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A comfortable graph structure for Grover walk

Yusuke Higuchi Department of Mathematics, Gakushuin University, Tokyo 171-8588, Japan Mohamed Sabri and Etsuo Segawa Graduate School of Information Sciences, Tohoku University, Aoba, Sendai 980-8579, Japan Graduate School of Environment and Information Sciences, Yokohama National University, Hodogaya, Yokohama 240-8501, Japan

Abstract. We consider a Grover walk model on a finite internal graph, which is connected with a finite number of semi-infinite length paths and receives the alternative inflows along these paths at each time step. After the long time scale, we know that the behavior of such a Grover walk should be stable, that is, this model has a stationary state. In this paper our objectives are to give some characterization upon the scattering of the stationary state on the surface of the internal graph and upon the energy of this state in the interior. For the scattering, we concretely give a scattering matrix, whose form is changed depending on whether the internal graph is bipartite or not. On the other hand, we introduce a comfortability function of a graph for the quantum walk, which shows how many quantum walkers can stay in the interior, and we succeed in showing the comfortability of the walker in terms of combinatorial properties of the internal graph.

Keywords: Quantum walk, (pseudo-)Kirchhoff’s laws, (signless-)Laplacian, Comfortability

1 Introduction

As discrete-time quantum walks on graphs are being studied, many interesting behaviors of quantum walks, which cannot be observed in the classical random walk setting, have become apparent: the accomplishment of quantum speed up in quantum algorithm, linear spreading, localization, periodicity, pseudo-perfect state transfer and so on (see [18, 20, 22] and references therein). Among them it is revealed that behaviors of quantum walks are closely related to geometric features of graphs; for examples, cycle geometry of graphs gives the localization [14, 16], a three-edge-coloring induces an eigenfunction of some quantum walks [21], and the rotation systems and 1-factorizations of graphs are reflected on the mixing time of quantum walks [12]. For a family of graphs with fractal structure, the spectral analysis reflects on its property for the hopping matrix [7] and the continuous-time quantum walk [3] are studied. In [9], the scattering for a tree with two semi-infinite lines appended is discussed under the time evolution of continuous-time quantum walk to investigate the quantum exponentially speed up of the hitting time to the marked vertex. In this paper, we observe what property of general graphs are extracted from the structure of the stationary state of discrete-time quantum walks. The stationary state, which is a kind of the long time behavior of quantum walk, has been discussed in [10, 11, 17].

Let us explain our setting. For a connected and locally finite graph G=(V,E)G=(V,E), which may be infinite, let us define the set of symmetric arcs AA induced by EE as follows: for any undirected edge eEe\in E with end vertices u,vVu,v\in V, the induced arcs are aa and a¯\overline{a}, which are arcs from uu to vv and vv to uu along the edge eEe\in E, respectively. The terminus and origin vertices of aAa\in A are denoted by t(a),o(a)Vt(a),\,o(a)\in V, respectively. The total space of quantum walk treated here is denoted by A\mathbb{C}^{A}, which is a vector space whose standard basis are labeled by AA. In other words, A{ψ:A}\mathbb{C}^{A}\cong\{\psi:A\to\mathbb{C}\} is spanned by {δa:aA}\left\{\delta_{a}:a\in A\right\}, where δa()\delta_{a}(\cdot) is the delta function such that

δa(a)={1if a=a,0otherwise.\delta_{a}(a^{\prime})=\begin{cases}1&\text{if }a=a^{\prime},\\ 0&\text{otherwise}.\end{cases}

The time evolution operator of a quantum walk U:AAU:\mathbb{C}^{A}\to\mathbb{C}^{A} is defined by

U=SC.U=SC.

Here Sδa=δa¯S\delta_{a}=\delta_{\overline{a}} and C=vVCvC=\bigoplus_{v\in V}C_{v}, where CvC_{v} is a local coin operator assigned at vertex vv which is a deg(v)\text{deg}(v)-dimensional unitary matrix on span{δa:t(a)=v}\text{span}\left\{\delta_{a}:t(a)=v\right\}. Note that since the time evolution operator UU is unitary in the sense that UU=IUU^{*}=I, then for any ψA\psi\in\mathbb{C}^{A}, the l2l^{2} norm is preserved, that is, Uψ2=ψ2=aA|ψ(a)|2||U\psi||^{2}=||\psi||^{2}=\sum_{a\in A}|\psi(a)|^{2} for the standard inner product. Let ψn\psi_{n} be the nn-th iteration of UU with the initial state ψ0\psi_{0}; that is, ψn+1=Uψn\psi_{n+1}=U\psi_{n}. We call ψ(a):=limnψn(a)\psi_{\infty}(a):=\lim_{n\to\infty}\psi_{n}(a) for any aAa\in A, if it exists, the stationary state. Another type of stationarity of quantum walks, the stationary measure, is discussed in [19], whose form is as follows: μ(ψn+1)(u):=t(a)=u|ψn+1(a)|2=t(a)=u|ψn(a)|2\mu(\psi_{n+1})(u):=\sum_{t(a)=u}|\psi_{n+1}(a)|^{2}=\sum_{t(a)=u}|\psi_{n}(a)|^{2} for any time step n=0,1,2,n=0,1,2,\dots and vertex uVu\in V.

We are interested in how the state converges to the stationary state as a fixed point of a dynamical system. In general, the state of the walker in a quantum walk on a finite graph G0=(V0,A0)G_{0}=(V_{0},A_{0}) does not necessarily converge because the time evolution operator is unitary and hence the absolute values of the eigenvalues of the operator are equal to 11. Then we connect semi-infinite paths, say tails, to a subset of vertices of δVV0\delta V\subset V_{0} in the finite graph G0G_{0}, which is called the surface. Moreover the ll^{\infty}-initial state Φ0\Phi_{0} is set so that a quantum walker along the tails come into the internal and a quantum walker goes out from that at every time step. See (1) for the detailed definition of the initial state and also Fig. 1. It is expected that the outflow balances the inflow in the long time limit and this quantum walk on this graph goes to a steady state. Indeed the convergence of states, in this sense, is shown in [10, 11, 17]. In this setup, the state space of the walk is described by ll^{\infty}-space. This quantum walk always converges to a stationary state whenever we choose vertices of the internal, to which the tails are connected.

In [15, 17], the Grover walk whose quantum coins are expressed by the Grover matrices {Gr(deg(u)}uV\left\{\mathrm{Gr}(\text{deg}(u)\right\}_{u\in V} was applied with the constant inflow. Then the scattering on the surface δV\delta V, which is a set of vertices connected to tails, in the long time limit recovers the local scattering at each vertex; that is, the scattering matrix is Gr(r)\mathrm{Gr}(r), where rr is the number of tails. This implies that we can obtain the global scattering by only some information on the surface, especially the number rr, whereas this also implies that we cannot obtain any information on the structure of the internal graph G0G_{0}.

Our motivation is to investigate how the geometrical structure of the internal graph affects the behavior of quantum walkers in terms of their stationary states. For this purpose, we must change something in our setting; for example, we replace a constant flow, which was studied in [15, 17], with some oscillated one. In this study, as a first trial, we focus on the simplest oscillation of the inflow with alternating signs which is a coarse graining of alternating current input. Our interest is to find a graph geometry from the scattering on the surface and also the energy in the interior in the long time limit. See Figure 1.

Refer to caption
Figure 1: The time evolution of the quantum walk with the alternative inflow: The surface of the internal graph is the set of vertices in K3K_{3} connected to the tails. At time n=0n=0, we set the initial state Φ0\Phi_{0} with alternating signs on the tails represented by 𝜶=[α1,α2]=[9,9]\bm{\alpha}=[\alpha_{1},\alpha_{2}]^{\top}=[9,9]^{\top}. The time evolution is determined by the Grover matrices assigned at each vertex: for each vertex uu and each time step, the transmitting weight is 2/deg(u)2/\deg(u) while the reflection weight is 2/deg(u)12/\deg(u)-1. Then on the tails, the dynamics is free because deg(u)=2\deg(u)=2 while in the internal graph, the time evolution is not trivial. For example, see the middle of the figure, at time n=1n=1 and at the vertex connecting to the left tail, say uu_{*}, a quantum walker who came from the tail with the amplitude “99” at the previous time is transmitted to the internal graph with the amplitude 9×2/deg(u)=9×2/3=69\times 2/\deg(u_{*})=9\times 2/3=6, while reflected to the tails with the amplitude 9×(2/deg(u)1)=39\times(2/\deg(u_{*})-1)=-3. Because of the setting of the initial state and free dynamics on the tails, the internal graph receives the inflow from the tails and also radiates the outflow to the tails at every time step.

The scattering on the surface gives how the inflows map to the outflows on the surface. This answer is given in following.

Theorem 1.

Let 𝛂\bm{\alpha} and 𝛃\bm{\beta} represent the inflow and outflow of the stationary state on the surface δV\delta V. Their details can be seen in Definition 1. Then we have

𝜷=σ𝜶,\bm{\beta}=\sigma\bm{\alpha},

where the scattering matrix σ\sigma is unitary and described by

σ={IG0 is non-bipartite,τG0 is bipartite.\sigma=\begin{cases}I&\text{: $G_{0}$ is non-bipartite,}\\ \tau&\text{: $G_{0}$ is bipartite.}\end{cases}

Here II is the identity matrix and τ\tau is described as in Section 3.

From this theorem, if the interior is non-bipartite, the scattering is the perfect reflection, while if the interior is bipartite, the scattering is described by the Grover matrix. Then in contrast with results in [15, 17], we succeed in extracting a structure of the internal graph from the scattering on the surface; that is, the bipartiteness.

Next it is natural to ask what happens to the interior. To answer this question, we introduce the idea of comfortability. The comfortability is the function of the interior, and gives how quantum walkers accumulate in the internal graph in the long time limit, which is the energy stored the interior. The notion of the comfortability for quantum walk is an analogy of the energy in the electric circuit [8]. The detailed definition of the comfortability QW(G0)\mathcal{E}_{QW}(G_{0}) is described in Definition 2. We obtain that when only two tails are connected, the comfortability of the graph can be expressed in terms of the geometric information of the graph as follows.

Theorem 2.

Assume the number of tails is 22, and the inflow 𝛂=[α1,α2]=[1,0]\bm{\alpha}=[\alpha_{1},\alpha_{2}]^{\top}=[1,0]^{\top}. Then the comfortability εQW(G0)\mbox{\Large$\varepsilon$}_{QW}(G_{0}) of the quantum walk is given by

εQW(G0)={(χ2(G0)/χ1(G0)+|E0|)/4 : G0 is bipartite,ι2(G0)/ι1(G0) : G0 is non-bipartite.\mbox{\Large$\varepsilon$}_{QW}(G_{0})=\begin{cases}\left(\chi_{2}(G_{0})/\chi_{1}(G_{0})+|E_{0}|\right)/4&\text{ : }G_{0}\text{ is bipartite,}\\ \iota_{2}(G_{0})/\iota_{1}(G_{0})&\text{ : }G_{0}\text{ is non-bipartite.}\end{cases}

Here |E0||E_{0}| is the number of edges of G0G_{0}, χ1(G0)\chi_{1}(G_{0}) is the number of the spanning trees of G0G_{0} and χ2(G0)\chi_{2}(G_{0}) is the number of the spanning forests of G0G_{0} with two components in which one contains u1u_{1} and the other contains unu_{n}. The geometric quantities of ι1(G0)\iota_{1}(G_{0}) and ι2(G0)\iota_{2}(G_{0}) are defined in Section 3.

Under the condition where the outflow balances the inflow whose amplitude is constant, we can control the amount of energy stored in the interior by tuning the oscillation of the inflow according to the geometric structure. We illustrate the ranking of the comfortability among all of the graphs with four vertices in Section 4. To show the above theorem, we first define a (pseudo-)current function [15, 17] on a (non-)bipartite graph and we show that these functions satisfy the (pseudo-)Kirchhoff laws in Theorems 5 and 6, respectively. Secondly, we obtain a potential function with respect to these (pseudo-)current functions, in terms of (signless-)Laplacian matrix in [2, 4] and Theorem 8, respectively. Thirdly, using these expressions of the potential functions, we obtain expressions for the comfortability in terms of the (non-)oriented incident matrix and the (signless-)Laplacian matrix, when the graph is (non-)bipartite in Propositions 1 and 2. Finally, using the speciality of the setting, we obtain the expression of the comfortability using the graph factors induced quantum walks defined by Definition 3.

We arrange the rest of this paper in the following way. In section 2, we introduce the setting, notations and some previously known results in our model. Moreover, we define the scattering matrix of the walk and a comfortability function of the graph. In section 3, we introduce our main results of the paper: Theorems 1 and 2 which tells information on the scattering matrix and the comfortability function, respectively. These theorems depends on the bipartiteness of graphs. Furthermore, we note that the theorem on the comfortability gives a method to compute the comfortability of the graph in terms of the geometric information of the graph. In section 4, we give an example to illustrate the comfortability of the graphs with 4 vertices and we show the comparison of the comfortabilities. In section 5, we give the proofs of our main theorems, Theorems 1 and 2.

2 Setting

Let G0=(V0,E0)G_{0}=(V_{0},E_{0}) be a finite connected graph and A0A_{0} be the symmetric arc set induced by E0E_{0}. We choose the boundary(surface) of G0G_{0}, δVV0\emptyset\neq\delta V\subset V_{0} with δV={v1(0),,vr(0)}\delta V=\{v_{1}^{(0)},...,v_{r}^{(0)}\}, where vi(0)vj(0)v_{i}^{(0)}\neq v_{j}^{(0)} if and only if iji\neq j. Let {j:j=1,,r}\left\{\mathbb{P}_{j}:j=1,...,r\right\} be the set of semi-infinite length paths called the tails each end vertex of which is denoted by vj(0)v_{j}^{(0)}, and each of which is connected to the finite graph G0G_{0} such that V(j)={vj(0)vj(1)vj(2)}V(\mathbb{P}_{j})=\left\{v_{j}^{(0)}\sim v_{j}^{(1)}\sim v_{j}^{(2)}\sim...\right\}. Here uvu\sim v means that the vertices uu and vv are adjacent. We denote the constructed graph by G~=(V~,E~)\tilde{G}=(\tilde{V},\tilde{E}). We also denote the arc set induced by E0E_{0} and E~\tilde{E} by A0A_{0} and A~\tilde{A} respectively. For any arc a=(u,v)A~a=(u,v)\in\tilde{A}, we write a¯=(v,u)\overline{a}=(v,u), o(a)=uo(a)=u and t(a)=vt(a)=v. Remark that o(a¯)=t(a)o(\overline{a})=t(a) and t(a¯)=o(a)t(\overline{a})=o(a).

For a discrete set Ω\Omega, let us define Ω\mathbb{C}^{\Omega} by the vector space whose basis are labeled by each element of Ω\Omega. The total state space associated with the quantum walk treated here is A~\mathbb{C}^{\tilde{A}}. We define the time evolution operator WW on A~\mathbb{C}^{\tilde{A}} in the matrix form by

(W)a,b={(2deg(o(a))δab¯) if o(a)=t(b),0otherwise,(W)_{a,b}=\begin{cases}\left(\dfrac{2}{\text{deg}(o(a))}-\delta_{a\overline{b}}\right)&\text{ if }\,o(a)=t(b),\\ 0&\text{otherwise},\end{cases}

which is so called the Grover walk. Note that the walk becomes “free” on the tails; that is,

(W)a,b={1o(a)=t(b)ab¯,0: otherwise(W)_{a,b}=\begin{cases}1&\text{: $o(a)=t(b)$, $a\neq\overline{b}$,}\\ 0&\text{: otherwise}\end{cases}

for any o(a)V0o(a)\notin V_{0}. We set the ll^{\infty}-initial state by using a complex value zz with |z|=1|z|=1:

Φ0(a)={zdist(vj(0),t(a))αjif o(a)=vj(s+1),t(a)=vj(s),s=0,1,2,,j=1,2,,r,0otherwise.\Phi_{0}(a)=\begin{cases}z^{-\text{dist}(v_{j}^{(0)},\;t(a))}\alpha_{j}&\text{if }o(a)=v_{j}^{(s+1)},\,t(a)=v_{j}^{(s)},\,s=0,1,2,...,\,j=1,2,...,r,\\ \\ 0&\text{otherwise}.\end{cases} (1)

Here αj\alpha_{j}\in\mathbb{C} (j=1,,r)(j=1,\dots,r) and dist(u,v)\mathrm{dist}(u,v) is the shortest length of paths in G~\tilde{G} between vertices uu and vv. Then quantum walkers inflows into the internal graph G0G_{0} at every time step nn from the tails. On the other hand, a quantum walker outflows towards the tails from the internal graph.

Let ΦnA~\Phi_{n}\in\mathbb{C}^{\tilde{A}} be the nn-th iteration of the quantum walk such that Φn+1=WΦn\Phi_{n+1}=W\Phi_{n}. Since the inflow oscillates with respect to the time step, the total state does not converge in the long time limit. We put Φn:=znΦn\Phi_{n}^{\prime}:=z^{n}\Phi_{n}, which satisfies Φn+1=zWΦn\Phi_{n+1}^{\prime}=zW\Phi_{n}^{\prime}. The convergence of Φn\Phi_{n}^{\prime} is ensured by [17] as follows.

Theorem 3 ([17]).

Φ:=limnΦn\Phi_{\infty}^{\prime}:=\lim\limits_{n\rightarrow\infty}\Phi_{n}^{\prime} exists; that is, WΦ=z1ΦW\Phi_{\infty}^{\prime}=z^{-1}\Phi_{\infty}^{\prime}.

Let us focus on the dynamics restricted to the internal graph. To this end, we define the boundary operator of A0A_{0}, χ:A~A0\chi:\mathbb{C}^{\tilde{A}}\rightarrow\mathbb{C}^{A_{0}} by,

(χf~)(a)=f~(a),aA0(\chi\tilde{f})(a)=\tilde{f}(a),a\in A_{0}

for any f~A~\tilde{f}\in\mathbb{C}^{\tilde{A}}. The adjoint χ:A0A~\chi^{*}:\mathbb{C}^{A_{0}}\rightarrow\mathbb{C}^{\tilde{A}} is described by

(χf)(a)={f(a)ifaA0,0otherwise(\chi^{*}f)(a)=\begin{cases}f(a)&\text{if}\,a\in A_{0},\\ 0&\text{otherwise}\end{cases}

for any fA0f\in\mathbb{C}^{A_{0}}. Remark that χχ:A0A0\chi\chi^{*}:\mathbb{C}^{A_{0}}\rightarrow\mathbb{C}^{A_{0}} is the identity operator on A0\mathbb{C}^{A_{0}} and χχ:A~A~\chi^{*}\chi:\mathbb{C}^{\tilde{A}}\rightarrow\mathbb{C}^{\tilde{A}} is the projection operator on A~\mathbb{C}^{\tilde{A}} with respect to A0A_{0}. Putting χΦn=:ϕn\chi\Phi_{n}^{\prime}=:\phi_{n}^{\prime} and χWχ=:E\chi W\chi^{*}=:E, we have

ϕn+1=zEϕn+ρ,ϕ0=0,\phi_{n+1}^{\prime}=zE\phi_{n}^{\prime}+\rho,\;\phi_{0}^{\prime}=0, (2)

where ρ=χWΦ0\rho=\chi W\Phi_{0}. The dynamical system given by (2) can be also accomplished by the restriction to the internal graph of the following alternating quantum walk for z=1z=-1 case: the time evolution operator of the alternating quantum walk U:A~A~U:\mathbb{C}^{\tilde{A}}\rightarrow\mathbb{C}^{\tilde{A}} is defined in the matrix form by

(U)a,b={(1)𝟙V0(o(a))(2deg(o(a))δab¯) if o(a)=t(b),0otherwise,(U)_{a,b}=\begin{cases}(-1)^{\mathbbm{1}_{V_{0}}(o(a))}\left(\dfrac{2}{\text{deg}(o(a))}-\delta_{a\overline{b}}\right)&\text{ if }\,o(a)=t(b),\\ 0&\text{otherwise},\end{cases} (3)

where 𝟙V0\mathbbm{1}_{V_{0}} is the characteristic function of V0V_{0}. The quantum coin assigned at every vertex in the internal graph is the “signed” Grover matrix. The initial state of the walker is

Ψ0(a)={αjifo(a)=vj(s+1),t(a)=vj(s),s=0,1,2,,j=1,,r,0otherwise.\Psi_{0}(a)=\begin{cases}\alpha_{j}&\text{if}\,o(a)=v_{j}^{(s+1)},\,t(a)=v_{j}^{(s)},\,s=0,1,2,...,\,j=1,...,r,\\ \\ 0&\text{otherwise}.\end{cases} (4)

In this paper, we consider Ψn+1=UΨn\Psi_{n+1}=U\Psi_{n} instead of Φn\Phi_{n}. Since Ψ:=limnΨn\Psi_{\infty}:=\lim_{n\to\infty}\Psi_{n} exists and UΨ=ΨU\Psi_{\infty}=\Psi_{\infty} holds according to [17], we compute its stationary state. In particular, we focus on the following quantities.

Definition 1.

Scattering matrix: Let 𝛂:=[α1,α2,,αr]T,𝛃:=[β1,β1,,βr]T\bm{\alpha}:=\left[\alpha_{1},\alpha_{2},...,\alpha_{r}\right]^{T},\,\bm{\beta}:=\left[\beta_{1},\beta_{1},...,\beta_{r}\right]^{T}, where βj=Ψ(a)\beta_{j}=\Psi_{\infty}(a) with o(a)=vj(0),t(a)=vj(1)o(a)=v_{j}^{(0)},\,t(a)=v_{j}^{(1)}. The scattering matrix σ\sigma, which is a rr-dimensional unitary matrix, is defined by

𝜷=σ𝜶.\bm{\beta}=\sigma\bm{\alpha}.

The existence of such a unitary matrix is ensured by [10, 11]. If z=1z=1, the scattering matrix gives us only the information of the number of tails because the scattering matrix for z=1z=1 is expressed by Gr(r)\mathrm{Gr}(r) [15]. In this paper, if z=1z=-1, we obtain the information on the internal graph in Theorem 1. We are also interested in the stationary state in the internal graph, especially how many quantum walkers exist; that is, how quantum walkers feel comfortable to the graph.

Definition 2.

The comfortability of G0G_{0} with 𝛂\bm{\alpha} with δV\delta V for quantum walker :

QW(G0;𝜶,δV)=12aA0|Ψ(a)|2.\mathcal{E}_{QW}(G_{0};\bm{\alpha},\delta V)=\frac{1}{2}\sum_{a\in A_{0}}|\Psi_{\infty}(a)|^{2}.

We extract some geometric graph structures from these quantities which derive from some quantum effects in Theorem 2.

3 Main Results

Now we state a main theorem concerning the scattering on the surface, which gives a quantum walk method to characterize bipartite graphs. For example, let us consider a graph with two tails and we set the inflow one of them. According to the following theorem, we can determine the bipartiteness of the graph from the scattering way: if the scattering is the perfect reflecting, then the graph is non-bipartite, while the scattering is the perfect transmitting, then the graph is bipartite.

Theorem 1.

(Scattering on the surface) Assume the time evolution operator is described by (3). For the stationary state Ψ\Psi_{\infty}, let 𝛂:=[α1,α2,,αr]T\bm{\alpha}:=\left[\alpha_{1},\alpha_{2},...,\alpha_{r}\right]^{T} and 𝛃:=[β1,β1,,βr]T\bm{\beta}:=\left[\beta_{1},\beta_{1},...,\beta_{r}\right]^{T}, where αj\alpha_{j} is the inflow described by (4) and βj\beta_{j} is the outflow described by βj=Ψ(a)\beta_{j}=\Psi_{\infty}(a) with o(a)=vj(0),t(a)=vj(1)o(a)=v_{j}^{(0)},\,t(a)=v_{j}^{(1)}. Then the scattering matrix, which is an rr-dimensional unitary matrix; 𝛃=σ𝛂\bm{\beta}=\sigma\bm{\alpha}, is expressed as follows.

σ={IG0 is non-bipartite,τG0 is bipartite.\sigma=\begin{cases}I&\text{: $G_{0}$ is non-bipartite,}\\ \tau&\text{: $G_{0}$ is bipartite.}\end{cases}

Here II is the identity matrix and τ\tau is described as follows:

τ=[Ik00Irk]Gr(r)[Ik00Irk]\tau=-\begin{bmatrix}I_{k}&0\\ 0&-I_{r-k}\end{bmatrix}\mathrm{Gr}(r)\begin{bmatrix}I_{k}&0\\ 0&-I_{r-k}\end{bmatrix}

with the computational basis labeled by {v1(0),,vk(0),vk+1(0),,vr(0)}\left\{v_{1}^{(0)},...,v_{k}^{(0)},v_{k+1}^{(0)},...,v_{r}^{(0)}\right\}, where v1(0),,vk(0)XδVv_{1}^{(0)},...,v_{k}^{(0)}\in X\cap\delta V and vk+1(0),,vr(0)YδVv_{k+1}^{(0)},...,v_{r}^{(0)}\in Y\cap\delta V.

Remark 1.

If we set the inflow to a bipartite graph with the partite sets XX and YY so that total inflow to the partite set XX coincides with that to the partite set YY, the perfect reflecting happens. This means that we can not detect the bipartiteness of graph with such an initial state. These inputs vectors are described by eigenvectors of the scattering matrix τ\tau with the eigenvalue 11.

We express the total energy in the interior, comfortability, in terms of the combinatorial geometry of finite graphs. To state our claim, we prepare some notations and settings. The odd unicyclic graph is the graph which contains only one cycle whose length is odd. Now let us consider the case where two tails are connected to G0G_{0} at u1u_{1} and unu_{n} (u1un)(u_{1}\neq u_{n}), and set the inflow to u1u_{1} and unu_{n} by 𝜶=(α1,α2)=(1,0)\bm{\alpha}=(\alpha_{1},\alpha_{2})=(1,0). Here nn is the number of vertices of G0G_{0}. In this case, the comfortability is described by QW(G0;𝜶,δV)=:QW(G0;u1,un)\mathcal{E}_{QW}(G_{0};\bm{\alpha},\delta V)=:\mathcal{E}_{QW}(G_{0};u_{1},u_{n}).

Definition 3.

(Important graph factors) Let χ1(G0)\chi_{1}(G_{0}) be the number of spanning trees of G0G_{0} and χ2(G0;u1,un)\chi_{2}(G_{0};u_{1},u_{n}) be the number of spanning forests of G0G_{0} with exactly two components, one containing u1u_{1} and the other containing unu_{n}. Here the isolate vertex is regarded as a tree. Define the set of odd unicyclic spanning subgraphs of G0G_{0} by

𝒞o=r1{{C1,,Cr}|Ci’s are odd unicyclic graphs}\mathcal{C}_{o}=\cup_{r\geq 1}\;\{\{C_{1},...,C_{r}\}\;|\;C_{i}\text{'s are odd unicyclic graphs}\}

and the set of spanning subgraphs of G0G_{0} whose one component is a tree TT which contains u1u_{1} and the remaining components are odd unicyclic graphs by

𝒯𝒞o=k0{{T,C1,,Ck}|u1T where T is a tree and Ci’s are odd unicyclic graphs}.\mathcal{T}\mathcal{C}_{o}=\cup_{k\geq 0}\;\{\{T,C_{1},...,C_{k}\}\;|\;u_{1}\in T\text{ where }T\text{ is a tree and }C_{i}\text{'s are odd unicyclic graphs}\}.

Here the situation k=0k=0 is the set of spanning trees of G0G_{0}. See Figure 2. Now define the functions ι1\iota_{1} and ι2\iota_{2} by

ι1(G0)=H𝒞o4ω(H)\iota_{1}(G_{0})=\sum\limits_{H\in\mathcal{C}_{o}}4^{\omega(H)}

and

ι2(G0;u1)=H𝒯𝒞o4ω(H)1,\iota_{2}(G_{0};u_{1})=\sum\limits_{H\in\mathcal{T}\mathcal{C}_{o}}4^{\omega(H)-1},

where ω(H)\omega(H) is the number of components in HH.

Then with the above notations, we have another main theorem of our paper as follows.

Theorem 2.

(Comfortability in the interior) Assume the number of tails is 22, and the inflow 𝛂=(α1,α2)=(1,0)\bm{\alpha}=(\alpha_{1},\alpha_{2})=(1,0) at u1u_{1} and unu_{n}, respectively. Then the comfortability of the quantum walk (3) is given by

εQW(G0;u1,un)={14(χ2(G0;u1,un)χ1(G0)+|E0|) if G0 is bipartite,ι2(G0;u1)ι1(G0) if G0 is non-bipartite.\mbox{\Large$\varepsilon$}_{QW}(G_{0};u_{1},u_{n})=\begin{cases}\dfrac{1}{4}\left(\dfrac{\chi_{2}(G_{0};u_{1},u_{n})}{\chi_{1}(G_{0})}+|E_{0}|\right)&\text{ if }G_{0}\text{ is bipartite,}\\ \\ \dfrac{\iota_{2}(G_{0};u_{1})}{\iota_{1}(G_{0})}&\text{ if }G_{0}\text{ is non-bipartite.}\end{cases}

Then we can determine how much quantum walker feels comfortable to the given graph by listing up the spanning subgraphs in Definition 3 of this graph. We will demonstrate it for the graphs with four vertices in the next section.

Theorem 4.

Assume the number of tails is 22, and the inflow 𝛂=(α1,α2)=(1,0)\bm{\alpha}=(\alpha_{1},\alpha_{2})=(1,0) at u1u_{1} and unu_{n}, respectively. Then the comfortability of the quantum walk discussed in [17], that is, z=1z=1 in (2), is given by

εQW(G0;u1,un)=14(χ2(G0;u1,un)χ1(G0)+|E0|).\mbox{\Large$\varepsilon$}_{QW}(G_{0};u_{1},u_{n})=\dfrac{1}{4}\left(\dfrac{\chi_{2}(G_{0};u_{1},u_{n})}{\chi_{1}(G_{0})}+|E_{0}|\right).

The proof of Theorem 4 is essentially same as in the proof of Theorem 2 for the bipartite graph case.

Refer to caption
Figure 2: 𝒞o\mathcal{C}_{o}-factor and 𝒯𝒞o\mathcal{T}\mathcal{C}_{o}-factor of K4K_{4}: The white colored vertex in the complete graph K4K_{4} corresponds to u1u_{1}. The left figure depicts the list of the family of odd unicyclic factor of K4K_{4} and the right figure depicts the list of the 𝒯𝒞o\mathcal{T}\mathcal{C}_{o}-factor of K4K_{4}. Note that the isolated vertex is regards as a tree and the family of the spanning tree is included in 𝒯𝒞o\mathcal{T}\mathcal{C}_{o}.

4 Example

As an example to our result in Theorem Theorem 2, we consider the connected graphs with 44 vertices labeled by {u1,u2,u3,u4}\left\{u_{1},u_{2},u_{3},u_{4}\right\}. The setting is the same as in Theorem Theorem 2 and we choose the inflow 𝜶=(1,0)\bm{\alpha}=(1,0) at the vertices u1u_{1} and u4u_{4}, respectively. We classify these graphs into 1010 classes based on the number of edges, bipartiteness and the configuration of u1u_{1} and u4u_{4} and the numbers of important factors state in Definition 3. Note that the comfortability for the non-bipartite case depends only on u1u_{1}. We conclude that every graph with 44 vertices belongs to exactly one of the following classes (see Fig 3):

𝒢1={K4},\mathcal{G}_{1}=\left\{K_{4}\right\},
𝒢2={K4u1uj:j=2,3,4},\mathcal{G}_{2}=\left\{K_{4}-u_{1}u_{j}:j=2,3,4\right\},
𝒢3={K4uiuj:i,j=2,3,4},\mathcal{G}_{3}=\left\{K_{4}-u_{i}u_{j}:i,j=2,3,4\right\},
𝒢4={C4:u1u4},\mathcal{G}_{4}=\left\{C_{4}:u_{1}\sim u_{4}\right\},
𝒢5={C4:u1u4},\mathcal{G}_{5}=\left\{C_{4}:u_{1}\nsim u_{4}\right\},
𝒢6={G:G is constructed by joining u1 to exactly one vertex in the cycle u2,u3,u4},\mathcal{G}_{6}=\left\{G:G\text{ is constructed by joining }u_{1}\text{ to exactly one vertex in the cycle }u_{2},u_{3},u_{4}\right\},
𝒢7={G:G is constructed by joining ui(i=2,3,4) to exactly one vertex in the cycle u1,uk,ul(k,l1,i)},\mathcal{G}_{7}=\left\{G:G\text{ is constructed by joining }u_{i}(i=2,3,4)\text{ to exactly one vertex in the cycle }u_{1},u_{k},u_{l}(k,l\neq 1,i)\right\},
𝒢8={T:T is a tree with dist(u1,u4)=1},\mathcal{G}_{8}=\left\{T:T\text{ is a tree with }\text{dist}(u_{1},u_{4})=1\right\},
𝒢9={T:T is a tree with dist(u1,u4)=2},\mathcal{G}_{9}=\left\{T:T\text{ is a tree with }\text{dist}(u_{1},u_{4})=2\right\},
𝒢10={T:T is a tree with dist(u1,u4)=3}.\mathcal{G}_{10}=\left\{T:T\text{ is a tree with }\text{dist}(u_{1},u_{4})=3\right\}.

Here K4uiujK_{4}-u_{i}u_{j} is the graph obtained by removing the edge uiuju_{i}u_{j} from K4K_{4}. Now for Gi𝒢i(i=1,,10)G_{i}\in\mathcal{G}_{i}(i=1,...,10), we compute εQW\mbox{\Large$\varepsilon$}_{QW} as follows. When G0=K4G_{0}=K_{4} for example, since K4K_{4} is non-bipartite, we have

εQW(K4)=ι2(K4;u1)ι1(K4).\mbox{\Large$\varepsilon$}_{QW}(K_{4})=\dfrac{\iota_{2}(K_{4};u_{1})}{\iota_{1}(K_{4})}.

To compute ι1(K4)\iota_{1}(K_{4}), we need to find the number of odd unicyclic subgraphs which span K4K_{4}. The only such possible subgraph is a 33-cycle with an additional edge. It is clear that there are 44 ways to choose a 33-cycle and for a chosen 33-cycle, there are 33 ways to choose an edge which connects the remaining vertex to the cycle. Hence, altogether there are 1212 such subgraphs, which are shown as 𝒞o\mathcal{C}_{o} in the figure 2. Observe that each subgraph has only one component and hence we have

ι1(K4)=48.\iota_{1}(K_{4})=48.

Now to compute ι2(K4;u1)\iota_{2}(K_{4};u_{1}), we have to find the spanning subgraphs which contains a tree with u1u_{1} and the remaining are odd cycles, which are possibly empty. There are two types of such subgraphs, as shown as 𝒯𝒞o\mathcal{T}\mathcal{C}_{o} in the figure 2 one being the spanning trees and the other being the cycle u2,u3,u4u_{2},u_{3},u_{4} along with the single vertex u1u_{1}. In the first case the number of spanning trees is the tree number, or the complexity, of K4K_{4}, given by 442=424^{4-2}=4^{2} (e.g., [2]) while in the second case there is only one such subgraph which has two components. So it follows that

ι2(K4;u1)=20\iota_{2}(K_{4};u_{1})=20

and hence

εQW(K4)=2048=512.\mbox{\Large$\varepsilon$}_{QW}(K_{4})=\dfrac{20}{48}=\dfrac{5}{12}.

Now consider G4𝒢4G_{4}\in\mathcal{G}_{4}. This graph is a 44-cycle with u1u4u_{1}\sim u_{4}. This graph is bipartite and hence

εQW(G4)=14(χ2(G4;u1,u4)χ1(G4)+|E0|).\mbox{\Large$\varepsilon$}_{QW}(G_{4})=\dfrac{1}{4}\left(\dfrac{\chi_{2}(G_{4};u_{1},u_{4})}{\chi_{1}(G_{4})}+|E_{0}|\right).

Since χ1(G4)\chi_{1}(G_{4}) is the tree number of G4G_{4} and by removing an edge from G4G_{4} we can get a spanning tree, we have

χ1(G4)=4.\chi_{1}(G_{4})=4.

To compute χ2(G4;u1,u4)\chi_{2}(G_{4};u_{1},u_{4}), we have to find the number of forests with exactly two components, one containing u1u_{1} and the other containing u4u_{4}. To find such a forest, the edge {u1,u4}\left\{u_{1},u_{4}\right\} has to be removed, and other than that, one of the remaining edges has to be removed. So we have

χ2(G4;u1,u4)=3\chi_{2}(G_{4};u_{1},u_{4})=3

and hence

εQW(G4)=14(34+4)=1916.\mbox{\Large$\varepsilon$}_{QW}(G_{4})=\dfrac{1}{4}(\dfrac{3}{4}+4)=\dfrac{19}{16}.

Similarly, we compute the comfortability on the remaining classes of graphs and we tabulate these values as shown in Table 1.

𝒢i\mathcal{G}_{i} 𝒢1\mathcal{G}_{1} 𝒢2\mathcal{G}_{2} 𝒢3\mathcal{G}_{3} 𝒢4\mathcal{G}_{4} 𝒢5\mathcal{G}_{5} 𝒢6\mathcal{G}_{6} 𝒢7\mathcal{G}_{7} 𝒢8\mathcal{G}_{8} 𝒢9\mathcal{G}_{9} 𝒢10\mathcal{G}_{10}
εQW(𝒢i)\mbox{\Large$\varepsilon$}_{QW}(\mathcal{G}_{i}) 5/125/12 3/43/4 1/21/2 19/1619/16 5/45/4 7/47/4 3/43/4 11 5/45/4 3/23/2
Scattering R R R T T R R T T T
Bipartiteness - - - \circ \circ - - \circ \circ \circ
|E||E| 66 55 55 44 44 44 44 33 33 33
Table 1: The comfortability of quantum walker to graphs with four vertices: The comfortability to each graph class is described by QW\mathcal{E}_{QW}. The symbols of “R” and “T” mean the perfectly reflection and transmitting, respectively. The best graph with four vertices of the comfortability is the 𝒢6\mathcal{G}_{6} and the worst graph is the complete graph.

Based on εQW\mbox{\Large$\varepsilon$}_{QW}, we have the following ordering of graphs:

G6QWG10QWG5,G9QWG4QWG8QWG2,G7QWG3QWG1.G_{6}\succ_{QW}G_{10}\succ_{QW}G_{5},G_{9}\succ_{QW}G_{4}\succ_{QW}G_{8}\succ_{QW}G_{2},G_{7}\succ_{QW}G_{3}\succ_{QW}G_{1}.
Refer to caption
Figure 3: The Hasse diagram of comfortability of graphs: The most comfortable graph for this quantum walker is 𝒢6\mathcal{G}_{6} which is a non-bipartite graph, while the most uncomfortable graph is 𝒢1\mathcal{G}_{1} (the complete graph). Here the entrance vertex u1u_{1} is indicated by the white vertex, and the exit vertex uNu_{N} is indicated by the black vertex for the bipartite case. Note that in the non-bipartite case, the comfortability for this quantum walker is independent of the position of the exit vertex (see Theorem 2).

Furthermore, we remark that for a tree TT in general, with nn vertices, the comfortability of the quantum walker is given by

εQW(T)=14(dist(u1,un)+(n1)).\mbox{\Large$\varepsilon$}_{QW}(T)=\dfrac{1}{4}\left(\text{dist}(u_{1},u_{n})+(n-1)\right).

Finally we also consider the comparison between the comfortability for z=1z=1 and z=1z=-1 cases. Any graph of four vertices is isomorphic to one of the following graphs Γ1\Gamma_{1},Γ2\Gamma_{2},…, Γ6\Gamma_{6} as in Figure 4:

Γ1\Gamma_{1}Γ2\Gamma_{2}Γ3\Gamma_{3}Γ4\Gamma_{4}Γ5\Gamma_{5}Γ6\Gamma_{6}
Figure 4: Non-isomorphic graphs Γj\Gamma_{j}’s for four vertices

We set

Comf(G)=maxu1,unV(G)QW(G;𝜶,u1,un)\mathrm{Comf}(G)=\max_{u_{1},u_{n}\in V(G)}\mathcal{E}_{QW}(G;\bm{\alpha},u_{1},u_{n})

for 𝜶=(1,0)\bm{\alpha}=(1,0) at u1u_{1} and unu_{n}. Then we have the comfortabilities for each graph and z=±1z=\pm 1 as in Table 2.

Γ1\Gamma_{1} Γ2\Gamma_{2} Γ3\Gamma_{3} Γ4\Gamma_{4} Γ5\Gamma_{5} Γ6\Gamma_{6}
z=1z=-1 5/45/4 3/23/2 5/45/4 7/47/4 3/43/4 5/125/12
z=1z=1 5/45/4 3/23/2 5/45/4 17/1217/12 3/23/2 13/813/8
Table 2: Comf(Γj)\mathrm{Comf}(\Gamma_{j}) for z=±1z=\pm 1

Thus we have, for z=1z=-1,

Comf(Γ6)<Comf(Γ5)<Comf(Γ1)=Comf(Γ3)<Comf(Γ2)<Comf(Γ4),\mathrm{Comf}(\Gamma_{6})<\mathrm{Comf}(\Gamma_{5})<\mathrm{Comf}(\Gamma_{1})=\mathrm{Comf}(\Gamma_{3})<\mathrm{Comf}(\Gamma_{2})<\mathrm{Comf}(\Gamma_{4}),

and for z=1z=1,

Comf(Γ1)=Comf(Γ3)<Comf(Γ4)<Comf(Γ2)=Comf(Γ5)<Comf(Γ6).\mathrm{Comf}(\Gamma_{1})=\mathrm{Comf}(\Gamma_{3})<\mathrm{Comf}(\Gamma_{4})<\mathrm{Comf}(\Gamma_{2})=\mathrm{Comf}(\Gamma_{5})<\mathrm{Comf}(\Gamma_{6}).

5 Proof of main theorems

5.1 Proof of Theorem 1

To prove this theorem we use the following lemma.

Lemma 1.

For a given uV0u\in V_{0}, Ψ(a)Ψ(a¯)\Psi_{\infty}(a)-\Psi_{\infty}(\overline{a}) is constant for all aA~a\in\tilde{A} with o(a)=uo(a)=u.

Proof.

It follows from the dynamics of the walk that, in the stationary state, for aA~a\in\tilde{A} such that o(a)V0o(a)\in V_{0},

Ψ(a)=b:t(b)=o(a)2deg(o(a))Ψ(b)+Ψ(a¯).\Psi_{\infty}(a)=-\sum\limits_{b:t(b)=o(a)}\dfrac{2}{\text{deg}(o(a))}\Psi_{\infty}(b)+\Psi_{\infty}(\overline{a}).

It follows that

Ψ(a)Ψ(a¯)=2deg(u)b:t(b)=uΨ(b),o(a)=u.\Psi_{\infty}(a)-\Psi_{\infty}(\overline{a})=-\dfrac{2}{\text{deg}(u)}\sum\limits_{b:t(b)=u}\Psi_{\infty}(b),\;o(a)=u.

Observe that for a given uV0u\in V_{0}, the right hand side of the equation is a constant. ∎

Now we give the proof of Theorem 1 as follows. By Lemma 1, the measure Ψ(a)Ψ(a¯)\Psi_{\infty}(a)-\Psi_{\infty}(\overline{a}) is a measure on uV0u\in V_{0} such that o(a)=uo(a)=u. We denote this measure by ρ(u)\rho(u). That is, Ψ(a)Ψ(a¯)=ρ(u),o(a)=u\Psi_{\infty}(a)-\Psi_{\infty}(\overline{a})=\rho(u),o(a)=u. It follows that if uvu\sim v then ρ(u)=ρ(v)\rho(u)=-\rho(v).

Suppose G0G_{0} is non-bipartite. Then there is an odd cycle C=(u1,,u2l1),uiV0C=(u_{1},...,u_{2l-1}),u_{i}\in V_{0} in G0G_{0}. Then we have ρ(u1)=ρ(u2)=ρ(u3)==ρ(u2l1)=ρ(u1)\rho(u_{1})=-\rho(u_{2})=\rho(u_{3})=...=\rho(u_{2l-1})=-\rho(u_{1}) which implies ρ(u1)=0\rho(u_{1})=0. Since G0G_{0} is connected, ρ(u)=0\rho(u)=0 for any uV0u\in V_{0}. Thus Ψ(a)Ψ(a¯)=0\Psi_{\infty}(a)-\Psi_{\infty}(\overline{a})=0 for any aA0a\in A_{0} such that o(a)V0o(a)\in V_{0}. In particular βi=αi\beta_{i}=\alpha_{i} for any ii. Hence 𝜷=𝜶\bm{\beta}=\bm{\alpha}.

Now suppose G0G_{0} is bipartite with V0=XYV_{0}=X\sqcup Y. Observe that

s:=ρV0(v)s:=\rho_{V_{0}}(v) (5)

is constant for all vXv\in X. Then

ρV0(v)=s\rho_{V_{0}}(v)=-s (6)

for all vYv\in Y. Define

AinX\displaystyle A^{X}_{in} ={aA~:o(a)V~V0,t(a)X},\displaystyle=\left\{a\in\tilde{A}:o(a)\in\tilde{V}\setminus V_{0},t(a)\in X\right\},
AoutX\displaystyle A^{X}_{out} ={aA~:a¯AinX},\displaystyle=\left\{a\in\tilde{A}:\overline{a}\in A^{X}_{in}\right\},
AinY\displaystyle A^{Y}_{in} ={aA~:o(a)V~V0,t(a)Y},\displaystyle=\left\{a\in\tilde{A}:o(a)\in\tilde{V}\setminus V_{0},t(a)\in Y\right\},
AoutY\displaystyle A^{Y}_{out} ={aA~:a¯AinY}.\displaystyle=\left\{a\in\tilde{A}:\overline{a}\in A^{Y}_{in}\right\}.

Then by applying the time evolution operator once, we get

U(Ψ|AinX+Ψ|AinY+Ψ|A0)=Ψ|AoutX+Ψ|AoutY+Ψ|A0.U\left(\Psi_{\infty}|_{A^{X}_{in}}+\Psi_{\infty}|_{A^{Y}_{in}}+\Psi_{\infty}|_{A_{0}}\right)=\Psi_{\infty}|_{A^{X}_{out}}+\Psi_{\infty}|_{A^{Y}_{out}}+\Psi_{\infty}|_{A_{0}}.

By taking the squared norm, we get

Ψ|AinX2+Ψ|AinY2+Ψ|A02=Ψ|AoutX2+Ψ|AoutY2+Ψ|A02.\left\lVert\Psi_{\infty}|_{A^{X}_{in}}\right\rVert^{2}+\left\lVert\Psi_{\infty}|_{A^{Y}_{in}}\right\rVert^{2}+\left\lVert\Psi_{\infty}|_{A_{0}}\right\rVert^{2}=\left\lVert\Psi_{\infty}|_{A^{X}_{out}}\right\rVert^{2}+\left\lVert\Psi_{\infty}|_{A^{Y}_{out}}\right\rVert^{2}+\left\lVert\Psi_{\infty}|_{A_{0}}\right\rVert^{2}.

It follows from (5) and (6) that

Ψ|AinX2+Ψ|AinY2\displaystyle\left\lVert\Psi_{\infty}|_{A^{X}_{in}}\right\rVert^{2}+\left\lVert\Psi_{\infty}|_{A^{Y}_{in}}\right\rVert^{2} =s𝟙AinX+SΨ|AinX2+s𝟙AinY+SΨ|AinY2\displaystyle=\left\lVert s\mathbbm{1}_{A^{X}_{in}}+S\Psi_{\infty}|_{A^{X}_{in}}\right\rVert^{2}+\left\lVert-s\mathbbm{1}_{A^{Y}_{in}}+S\Psi_{\infty}|_{A^{Y}_{in}}\right\rVert^{2}
=s2|XδV|+2saAinXΨ(a)+Ψ|AinX2\displaystyle=s^{2}|X\cap\delta V|+2s\sum\limits_{a\in A^{X}_{in}}\Psi_{\infty}(a)+\left\lVert\Psi_{\infty}|_{A^{X}_{in}}\right\rVert^{2}
+s2|YδV|2saAinYΨ(a)+Ψ|AinY2.\displaystyle\qquad\qquad+s^{2}|Y\cap\delta V|-2s\sum\limits_{a\in A^{Y}_{in}}\Psi_{\infty}(a)+\left\lVert\Psi_{\infty}|_{A^{Y}_{in}}\right\rVert^{2}.

Here SS is the shift operator: Sδa=δa¯S\delta_{a}=\delta_{\bar{a}} for any aA~a\in\tilde{A}. Hence if s0s\neq 0,

s=2|δV|(aAinXΨ(a)aAinYΨ(a)).s=-\dfrac{2}{|\delta V|}\left(\sum\limits_{a\in A^{X}_{in}}\Psi_{\infty}(a)-\sum\limits_{a\in A^{Y}_{in}}\Psi_{\infty}(a)\right). (7)

Then it follows that

βi=αi2|δV|(aAinXΨ(a)aAinYΨ(a)),\beta_{i}=\alpha_{i}-\dfrac{2}{|\delta V|}\left(\sum\limits_{a\in A^{X}_{in}}\Psi_{\infty}(a)-\sum\limits_{a\in A^{Y}_{in}}\Psi_{\infty}(a)\right),

where βi=Ψ(a)\beta_{i}=\Psi_{\infty}(a) for some aAoutXa\in A^{X}_{out} and

βi=αi2|δV|(aAinYΨ(a)aAinXΨ(a)),\beta_{i}=\alpha_{i}-\dfrac{2}{|\delta V|}\left(\sum\limits_{a\in A^{Y}_{in}}\Psi_{\infty}(a)-\sum\limits_{a\in A^{X}_{in}}\Psi_{\infty}(a)\right),

where βi=Ψ(a)\beta_{i}=\Psi_{\infty}(a) for some aAoutYa\in A^{Y}_{out}. Hence 𝜷=τ𝜶\bm{\beta}=\tau\bm{\alpha} where

τ=(2|δV|+12|δV|2|δV|2|δV|+1...2|δV|2|δV|2|δV|+12|δV|2|δV|2|δV|+1...).\tau=\left(\begin{array}[]{@{}c|c@{}}\begin{matrix}-\dfrac{2}{|\delta V|}+1&-\dfrac{2}{|\delta V|}&...\\ -\dfrac{2}{|\delta V|}&-\dfrac{2}{|\delta V|}+1&...\\ .&&\\ .&&\\ .&&\\ \end{matrix}&\mbox{\Large$\dfrac{2}{|\delta V|}$}\\ \hline\cr\mbox{\Large$\dfrac{2}{|\delta V|}$}&\begin{matrix}-\dfrac{2}{|\delta V|}+1&-\dfrac{2}{|\delta V|}&...\\ -\dfrac{2}{|\delta V|}&-\dfrac{2}{|\delta V|}+1&...\\ .&&\\ .&&\\ .&&\\ \end{matrix}\end{array}\right).

It is clear that, to satisfy the condition s0s\neq 0, we must have aAinYΨ(a)aAinXΨ(a)\sum_{a\in A^{Y}_{in}}\Psi_{\infty}(a)\neq\sum_{a\in A^{X}_{in}}\Psi_{\infty}(a). Now let us see that s=0s=0 if and only if κX=κY\kappa_{X}=\kappa_{Y}, where κX=aAinXΨ(a)\kappa_{X}=\sum_{a\in A^{X}_{in}}\Psi_{\infty}(a) and κY=aAinYΨ(a)\kappa_{Y}=\sum_{a\in A^{Y}_{in}}\Psi_{\infty}(a).

By Lemma 1 we have, s=0s=0 if and only if

Ψ(a)=Ψ(a¯),aA0 with o(a)V0\Psi_{\infty}(a)=\Psi_{\infty}(\overline{a}),\;a\in A_{0}\text{ with }o(a)\in V_{0}

and

b:t(b)=uΨ(b)=0,uV0.\sum\limits_{b:t(b)=u}\Psi_{\infty}(b)=0,\;u\in V_{0}.

Thus we have

0\displaystyle 0 =a:t(a)YΨ(a)\displaystyle=\sum\limits_{a:t(a)\in Y}\Psi_{\infty}(a)
=a:t(a)Y,o(a)XΨ(a)+κY\displaystyle=\sum_{a:\;t(a)\in Y,\;o(a)\in X}\Psi_{\infty}(a)+\kappa_{Y}
=a:o(a)X,t(a)YΨ(a)+κY\displaystyle=\sum_{a:\in o(a)\in X,\;t(a)\in Y}\Psi_{\infty}(a)+\kappa_{Y}
=a:o(a)XΨ(a)κX+κY\displaystyle=\sum\limits_{a:o(a)\in X}\Psi_{\infty}(a)-\kappa_{X}+\kappa_{Y}
=κX+κY,\displaystyle=-\kappa_{X}+\kappa_{Y},

which implies κX=κY\kappa_{X}=\kappa_{Y} if s=0s=0.

If 𝜶\bm{\alpha} satisfies κX=κY\kappa_{X}=\kappa_{Y}, then s=0s=0 and the same argument as in the case where G0G_{0} is non-bipartite is valid; 𝜷=𝜶\bm{\beta}=\bm{\alpha} holds. It is easy to check that 𝜷=τ𝜶\bm{\beta}=\tau\bm{\alpha} holds for τ\tau defined above. \hfill\square

11
Figure 5: Example of the inflow of s0s\neq 0.

5.2 Proof of Theorem 2

The proof of Theorem 2 is divided into three parts with bipartite and non-bipartite cases as follows. First, we will show that the stationary state is expressed by a (pseudo-)current function. Secondly, we will determine the potential functions with respect to the (pseudo-)current function which satisfies the Poisson equation in terms of the (signless-)Laplacian. Finally, we will express the comfortability in terms of geometries of the graph using the above.

5.2.1 Kirchhoff and pseudo-Kirchhoff laws

In this subsection we will see that the stationary state can represent a kind of current function [15, 17] on the underlying bipartite and non-bipartite graphs.

Bipartite case
First, we will introduce a current function induced by the stationary state Ψ\Psi_{\infty} on a bipartite graph. Let G0G_{0} be a bipartite graph with V0=XYV_{0}=X\sqcup Y. Define the function f()f(\cdot) such that

f(a)={0if t(a)X,1if t(a)Y.f(a)=\begin{cases}0&\text{if }\,t(a)\in X,\\ \\ 1&\text{if }\,t(a)\in Y.\end{cases}

Then by Lemma 1, it follows that the measure 12((1)f(a)Ψ(a)+(1)f(a¯)Ψ(a¯))\dfrac{1}{2}\left((-1)^{f(a)}\Psi_{\infty}(a)+(-1)^{f(\overline{a})}\Psi_{\infty}(\overline{a})\right) is a constant for any aA0a\in A_{0}. We denote this constant by ρ\rho. Then the following theorem holds.

Theorem 5.

Let the setting be the same as the above and define j(a)=(1)f(a)Ψ(a)ρj(a)=(-1)^{f(a)}\Psi_{\infty}(a)-\rho. Then j()j(\cdot) satisfies the Kirchhoff’s current and voltage laws:

  1. 1.

    (Kirchhoff current law) For any uV~u\in\tilde{V} and aA~a\in\tilde{A},

    bA~:t(b)=uj(b)=0,j(a)+j(a¯)=0.\sum\limits_{b\in\tilde{A}:t(b)=u}j(b)=0,\,j(a)+j(\overline{a})=0.
  2. 2.

    (Kirchhoff voltage law) For any cycle c=(a1,,as)c=(a_{1},...,a_{s}) with t(a1)=o(a2),,t(as1)=o(as),t(as)=o(a1)t(a_{1})=o(a_{2}),...,t(a_{s-1})=o(a_{s}),t(a_{s})=o(a_{1}) in G0G_{0}, it holds

    k=1sj(ak)=0.\sum\limits_{k=1}^{s}j(a_{k})=0.
Proof.

We can rewrite the expression in Lemma 1 as

12((1)f(a)Ψ(a)+(1)f(a¯)Ψ(a¯))=1deg(u)b:t(b)=u(1)f(b)Ψ(b),o(a)=u,\dfrac{1}{2}\left((-1)^{f(a)}\Psi_{\infty}(a)+(-1)^{f(\overline{a})}\Psi_{\infty}(\overline{a})\right)=\dfrac{1}{\text{deg}(u)}\sum\limits_{b:t(b)=u}(-1)^{f(b)}\Psi_{\infty}(b),\,o(a)=u,

in which the right hand side is a measure on the vertex uu, denoted by ρ(u)\rho(u). Then it follows that if uvu\sim v then ρ(u)=ρ(v)\rho(u)=\rho(v). Since G0G_{0} is connected, it follows that ρ(u)\rho(u) is a constant for all uV0u\in V_{0}, which is denoted by ρ\rho. Summarizing the above, we have

1deg(u)b:t(b)=u(1)f(b)Ψ(b)=ρ,uV0.\dfrac{1}{\text{deg}(u)}\sum\limits_{b:t(b)=u}(-1)^{f(b)}\Psi_{\infty}(b)=\rho,u\in V_{0}. (8)

Then from (7) we can express

ρ=1|δV|(aAinXΨ(a)aAinYΨ(a)).\rho=\dfrac{1}{|\delta V|}\left(\sum\limits_{a\in A^{X}_{in}}\Psi_{\infty}(a)-\sum\limits_{a\in A^{Y}_{in}}\Psi_{\infty}(a)\right).

Equation (8) implies

a:t(a)=uj(a)=a:t(a)=u(1)f(a)Ψ(a)ρdeg(u)=0,uV0.\sum\limits_{a:t(a)=u}j(a)=\sum\limits_{a:t(a)=u}(-1)^{f(a)}\Psi_{\infty}(a)-\rho\text{deg}(u)=0,\;u\in V_{0}.

Also we have

j(a)+j(a¯)=(1)f(a)Ψ(a)+(1)f(a¯)Ψ(a¯)2ρ=0,aA0.j(a)+j(\overline{a})=(-1)^{f(a)}\Psi_{\infty}(a)+(-1)^{f(\overline{a})}\Psi_{\infty}(\overline{a})-2\rho=0,\;a\in A_{0}.

Now suppose C=(a1,,a2l),aiA0C=(a_{1},...,a_{2l}),a_{i}\in A_{0} is an even cycle in G0G_{0}. Assume that o(a1)Xo(a_{1})\in X. Define φ\varphi such that

φ(a)={1if a=a2k+1 or a¯2k+1,1if a=a2k or a¯2k,0otherwise.\varphi(a)=\begin{cases}1&\text{if }a=a_{2k+1}\text{ or }\overline{a}_{2k+1},\\ -1&\text{if }a=a_{2k}\text{ or }\overline{a}_{2k},\\ 0&\text{otherwise.}\end{cases}

Let ψ:=χΨ\psi_{\infty}:=\chi^{*}\Psi_{\infty}. Since φ\varphi is a centered eigenvector of χUχ\chi U\chi^{*} whose eigenvalue is 1-1, by [17, Lemmas 3.4 and 3.5] we have ψφ\psi_{\infty}\perp\varphi and it follows that

0\displaystyle 0 =ψ|φ\displaystyle=\left\langle\psi_{\infty}|\varphi\right\rangle
=aA0ψ(a)φ(a)\displaystyle=\sum\limits_{a\in A_{0}}\psi_{\infty}(a)\varphi(a)
=k=1l[[ψ(a2k1)+ψ(a¯2k1)][ψ(a2k1)+ψ(a¯2k1)]]\displaystyle=\sum\limits_{k=1}^{l}\left[\left[\psi_{\infty}(a_{2k-1})+\psi_{\infty}(\overline{a}_{2k-1})\right]-\left[\psi_{\infty}(a_{2k-1})+\psi_{\infty}(\overline{a}_{2k-1})\right]\right]
=k=1l[(1)f(a2k1)(j(a2k1)j(a¯2k1))(1)f(a2k)(j(a2k)j(a¯2k))]\displaystyle=\sum\limits_{k=1}^{l}\left[(-1)^{f(a_{2k-1})}\left(j(a_{2k-1})-j(\overline{a}_{2k-1})\right)-(-1)^{f(a_{2k})}\left(j(a_{2k})-j(\overline{a}_{2k})\right)\right]
=k=12l[j(ak)j(a¯k)].\displaystyle=\sum\limits_{k=1}^{2l}\left[j(a_{k})-j(\overline{a}_{k})\right].

Since j(a)+j(a¯)=0j(a)+j(\overline{a})=0, it follows that

k=1sj(ak)=0.\sum\limits_{k=1}^{s}j(a_{k})=0.

Non-bipartite case
Next let us investigate the property of ψ\psi_{\infty} in the case non-bipartite case. We can see that the stationary state itself has similar properties to the electrical current flow as follows.

Theorem 6.

Let ΨA~\Psi_{\infty}\in\tilde{A} be the stationary state on the non-bipartite graph. Then Ψ\Psi_{\infty} satisfies the following properties:

  1. 1.

    (Pseudo-Kirchhoff current law) For each uV0u\in V_{0} and aA0~a\in\tilde{A_{0}},

    b:t(a)=uΨ(a)=0,Ψ(a)=Ψ(a¯).\sum_{b:t(a)=u}\Psi_{\infty}(a)=0,\;\Psi_{\infty}(a)=\Psi_{\infty}(\bar{a}).
  2. 2.

    (Pseudo-Kirchhoff voltage law) For any even closed walk (e1,,e2s)(e_{1},\dots,e_{2s}) such that
    t(e1)=o(e2),,t(e2s1)=o(e2s)t(e_{1})=o(e_{2}),\dots,t(e_{2s-1})=o(e_{2s}) and t(e2s)=o(e1)t(e_{2s})=o(e_{1}),

    k=12s(1)kΨ(ek)=0.\sum_{k=1}^{2s}(-1)^{k}\Psi_{\infty}(e_{k})=0.
Proof.

The proof of the first part is obtained from the proof of Theorem 1. The second part shows ψ,ϕ=0\langle\psi_{\infty},\phi\rangle=0 for any ϕ\phi which is an eigenfunction of the eigenvalue 1-1 in the birth part. For details, refer to [14, Definition 4 and Theorem 1]. Moreover it has been shown by [17] that the stationary state is orthogonal to the birth part. Then we have the desired conclusion. ∎

5.2.2 Laplacian and signless Laplacian

We have seen that the stationary state has the properties of current function or pseudo-current function in A~\mathbb{C}^{\tilde{A}}. Then it is natural to determine the potential function in V0\mathbb{C}^{V_{0}} with respect to the current. In this subsection, we characterize the potential function using the Laplacian and the signless Laplacian for the cases of bipartite and non-bipartite graphs, respectively.

Bipartite case
Let MM be the adjacency matrix and DD be the degree matrix of G0G_{0}. The Laplacian matrix of G0G_{0} is denoted by L=MDL=M-D. Using the Laplacian matrix with the Poisson equation, we can characterise the current function j()j(\cdot) on A0A_{0} in terms of a potential function on V0V_{0} ([2] and [4]) in which exists under the Kirchhoff current and voltage laws.

Theorem 7 (see e.g., [2, 4]).

Let the setting be the same as in Theorem 5. Let G0G_{0} be a bipartite graph and LL be the Laplacian matrix of G0G_{0}. Then there exists a potential function ϕV0\phi\in\mathbb{C}^{V_{0}} such that j(a)=ϕ(o(a))ϕ(t(a))j(a)=\phi(o(a))-\phi(t(a)). Here ϕ\phi satisfies the following equation.

Lϕ=q,L\phi=-q,

where q(u)=aA~A0:t(a)=uj(a),uV0.q(u)=\sum\limits_{a\in\tilde{A}\setminus A_{0}:t(a)=u}j(a),\;u\in V_{0}.

Here we should remark that q(ker(L))q\in(\text{ker}(L))^{\perp}: actually we have

0=aA:t(a)V0j(a)aAA0:t(a)V0j(a)=vV0q(v)𝟏(v),0=\sum_{a\in A:t(a)\in V_{0}}j(a)\equiv\sum_{a\in A\setminus A_{0}:t(a)\in V_{0}}j(a)=\sum_{v\in V_{0}}q(v)\bm{1}(v),

where 𝟏()\bm{1}(\cdot) is the constant function whose value is 11 and ker(L)={c𝟏:c}\text{ker}(L)=\left\{c\bm{1}:c\in\mathbb{C}\right\}.

Non-bipartite case
When the underlying graph is non-bipartite, the following two properties hold for the stationary sate Ψ\Psi_{\infty} by Theorem 6:

Ψ(a)\displaystyle\Psi_{\infty}(a) =Ψ(a¯),for anyaA~,o(a)V0;\displaystyle=\Psi_{\infty}(\overline{a}),\;\text{for any}\;a\in\tilde{A},o(a)\in V_{0}; (9)
aA~:t(a)=uΨ(a)\displaystyle\sum\limits_{a\in\tilde{A}:t(a)=u}\Psi_{\infty}(a) =0,for anyuV0.\displaystyle=0,\;\text{for any}\;u\in V_{0}. (10)

As an analogy to the current function in the bipartite case, we represent these properties in term of the non-oriented incidence matrix and the signless Laplacian matrix in the following theorem. Here the sighless Laplacian matrix of G0G_{0} is denoted by Q=M+DQ=M+D.

Theorem 8.

Let Ψ\Psi_{\infty} be the stationary state of quantum walk such that Ψ(a)=limn(UnΨ0)(a)\Psi_{\infty}(a)=\lim_{n\to\infty}(U^{n}\Psi_{0})(a) for any aA~a\in\tilde{A}. Let G0G_{0} be a non-bipartite graph and QQ be the signless Laplacian matrix of G0G_{0}. Then there uniquely exists ϕV0\phi\in\mathbb{C}^{V_{0}} such that Ψ(a)=ϕ(o(a))+ϕ(t(a))\Psi_{\infty}(a)=\phi(o(a))+\phi(t(a)) for any aA0a\in A_{0}. Here ϕ\phi satisfies the following equation.

Qϕ=qQ\phi=-q

where q(u)=aA~A0:t(a)=uΨ(a),uV0.q(u)=\sum\limits_{a\in\tilde{A}\setminus A_{0}:t(a)=u}\Psi_{\infty}(a),\;u\in V_{0}.

Proof.

Denote the non-oriented incidence matrix on the set of arcs by C~:A0V0\tilde{C}:\mathbb{C}^{A_{0}}\rightarrow\mathbb{C}^{V_{0}} which satisfies

(C~ψ)(vi)=j=1mcijψ(aj), 1in,(\tilde{C}\psi)(v_{i})=\sum\limits_{j=1}^{m}c_{ij}\psi(a_{j}),\;1\leq i\leq n,

where m=|A0|m=|A_{0}| and

cij={1if t(aj)=vi or o(aj)=vi,0otherwise. c_{ij}=\begin{cases}1&\text{if }\,t(a_{j})=v_{i}\text{ or }o(a_{j})=v_{i},\\ 0&\text{otherwise. }\end{cases}

Then we have

(C~ψ)(v)=aA0:t(a)=vψ(a)+aA0:o(a)=vψ(a),vV0.(\tilde{C}\psi)(v)=\sum\limits_{a\in A_{0}:t(a)=v}\psi(a)+\sum\limits_{a\in A_{0}:o(a)=v}\psi(a),\;v\in V_{0}.

The adjoint operator C~:V0A0\tilde{C}^{*}:\mathbb{C}^{V_{0}}\rightarrow\mathbb{C}^{A_{0}} is given by

(C~f)(a)=f(t(a))+f(o(a)).(\tilde{C}^{*}f)(a)=f(t(a))+f(o(a)).

Then we have C~C~\tilde{C}\tilde{C}^{*} in terms of the signless Laplacian matrix QQ

C~C~=2(D+M)=2Q.\tilde{C}\tilde{C}^{*}=2(D+M)=2Q.

Let ψ=χΨ\psi_{\infty}=\chi^{*}\Psi_{\infty}. Now let us see that if there exists a potential function ϕV0\phi\in\mathbb{C}^{V_{0}} such that ψ(a)=(C~ϕ)(a)\psi_{\infty}(a)=(\tilde{C}^{*}\phi)(a) for any aA0a\in A_{0}, then ϕ\phi must satisfy

Qϕ=q.Q\phi=-q.

Remark that for any aA0a\in A_{0}

ψ(a)=(C~ϕ)(a)=ϕ(t(a))+ϕ(o(a))=ϕ(t(a¯))+ϕ(o(a¯))=(C~ϕ)(a¯)=ψ(a¯)\psi_{\infty}(a)=(\tilde{C}^{*}\phi)(a)=\phi(t(a))+\phi(o(a))=\phi(t(\overline{a}))+\phi(o(\overline{a}))=(\tilde{C}^{*}\phi)(\overline{a})=\psi_{\infty}(\overline{a})

and hence the first property of (9) can be easily confirmed. We have, by the property (9),

aA~:o(a)=uΨ(a)\displaystyle\sum\limits_{a\in\tilde{A}:o(a)=u}\Psi_{\infty}(a) =aA~:o(a)=uΨ(a¯)\displaystyle=\sum\limits_{a\in\tilde{A}:o(a)=u}\Psi_{\infty}(\overline{a})
=aA~:t(a)=uΨ(a),\displaystyle=\sum\limits_{a\in\tilde{A}:t(a)=u}\Psi_{\infty}(a),

and by the property (10),

aA~:t(a)=uΨ(a)+aA~:o(a)=uΨ(a)=0,uV0.\sum\limits_{a\in\tilde{A}:t(a)=u}\Psi_{\infty}(a)+\sum\limits_{a\in\tilde{A}:o(a)=u}\Psi_{\infty}(a)=0,\;u\in V_{0}.

Let us divide the summation in the above equation by

aA0:t(a)=uΨ(a)+aA~A0:t(a)=uΨ(a)+aA0:o(a)=uΨ(a)+aA~A0:o(a)=uΨ(a)=0.\sum\limits_{a\in A_{0}:t(a)=u}\Psi_{\infty}(a)+\sum\limits_{a\in\tilde{A}\setminus A_{0}:t(a)=u}\Psi_{\infty}(a)+\sum\limits_{a\in A_{0}:o(a)=u}\Psi_{\infty}(a)+\sum\limits_{a\in\tilde{A}\setminus A_{0}:o(a)=u}\Psi_{\infty}(a)=0.

Then it follows that

(C~ψ)(u)=2aA~A0:t(a)=uΨ(a).(\tilde{C}\psi_{\infty})(u)=-2\sum\limits_{a\in\tilde{A}\setminus A_{0}:t(a)=u}\Psi_{\infty}(a).

which implies

Qϕ=12C~C~ϕ=q,Q\phi=\dfrac{1}{2}\tilde{C}\tilde{C}^{*}\phi=-q,

where q(u)=aA~A0:t(a)=uΨ(a),uV0.q(u)=\sum_{a\in\tilde{A}\setminus A_{0}:t(a)=u}\Psi_{\infty}(a),u\in V_{0}. Although the existence of the potential function ϕ\phi is ensured by the pseudo-Kirchhoff voltage law in Theorem 6, we prove here directly as follows. To show the existence of ϕ\phi, it is enough to show that

ψRange(C~)=ker(C~).\psi_{\infty}\in\text{Range}(\tilde{C}^{*})=\text{ker}(\tilde{C})^{\perp}.

Let us see that

ker(C~)=(ker(d)+),\text{ker}(\tilde{C})=(\text{ker}(d)\cap\mathcal{H}_{+})\oplus\mathcal{H}_{-}, (11)

where +={ψA0:ψ(a)=ψ(a¯) for any aA0}\mathcal{H}_{+}=\left\{\psi\in\mathbb{C}^{A_{0}}:\psi(a)=\psi(\overline{a})\text{ for any }a\in A_{0}\right\}, ={ψA0:ψ(a)=ψ(a¯) for any aA0}\mathcal{H}_{-}=\left\{\psi\in\mathbb{C}^{A_{0}}:\psi(a)=-\psi(\overline{a})\text{ for any }a\in A_{0}\right\} and

(dφ)(u)=1degG0(u)aA0:t(a)=uφ(a).(d\varphi)(u)=\dfrac{1}{\sqrt{\deg_{G_{0}}(u)}}\sum\limits_{a\in A_{0}:t(a)=u}\varphi(a).

Note that +=A0\mathcal{H}_{+}\oplus\mathcal{H}_{-}=\mathbb{C}^{A_{0}} and +=ker(1S0)\mathcal{H}_{+}=\ker(1-S_{0}), =ker(1+S0)\mathcal{H}_{-}=\ker(1+S_{0}), where S0S_{0} is the shift operator on A0\mathbb{C}^{A_{0}} such that S0δa=δa¯S_{0}\delta_{a}=\delta_{\bar{a}} for any aA0a\in A_{0}. The operator C~\tilde{C} can be rewritten by

(C~ψ)(u)=degG0(u)((dψ)(u)+(dS0ψ)(u)).(\tilde{C}\psi)(u)=\sqrt{\deg_{G_{0}}(u)}\left((d\psi)(u)+(dS_{0}\psi)(u)\right).

Then we have

kerC~=ker(d(1+S0)),\ker\tilde{C}=\ker(d(1+S_{0})),

which implies (11). By [17] it follows that

ψ(ker(d)+).\psi_{\infty}\in(\text{ker}(d)\cap\mathcal{H}_{+})^{\perp}.

Since ψ(a¯)=ψ(a)\psi_{\infty}(\bar{a})=\psi_{\infty}(a), then ψ+=\psi_{\infty}\in\mathcal{H}_{+}=\mathcal{H}_{-}^{\perp}. Therefore ψker(C~)=Range(C~)\psi_{\infty}\in\ker(\tilde{C})^{\perp}=\text{Range}(\tilde{C}^{*}).

Furthermore, since G0G_{0} is non-bipartite, the least eigenvalue of QQ is positive (cf. [6]) and hence QQ is invertible. This completes the proof of the existence and the uniqueness of ϕ\phi. ∎

5.2.3 Comfortability

We have seen that relations between stationary state and the (signless-)Laplacian, which contains information of the geometry of the graph. Let us express the comfortability in terms of some graph geometrical properties. This leads the completion of the proof of Theorem 2.

Bipartite case
Now we introduce the energy of the electric circuit εEC(G0)\mbox{\Large$\varepsilon$}_{EC}(G_{0}) which is given by,

εEC(G0)=12j2=12aA0|j(a)|2.\mbox{\Large$\varepsilon$}_{EC}(G_{0})=\dfrac{1}{2}\left\lVert j\right\rVert^{2}=\frac{1}{2}\sum_{a\in A_{0}}|j(a)|^{2}.

To give the next proposition, we prepare the following notion. Note that the Laplacian matrix LL is singular and hence ϕ\phi is not determined uniquely. We impose the ground condition ϕ(n)=0\phi(n)=0 which reduces the equation in Theorem 7 to L(n)ϕ(n)=q(n)L^{(n)}\phi^{(n)}=-q^{(n)}, where L(n)L^{(n)} is the matrix obtained by removing the nn-th row and the nn-th column of the Laplacian matrix, ϕ(n)\phi^{(n)} and q(n)q^{(n)} are the vectors obtained by removing the nn-th element from ϕ\phi and qq respectively. Here, det(L(n))\text{det}(L^{(n)}) is the number of spanning trees of G0G_{0} (see [2]), and hence non-zero. So ϕ(n)\phi^{(n)} is determined uniquely by ϕ(n)=(L(n))1q(n)\phi^{(n)}=-(L^{(n)})^{-1}q^{(n)}. More over, we denote by BB and B~\tilde{B}, the usual incidence matrix and the non-oriented incidence matrix on the set of edges respectively. More precisely, we fix an orientation of each eE0e\in E_{0} and denote it by E0\vec{E}_{0}:

A0=E0(E0)¯,A_{0}=\vec{E}_{0}\cup\overline{(\vec{E}_{0})},

where (E0)¯={aA0|a¯E0}\overline{(\vec{E}_{0})}=\{a\in A_{0}\;|\;\bar{a}\in\vec{E}_{0}\}. Then B:E0V0B:\mathbb{C}^{\vec{E}_{0}}\to\mathbb{C}^{V_{0}} is denoted by

(Bψ)(u)\displaystyle(B\psi)(u) =t(a)=uψ(a)o(a)=uψ(a);\displaystyle=\sum_{t(a)=u}\psi(a)-\sum_{o(a)=u}\psi(a);

then its adjoint is expressed by

(Bf)(a)=f(t(a))f(o(a)).(B^{*}f)(a)=f(t(a))-f(o(a)).

The non-oriented incidence matrix B~:E0V0\tilde{B}:\mathbb{C}^{\vec{E}_{0}}\to\mathbb{C}^{V_{0}} is denoted by

(B~ψ)(u)\displaystyle(\tilde{B}\psi)(u) =t(a)=uψ(a)+o(a)=uψ(a);\displaystyle=\sum_{t(a)=u}\psi(a)+\sum_{o(a)=u}\psi(a);

then its adjoint is expressed by

(B~f)(a)=f(t(a))+f(o(a)).({\tilde{B}}^{*}f)(a)=f(t(a))+f(o(a)).

Now we give the following proposition.

Proposition 1.

The electrical energy of the circuit is given by

εEC(G0)=1det(L(n))i,j=1n1(1)i+jq(n)(i)q(n)(j)HG0|E(H)|=n2det(BH(n,j))det((BH(n,i))),\mbox{\Large$\varepsilon$}_{EC}(G_{0})=\dfrac{1}{\mathrm{det}(L^{(n)})}\sum\limits_{i,j=1}^{n-1}(-1)^{i+j}q^{(n)}(i)q^{(n)}(j)\sum\limits_{\begin{subarray}{c}H\subset G_{0}\\ |E(H)|=n-2\end{subarray}}\mathrm{det}(B_{H}^{(n,j)})\mathrm{det}((B_{H}^{(n,i)})^{*}),

where BH(n,j)B_{H}^{(n,j)} is the matrix obtained by choosing the columns corresponding to the edges in HH and removing the jj-th and nn-th rows in the oriented incidence matrix of G0G_{0}.

Proof.

By definition, we can write the electrical energy in terms of the Laplacian matrix as follows.

εEC(G0)\displaystyle\mbox{\Large$\varepsilon$}_{EC}(G_{0}) =12j2=Bϕ,Bϕ=ϕ,Lϕ=ϕ,q=ϕ(n),q(n)=(L(n))1q(n),q(n)\displaystyle=\dfrac{1}{2}\left\lVert j\right\rVert^{2}=\left\langle B^{*}\phi,B^{*}\phi\right\rangle=\left\langle\phi,L\phi\right\rangle=-\left\langle\phi,q\right\rangle=-\left\langle\phi^{(n)},q^{(n)}\right\rangle=\left\langle(L^{(n)})^{-1}q^{(n)},q^{(n)}\right\rangle
=i=1n1q(n)(i)((L(n))1q(n))(i).\displaystyle=\sum\limits_{i=1}^{n-1}q^{(n)}(i)((L^{(n)})^{-1}q^{(n)})(i).

By Cramer’s rule, we get,

εEC(G0)=1det(L(n))i=1n1q(n)(i)det(Li(n)),\mbox{\Large$\varepsilon$}_{EC}(G_{0})=\dfrac{1}{\text{det}(L^{(n)})}\sum\limits_{i=1}^{n-1}q^{(n)}(i)\text{det}(L^{(n)}_{i}),

where det(Li(n))\text{det}(L^{(n)}_{i}) is the matrix obtained by replacing the ii-th column of L(n)L^{(n)} by q(n)q^{(n)}. Now by expanding along the ii-th column, we get

εEC(G0)=1det(L(n))i,j=1n1q(n)(i)q(n)(j)det(L(j,i)(n)),\mbox{\Large$\varepsilon$}_{EC}(G_{0})=\dfrac{1}{\text{det}(L^{(n)})}\sum\limits_{i,j=1}^{n-1}q^{(n)}(i)q^{(n)}(j)\text{det}(L^{(n)}_{(j,i)}),

where L(j,i)(n)L^{(n)}_{(j,i)} is the matrix obtained by removing the jj-th row and the ii-th column from L(n)L^{(n)}. Note that L(j,i)(n)=B(j,n)(B(i,n))L^{(n)}_{(j,i)}=B^{(j,n)}(B^{(i,n)})^{*} and hence

εEC(G0)=1det(L(n))i,j=1n1q(n)(i)q(n)(j)det(B(j,n)(B(i,n))),\mbox{\Large$\varepsilon$}_{EC}(G_{0})=\dfrac{1}{\text{det}(L^{(n)})}\sum\limits_{i,j=1}^{n-1}q^{(n)}(i)q^{(n)}(j)\text{det}(B^{(j,n)}(B^{(i,n)})^{*}),

where B(j,n)B^{(j,n)} is the matrix obtained by removing the jj-th and the nn-th rows from BB. Now by Binet-Cauchy theorem (see [6]), it follows that

εEC(G0)=1det(L(n))i,j=1n1(1)i+jq(n)(i)q(n)(j)HG0|E(H)|=n2det(BH(n,j))det((BH(n,i))),\mbox{\Large$\varepsilon$}_{EC}(G_{0})=\dfrac{1}{\text{det}(L^{(n)})}\sum\limits_{i,j=1}^{n-1}(-1)^{i+j}q^{(n)}(i)q^{(n)}(j)\sum\limits_{\begin{subarray}{c}H\subset G_{0}\\ |E(H)|=n-2\end{subarray}}\text{det}(B_{H}^{(n,j)})\text{det}((B_{H}^{(n,i)})^{*}),

where BH(n,j)B_{H}^{(n,j)} is the matrix obtained by choosing the columns corresponding to the edges in HH from B(j,n)B^{(j,n)}. ∎

Now we give the proof of Theorem 2 by applying Proposition 1. Observing that ρ=1/2\rho=1/2, q(n)(1)=1ρ=1/2q^{(n)}(1)=1-\rho=1/2 and q(n)(i)=0q^{(n)}(i)=0 (i=2,,n1)(i=2,\dots,n-1) in our setting of Theorem 2 and applying Proposition 1 to our setting, we get

εEC(G0)=141det(L(n))HG0|E(H)|=n2(det(BH(n,1)))2.\mbox{\Large$\varepsilon$}_{EC}(G_{0})=\frac{1}{4}\dfrac{1}{\text{det}(L^{(n)})}\sum\limits_{\begin{subarray}{c}H\subset G_{0}\\ |E(H)|=n-2\end{subarray}}\left(\text{det}(B_{H}^{(n,1)})\right)^{2}.

Focusing on the linearly dependence on the column vectors of the incidence matrix of the spanning subgraph HG0H\subset G_{0}, we obtain the following:

  1. 1.

    If HH contains a cycle, then det(BH(n,1))=0\text{det}(B_{H}^{(n,1)})=0;

  2. 2.

    If HH contains a connected component including both u1u_{1} and unu_{n}, then det(BH(n,1))=0\text{det}(B_{H}^{(n,1)})=0.

Hence, it implies that if det(BH(n,1))0\text{det}(B_{H}^{(n,1)})\neq 0. Then HH is a spanning forest which contains exactly two components, one containing u1u_{1} and the other one containing unu_{n}. On the other hand, if HH is a spanning forest which contains exactly two components, one containing u1u_{1} and the other one containing unu_{n}, BH(n,1)B_{H}^{(n,1)} is of the form

BH(n,1)=(BT100BT2)B_{H}^{(n,1)}=\left(\begin{array}[]{@{}c|c@{}}B_{T_{1}}&0\\ \hline\cr 0&B_{T_{2}}\end{array}\right)

for the trees T1T_{1} and T2T_{2} in the forest which are the spanning trees of the two components of the forest. Now by [2], det(BT1)=det(BT2)=±1\text{det}(B_{T_{1}})=\text{det}(B_{T_{2}})=\pm 1 and hence (det(BH(n,1)))2=1\left(\text{det}(B_{H}^{(n,1)})\right)^{2}=1. Thus, (det(BH(n,1)))2=1\left(\text{det}(B_{H}^{(n,1)})\right)^{2}=1 if and only if HH is a spanning forest which contains exactly two components, one containing u1u_{1} and the other one containing unu_{n}. Hence it follows that

4εEC(G0)=χ2(G0;u1,un)χ1(G0),4\mbox{\Large$\varepsilon$}_{EC}(G_{0})=\dfrac{\chi_{2}(G_{0};u_{1},u_{n})}{\chi_{1}(G_{0})},

where χ1(G0)\chi_{1}(G_{0}) is the tree number of G0G_{0} and χ2(G0;u1,un)\chi_{2}(G_{0};u_{1},u_{n}) is the number of spanning forests of G0G_{0} with exactly two components, one containing u1u_{1} and the other containing unu_{n}.

Let G0G_{0} be a bipartite graph. Then

εQW\displaystyle\mbox{\Large$\varepsilon$}_{QW} =12aA0|Ψ(a)|2=12aA0|j(a)+ρ|2=12aA0j(a)2+ρaA0j(a)+ρ2|E0|\displaystyle=\dfrac{1}{2}\sum\limits_{a\in A_{0}}|\Psi_{\infty}(a)|^{2}=\dfrac{1}{2}\sum\limits_{a\in A_{0}}|j(a)+\rho|^{2}=\dfrac{1}{2}\sum\limits_{a\in A_{0}}j(a)^{2}+\rho\sum\limits_{a\in A_{0}}j(a)+\rho^{2}|E_{0}|
=εEC(G0)+ρ2|E0|.\displaystyle=\mbox{\Large$\varepsilon$}_{EC}(G_{0})+\rho^{2}|E_{0}|. (12)

Note that in our setting with only two tails, we have ρ2=14\rho^{2}=\dfrac{1}{4}, which leads to the formula in the theorem.

Non-bipartite case

Now let G0G_{0} be non-bipartite. Then we have ψ(e)=ψ(e¯)=(B~ϕ)(e)\psi_{\infty}(e)=\psi_{\infty}(\bar{e})=(\tilde{B}^{*}\phi)(e) for any eE0e\in\vec{E}_{0} and Qϕ=qQ\phi=-q. Now since G0G_{0} is non-bipartite, it follows that QQ is invertible (for example, see [6, Theorem 7.8.1]) and hence ϕ=Q1q\phi=-Q^{-1}q. By a similar argument, it follows that

εQW(G0)\displaystyle\mbox{\Large$\varepsilon$}_{QW}(G_{0}) =12aA0|Ψ(a)|2=12ψ,ψ=B~ϕ,B~ϕ=ϕ,Qϕ=Q1q,q.\displaystyle=\dfrac{1}{2}\sum\limits_{a\in A_{0}}|\Psi_{\infty}(a)|^{2}=\dfrac{1}{2}\left\langle\psi_{\infty},\psi_{\infty}\right\rangle=\left\langle\tilde{B}^{*}\phi,\tilde{B}^{*}\phi\right\rangle=\left\langle\phi,Q\phi\right\rangle=\left\langle Q^{-1}q,q\right\rangle.

By using the Cramer’s rule and the Binet-Cauchy theorem, we can derive a similar expression for the non-bipartite graphs as follows:

Proposition 2.

Let G0G_{0} be a non-bipartite graph with arbitrary boundary . Then we have

εQW(G0)=1det(Q)i,j=1n(1)i+jq(i)q(j)HG0|E(H)|=n1det(B~H(i))det(B~H(j)),\mbox{\Large$\varepsilon$}_{QW}(G_{0})=\dfrac{1}{\text{det}(Q)}\sum\limits_{i,j=1}^{n}(-1)^{i+j}q(i)q(j)\sum\limits_{\begin{subarray}{c}H\subset G_{0}\\ |E(H)|=n-1\end{subarray}}\text{det}(\tilde{B}_{H}^{(i)})\text{det}(\tilde{B}_{H}^{(j)}),

where B~H(j)\tilde{B}_{H}^{(j)} is the matrix obtained by choosing the columns corresponding to the edges in HH and removing the jj-th row from B~\tilde{B}.

Observing that q(j)=δ1(j)q(j)=\delta_{1}(j) in our setting and applying Proposition 2, then we have

εQW(G0)=1det(Q)HG0|E(H)|=n1(det(B~H(1)))2.\mbox{\Large$\varepsilon$}_{QW}(G_{0})=\dfrac{1}{\text{det}(Q)}\sum\limits_{\begin{subarray}{c}H\subset G_{0}\\ |E(H)|=n-1\end{subarray}}\left(\text{det}(\tilde{B}_{H}^{(1)})\right)^{2}.

Focusing on the linearly dependence on the column vectors of the incidence matrix of the spanning subgraph HG{u1}H\subset G\setminus\{{u_{1}}\}, we obtain the following:

  1. 1.

    If HH has an even cycle, then det(B~H(1))=0\text{det}(\tilde{B}_{H}^{(1)})=0;

  2. 2.

    If HH has a connected component having at least two odd cycles, then det(B~H(1))=0\text{det}(\tilde{B}_{H}^{(1)})=0;

  3. 3.

    If HH has a connected component having an odd cycle and u1u_{1}, then det(B~H(1))=0\text{det}(\tilde{B}_{H}^{(1)})=0.

Hence it implies that if det(B~H(1))0\text{det}(\tilde{B}_{H}^{(1)})\neq 0 then H=TC1CkH=T\cup C_{1}\cup...\cup C_{k} where u1Tu_{1}\in T where TT is a tree and C1CkC_{1}\cup...\cup C_{k} are odd unicycles. Now if H=TC1CkH=T\cup C_{1}\cup...\cup C_{k} where u1Tu_{1}\in T where TT is a tree and C1CkC_{1}\cup...\cup C_{k} are odd unicycles, then B~H(1)\tilde{B}_{H}^{(1)} is of the form

B~H(1)=(B~T000B~C10...00B~Ck).\tilde{B}_{H}^{(1)}=\left(\begin{array}[]{@{}c|c|c|c@{}}\tilde{B}_{T}&0&...&0\\ \hline\cr 0&\tilde{B}_{C_{1}}&...&0\\ \hline\cr.&&&\\ .&&&\\ .&&&\\ \hline\cr 0&...&0&\tilde{B}_{C_{k}}\end{array}\right).

Note that det(B~T)=±1\text{det}(\tilde{B}_{T})=\pm 1 (by [2]) and by expanding the determinant of the non-oriented incidence matrix of a cycle, we can show that det(B~Ci)=±2\text{det}(\tilde{B}_{C_{i}})=\pm 2 (see Appendix for the details), which implies that

(det(B~H(1)))2=4ω(H)1,\left(\text{det}(\tilde{B}_{H}^{(1)})\right)^{2}=4^{\omega(H)-1},

where ω(H)\omega(H) is the number of components in HH. Furthermore, let

𝒞o={{C1,,Cr}|Ci’s are odd unicyclic graphs}\mathcal{C}_{o}=\{\{C_{1},...,C_{r}\}|C_{i}\text{'s are odd unicyclic graphs}\}

and

𝒯𝒞o={{T,C1,,Ck}|u1T where T is a tree and Ci’s are odd unicyclic graphs}.\mathcal{T}\mathcal{C}_{o}=\{\{T,C_{1},...,C_{k}\}|u_{1}\in T\text{ where }T\text{ is a tree and }C_{i}\text{'s are odd unicyclic graphs}\}.

Then by [5],

det(Q)=HC04ω(H)\text{det}(Q)=\sum\limits_{H\in C_{0}}4^{\omega(H)}

and it follows that

εQW=H𝒯𝒞o4ω(H)1HC04ω(H)\mbox{\Large$\varepsilon$}_{QW}=\dfrac{\sum\limits_{H\in\mathcal{T}\mathcal{C}_{o}}4^{\omega(H)-1}}{\sum\limits_{H\in C_{0}}4^{\omega(H)}}

which completes the proof. \hfill\square

Appendix

Let CiC_{i} be an odd unicyclic graph and let B~Ci\tilde{B}_{C_{i}} be the non-oriented incident matrix of CiC_{i}. Let us prove that det(B~Ci)=±2\text{det}(\tilde{B}_{C_{i}})=\pm 2. Since an odd unicyclic graph is obtained by connecting trees to an odd cycle (say CC^{\prime}), If we expand det(B~Ci)\text{det}(\tilde{B}_{C_{i}}) along the row corresponding to a leaf of a tree connected to CC^{\prime}, it reduces to a similar determinant. By expanding recursively along the rows corresponding to the leaves of the graph, we obtain

det(B~Ci)=±det(BC).\text{det}(\tilde{B}_{C_{i}})=\pm\text{det}(B_{C^{\prime}}).

Now it is enough to show that det(BC)=2\text{det}(B_{C^{\prime}})=2. Observe that we have

det(BC)=det(10011100......0011).\text{det}(B_{C^{\prime}})=\text{det}\left(\begin{array}[]{@{}c c c c c@{}}1&0&...&0&1\\ 1&1&0&...&0\\ .&.&&&\\ .&&.&&\\ .&&&.&\\ 0&...&0&1&1\end{array}\right).

Expanding the determinant in the right hand side along the first row, since there are odd number of rows and columns, we get

det(BC)=det(10001100......0011)+det(11000110......0001).\text{det}(B_{C^{\prime}})=\text{det}\left(\begin{array}[]{@{}c c c c c@{}}1&0&...&0&0\\ 1&1&0&...&0\\ .&.&&&\\ .&&.&&\\ .&&&.&\\ 0&...&0&1&1\end{array}\right)+\text{det}\left(\begin{array}[]{@{}c c c c c@{}}1&1&...&0&0\\ 0&1&1&...&0\\ .&.&&&\\ .&&.&&\\ .&&&.&\\ 0&...&0&0&1\end{array}\right).

Now to compute the two determinants in the right hand side, we expand the first determinant recursively along the rows and the second along the columns and we obtain that these two determinants are equal to 11. Hence we conclude that

det(B~Ci)=±2.\text{det}(\tilde{B}_{C_{i}})=\pm 2.

Acknowledgments: Yu.H. acknowledges financial supports from the Grant-in-Aid of Scientific Research (C) Japan Society for the Promotion of Science (Grant No. 18K03401). M.S. acknowledges financial supports from JST SPRING (Grant No. JPMJSP2114). E.S. acknowledges financial supports from the Grant-in-Aid of Scientific Research (C) Japan Society for the Promotion of Science (Grant No. 19K03616) and Research Origin for Dressed Photon. We would like to thank prof. Hajime Tanaka for the invaluable discussions and the comments which were very useful to carry out this study.

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