Perfect state transfer in NEPS of some graphs

ABSTRACT

Let G be a graph with adjacency matrix $A_G$. The transition matrix of G corresponding to $A_G$ is denoted as $H_{A_G}\left(t\right):=\exp (-\mathrm{i}t{A}_G)(t\in \mathbb{R},\mathrm{i}=\sqrt{-1})$. If there is some time $\tau \in \mathbb{R}$ such that $H_{A_G}(\tau {)}_{u,v}$ has unit modulus, where u and v are distinct vertices in G, then we say that G admits perfect state transfer from u to v. In this paper, we first show that a non-complete extended p-sum (NEPS) with badly decomposed factors has no perfect state transfer. And then, we prove that NEPS of a cube with odd distance has perfect state transfer when the sum of elements in its basis is not zero and that NEPS of a cube with even distance exhibits perfect state transfer if and only if there is a tuple in the basis such that it has exact one coordinate which is valued 1.

Keywords:

• Perfect state transfer
• NEPS of graphs
• badly decomposed graph
• cube

2010 Mathematics Subject Classifications:

• 05C50
• 81P68
• 15A18

Acknowledgments

We greatly appreciate the anonymous referees for their comments and suggestions.

Disclosure statement

No potential conflict of interest was reported by the authors.

Notes

1 Indeed, let $F_i$ for i=1,2,3 be as in the proof of Proposition 2.12. Then by (2(2) $H_{A_G}\left(t\right)=\sum _{k\ge 0}\frac{(-\mathtt{i}{\mathtt{)}}^{\mathtt{k}}{\mathtt{A}}_{\mathtt{G}}^{\mathtt{k}}{\mathtt{t}}^{\mathtt{k}}}{k!}=\sum _{k\ge 0}\frac{(-\mathtt{i}{\mathtt{)}}^{\mathtt{k}}\left(\sum _{\mathtt{r}\mathtt{=}\mathtt{1}}^{\mathtt{n}}{\lambda }_{\mathtt{r}}^{\mathtt{k}}{\mathtt{E}}_{\mathtt{r}}\right){\mathtt{t}}^{\mathtt{k}}}{k!}=\sum _{r=1}^n\exp (-\mathtt{i}\mathtt{t}{\lambda }_{\mathtt{r}}\mathtt{)}{\mathtt{E}}_{\mathtt{r}}\mathtt{.}$(2) ), we have $H_{A_{K_{m,n}}}(t{)}_{u,v}=\exp (-\mathrm{i}t\sqrt{mn}\left)\right(F_1{)}_{u,v}+\exp \left(\mathrm{i}t\sqrt{mn}\right)(F_2{)}_{u,v}+(F_3{)}_{u,v}.$ If m=2 and u=1, v=2, then $\begin{array}{rl}{H}_{A_{K_{2,n}}}(t{)}_{1,2}& =\frac{1}{4}\exp (-\mathrm{i}t\sqrt{2n})+\frac{1}{4}\exp \left(\mathrm{i}t\sqrt{2n}\right)-\frac{1}{2}\\ & =\frac{1}{2}\cos \left(t\sqrt{2n}\right)-\frac{1}{2},\end{array}$ which has unit modulus at $t=\pi /\sqrt{2n}$. Thus, $K_{2,n}$ has perfect state transfer at $t=\pi /\sqrt{2n}$. Note that by [5, Corollary 8] we can also get a sufficient condition for $K_{2,n}$ to have perfect state transfer.

Additional information

Funding

Supported by National Natural Science Foundation of China (NSFC) (Nos. 11571135, 11671320 and 11601431) and the China Postdoctoral Science Foundation (No. 2016M600813), the Natural Science Foundation of Shaanxi Province (No. 2017JQ1019) and the Fundamental Research Funds for the Central Universities (No. 3102018ZY038).