We would like to introduce a useful tool. For more expansive treatment, the refer is referred to [10].

Let *G* be a graph and *π* = {*C*_{1}, *C*_{2},⋯, *C*_{r}} be a partition of *V*(*X*). We say *π* is equitable if every vertex in *C*_{i} has the same numbers *b*_{ij} of neighbors in *C*_{j}. The quotient graph of *G* induced by *π*, denoted by *G*/*π*, is a directed graph with vertex set *π* and *b*_{ij} arcs from the *ith* to *jth* cells of *π*. The entries of the adjacency matrix of *G*/*π* are given by *A*(*G*/*π*)_{ij} = *b*_{ij}. We can symmetrize *A*(*G*/*π*) to *B* by letting
$\begin{array}{}{B}_{ij}=\sqrt{{b}_{ij}{b}_{ji}}.\end{array}$
We call the weighted graph with adjacency matrix *B*, the symmetrized quotient graph. In the following, we always use *B* to denote the symmetrized form of the matrix *A*(*G*/*π*). We list some known results which will be used within this paper.

#### Lemma 2.1

([3]). *Let G be a graph and let u*, *v be two vertices of G. Then there is perfect state transfer between u and v at time t with phase γ if and only if all of the following conditions hold*.

*Vertices u and v are strongly cospectral*.

*There are integers a*, △ where △ *is square*-*free so that for each eigenvalue* λ *in supp*_{G}(*u*):

$\begin{array}{}\lambda =\frac{1}{2}(a+{b}_{\lambda}\sqrt{\mathrm{\u25b3}}),\end{array}$
*for some integer b*_{λ}.

$\begin{array}{}{e}_{u}^{T}{E}_{\lambda}(G){e}_{v}\end{array}$
*is positive if and only if*
$\begin{array}{}(\rho (G)-\lambda )/g\sqrt{\mathrm{\u25b3}}\end{array}$
*is even*, *where*

$$\begin{array}{}{\displaystyle g:=gcd(\{\frac{\rho (G)-\lambda}{\sqrt{\mathrm{\u25b3}}}\}:\lambda \in Sup{p}_{G}(u))}\end{array}$$

*Moreover*, *if the above conditions hold, then the following also hold*.

*There is a minimum time of perfect state transfer between u and v given by*

$$\begin{array}{}{\displaystyle {t}_{0}:=\frac{\pi}{g\sqrt{\mathrm{\u25b3}}}}\end{array}$$

*The time of perfect state transfer t is an odd multiple of t*_{0}.

*The phase of perfect state transfer is given by γ* = *e*^{–itρ(G)}.

#### Lemma 2.2

([8). *Let G be a graph with perfect state transfer between vertices u and v*. *Then*, *for each automorphism τ* ∈ *Aut*(*G*), *τ*(*u*) = *u if and only if τ*(*v*) = *v*.

#### Lemma 2.3

([10]). *Suppose G has pretty good state transfer between vertices u and v. Then u and v are strongly cospectral*, *and each automorphism fixing u must fix v*.

#### Lemma 2.4

([2]). *Let X be a graph with an equitable π and assume* {*a*} *and* {*b*} *are singleton cells of π. Let B denote the adjacency matrix of the symmetrized quotient graph relative π. Then for any time t*,

$$\begin{array}{}{\displaystyle ({e}^{-itA(X)}{)}_{a,b}=({e}^{-itB}{)}_{\{a\},\{b\}}}\end{array}$$

*and therefore X has perfect state transfer from a to b at time t if and only if the symmetrized quotient graph has perfect state transfer from* {*a*} *to* {*b*}.

Let *S*_{k+1} and *S*_{l+1} be the two star graphs with *k* + 1 and *l* + 1 edges respectively and *w* be a vertex of degree one in *S*_{k+1} and *S*_{l+1}. Then the 1-sum of *S*_{k+1} and *S*_{l+1}, denoted by *F*_{k,l}, is obtained by merging the vertex *w* of *S*_{k+1} with the vertex *w* of *S*_{l+1} and no edge joins a vertex of *S*_{k+1}∖ *w* with a vertex of *S*_{l+1}∖ *w*.

#### Lemma 2.5

*There is no perfect state transfer from u*_{1} *to u*_{2} *on F*_{2,l} *(Fig. 1)*.

#### Proof

Let *F*_{2,l} be a graph shown in Fig. 1. Let *π* be the equitable partition with cells

$$\begin{array}{}{\displaystyle \{\{{u}_{1}\},\{{u}_{2}\},\{u\},\{w\},\{v\},N(v)\mathrm{\setminus}\{w\}\}.}\end{array}$$

Then the adjacency matrix *B* of the corresponding symmetrized quotient graph induced by *π* is

$$\begin{array}{}{\displaystyle \left(\begin{array}{cccccc}0& 0& 1& 0& 0& 0\\ 0& 0& 1& 0& 0& 0\\ 1& 1& 0& 1& 0& 0\\ 0& 0& 1& 0& 1& 0\\ 0& 0& 0& 1& 0& \sqrt{l}\\ 0& 0& 0& 0& \sqrt{l}& 0\end{array}\right).}\end{array}$$

The eigenvalue of *B* are

*θ*_{1} = 0 with multiplicity 2,

$\begin{array}{}{\theta}_{2}=\frac{\sqrt{2}}{2}\sqrt{l+4+\sqrt{{l}^{2}-4l+8}},\end{array}$

$\begin{array}{}{\theta}_{3}=-\frac{\sqrt{2}}{2}\sqrt{l+4+\sqrt{{l}^{2}-4l+8}},\end{array}$

$\begin{array}{}{\theta}_{4}=\frac{\sqrt{2}}{2}\sqrt{l+4-\sqrt{{l}^{2}-4l+8}},\end{array}$

$\begin{array}{}{\theta}_{5}=-\frac{\sqrt{2}}{2}\sqrt{l+4-\sqrt{{l}^{2}-4l+8}}.\end{array}$

If there is perfect state transfer from {*u*_{1}} to {*u*_{2}} on *F*_{2,l}/*π*, by Lemma 2.1,

$$\begin{array}{}{\displaystyle \frac{{\theta}_{2}-{\theta}_{3}}{{\theta}_{4}-{\theta}_{5}}}\end{array}$$

is rational. Therefore

$$\begin{array}{}{\displaystyle \frac{({\theta}_{2}-{\theta}_{3}{)}^{2}}{({\theta}_{4}-{\theta}_{5}{)}^{2}}=\frac{l+4+\sqrt{{l}^{2}-4l+8}}{l+4-\sqrt{{l}^{2}-4l+8}}=\frac{{l}^{2}-2l+12}{6l+4}+\frac{(l+4)\sqrt{{l}^{2}-4l+8})}{6l+4}}\end{array}$$

is rational, and hence *l*^{2}–4*l*+8 is a perfect square. This means that there exists a integer *p* such that *l*^{2}–4*l*+8 = (*l*–2)^{2} + 4 = *p*^{2}. Note that both *l*–2 and *p* are integers. This can occur only if *l* = 2. But in this case (*θ*_{2}–*θ*_{3})/(*θ*_{4}–*θ*_{5}) =
$\begin{array}{}\sqrt{2}\end{array}$
is irrational. This is a contradiction. Hence there is no perfect state transfer from {*u*_{1}} to {*u*_{2}} on *F*_{2, l}/*π* and by Lemma 2.4, there is no perfect state transfer from *u*_{1} to *u*_{2} on *F*_{2,l}.□

Next we start to study perfect state transfer between two vertices *u* and *v* on *F*_{k,k} (see Fig. 2).

#### Lemma 2.6

*There is no perfect state transfer from u to v on F*_{k,k}.

#### Proof

Suppose *F*_{k,k} be a graph shown in Fig.2. Let *π* be the equitable partition with cells

$$\begin{array}{}{\displaystyle \{N(u)\setminus \{w\},\{u\},\{w\},\{v\},N(v)\setminus \{w\}\}.}\end{array}$$

Then the adjacency matrix *B* of the corresponding symmetrized quotient graph induced by *π* is

$$\begin{array}{}{\displaystyle \left(\begin{array}{ccccc}0& \sqrt{k}& 0& 0& 0\\ \sqrt{k}& 0& 1& 0& 0\\ 0& 1& 0& 1& 0\\ 0& 0& 1& 0& \sqrt{k}\\ 0& 0& 0& \sqrt{k}& 0\end{array}\right).}\end{array}$$

The eigenvalue of *B* are

$$\begin{array}{}{\displaystyle {\theta}_{1}=0,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\theta}_{2}=\sqrt{k},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\theta}_{3}=-\sqrt{k},}\\ {\displaystyle {\theta}_{4}=\sqrt{k+2},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\theta}_{5}=-\sqrt{k+2}.}\end{array}$$

The corresponding eigenvectors are the columns of the following matrix:

$$\begin{array}{}{\displaystyle \left(\begin{array}{ccccc}1& 1& 1& \sqrt{k}& \sqrt{k}\\ 0& 1& -1& \sqrt{k+2}& -\sqrt{k+2}\\ -\sqrt{k}& 0& 0& 2& 2\\ 0& -1& 1& \sqrt{k+2}& -\sqrt{k+2}\\ 1& -1& -1& \sqrt{k}& \sqrt{k}\end{array}\right).}\end{array}$$

It is easy to check that *θ*_{i} ∈ *Supp*_{G}({*u*}) for *i* = 2,3,4,5. If there is perfect state transfer from {*u*} to {*v*} on *F*_{k,k}/*π*, by Lemma 2.1, there exist integers *m*,*n* and a square-free integer *p* such that *θ*_{2}–*θ*_{3} = *m*
$\begin{array}{}\sqrt{p}\end{array}$
and *θ*_{4}–*θ*_{5} = *n*
$\begin{array}{}\sqrt{p}\end{array}$. Then *n* > *m* ≥ 1. Now

$$\begin{array}{}{\displaystyle ({n}^{2}-{m}^{2})p={n}^{2}p-{m}^{2}p=(k+2)-k=2.}\end{array}$$

This is impossible. Hence there is no perfect state transfer from {*u*} to {*v*} on *F*_{k,k}/*π* and by Lemma 2.4, there is no perfect state transfer from *u* to *v* on *F*_{k,k}.□

#### Lemma 2.7

*w is not strongly cospectral with the other vertices*.

#### Proof

If *w* is strongly cospectral with the vertex *a*, then they must have the same degree. Thus *a* ≠ *u*_{i} (*i* = 1,2, ⋯, *k*) and *a* ≠ *v*_{j} (*i* = 1,2, ⋯, *k*). Without loss of generality, suppose *a* = *u*. Then *k* = 1. Note that *F*_{1,l}∖{*u*} = *K*_{1}∪ *S*_{l+1} and *F*_{1,l}∖ \{*w*} = *K*_{2}∪ *S*_{l}. TThese are not cospectral, hence it follows they are not strongly cospectral which in itself provides a contradiction.□

#### Theorem 2.8

*There is no perfect state transfer on F*_{k,l}.

#### Proof

We divide into four cases to discuss:

Note that if there exists perfect state transfer from *a* to *b*, then *a* and *b* must be strongly cospectral. Since *w* is not strongly cospectral with the other vertices, there is no perfect state transfer from *w* to other vertices.

If there exists perfect state transfer from *u* to *v*, then *u* and *v* must have the same degree. This means that *k* = *l*. By Lemma 2.7, there is no perfect state transfer from *u* to *v* on *F*_{k,k}.

If there exists perfect state transfer from *u*_{i} to *u*_{j}, by Lemma 2.2, *k* = 2. By Lemma 2.5, there is no perfect state transfer from *u*_{1} to *u*_{2} on *F*_{2,l}. A similar argument will show that there is no perfect state transfer from *v*_{i} to *v*_{j}.

If there exists perfect state transfer from *u*_{i} to *v*_{j}, by Lemma 2.2, *k* = *l* = 1. Then the graph is *P*_{5}. We know that there is no perfect state transfer in *P*_{5} from [5].□

## Comments (0)

General note:By using the comment function on degruyter.com you agree to our Privacy Statement. A respectful treatment of one another is important to us. Therefore we would like to draw your attention to our House Rules.