_3\\ \end{bmatrix} , \quad H^{(1)} = \begin{bmatrix} 0 &{} 0 &{} v_2 \\ 0 &{} 0 &{} 0 \\ v_2&{} 0 &{} 0\\ \end{bmatrix} \end{aligned}$$

(53)

$v_1$ and $v_2$ are negative numbers and $v_3$ is a non-positive number where $v_1/v_2 \ge 1$. Let the eigenvectors basis of $H^{(0)}$ be $(|\omega \rangle , |e_1\rangle , |e_2\rangle )$. Let $\kappa = \frac{v_3}{v_1} \ge 0$ and $\beta _\pm =\frac{\kappa \pm \sqrt{\kappa ^2 + 4}}{2}$, then eigenvectors

$$\begin{aligned} |e_1\rangle = \frac{(|b_1\rangle + \beta _+ |b_2\rangle )}{\sqrt{1 + \beta _+^2}}, \quad |e_2\rangle = \frac{(|b_1\rangle + \beta _- |b_2\rangle )}{\sqrt{1 + \beta _-^2}} \end{aligned}$$

(54)

with corresponding eigenvalues $\lambda _\pm = v_1 \beta _\pm$. H expressed in the $(|\omega \rangle , |e_1\rangle , |e_2\rangle )$ eigenbasis is

$$\begin{aligned} H_{\mathrm{seek}} = \begin{bmatrix} -1 &{} v_2 \frac{\beta _+}{\sqrt{\beta _+^2 +1}} &{} v_2 \frac{\beta _-}{\sqrt{\beta _-^2 +1}} \\ \frac{v_2 \sqrt{\beta _+^2 + 1 }}{\beta _+ - \beta _-} &{} \lambda _+ &{} 0 \\ \frac{-v_2 \sqrt{\beta _-^2 + 1 }}{\beta _+ - \beta _-} &{} 0 &{} \lambda _-\\ \end{bmatrix} \end{aligned}$$

(55)

Proof

Let $|e'\rangle = |b_1\rangle +\beta |b_2\rangle$ be an eigenvector of $H^{(0)}$ with eigenvalue $\lambda$. After some calculation, we obtain $\lambda _{\pm } = {\beta _{\pm } v_1}$ where

$$\begin{aligned} \beta _{\pm } = \frac{\kappa \pm \sqrt{\kappa ^2 + 4}}{2}, \quad \kappa = \frac{v_3}{v_1}. \end{aligned}$$

(56)

Let eigenvector $|e'_1\rangle = |b_1\rangle +\beta _+ |b_2\rangle$ with eigenvalue $\lambda _+$ and eigenvector $|e'_2\rangle = |b_1\rangle +\beta _- |b_2\rangle$ with eigenvalue $\lambda _-$, by renormalization we have

$$\begin{aligned} {} |e_1\rangle = \frac{|e'_1\rangle }{\sqrt{\beta _+^2 +1}}, \quad |e_2\rangle = \frac{|e'_2\rangle }{\sqrt{\beta _-^2 +1}}. \end{aligned}$$

(57)

By basis change and Eq. (53) with $H^{(1)}$ in the $(|\omega \rangle , |e_1\rangle , |e_2\rangle )$ eigenbasis, we obtain $H^{(1)}|\omega \rangle$ as

$$\begin{aligned} v_2|b_2\rangle = \frac{v_2 \sqrt{\beta _+^2 +1}}{\beta _+ -\beta _-}|e_1\rangle + \frac{-v_2 \sqrt{\beta _-^2 +1}}{\beta _+ -\beta _-}|e_2\rangle . \end{aligned}$$

(58)

Hence, the Hamiltonian $H_{\mathrm{seek}}$ can be expressed as shown in Eq. (55). $\square$

Lemma 1

[43] (Computational basis to eigenbasis) Given a derived reduced Hamiltonian $H_{\mathrm{seek}}$ written in the $(|\omega \rangle , |e_1\rangle , |e_2\rangle )$ basis as shown in Theorem 2, we then know that (a) Hamiltonian $H_{\mathrm{seek}}$ is symmetric and (b) $\beta _+> 0 > \beta _-$ and $\lambda _+ <0, \lambda _- > 0$.

Proof

With the value of $\beta _\pm$ as shown in Theorem 2, we know $\beta _+ \beta _- = -1$, we know

$$\begin{aligned} \beta _+ (\beta _+ - \beta _-) = \beta _+^2 +1,\quad \beta _-(\beta _+ - \beta _-) = -(1+ \beta _-^2), \end{aligned}$$

(59)

$$\begin{aligned} v_2 \frac{\beta _+}{\sqrt{\beta _+^2 +1}} = \frac{v_2 \sqrt{\beta _+^2 + 1 }}{\beta _+ - \beta _-}, \quad v_2 \frac{\beta _-}{\sqrt{\beta _-^2 +1}} = \frac{-v_2 \sqrt{\beta _-^2 + 1 }}{\beta _+ - \beta _-}. \end{aligned}$$

(60)

Hence, H is symmetric. For property (b), it is immediate since $\sqrt{\kappa ^2 + 4}> \kappa > 0$ and $\lambda _{\pm } = v_1\beta _{\pm }$. $\square$

By Lemma 1, $H_{\mathrm{seek}}$ can be written in the $(|\omega \rangle , |e_1\rangle ,|e_2\rangle )$ basis as

$$\begin{aligned} H_{\mathrm{seek}} = \begin{bmatrix} -1 &{} \delta _1 &{} \delta _2 \\ \delta _1&{} \lambda _+ &{} 0 \\ \delta _2 &{} 0 &{} \lambda _-\\ \end{bmatrix} . \end{aligned}$$

(61)

where

$$\begin{aligned} \delta _1 = v_2 \frac{\beta _+}{\sqrt{\beta _+^2 +1}}, \quad \delta _2 = v_2 \frac{\beta _-}{\sqrt{\beta _-^2 +1}}. \end{aligned}$$

(62)

Appendix B: Low-state transport escape rate

Theorem 3

[43] Given a Hamiltonian $H_{\mathrm{seek}}$ in the form shown in Lemma 1, it is desirable to have $\lambda _+ = -1$ such that $|\omega \rangle$ and $|e_1\rangle$ form the basis for the two states of the lowest eigenvalue. If $v_1= -\gamma N(\sqrt{\alpha (1- \alpha )})$, $v_2= -\gamma (\sqrt{N (1-\alpha )})$ then the degeneracy between site energies of $|\omega \rangle$ and $|e_1\rangle$ facilitates transport between these two low energy states, hence $\gamma = (N\sqrt{\alpha (1- \alpha )}\beta _+)^{-1}$. The transport between $|\omega \rangle$ and $|e_2\rangle$ is negligible since $\delta _2$ is much smaller than $\lambda _-$.

Proof

Since we desire to have faster transport between the lowest eigen energy states, we need to set

$$\begin{aligned} \lambda _+ = v_1\beta _+ = -1. \end{aligned}$$

(63)

With the fact that $v_1 = -\gamma N(\sqrt{\alpha (1- \alpha )})$, we need to set

$$\begin{aligned} \gamma = (N\sqrt{\alpha (1- \alpha )}\beta _+)^{-1} \end{aligned}$$

(64)

From Eq. (62) and $\lambda _- = v_1\beta _{-}$, we know $\delta _2$ is much smaller than $\lambda _-$ because

$$\begin{aligned} \frac{\delta _2}{\lambda _-} = \frac{v_2}{v_1 \sqrt{(\beta _-^2 +1)}}=\frac{1}{\sqrt{\alpha N (\beta _-^2 +1)}}< \frac{1}{\sqrt{\alpha N}}. \end{aligned}$$

(65)

$\square$