A note on perfect quantum state transfers on trees

TL;DR

This paper proves that, aside from the simple paths P2 and P3, no other tree structures support perfect quantum state transfers, clarifying the limitations of quantum communication on tree graphs.

Contribution

The paper establishes a negative result showing that only P2 and P3 trees admit perfect quantum state transfers, resolving a previously open question.

Findings

Only P2 and P3 trees admit perfect state transfers.

Other trees do not support perfect quantum state transfers.

The result narrows the class of graphs suitable for quantum communication.

Abstract

It has been asked whether there are trees other than $P_2$ and $P_3$ which can admit perfect state transfers. In this note we show that the answer is negative.

Equations

$$H_X(t)=\exp (iAt),\text{for any }t.$$

Full text

††journal: Discrete Mathematics

1 Introduction

For any simple graph $X$ with the adjacency matrix $A$ and with $|V(X)|=n$, we define the function $H_{X}(t)$ as

[TABLE]

If $X$ is clear from the context, we may just write $H(t)$. We say there is a perfect state transfer or a PST between distinct vertices $u$ and $v$ of $X$ at time $\tau$, if $|H(\tau)_{u,v}|=1$. For the motivation of this definition in designing quantum communication networks and a survey of important results, the reader may refer to [1, 2, 3].

Godsil provides a proof in [3] that there is a PST between the endpoints of the paths $P_{2}$ and $P_{3}$. Also, the following has been proved in [1].

Proposition 1.1**.**

The path $P_{n}$ has no PST for any $n\geq 4$. ∎

Therefore, Godsil asks in [3] whether there are any trees besides $P_{2}$ and $P_{3}$ on which a PST can occur. We prove that the answer is no. The main tool to do this is the following result also from [3]. Given any vertex $u$ from a graph $X$, we denote by $\Delta_{u}$ the distance partition of the vertices of $X$ with respect to $u$.

Proposition 1.2**.**

Let $u$ and $v$ be vertices in $X$. If there is perfect state transfer from $u$ to $v$, then $\Delta_{u}=\Delta_{v}$.∎

2 There is no PST on trees

In this section we prove the main result of the note.

Theorem 2.1**.**

If $T\neq P_{2},P_{3}$ is a tree, then there is no PST on $T$.

Proof.

Suppose that there is a PST on $T$ between two distinct vertices $u$ and $v$. Assume $P:u=w_{0}-w_{1}-w_{2}-\cdots-w_{r}-w_{r+1}=v$ is the unique path between $u$ and $v$. First we show that both $u$ and $v$ must be leaves. If $u$ is adjacent to a vertex $z\neq w_{1}$, then $w_{1}$ and $z$ belong to the same cell of the distance partition $\Delta_{u}$, while they are in distinct cells of the partition $\Delta_{v}$. Therefore $\Delta_{u}\neq\Delta_{v}$ which, according to Proposition 1.2, is a contradiction. Hence $u$ is a leaf and with the same argument, $v$ is a leaf as well. Then we show that indeed $T=P$. To do this, suppose (for contrary) that there is an $i\in\{1,\ldots,r\}$ such that $w_{i}$ has a neighbour $z$ other than $w_{i-1}$ and $w_{i+1}$. Then $w_{i-1}$ and $z$ belong to the same cell of $\Delta_{v}$ while they belong to distinct cells of $\Delta_{u}$. This, similarly, is a contradiction. Thus the claim is proved; that is, $T$ is a path and since $T\neq P_{2},P_{3}$, according to Proposition 1.1, $T$ cannot have a PST. ∎

Acknowledgment

The first author would like to thank the financial support the Iranian National Elites’ Foundation.