Exceptional Configurations of Quantum Walks with Grover’s Coin

Abstract

We study search by quantum walk on a two-dimensional grid using the algorithm of Ambainis, Kempe and Rivosh [AKR05]. We show what the most natural coin transformation — Grover’s diffusion transformation — has a wide class of exceptional configurations of marked locations, for which the probability of finding any of the marked locations does not grow over time. This extends the class of known exceptional configurations; until now the only known such configuration was the “diagonal construction” by [AR08].

N. Nahimovs is supported by EU FP7 project QALGO, A. Rivosh is supported by ERC project MQC.

A General Graphs

In this appendix, we overview the stationary states of quantum walks with Grover’s coin for general graphs.

Quantum Walks on a General Graph

Consider a graph $G = (V, E)$ with a set of vertices V and a set of edges E. Let $n = |V|$ and $m = |E|$. Let N(x) be a neighbourhood of a vertex x, that is a set of vertices x is adjacent to. We define a location register with n basis states $| i \rangle$ for $i \in \{1,\dots ,n\}$ and a direction or coin register, which for a vertex $v_i$ has $d_i = \deg (v_i)$ basis states $| j \rangle$ for $j \in N(v_i)$. The state of the quantum walk is given by:

$$\begin{aligned} | \psi (t) \rangle = \sum _{i=1}^{n} \sum _{j \in N(v_i)} \alpha _{i,j} | i,j \rangle . \end{aligned}$$

A step of the quantum walk is performed by first applying $I \otimes C$, where C is a unitary transformation on the coin register. The usual choice of transformation on the coin register is Grover’s diffusion transformation D. Then, we apply the shift transformation S:

$$\begin{aligned} S = \sum _{i=1}^{n} \sum _{j \in N(v_i)} | j,i \rangle \langle i,j | , \end{aligned}$$

which for each pair of connected vertices i, j swaps an amplitude of vertex i pointing to j with an amplitude of vertex j pointing to i.

We start the quantum walk in the equal superposition over all pairs vertex-direction:

$$\begin{aligned} | \psi _0 \rangle = \frac{1}{\sqrt{n \cdot \deg (G)}} \sum _{i=1}^{n} \sum _{j \in N(v_i)} | i,j \rangle , \end{aligned}$$

where $\deg (G) = \sum _{i} \deg (v_i)$. It can be easily verified that the state of the walk stays unchanged, regardless of the number of steps.

To use the quantum walk as a tool for search, we mark some vertices. For the unmarked vertices, we apply the same transformations as above. For the marked vertices, we apply $-I$ instead of D as the coin flip transformation. The shift transformation remains the same for both the marked and unmarked vertices.

Another way to look at a step of the algorithm is that we first perform a query Q transformation, which flips signs of amplitudes of marked vertices, then conditionally perform the coin transformation (I or D depending on whether a vertex is marked or not) and then perform the shift transformation S. In case of the Grover’s coin the step of the algorithm is the query Q followed by D followed by S.

Stationary States of the Quantum Walk with Grover’s Coin for General Graphs

Consider a graph $G = (V, E)$ with two marked vertices $v_i$ and $v_j$. Let vertices be connected and let each of them be connected to some other k vertices. Let $| \phi _{stat}^a \rangle$ be a state having amplitudes of all basis states except $| i,j \rangle$ and $| j,i \rangle$ equal to a and amplitudes of basis states $| i,j \rangle$ and $| j,i \rangle$ equal to $-ka$ (see Fig. 5). Then this state is not changed by a step of the algorithm with Grover’s coin.

Fig. 5.

Symmetric stationary state for 2 marked vertices.

Theorem 2

Let $G = (V,E)$ be a graph with two marked vertices i and j; let $(v_i,v_j) \in E$ and $N(v_i) = N(v_j) = k + 1$; and let

$$\begin{aligned} | \phi _{stat}^a \rangle = \sum _{i=1}^{n} \sum _{j \in N(v_i)} | i,j \rangle - (k+1)a (| i,j \rangle - | j,i \rangle ). \end{aligned}$$

Then, $| \phi _{stat}^a \rangle$ is an eigenstate of a step of the quantum walk on G with Grover’s coin.

Fig. 6.

Symmetric stationary state for 3 marked vertices.

Proof

Consider the effect of a step of the algorithm on $| \phi _{stat}^a \rangle$. The query transformation changes the signs of all amplitudes of the marked vertices. The coin flip performs an inversion above the average: for unmarked vertices it does nothing as all amplitudes are equal to a; for marked vertices the average is 0, so the inversion results in sign flip. Thus, CQ does nothing for amplitudes of the unmarked vertices and twice flips the sign of amplitudes of the marked vertices. Therefore, we have

$$\begin{aligned} CQ| \phi _{stat}^a \rangle = | \phi _{stat}^a \rangle . \end{aligned}$$

The shift transformation swaps amplitudes of adjacent vertices. For $| \phi _{stat}^a \rangle$, it swaps a with a and $-ka$ with $-ka$. Thus, we have

$$\begin{aligned} SCQ| \phi _{stat}^a \rangle = | \phi _{stat}^a \rangle . \end{aligned}$$

   $\square$

The initial state of the algorithm $| \psi _0 \rangle$ can be written as

$$\begin{aligned} | \psi _0 \rangle = \phi _{stat}^a + (k+1)a(| i,j \rangle + | j,i \rangle ), \end{aligned}$$

for $a=1/\sqrt{n \cdot \deg (G)}$. Therefore, the only part of the initial state, which is changed by a step of the algorithm, is

$$\begin{aligned} \frac{k+1}{\sqrt{n \cdot \deg (G)}}(| i,j \rangle + | j,i \rangle ). \end{aligned}$$

Next figures show similar constructions for three (Fig. 6) and four (Fig. 7) marked vertices. We give them without a proof (which is similar to the two marked vertex case). It is easy to see how one can extend the construction to any number of marked vertices.

Fig. 7.

Symmetric stationary state for 4 marked vertices.

Fig. 8.

Generic stationary state for 3 marked vertices.

The above constructions are symmetric in the sense that each of the marked vertices has the same number of neighbours. One can also construct a stationary state without this restriction. The Fig. 8 shows the general stationary state of three marked locations. The parameters of the construction (number of adjacent vertices) are restricted by Eq. 1.

$$\begin{aligned} {\left\{ \begin{array}{ll} l_{12} + l_{12} = m_1 \\ l_{21} + l_{23} = m_2 \\ l_{31} + l_{32} = m_3 \\ l_{12} = l_{21} \\ l_{23} = l_{32} \\ l_{31} = l_{13} \end{array}\right. }. \end{aligned}$$

(1)

For example, for $l_{12} = l_{21} = 1$, $l_{23} = l_{32} = 2$ and $l_{31} = l_{13} = 3$ we will have $m_1 = 4$, $m_2 = 3$ and $m_3 = 5$.

Again, it is easy to see how one can extend the construction to any number of marked vertices.