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Abstract
The focus of this study is scattering matrices in the framework of quantum graphs,more precisely the matrices which describe equi-transmission. They are unitary andHermitian and are independent of the energies of the associated system. In the firstarticle it is shown that in the case where reflection does not occur, such matrices existonly in even dimensions. A complete description of the matrices in dimensions 2, 4,and 6 is given. In dimension 6, 60 five-parameter families are obtained. The 60 matri-ces yield a combinatorial bipartite graph K2
6 . In the second article it is shown that whenreflection is allowed, the standard matching conditions matrix is equi-transmitting forany dimension n. All equi-transmitting matrices up to order 6 are described. For oddn (3 and 5), the standard matching conditions matrix is the only equi-transmitting ma-trix. For even n (2, 4 and 6) there exists other equi-transmitting matrices apart fromthose equivalent to the standard matching conditions. All such additional matriceshave zero trace.
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Contents
Abstract v
Acknowledgements ix
I Basic concepts and introduction 11.1 Quantum graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Matching conditions . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Vertex conditions and vertex scattering matrix . . . . . . . . . . . . . 7
References 11
II Research Articles 13
Paper I On Reflectionless Equi-transmitting Matrices 15
Paper II On Equi-transmitting Matrices 31
III Appendices 55
A Reflectioless Equi-transimtting Matrices of Order 6 57
B Equi-transmitting Matrices of Order 6 81
Acknowledgements
My first appreciation goes to my advisor Pavel Kurasov for introducing me to thesubject and guiding me through the research. His patience, motivation and support areduly recognised.
I would also like to appreciate the students and staff at the Mathematics Depart-ment, Stockholm University who have extended a right hand of fellowship in one wayor the other. In particular I thank Muhammad Usman and Marlena Nowaczyk for theuseful discussions we had and Jared Ongaro for helping me with matters of LaTeXwhenever I got into miry grounds.
Many thanks to ISP for the financial support for my studies, Mathematics Depart-ment of Stockholm University for hosting me and University of Nairobi for releasingme for studies.
Heartfelt gratitude to my wife Betty and the children for their love and patienceespecially given that I have had to be away from them for long periods.
!
Part I
Basic concepts and introduction
1
1.1 Quantum graphs
A quantum graph is a metric graph on which a differential operator is defined. Todefine the operator as a self-adjoint operator one needs to introduce certain vertexconditions. Below we give a description of the three components of a quantum graph.
Metric graph
A discrete or combinatorial graph is an ordered pair (VVV ,EEE) of the set VVV = {xi} ofvertices and the set EEE of edges connecting the vertices. The set of edges can be viewedas the set of 2-element subsets of VVV . In combinatorial graphs, the focus is on thevertices with edges serving the purpose of indicating connectivity. A metric graph isobtained by assigning positive length to the edges and the attention shifts to the edges.Consider N compact and semi-infinite edges En, each considered as a subset of anindividual copy of R
En =
{[x2n−1,x2n] , n = 1, . . . ,Nc
[xn,∞) , n = Nc +1, . . . ,Nc +Ni = N,
where Nc (respectively Ni) denotes the number of compact (respectively semi-infinite)intervals. Let VVV = {xi} be the set of all endpoints of the intervals En. Its arbitrarypartition into M equivalent classes Vm, m = 1, . . . ,M gives the set of vertices.
Definition 1.1. A metric graph Γ is the union of the edges En with the end pointsbelonging to the same vertex identified
Γ = ∪Nn=1En/x∼y,
where the equivalence relation is defined as
x∼ y⇐⇒[
x = yx 6= y⇒∃Vm : x,y ∈Vm.
The number vm of elements in the class Vm is called the valence of Vm. Note thatin a metric graph, intermediate points in an edge are also considered as points on thegraph. The distance between any two points on the graph is therefore the shortestmetric distance between them. This means that for any two points belonging to thesame edge and are close to each other, the distance between them is the usual length
3
of the corresponding interval. A finite graph is compact if Ni = 0. Each compact edgehas the length ln = x2n−x2n−1, and the total length of a compact metric graph is givenas
L=N
∑n=1
ln.
With edges of a metric graph so described, one can then define functions on Γ.The inclusion of intermediate points on the edges enables one to define the Lebesguemeasure dx on the graph. Some function spaces that can be defined on metric graphsinclude the Hilbert space L2 (Γ) = ⊕En∈EEEL2 (En) and the Sobolev space W 2
2 (Γ) =⊕En∈EEEW 2
2 (En). The boundary values of functions defined on the edges are given by
u(x j) = limx→x j
u(x).
The boundary values of the normal derivatives are given by
∂nu(x j) =
limx→x j
ddx
u(x) x j is the left endpoint
− limx→x j
ddx
u(x) x j is the right endpoint.
while that of the extended normal derivatives are given by
∂nu(x j) =
limx→x j
(ddx
u(x)− ia(x)u(x)), x j is the left endpoint
− limx→x j
(ddx
u(x)− ia(x)u(x)), x j is the right endpoint.
where a(x) is the magnetic potential of the magnetic Schrödinger operator. In all theselimits, x approaches x j from inside the corresponding edge. The normal derivativesare independent of the direction in which the edge is parameterized.
Differential operators
Differential operators play the role of describing the wave dynamics along the edgesin a metric graph. The differential operators usually used on quantum graphs are
• Laplace operatorL =− d2
dx2 ;
• Schrödinger operatorLq =− d2
dx2 +q(x) ;
• magnetic Schrödinger operator
Lq,a =(i d
dx +a(x))2
+q(x) .
4
The functions a(x) and q(x) are the magnetic and electric potentials respectively bothof which are real-valued. In addition the magnetic potential is continuously differ-entiable and the electric potential is locally square integrable and decays on infiniteedges. Note that the differential operators described here are not self-adjoint unlessappropriate vertex conditions at the vertices are introduced. The most commonly usedoperator, for simplicity, is the Laplace operator. In this study we restrict ourselves tothe Laplace operator unless stated otherwise.
Vertex conditions
Matching conditions serve two main purposes. One they indicate how the edges arecoupled together thereby describing the topology of the metric graph. Secondly theyare necessary to ensure that the differential operator is self-adjoint. Since vertex con-ditions are the main focus of our studies, the whole of the next section is devoted todiscussing them. We now give formal definition of a quantum graph.
Definition 1.2. A quantum graph is a metric graph furnished with a differential oper-ator which acts on functions defined on the edges. The functions satisfy certain vertexconditions so as to make the differential operator self adjoint.
1.2 Matching conditions
Vertices of valence one are called boundary vertices while those with valency greaterthan one are called internal vertices. Vertex conditions defined at boundary verticesare therefore called boundary conditions while those defined on internal vertices arecalled matching conditions.
The study of vertex conditions of a quantum graph is generally localised to ver-tices. It thus suffices in such studies to consider a star graph. The results obtained canthen be extended to the whole quantum graph.
Suppose for some vertex conditions a vertex V can be partitioned into vertices V ′
and V ′′ such that V = V ′ ∪V ′′ and V ′ ∩V ′′ = ∅ then such vertex conditions are notproperly connecting. If such a partitioning is not possible then the vertex conditionsare properly connecting. This study deals with properly connecting vertex conditions.
Self-adjoint vertex conditions parameterization
Various self-adjoint vertex conditions parameterization have emerged in the develop-ment of the field of quantum graphs. Below we consider some of them. Supposethe domain of the Laplacian defined on Γ is the Hilbert space L2 (Γ) =⊕En∈EEEL2 (En).Then the solution of the differential equation Lu(x) = λu(x), λ = k2 is given by
u(x) = aeikx +be−ikx.
Since the differential operator is of second order, it follows from the theory of ODEthat two boundary conditions are necessary to connect the functional values and thederivatives. In [8] Kostrykin and Schrader introduced the parameterization of vertexconditions at a vertex V with valence v as
5
A~u(V )+B~u′ (V ) = 0,
where ~u(V ) is the vector of the boundary values{
u(x j)}
x j∈V , ~u′ (V ) is the vector of
the boundary normal derivatives{
∂nu(x j)}
x j∈V and A and B are two v× v matrices.This parameterization results in a self-adjoint vertex conditions if and only if the v×2vmatrix (A,B) has maximal rank and the matrix AB∗ is Hermitian. The proof of thisresult can be found in [1; 9]. This parameterization fails to be unique, for supposewe multiply the above equation to the left by some arbitrary invertible v× v matrix C.Then we obtain the condition CAu(V )+CBu′ (V ) = 0 where the matrix (CA,CB) hasmaximal rank and the matrix CA(CB)∗ is Hermitian.
To address the problem of uniqueness, Harmer in [5] defined another self-adjointboundary parameterization which can be formulated as follows. Instead of matricesA and B, consider any v× v unitary matrix U . Then it is required that the boundaryvalues of u satisfy
i(U− I)~u(V ) = (U + I)~u′ (V ) . (1.1)
In [10] Kurasov and Nowaczyk gave an equivalent parameterization described asfollows. Consider a star graph with v vertices at the central vertex parameterized byΓ =∪v
j=1 [0,∞). To the Laplace operator defined on Γ the corresponding maximal andminimal operators are given by the same formal expression as L but their domains aredefined as Dom(Lmax) =⊕v
j=1W 22 ((0,∞)) and Dom(Lmin) =⊕v
j=1C∞0 ((0,∞)) respec-
tively.
Theorem 1. The family of all self-adjoint extensions of the minimal operator Lmin canbe uniquely parameterized by an arbitrary v×v unitary matrix S, so that the operatorLS is the restriction of Lmax = L∗min to the set of functions satisfying the matchingconditions
i(S− I)~u(V ) = (S+ I)~u′ (V ) . (1.2)
This parameterization is identical to (1.1) provided S =U . However the latter hasa clear physical interpretation, namely, when the value of the energy k = 1 then thematrix S coincide with the vertex scattering matrix1. In this study we will use theself-adjoint parameterization (1.2)
Let us consider some examples of the matching conditions that appear in the liter-ature.
1. Standard matching conditions(SMC): The most commonly used matching/boundaryconditions, called standard, (Neumann, or Kirchhoff) matching conditionsare
u is continuous at every vertex Vm
∑x j∈Vm
∂nu(x j) = 0.
1See definition 1.3
6
The derivatives are taken in the direction moving away from the vertex, i.e.they are normal derivatives. One observes that in the case of a boundary vertex,that is a vertex with valence one, the conditions reduce to a single Neumanncondition ∂nu(x j) = 0. For the vertex of valence two, the standard matchingconditions imply that both the function and its derivatives are continuous at Vm.Hence the vertex can be removed and the two edges substituted with a singleedge whose length is the sum of the two original edges.
2. δ−interaction: This matching condition is described by the following equa-tions
u is continuous at every vertex Vm
∑x j∈Vm
∂nu j (V ) = α ·u(V ), α ∈ R.
If α = ∞ then u(V ) = 0 and from the continuity of the function at V this match-ing condition becomes the Dirichlet condition.
3. δ ′−interaction: This is viewed as the dual of the δ−interaction vertex condi-tion. It is described by the following equations
∂nui (V ) = ∂nu j (V ) , i 6= j
∑x j∈Vm
u j (V ) = β ·∂nu(V ), β ∈ R.
If β = ∞ then ∂nu(V ) = 0 and from the first condition this matching conditionbecomes the Neumann condition.
1.3 Vertex conditions and vertex scattering matrix
Related to the study of coupling of edges at a vertex is the concept of scattering ofwaves. The scattering of waves at a vertex is described by means of the vertex scatter-ing matrix. For the star graph Γ, consider arbitrary solution to the Helmholtz equation−u′′ = k2u given by
u(x) = a jeik(x−x j) +b je−ik(x−x j), x ∈ E j. (1.3)
Let the amplitude of the incomming wave along edge E j be b j while that of theoutgoing wave along edge Ei be ai with the corresponding vectors given by~b and ~arespectively.
Definition 1.3. The v× v unitary matrix Sv (k) such that
~a = Sv (k)~b, (1.4)
7
is the vertex scattering matrix.
We now discuss the relation between the scattering matrix and the self-adjointvertex conditions parameterization. The vertex value of the solution (1.3) at the vertexV in the star graph Γ=∪v
j=1 [0,∞) is~u(0) =~a+~b. The value of the normal derivatives
at the vertex is ∂n~u(0) = ik(~a−~b
). Substituting these values into equation (1.2) we
obtain that
(S− I)(~a+~b
)= k (S+ I)
(~a−~b
).
Substituting the value of ~a from (1.4) into this equation and simplifying one obtainsthe following
SV (k) = S(k+1)+(k−1)IS(k−1)+(k+1)I . (1.5)
It can be observed that for k = 1, the arbitrary unitary matrix S is precisely thevertex scattering matrix, that is SV (1) = S. This makes the parameterization (1.2)special in that it has relevant interpretation.
Since matrix S is unitary, it possesses the following spectral decomposition
S = ∑vn=1 eiθn〈~en, ·〉~en,
where eiθn , θ ∈ [−π,π] are the eigenvalues of S corresponding to the eigenvectors~enwhich are chosen orthonormal. Plugging this representaion of S into equation (1.5)the following spectral decomposition of SV (k) is obtained
SV (k) = ∑vn=1
k(eiθn+1)+(eiθn−1)k(eiθn+1)−(eiθn−1)
〈~en, ·〉~en.
From this equation it can be seen that both S and SV (k) have the same eigenvec-tors. The eigenvalues of SV (k) depend on the energy k. For the eigenvalues 1 and−1 of S the corresponding eigenvalues of SV (k) are also 1 and −1 and the quotientk(eiθn+1)+(eiθn−1)k(eiθn+1)−(eiθn−1)
independent of k. These eigenvalues are stationary while all the
other eigenvalues tend to 1 as k→ ∞.It follows that if S is chosen such that the spectrum σ (S) = {±1}, then the corre-
sponding vertex scattering matrix is independent of the energy. For this to happen it isnecessary that S should not only be unitary, but also Hermitian. It is easy to see that ifS is Hermitian then SV (k) is Hermitian as well.
The vertex scattering matrix can be seen as the unitary matrix which describesthe probabilities that the waves penetrate the vertex from one edge to another. Inquantum mechanics such probabilities are given by the squared absolute value of thecorresponding matrix entries
∣∣si j∣∣2. One may say that all edges are equivalent if and
only if
8
{∣∣si j∣∣2 = |slm|2 , i 6= j, l 6= m
|sii|2 =∣∣s j j∣∣2 , i 6= j.
Moreover we shall require that the matrices are independent of the energy. Weare going to call such matrices equi-transmitting. Below we give the definition of theequi-transmitting matrices we have studied in this thesis.
Definition 1.4. An n× n matrix S is equi-transmitting (ET-matrix) if the followinghold
1. S = S∗ = S−1,
2. |sii|=∣∣s j j∣∣ , i 6= j,
3.∣∣si j∣∣= |slm| , i 6= j, l 6= m.
Let us denote the moduli of the reflection and transmission coefficients as follows
|sii|= r for all i and∣∣si j∣∣= |slm|= t, i 6= j, l 6= m.
When r = 0 the corresponding matrix is a reflectionless equi-transmitting matrix (RET-matrix).
Harrison, Smilansky and Winn have studied RET-matrices in [11]. The only re-quirement they set for such matrices is unitarity. They have been able to constructsuch matrices of order five which is not necessarily ET for all energies, but just for theone particular energy that they considered.
Turek and Cheon [13] have studied equi-transmitting matrices. In their study, theydescribe bounds for the ratio d = r/t. They also constructed several examples of ETmatrices in even dimension using famous Hadamard matrices.
Our goal is to describe all ET and RET-matrices up to dimension 6. The ET-matrices can be represented as S = D∗θCDθ where
C =
r t t · · · tt r ta23 · · · ta2nt ta23 r ta3n...
.... . .
...t ta2n ta3n · · · r
,
Dθ = diag{
eiθ1 ,eiθ2 , . . . ,eiθn}, θi ∈ [0,2π) , i = 1,2, . . . ,n,
ai j ∈{C :∣∣ai j∣∣= 1, for all i = 2, . . . ,n−1, j = 3, . . . ,n
}.
The matrices S and C are equal up to the phases defined by Dθ and the equi-transmission properties of S are preserved in C. We therefore narrow down our studyof ET-matrices to matrix C and in particular describe the results in terms of matrix C.
In chapter 2 we study RET-matrices, that is when r = 0, where we construct suchmatrices for orders 2, 4 and 6. It is proved that such matrices exist only for evendimensions.
9
Chapter 3 deals with ET matrices when r is in the interval (0,1). It is shown thatstandard matching conditions determine ET-matrices in any dimension and there existother ET-matrices generated by them. We have also shown that for n = 3 and 5 onlysuch ET-matrices exist. Our conjecture is that this remains true for any odd n≥ 7.
For even n ∈ {2,4,6} and when the trace of the matrix C is zero, there exist otherET-matrices apart from those generated by the SMC-matrices as follows
1. In all the three cases there is a one-parameter family matrices.
2. When n = 6 there is a unique real matrix equivalent to the SMC-matrix and atwo-parameter family matrices.
Again, our conjecture is that this remains true for even n ≥ 8, that is, there exitsother ET-matrices not generated by the standard matching conditions. This study hastherefore given a complete description of all possible ET-matrices for n≤ 6.
10
References
[1] G. Berkolaiko and P. Kuchment. Introduction to quantum graphs, volume 186 of Mathematical Surveys and Mono-graphs. American Mathematical Society, Providence, RI, 2013. 6
[2] V. Caudrelier and E. Ragoucy. Direct computation of scattering matrices for general quantum graphs. Nuclear Phys.B, 828(3):515–535, 2010.
[3] R. Diestel. Graph theory, volume 173 of Graduate Texts in Mathematics. Springer, Heidelberg, fourth edition, 2010.
[4] A. Friedman. Foundations of modern analysis. Dover Publications Inc., New York, 1982. Reprint of the 1970original.
[5] M. Harmer. Hermitian symplectic geometry and extension theory. J. Phys. A, 33(50):9193–9203, 2000. 6
[6] M. Harmer. Hermitian symplectic geometry and the factorization of the scattering matrix on graphs. J. Phys. A,33(49):9015–9032, 2000.
[7] R. B. Holmes and V. I. Paulsen. Optimal frames for erasures. Linear Algebra Appl., 377:31–51, 2004.
[8] V. Kostrykin and R. Schrader. Kirchhoff’s rule for quantum wires. II. The inverse problem with possible applicationsto quantum computers. Fortschr. Phys., 48(8):703–716, 2000. 5
[9] P. Kurasov. Quantum graphs: Spectral theory and inverse problems, 2014. monograph in preparation. 6
[10] P. Kurasov and M. Nowaczyk. Geometric properties of quantum graphs and vertex scattering matrices. OpusculaMath., 30(3):295–309, 2010. 6
[11] Harrison J. M., Smilansky U., and Winn B. Quantum graph where back-scattering is prohibited. Journal of physicsA: Mathematical and theoritical, 40:14181–14193, 2007. 9
[12] C. Meyer. Matrix analysis and applied linear algebra. Society for Industrial and Applied Mathematics (SIAM),Philadelphia, PA, 2000. With 1 CD-ROM (Windows, Macintosh and UNIX) and a solutions manual (iv+171 pp.).
[13] O. Turek and T. Cheon. 9
11
Part II
Research Articles
13
ON REFLECTIONLESS EQUI-TRANSMITTING MATRICES
PAVEL KURASOV, RAO OGIK, AND AMAR RAUF
ABSTRACT. Reflectionless equi-transmitting unitary matrices are studied in con-nection to matching conditions in quantum graphs. All possible such matrices ofsize 6 are described explicitly. It is shown that such matrices form 30 six-parameterfamilies intersected along 12 five-parameter families closely connected to conferencematrices.
1 IntroductionUnitary and Hermitian matrices are used to describe different problems in physicsand are well-studied in mathematics. It might appear that all relevant questions havebeed studied and there is completely nothing to add, but modern fields of researchpose new questions and may shed light on old ones. Our interest in such matrices isrelated to quantum graphs - differential operators on metric graphs - an area of mod-ern mathematical physics expanding in recent years, see for example [1, 9]. In thesemodels transmission of waves through the vertices is described by finite unitary ma-trices, in general energy dependent. The size of these matrices, to be denoted by v,coincides with the number of edges connected at each vertex. The vertex scatteringmatrices are energy independent if they are not only unitary but are Hermitian as well.This explains our interest in unitary Hermitian matrices. It is natural to ask the fol-lowing question: Which scattering matrices describe edge couplings where the edgesare indistinguishable? In quantum mechanics transition probabilities are given by themodulus squares of matrix elements. Therefore we shall be interested in matrices Spossessing modular permutation symmetry
{|sii|= |s j j|, i, j = 1, . . . ,v;
|si j|= |slm|; i 6= j, l 6= m, i, j, l,m = 1, . . . ,v,(1.1)
where v is the valency (degree) of the corresponding vertex, i.e. the size of S. In whatfollows matrices satisfying (1.1) will be called equi-transmitting. Interest in suchmatrices in connection to quantum graphs is explained by the fact that so called stan-dard matching conditions (the function is continuous and the sum of normal deriva-tives is zero, also known as Kirchhoff or Neumann conditions), widely used in the
2010 Mathematics Subject Classification. Primary 34L25, 81U40, Secondary 35P25, 81V99.Key words and phrases. Quantum graphs, vertex scattering matrix, equi-transmitting matrices.TO APPEAR IN OPUSCULA MATHEMATICA
1
area, exhibit a non-physical behavior, since the corresponding vertex scattering ma-trix has a dominated diagonal
Sst =
−1+2/v 2/v 2/v . . .2/v −1+2/v 2/v . . .2/v 2/v −1+2/v . . .
......
.... . .
−→v→∞
−I. (1.2)
This fact was first pointed out in [6], where it was suggested to study reflectionlessequi-transmitting matrices - equi-transmitting matrices with zero diagonal. Examplesof such symmetric matrices were constructed. It was noticed that the problem toobtain such matrices is related to the classical (not yet completely solved) problem ofdescribing all Hadamard matrices. Constructed examples determine vertex scatteringmatrices, which are reflectionless and equi-transmitting at a certain energy, but not forall values of the energy parameter. In order to obtain vertex scattering matrices whichare reflectionless and equi-transmitting for all energies one should require that suchmatrices are also Hermitian (see Theorem 1). Therefore it appears natural to includeHermiticity requirement into the definition of reflectionless equi-transmitting matrices(see Definition 1).
General equi-transmitting matrices (without necessarily requiring zero diagonal)have been studied in a series of papers by Turek and Cheon [10, 11]. It was noticedthat real reflectionless equi-transmitting matrices have already been studied under thename of conference matrices. These are matrices with zero diagonal and ±1 outsidethe diagonal, such that all rows and columns are mutually orthogonal. The main focusof [11] is on the relation between the absolute values of the diagonal and non-diagonalelements in an equi-transmitting matrix.
The goal of our paper is to study reflectionless equi-transmitting matrices system-atically. Trivially such matrices exists only in even dimensions (Theorem 1). Thedimensions 2 and 4 can easily be studied leading to a one-parameter and two three-parameter families respectively. The first interesting dimension is 6. It appears that allreflectionless equi-transmitting matrices form 30 six-parameter families intersectingalong 12 five-parameter families. The five parameters can be interpreted as magneticpotentials between different edges. Removing these parameters one gets 12 confer-ence matrices of size 6. The structure of this family provides a beautiful example,which may be used to study such matrices in higher dimensions.
2 Vertex scattering matricesIt is sufficient when discussing the matching conditions for a quantum graph, to con-sider a star graph. By a star graph Γ, we mean the union of v semi-infinite intervalsemenating from a vertex V , i.e. Γ = ∪v
j=1 [0,∞).
2
Figure 1. Star graph with 6 semi-infinite edges
On the star graph Γ, consider the operator L =− d2
dx2 defined on the vector function~u from the Sobolev space W 2
2 ([0,∞) ,Cv) sastisfying the matching conditions
i(S− I)~u = (S+ I)∂~u, (2.1)
where S is a unitary v× v matrix. Here the value of a function u and its normalderivative at the vertex V are denoted by ~u =~u(0) and ∂~u =~u′ (0) respectively. Letus calculate the corresponding vertex scattering matrix SV (k). Suppose the vectorsrepresenting the amplitudes of the incoming and outgoing waves are denoted by~b and~a respectively. The solution of the differential equation −u′′ = λu can be written asu(x) =~be−ikx +~aeikx, ~a,~b ∈ Cv, λ = k2. Then the values of the vectors ~u and ∂~u at Vare given by
~u =~b+~a and ∂~u =−ik~b+ ik~a. (2.2)
The vertex scattering matrix can be given in terms of the amplitutes of the incomingand outgoing waves and so take the form ~a = SV (k)~b. Substituting this into (2.2) weobtain
~u =~b+SV (k)~b and ∂~u =−ik~b+ ikSV (k)~b.
With these values of~u and ∂~u, (2.1) becomes
i(S− I)(I +SV (k))~b = ik (S+ I)(−I +SV (k))~b.
From this we obtain an expression for the vertex scattering matrix which in general isenergy dependent:
SV (k) = (k+1)S+(k−1)I(k−1)S+(k+1)I .
It is given in terms of the unitary matrix S used to parameterize the matching con-ditions.
3
When k = 1, then SV (k) coincides with S. Since S is unitary, it has the spectral repre-sentation given by
S = ∑vn=1 eiθn 〈~en, ·〉~en
where θn ∈ (−π,π], and~en ∈Cv are the eigenvectors of S which are chosen orthonor-mal. Substituting this spectral representation into (??), we obtain
SV (k) = ∑vn=1
k(eiθn+1)+(eiθn−1)k(eiθn+1)−(eiθn−1)
〈~en, ·〉~en. (2.3)
The unitary matrix SV (k) has the same eigenvectors as the matrix S, but the eigenval-ues in general depend on the energy. However eigenvalues ±1 which coincide withthe eigenvalues ±1 of S do not depend on k. All other eigenvalues tend to 1 as k→∞.The high energy limit of SV (k) exists [4, 5, 8] and is given by
SV (∞) = limk→∞ SV (k) =−P−1 +(I−P−1) = I−2P−1,
where P−1 is the spectral projection into the eigenspace of the eigenvlaue −1.
3 Definition of equi-transmitting matricesOur aim is to study matrices S leading to so-called reflectionless equi-transmittingvertex scattering matrices. Such matrices have zero diagonal. The following theoremproves that all such matrices have to be not only unitary but also Hermitian.
Theorem 1. Suppose S is a v× v unitary matrix and let (SV (k)) j j = 0, j = 1, . . . ,v.Then v is even and S is Hermitian.
Proof. Let the eigenvalues of S be denoted by λn and the corresponding eigenvectorsby~en = (zn
1,zn2, . . . ,z
nv). Then for j = 1, . . . ,v, we have
(SV (k)) j j (k) = ∑vn=1
k(λn+1)+(λn−1)k(λn+1)−(λn−1)
∣∣∣znj
∣∣∣2= 0. (3.1)
This gives us a v× v system of equations. For brevity let’s introduce the notationan = λn +1 and bn = λn−1. Then for j = 1, . . . ,v the system of equations becomes
∑vn=1
kan+bnkan−bn
∣∣∣znj
∣∣∣2= 0.
Since the eigenvectors {~en} form an orthonormal basis, we have |zn1|2+ · · ·+ |zn
v |2 = 1,n = 1, . . . ,v. So summing all the equations in the system we end up with the singleequation
∑vn=1
kan+bnkan−bn
= 0. (3.2)
It follows that4
∑vn=1
{(kan +bn)∏v
m = 1m 6= n
(kam−bm)
}= 0. (3.3)
Assume first that v is odd. The function in the above equation can be seen as apolynomial in k and it is identically zero only if all coefficients at different powers arezero. In particular, the coefficient in front of kv gives us
v∏vn=1 an = 0. (3.4)
Without loss of generality we take av = 0 which yields λv =−1. Substituting thisinto equation (3.2), the last term in that sum equals −1. Hence we obtain
∑v−1n=1
kan+bnkan−bn
= 1. (3.5)
Simplifying this equation we have
∑v−1n=1
{(kan +bn)∏v−1
m = 1m 6= n
(kam−bm)
}= ∏v−1
n=1 (kan−bn). (3.6)
Proceeding as before and comparing the coefficients of the highest power of k we seethat
(v−1)∏v−1n=1 an = ∏v−1
n=1 an ⇒ (v−2)∏v−1n=1 an = 0. (3.7)
Since v is odd, v− 2 6= 0 and so without loss of generality we take av−1 = 0 so thatλv−1 = −1. Continuing in a similar manner, we observe that at every step, the coef-ficient (v− l) in the second equation in (3.7) is such that l ∈ 2N. Therefore v− l 6= 0for all applicable l. This has the consequence that an = 0 for all n = 1, . . . ,v so thatλn =−1 for all n = 1, . . . ,v. But that is impossible since the trace of S is zero. Hencev is even.
For convenience of manipulation, we set v = 2N, N ∈ N. By the precceding cal-culations we see that a2N = a2N−1 = · · · = aN+1 = 0 which implies that λn = −1,n = N +1, . . . ,2N. Taking into account that the trace of S is zero we get:
TrS = ∑2Nn=1 λn = ∑N
n=1 λn +(−1)N = 0 ⇒ ∑Nn=1 λn = N.
Since |λn|= 1 for all n = 1, . . . ,2N we obtain that λn = 1, n = 1, . . . ,N. Therefore allthe eigenvalues of S are not only on the unit circle but are also real and precisely are−1 and 1. Hence S is also Hermitian. �
In [6] Harrison, Smilansky and Winn studied reflectionless equi-transmitting ma-trices. The definition used there does not require that the matrix be Hermitian. In viewof Theorem 1 it is natural to require that reflectionless equi-transmitting matrices arenot only unitary, but Hermitian as well. Such matrices lead to energy independent ver-tex scattering matrices as can be deduced from (2.3). We therefore give the followingdefinition
5
Definition 1. An n× n matrix S is called Reflectionless Equi-Transmitting (RET-matrix) if it is unitary Hermitian and it has zero diagonal while the off-diagonal ele-ments have the same absolute value, i.e.
S = S−1 = S∗,
s j j = 0, j = 1, . . . ,n,
∣∣si j∣∣= |slm| , i 6= j, l 6= m; i, j, l,m = 1, . . . ,n.
4 Reflectionless Equi-Transmitting matrices of sizes twoand four
For the sake of completeness we describe here all RET-matrices for n = 2,4. It is clearthat RET-matrices in dimension two have the form
S =
(0 eiθ
e−iθ 0
)= diag
{1,e−iθ}
(0 11 0
)diag
{1,eiθ}, (4.1)
where θ ∈ [0,2π). We have just one one-parameter family.
Every RET-matrix in dimension four possesses the representation
S = diag{
1,e−iθ1 ,e−iθ2 ,e−iθ3} 1√
3C diag
{1,eiθ1 ,eiθ2 ,eiθ3
}, (4.2)
where θn ∈ [0,2π) and
C =
0 1 1 11 0 a b1 a 0 c1 b c 0
. (4.3)
The numbers a,b,c ∈ C have absolute value one and are chosen such that the rows(and columns) are orthogonal. It should be observed that the rows (and columns) ofC are already normalized. Using the orthogonality conditions of the rows of C andsolving for a,b and c we obtain a = ±i, b = −a and c = −a. From these values ofthe parameters we obtain the following two matrices, which are complex conjugate ortranspose of each other:
C1 =
0 1 1 11 0 i −i1 −i 0 i1 i −i 0
C2 =
0 1 1 11 0 −i i1 i 0 −i1 −i i 0
. (4.4)
Hence the set of all RET-matrices in dimension 4 consists of two 3−parameter nonin-tersecting families.6
5 Reflectionless Equi-Transmitting matrices of size sixTo construct RET-matrices in dimension six we use the following lemma
Lemma 1. The sum of four complex numbers z1,z2,z3,z4 of equal magnitude equalszero if and only if at least one of the following cases occur:
z2 =−z1,z4 =−z3,
or
z3 =−z1,z4 =−z2,
or
z4 =−z1,z3 =−z2.
To see that the lemma holds one only needs to observe that the sum of the fourcomplex numbers z j, j = 1,2,3,4 which have the same absolute value equals zero ifand only if they form the sides of a rhombus.
Similar to dimension four, equi-transmitting matrices in dimension six take theform
S = D−1 1√5
C D, (5.1)
where D = diag{
1,eiθ1 ,eiθ2 ,eiθ3 ,eiθ4 ,eiθ5}
, θn ∈ [0,2π) and
C =
0 1 1 1 1 11 0 a b c d1 a 0 e f g1 b e 0 h l1 c f h 0 m1 d g l m 0
. (5.2)
The parameters a,b,c,d,e, f ,g,h, l,m ∈ C have absolute value one and are chosen sothat the rows (columns) of S are orthogonal. The orthogonality conditions yield thefollowing 15 equations in 10 unknowns:
7
a+b+ c+d = 0 (1)a+ e+ f +g = 0 (2)b+ e+h+ l = 0 (3)c+ f +h+m = 0 (4)d +g+ l +m = 0 (5)
1+be+ c f +dg = 0 (6)1+ae+ ch+dl = 0 (7)1+a f +bh+dm = 0 (8)1+ag+bl + cm = 0 (9)
1+ab+ f h+gl = 0 (10)1+ac+ eh+gm = 0 (11)1+ad + el + f m = 0 (12)
{1+bc+ e f + lm = 0 (13)1+bd + eg+hm = 0 (14)
{1+ cd + f g+hl = 0 (15).
(5.3)
Below we give just two examples illustrating how all RET-matrices of size 6 canbe obtained.
Example 1 (Case 1.1). We recall that the parameters in the matrix C are complex numbersof absolute value one. Therefore for any of these parameters x we have that xx = |x|2 = 1.Applying Lemma 1 to equation (1) in the system (5.3) we end up with the following threecases:
Case 1 : b =−a, d =−c,
Case 2 : c =−a, d =−b, (5.4)
Case 3 : d =−a, c =−b.
Let us consider the first possibility (cases 2 and 3 can be treated in a similar way). Substi-tuting the values of b and d into (5.3), we obtain the following system:
a+ e+ f +g = 0 (2)−a+ e+h+ l = 0 (3)c+ f +h+m = 0 (4)−c+g+ l +m = 0 (5)
1−ae+ c f − cg = 0 (6)1+ae+ ch− cl = 0 (7)1+a f −ah− cm = 0 (8)1+ag−al + cm = 0 (9)
f h+gl = 0 (10)1+ac+ eh+gm = 0 (11)1−ac+ el + f m = 0 (12)
{1−ac+ e f + lm = 0 (13)1+ac+ eg+hm = 0 (14)
{f g+hl = 0 (15).
(5.5)
Equation (15) can be eliminated since it is a multiple of equation (10). Application of Lemma1 to equation (2) yields the following cases:
Case 1.1 : e =−a, g =− f ,
Case 1.2 : f =−a, g =−e, (5.6)
Case 1.3 : g =−a, f =−e.
8
We pick the first case again, and substitute the corresponding values of e and g into (5.5) andobtain the following system:
−a−a+h+ l = 0 (3)c+ f +h+m = 0 (4)−c− f + l +m = 0 (5)
1+a2 + c f + c f = 0 (6)ch− cl = 0 (7)1+a f −ah− cm = 0 (8)1−a f −al + cm = 0 (9)
f h− f l = 0 (10)1+ac−ah− f m = 0 (11)1−ac−al + f m = 0 (12)
{1−ac−a f + lm = 0 (13)1+ac+a f +hm = 0 (14)
(5.7)
From equation (7) in the above system we have that l = h. Substituting this value of l intoequation (3) of the same system, we have that ℜ(a) = h. This means that h ∈R and since it hasabsolute value one it follows that h =±1. This in turn implies that l =±1 and a =±1. Addingequations (4) and (5) and then substituting the values of l and h, we obtain m=∓1. Substitutinga = ±1 into equation (6) yields c f = −1⇒ f = −c. The parameter c can be chosen arbitraryleading to the following matrices:
C1.1 =
0 1 1 1 1 11 0 ±1 ∓1 c −c1 ±1 0 ∓1 −c c1 ∓1 ∓1 0 ±1 ±1 c −c ±1 0 ∓11 −c c ±1 ∓1 0
. (5.8)
�
Remark. The case we have considered will be numbered Case 1.1. This means thatwe have applied Lemma 1 twice and that in each case we choose the first option. Thefirst application is to equation (1) of the initial system which yields three possibilities.We choose the first option from the three possibilities with the corresponding substi-tution giving us the second system. This system now has at most fourteen equations,equation (1) having been eliminated by the above substitution. The second applicationof Lemma 1 is now to eqution (2) of the second system which also yields three pos-sibilities. The second "1" in the notation means that we choose the first option againfrom the second three possibilities in order to determine the matrix.
Example 2 (Case 3.1.1). Suppose that now we consider the third case in (5.4). Substitutingthe values of d =−a and c =−b into the system (5.3) we obtain the following system:
a+ e+ f +g = 0 (2)b+ e+h+ l = 0 (3)−b+ f +h+m = 0 (4)−a+g+ l +m = 0 (5)
1+be−b f −ag = 0 (6)1+ae−bh−al = 0 (7)1+a f +bh−am = 0 (8)1+ag+bl−bm = 0 (9)
1+ab+ f h+gl = 0 (10)1−ab+ eh+gm = 0 (11)el + f m = 0 (12)
{e f + lm = 0 (13)1−ab+ eg+hm = 0 (14)
{1+ab+ f g+hl = 0 (15).
(5.9)
9
From this system, we see that equation (13) is a multiple of equation (12) and so can bediscarded. Applying Lemma 1 to equation (2) of the system (5.9) we obtain the following threecases:
Case 3.1 : e =−a, g =− f ,
Case 3.2 : f =−a, g =−e,
Case 3.3 : g =−a, f =−e.
We pick the case 3.1 and substitute the corresponding values of e and g into the system (5.9)and thus obtain the following system:
b−a+h+ l = 0 (3)−b+ f +h+m = 0 (4)−a− f + l +m = 0 (5)
1−ab−b f +a f = 0 (6)−bh−al = 0 (7)1+a f +bh−am = 0 (8)1−a f +bl−bm = 0 (9)
1+ab+ f h− f l = 0 (10)1−ab−ah− f m = 0 (11)−al + f m = 0 (12)
{1−ab+a f +hm = 0 (14)
{ab+hl = 0 (15)
(5.10)
We discard equation (15) because it is a multiple of equation(7). Here we need to apply Lemma1 once more, this time to equation (3) of the system (5.10).
Case 3.1.1 : b = a, l =−h
Case 3.1.2 : h = a, l =−b (5.11)
Case 3.1.3 : l = a, h =−b.
We then pick the first case. Substituting the values of b and l in the case 3.1.1 into the system(5.10) gives us the following system:
{−a+ f +h+m = 0 (4)−a− f −h+m = 0 (5)
−a f +a f = 0 (6)−ah+ah = 0 (7)1+a f +ah−am = 0 (8)1−a f −ah−am = 0 (9)
1+a2 + f h+ f h = 0 (10)1−a2−ah− f m = 0 (11)ah+ f m = 0 (12)
{1−a2 +a f +hm = 0 (14).
(5.12)
From equation (7) we see that a ∈ R, ⇒ a = ±1. Adding equations (4) and (5) and thensubstituting the values of a we have that ℜ(m) = ±1⇒ m = ±1. Substituting the values of ainto equation (10) we obtain that f h =−1, which implies that h =− f . From these we see thatthe corresponding matrix is10
C3.1.1 =
0 1 1 1 1 11 0 ±1 ± ∓1 ∓11 ±1 0 ∓1 f − f1 ±1 ∓1 0 − f f1 ∓1 f − f 0 ±11 ∓1 − f f ±1 0
. (5.13)
�
Continuing as in the above examples in all cases we obtain 30 (different) one-parameter families of matrices. A summary of all the cases can be seen in table 1.Below we list all the 30 different parameter dependent matrices where the subscriptdenotes the row (and column) which does not contain a parameter.
A2 =
0 1 1 1 1 11 0 ±1 ±1 ∓1 ∓11 ±1 0 ∓1 α −α1 ±1 ∓1 0 −α α1 ∓1 α −α 0 ±11 ∓1 −α α ±1 0
,B2 =
0 1 1 1 1 11 0 ±1 ∓1 ±1 ∓11 ±1 0 β ∓1 −β1 ∓1 β 0 −β ±11 ±1 ∓1 −β 0 β1 ∓1 −β ±1 β 0
,C2 =
0 1 1 1 1 11 0 ±1 ∓1 ∓1 ±11 ±1 0 γ −γ ∓11 ∓1 γ 0 ±1 −γ1 ∓1 −γ ±1 0 γ1 ±1 ∓1 −γ γ 0
,
A3 =
0 1 1 1 1 11 0 ±1 ∓1 α −α1 ±1 0 ±1 ∓1 ∓11 ∓1 ±1 0 −α α1 α ∓1 −α 0 ±11 −α ∓1 α ±1 0
,B3 =
0 1 1 1 1 11 0 ±1 β ∓1 −β1 ±1 0 ∓1 ±1 ∓11 β ∓1 0 −β ±11 ∓1 ±1 −β 0 β1 −β ∓1 ±1 β 0
,C3 =
0 1 1 1 1 11 0 ±1 γ −γ ∓11 ±1 0 ∓1 ∓1 ±11 γ ∓1 0 ±1 −γ1 −γ ∓1 ±1 0 γ1 ∓ ±1 −γ γ 0
,
A4 =
0 1 1 1 1 11 0 ±1 ∓1 α −α1 ±1 0 ∓1 −α α1 ∓1 ∓1 0 ±1 ±11 α −α ∓1 0 ∓11 −α α ±1 ∓1 0
,B4 =
0 1 1 1 1 11 0 β ±1 ∓1 −β1 β 0 ∓1 −β ±11 ±1 ∓1 0 ±1 ∓11 ∓1 −β ±1 0 β1 −β ±1 ∓1 β 0
,C4 =
0 1 1 1 1 11 0 γ ±1 −γ ∓11 γ 0 ∓1 ±1 −γ1 ±1 ∓1 0 ∓1 ±11 −γ ±1 ∓1 0 γ1 ∓1 −γ ±1 γ 0
,
A5 =
0 1 1 1 1 11 0 α −α ±1 ∓11 α 0 ± ∓1 −α1 −α ±1 0 ∓1 α1 ±1 ∓1 ∓1 0 ±11 ∓1 α α ±1 0
,B5 =
0 1 1 1 1 11 0 ±1 β ∓1 −β1 ±1 0 −β ∓1 β1 β −β 0 ±1 ∓11 ∓1 ∓1 ±1 0 ±11 −β β ∓1 ±1 0
,C5 =
0 1 1 1 1 11 0 γ ±1 ∓1 −γ1 γ 0 −γ ±1 ∓11 ±1 −γ 0 ∓1 γ1 ∓1 ±1 ∓1 0 ±11 −γ ∓1 γ ±1 0
,
A6 =
0 1 1 1 1 11 0 α −α ±1 ∓11 α 0 ∓1 −α ±11 −α ∓1 0 α ±11 ±1 −α α 0 ∓11 ∓1 ±1 ±1 ∓1 0
,B6 =
0 1 1 1 1 11 0 β ±1 −β ∓11 β 0 −β ∓1 ±11 ±1 −β 0 β ∓11 −β ∓ β 0 ±11 ∓1 ±1 ∓1 ±1 0
,C6 =
0 1 1 1 1 11 0 ±1 γ −γ ∓11 ±1 0 −γ γ ∓11 γ −γ 0 ∓1 ±11 −γ γ ∓1 0 ±11 ∓1 ∓1 ±1 ±1 0
.
The parameter dependent matrices are such that one row and one column with thesame indexing are without the parameter. The parameter occurs twice in each of theremaining four rows and columns. Since the first row and column are fixed, thereremains only four possibilities for taking up the two positions to be occupied by theparameter. This gives 6 matrices. Since there are five different ways in which theparameter free row (and column) can be taken up, we obtain the 6×5 = 30 matricesin agreement with the results obtained.
11
It is natural to classify the one parameter matrices according to which row (andcolumn) is parameter free. This gives us five families of one parameter matrices whichwe denote as follows:
{Ai,Bi,Ci} i = 2,3,4,5,6. (5.14)It is possible to obtain one matrix from another within a given family by permutations.
Assigning the values ±1 to the parameters yields 60 parameter free matrices. Weobserve that there are only 12 distinct such matrices obtained from the 60. Each ofthese 12 is an intersection of certain five (out of 60) families of matrices. Equation(5.15) illustrates how the intersections are obtained. Below we also list the parameterfree matrices.
D1 =
0 1 1 1 1 11 0 ±1 ±1 ∓1 ∓11 ±1 0 ∓1 ±1 ∓11 ±1 ∓1 0 ∓1 ±11 ∓1 ±1 ∓1 0 ±11 ∓1 ∓1 ±1 ±1 0
,D2 =
0 1 1 1 1 11 0 ±1 ±1 ∓1 ∓11 ±1 0 ∓1 ∓1 ±11 ±1 ∓1 0 ±1 ∓11 ∓1 ∓1 ±1 0 ±11 ∓1 ±1 ∓1 ±1 0
,D3 =
0 1 1 1 1 11 0 ±1 ∓1 ±1 ∓11 ±1 0 ±1 ∓1 ∓11 ∓1 ±1 0 ∓1 ±11 ±1 ∓1 ∓1 0 ±11 ∓1 ∓1 ±1 ±1 0
,
D4 =
0 1 1 1 1 11 0 ±1 ∓1 ∓1 ±11 ±1 0 ∓1 ±1 ∓11 ∓1 ∓1 0 ±1 ±11 ∓1 ±1 ±1 0 ∓11 ±1 ∓1 ±1 ∓1 0
,D5 =
0 1 1 1 1 11 0 ±1 ∓1 ∓1 ±11 ±1 0 ±1 ∓1 ∓11 ∓1 ±1 0 ±1 ∓11 ∓1 ∓1 ±1 0 ±11 ±1 ∓1 ∓1 ±1 0
,D6 =
0 1 1 1 1 11 0 ±1 ∓1 ±1 ∓11 ±1 0 ∓1 ∓1 ±11 ∓1 ∓1 0 ±1 ±11 ±1 ∓1 ±1 0 ∓11 ∓1 ±1 ±1 ∓1 0
.
12
n1 n2 n3 n4 Matrix
1 b =−a,d =−c
1 e =−a,g =− f A4
2 f =−a,g =−e
1 e = a, l =−h A3
2 h = a, l =−e A6
3 l = a,h =−e B2
3 g =−a, f =−e
1 e = a, l =−h A3
2 h = a, l =−e C2
3 l = a,h =−e A5
2 c =−a,d =−b
1 e =−a,g =− f
1 b = a, l =−h A2
2 h = a, l =−b B6
3 l = a,h =−b B3
2 f =−a,g =−e B5
3 g =−a, f =−e
1 e =−b, l =−h C4
2 h =−b, l =−e D1
3 l =−b,h =−e D4
3 d =−a,c =−b
1 e =−a,g =− f
1 b = a, l =−h A2
2 h = a, l =−b C3
3 l = a,h =−b C5
2 f =−a,g =−e
1 e =−b, l =−h B4
2 h =−b, l =−e D6
3 l =−b,h =−e
1 b =−a,m = e B2
2 e =−a,m = b C3
3 m = a,e =−b B4
3 g =−a, f =−e C6
Table 1. Summary of calculations:how the number of parameters is reduced.
The 12 parameter free matrices are in fact conference matrices.1 The intersectionsare formed by picking one and only one matrix from each family. We list below thetwelve intersections. In the notation used to denote the intersections, the superscriptindicates the choice made. For instance, Du
1 means that we have chosen the matrixcorresponding to the upper sign in matrix D1. In a similar manner, Dl
1 means that wehave chosen the matrix in D1 corresponding to the lower sign.
1A conference matrix is an n× n matrix C with diagonal entries 0 and off diagonal entries ±1 whichsatisfies CCt = (n−1) I.
13
Du1 = Au
2 (1) = Bu3 (1) =Cu
4 (1) =Cu5 (1) =Cu
6 (1) ,
Dl1 = Al
2 (−1) = Bl3 (−1) =Cl
4 (−1) =Cl5 (−1) =Cl
6 (−1) ,
Dl2 = Al
2 (1) =Cl3 (−1) = Bl
4 (−1) = Bl5 (−1) = Bl
6 (−1) ,
Du2 = Au
2 (−1) =Cu3 (1) = Bu
4 (1) = Bu5 (1) = Bu
6 (1) ,
Dl3 = Bl
2 (−1) = Al3 (−1) = Bu
4 (−1) = Al5 (−1) =Cl
6 (1) ,
Du3 = Bu
2 (1) = Au3 (1) = Bl
4 (1) = Au5 (1) =Cu
6 (−1) ,
Dl4 =Cl
2 (1) = Bl3 (1) = Al
4 (1) = Au5 (−1) = Bu
6 (−1) ,
Du4 =Cu
2 (−1) = Bu3 (−1) = Au
4 (−1) = Al5 (1) = Bl
6 (1) ,
Du5 =Cu
2 (1) = Au3 (−1) =Cl
4 (1) = Bu5 (−1) = Al
6 (1) ,
Dl5 =Cl
2 (−1) = Al3 (1) =Cu
4 (−1) = Bl5 (1) = Au
6 (−1) ,
Du6 = Bu
2 (−1) =Cu3 (−1) = Au
4 (1) =Cl5 (1) = Au
6 (1) ,
Dl6 = B1
2 (1) =Cl3 (1) = Al
4 (−1) =Cu5 (−1) = Al
6 (−1) .
(5.15)
We now discuss the observations made from the twelve intersections. Suppose weconsider these intersections as vertices of a discrete graph. An edge connecting any ofthe vertices is one and only one of the thirty matrices where one vertex corresponds tothe matrix obtained by assigning the value +1 to the parameter, while the other vertexis obtained by assigning the value−1 to the parameter. For example the edge connect-ing the vertices Du
1 and Du2 is the matrix A2 (α), where Du
1 = Au2 (1) and Du
2 = Au2 (−1).
Figure 2 is the graph obtained where we have assigned a distinct colour to each familyaccording to the classification in the notation (5.14). The graph obtained is bipartiteand 5-regular. In the figure below each edge represents a one-parameter family ( aloop). In fact each family is described by 6 parameters if one take into account theparameters θ1, . . . ,θ5 appearing in equation (5.1). The corresponding intersection aretherefore 5-parameter families corresponding to six-dimensional conference matrices.
14
Du1
Dl2
Dl3
Dl4
Du5
Du6
Dl1
Du2
Du3
Du4
Dl5
Dl6
Figure 2. Five- and six-parameter families of 6×6 RET-matrices
Conclusion. After our work was accomplished Ingemar Bengtsson pointed out, thatsimilar matrices appear in the coding theory [7]. In fact 4-dimensional RET-matrices(4.4) can be found there as well as the 6-dimensional matrix Dl
1. But no description ofall possible matrices is given there either.
Symmetric conference matrices have been discussed recently in [3], where an ex-ample of such matrices dimension 10 can be found. It is claimed that it is not knownwhether such matrix in dimension 8 exists or not.
We have constructed all RET-matrices in dimensions 2, 4 and 6. It is natural tocontinue studies by constructing RET-matrices for n = 8 and for arbitrary values of n.
Acknowledgement. The authors would like to thank Ingemar Bengtsson for the in-terest in their work. The work of PK was partially supported by the Swedish ResearchCouncil Grant #50092501, Institut Mittag-Leffler and ZiF (Zentrum für interdiszi-plinäre Forschung, Bielefeld). RO would like to thank ISP (International ScienceProgram, Uppsala University) for support.
15
References[1] G. Berkolaiko and P. Kuchment Introduction to quantum graphs, American Mathematical Society,
Rhode Island, 2013.[2] T. Cheon, Reflectionless and Equiscattering Quantum Graphs and Their Applications, Int. J. System &
Measurement, 5 (2012) 34-44.[3] T. Cheon, Reflectionless and Equiscattering Quantum Graphs, textitInt. J. Adv. Systems and Measure-
ments, 5 (2012), 18–22.[4] M. Harmer, Hermitian symplectic geometry and extension theory, Journal of Physics. A. Mathematical
and General, 33, (2000),9193-9203.[5] M. Harmer, Hermitian symplectic geometry and the factorization of the scattering matrix on graphs,
Journal of Physics. A. Mathematical and General, 33, (2000),9015–9032.[6] J.M. Harrison, U. Smilansky, and B. Winn, Quantum graphs where back-scattering is prohibited, Jour-
nal of Physics. A. Mathematical and Theoritical, 40, (2007), 14181-14193.[7] R.B. Holmes and V.I. Paulsen, Optimal frames for erasures, Linear Algebra Appl. 377 (2004), 31?51.[8] P. Kurasov and M. Nowaczyk, Geometric properties of quantum graphs and vertex scattering matrices,
Opuscula Mathematica, 30, (2010).[9] O. Post, Spectral analysis on graph-like spaces, Springer, 2010.[10] O. Turek and T. Cheon, Quantum graph vertices with permutation-symmetric scattering probabilities,
Phys. Lett. A 375 (2011), no. 43, 3775?3780.[11] O. Turek and T. Cheon, Hermitian unitary matrices with modular permutation symmetry,
arXiv:1104.0408.
DEPT. OF MATHEMATICS, STOCKHOLM UNIV. 106 91 STOCKHOLM, SWEDEN
16
ON EQUI-TRANSMITTING MATRICES
PAVEL KURASOV AND RAO OGIK
ABSTRACT. Equi-transmitting scattering matrices are studied. A complete descrip-tion of such matrices up to order six is given. It is shown that the standard matchingconditions matrix is essentially the only equi-transmitting matrix for orders 3 and 5.For orders 4 and 6, there exists other equi-transmitting but all such matrices havezero trace.
1 IntroductionFrom quantum physics it is known that, for objects with finite number of scatteringchannels, the scattering of waves at a node is described by unitary matrices. Theunitarity of the scattering matrix means that the system is conservative.
The transition probabilities between different channels are given by modulus squaresof the corresponding matrix elements
∣∣si j∣∣2. Therefore a vertex scattering matrix SV
determines a system where all channels are equivalent if and only if∣∣si j∣∣2 = |slm|2, i 6=
j, l 6= m. This implies that even diagonal elements have the same value |sii|2 =∣∣s j j∣∣2,
i 6= j. Such matrices are called equi-transmitting matrices.This classical problem has found renaissance in quantum graphs where the transi-
tion through vertices is described by vertex scattering matrix SV (k). The matrix hasthe following explicit energy dependence
SV (k) = (k+1)SV (1)+(k−1)I(k−1)SV (1)+(k+1)I . (1.1)
Here SV (1) is the vertex scattering matrix for k = 1 and can be considered as aparameter describing all possible vertex conditions [15, 14]. We are interested inequi-transmitting matrices SV (k) and so SV (1) has to be equi-transmitting. If SV (1)is not only unitary, but also Hermitian, then SV (k) does not depend on the energyk2. Therefore for SV (k) to be equi-transmitting it is sufficient that SV (1) is not onlyequi-transmitting and unitary but also Hermitian.
A widely used vertex condition is the so called standard (Kirchoff) matching con-ditions (SMC) on the functions at a vertex V are
u is continuous at every vertex V
∑x j∈Vm
∂nu(x j) = 0.
2010 Mathematics Subject Classification. Primary 34L25, 81U40, Secondary 35P25, 81V99.Key words and phrases. Quantum graphs, vertex scattering matrix, equi-transmitting matrices.PUBLISHED AS RESEARCH REPORTS IN MATHEMATICS ISSN: 1401-5617, 1(2014)
1
The corresponding vertex scattering matrix is
2v −1 2
v2v · · ·
2v
2v −1 2
v · · ·2v
2v
2v −1
......
. . .
,
where v is the degree of the vertex. We see that it is equi-transmitting.Turek and Cheon have studied equi-transmitting matrices in [19]. Their construc-
tion is by means of the Hadamard and conference matrices and they have given severalexamples of equi-transmitting matrices.
The aim of this study is to give a complete description of all equi-transmitting ma-trices up to dimension 6. In section two we describe the general structure of an equi-transmitting matrix and the various transformations that preserve equi-transmissionproperties. In section three we describe the possible values of r and t. It is shown thatr, t ∈ [0,1] with the interesting cases being r, t ∈ (0,1). In section four we consider sev-eral examples of equi-transmitting matrices including the one corresponding to SMC.Sections five and six deal with the construction of such matrices in dimensions 2 to 6for r, t ∈ (0,1). The following results are obtained:
• When n = 3 and n = 5, the only equi-transmitting matrix are those corre-sponding to the SMC-matrix.
• When n= 2,4,6 and Tr(S) = 0, other equi-transmitting matrices exist besidesthose generated by the SMC-matrix.
Definition 1. An n×n matrix S is equi-transmitting (ET-matrix)if the following hold
1. S = S∗ = S−1,2. |sii|=
∣∣s j j∣∣ , i 6= j,
3.∣∣si j∣∣= |slm| , i 6= j, l 6= m.
2 Formulation and transformationsOne of the important notion in quantum graphs is the vertex scattering matrix. Fora vertex V with valency n, the associated vertex scattering matrix SV (k) is an n× nunitary matrix that describes how waves are scattered at that vertex. If the matrixSV (1) is plugged into the matching conditions equation i(S− I)~u(V ) = (S+ I)~u′ (V ),then it is also possible to identify the kind of coupling it represents.
For the ET-matrix defined above, let |sii|= r, i = 1, . . . ,n and∣∣si j∣∣= |slm|= t, i 6= j,
l 6= m, i, j, l,m = 1, . . . ,n. The matrices have the representation given by
S = D∗θCDθ , (2.1)
where2
C =
r t t · · · tt r ta23 · · · ta2nt ta23 r ta3n...
.... . .
...t ta2n ta3n · · · r
,
Dθ = diag{
eiθ1 , . . . ,eiθn}, θi ∈ [0,2π) , i = 1,2, . . . ,n−1,
ai j ∈{C :∣∣ai j∣∣= 1, for all i = 2, . . . ,n−1, j = 3, . . . ,n
}.
It is easy to see that the matrix C has the property that∣∣ci j∣∣ =
∣∣si j∣∣, i, j = 1,2, . . . ,n.
Therefore it suffices in studying the equi-transmitting properties of S to consider thematrix C. The two matrices are equal up to the phases described by the matrix Dθ .
Having an ET-matrix S one can obtain another ET-matrix by means of appropriatematrix permutation. Below we discuss such transformations
1. Let D be a diagonal matrix whose diagonal entries comprise ±1. Then thetransformation
SD = D−1SD
has the effect that certain entries of S have their signs reversed. However signsof the diagonal entries remain unchanged because the reversal of their signs isdone. Such a transformation yields yet another ET-matrix since
∣∣sDi j
∣∣=∣∣si j∣∣,
i, j = 1,2, . . . ,n.
2. The transformation S 7→ −S has the effect of reversing the signs of all theentries of S. It is clear that the resulting matrix is again an ET-matrix.
3. Let Plm be a permutation matrix. The matrix
SP = PlmSPlm
is obtained by interchanging rows and columns i and j. Therefore∣∣sPi j
∣∣ =∣∣si j∣∣, i, j = 1,2, . . . ,n meaning that SP is also an ET-matrix.
From the foregoing it can be seen that the study of ET-matrices can be done up tothe above transformations.
3 Possible values of r and tFrom the definition of the ET-matrix, the numbers r and t are nonnegative. Since thematrix C is unitary, the rows form normalised vectors. Thus row 1 yields the equation
r2 +(n−1) t2 = 1 (3.1)3
which is key to the construction of matrix C. From the equation and the condition thatr and t are nonnegative, it follows that r, t ∈ [0,1].
When r = 1, then t = 0 and from the representation of C (see equation (2.1)) it canbe seen that the resulting matrix is a diagonal matrix whose diagonal entries comprises±1. This is a trivial class of ET-matrices. On the other hand when r = 0 then t = 1and one obtains a RET-matrix. These have been discussed up to order six in [16].
Since S is Hermitian and unitary its spectrum consists of ±1. Let n+ be the mul-tiplicity of +1 in the spectrum of S. Let ν+ be the number of positive diagonalentries of the matrix S. Then if the order of S is n, the numbers n+ and ν+ canassume any value in the set {0,1, . . . ,n}. The trace of S is given by the relationsTr(S) = 2n+−n = r (2ν+−n). This implies that
r = 2n+−n2ν+−n . (3.2)
This formula is applicable if and only if 2ν+−n 6= 0. Assuming this condition is takeninto consideration, it follows that r = 1 if and only if n+ = ν+, while r = 0 if and onlyif n+ = 1
2 n.It should be noted that cases when 2ν+−n = 0, which implies that ν+ = 1
2 n, onlyarise when n is even. In such a case the formula (3.2) cannot be used to determine thevalue of r. The only option in determining r and hence constructing S, if possible, isby using the definition of S and its corresponding representation. This case has turnedout to be significant despite our initial skepticism of what can be obtained from it.It is only in this case that an ET-matrix other than those generated by the standardmatching condition matrix can be constructed.
The figures below give a graphical depiction of the possible values of r. The firstfigure is for the case when n is odd, while the second figure is for the case when n iseven.
n
n
n+
ν+
n−12
n−12
(a)
n
n
n+
ν+
(b)
n2
n2
( n2 ,
n2
)r indeterminater = 1
r = 0r < 0r > 1r ∈ (0,1)
Figure 1. Possible values of r; (a) when n is odd and (b) when n is even
The diagonal white cells represent cases where r = 1 and for which the correspond-ing ET-matrix is a diagonal matrix. The orange cells represent cases where r > 1 andhence no ET-matrix can be constructed. The blue cells represent cases where r < 0 so4
that no ET-matrix can be constructed. The lime cells represent cases where r ∈ (0,1)and where there is the possibility of the existence of ET-matrices. In figure (b), thelight gray cells represent cases where r = 0 and where the corresponding ET-matricesare reflectionless. The red cells in figure (b) represent where r is not defined so thatno ET-matrix can be constructed. In the same figure, the middle violet cell repre-sent where r is indeterminate from formula (3.2) but where it is possible to constructET-matrices from the definition of S.
The various cases to be considered when investigating the existence of ET-matriceswill be denoted by the ordered pair (n+,ν+). It should be noted that the case(n−n+,n−ν+) yields the same values of r and t.
We now discuss the viable cases for the existence of an ET-matrix when r, t ∈ (0,1).The cases where n+ = ν+ are excluded since they give rise to r = 1. The exceptionto this is the case (n+,ν+) =
( n2 ,
n2
)when n is even because it is also viable for the
existence of the required ET-matrix. When n is odd then cases for which r, t ∈ (0,1)are given by
(n+,ν+) and (n−n+,n−ν+) , where
n+ ∈{
1,2, . . . , n−12
}
ν+ ∈{
0,1 . . . , n−32
}.
(3.3)
On the other hand when n is even then such cases are given by
(n+,ν+) ,(n−n+,n−ν+) and( n
2 ,n2
), where
n+ ∈{
1,2, . . . , n2 −1
}
ν+ ∈{
0,1 . . . , n2 −2
}.
(3.4)
4 Important general examples
Diagonal matricesFor all n, whenever r = 1, there exists an equi-transmitting diagonal matrix. Thismatrix is not interesting in the framework of quantum graphs since it describes com-pletely independent (not connected) edges. Such matrices will be called trivial ET-matrices, while those corresponding to r ∈ (0,1) will be called non trivial ET-matrices.
Reflectionless equi-transmitting matricesThese are obtained when r = 0. We have discussed these matrices in [16] where it hasbeen proved that such matrices exist only when n is even.
5
Standard matching conditions equi-transmitting matricesThis is a special class of ET-matrices for r ∈ (0,1) and it appears when we imposeSMC. For every incoming wave in a star graph with n infinite edges there are bothreflected waves along the same edge and transmitted waves along the other n− 1edges. This condition is described by the equation
u(x) =
e−ikx +Rieikx,x ∈ Ei
Tjieikx,x ∈ E j
(4.1)
where Ri is the reflection coefficient along edge Ei and Tji is the transmission coeffi-cient of waves transmitted from edge Ei into edge E j.
Suppose the coupling of the edges at the central vertex V is described by the SMC.Suppose also that the edges are parameterized as E j = [0,∞), j = 1, . . . ,n. Theseconditions are invariant under permutations. Therefore all reflection coefficients Riare equal and all transmission coefficients Tji are also equal. The requirement thatu j (0) = ul (0), j, l = 1 . . . ,n together with equation (4.1) yield the equation R+1 = T .On the other hand the conditions ∑∂nu(0) = 0 as well as equation (4.1) result in theequation ik (R−1+(n−1)T ) = 0. These two equations yield the values T = 2
n andR =−1+ 2
n . Thus the matrix S arising from the SMC is given by
s jl =
−1+ 2
n , j = l
2n , j 6= l.
(4.2)
It is easy to see that this matrix is Hermatian (symmetric in this case since all theentries are real). It is also easy to see that StS = SSt = I, i.e. S is orthogonal. Thus S isHermitian, unitary and equi-transmitting for all n≥ 2. Using the notations introducedin section 2, we see that
∣∣s jl∣∣=
r =∣∣−1+ 2
n
∣∣ , j = l
t = 2n , j 6= l
. (4.3)
This construction determines a Hermitian unitary equi-transmitting matrix S of ordern for all n≥ 3 and for r ∈ (0,1). We will refer to this matrix as the SMC-matrix.
Cases (1,0) and (n−1,n)
For any given n the case (n+,ν+)=(1,0) yields the equation −nr = −(n−2) whichimplies that r = 1− 2
n . When this value of r is substituted into the equation r2 +
(n−1) t2 = 1 and the positive value of t evaluated, it is obtained as t = 2n . Since in this
case all the diagonal entries are negative, the associated matrix C has the representation6
C1,0 =
−1+ 2n
2n
2n · · · 2
n2n −1+ 2
n2n a23 · · · 2
n a2n
2n
2n a23 −1+ 2
n2n a3n
......
. . .2n
2n a2n
2n a3n · · · −1+ 2
n
,
where ai j ∈{z ∈ C : |z|= 1, i = 2, . . . ,n−1, j = 3, . . . ,n, i < j}. Thus there are 12 (n−1)(n−2)
such parameters. The inner products of the rows yield certain 12 n(n−1) equations
since the rows are orthogonal. The orthogonality between the first row and all theother rows give the following n−1 equations
2r+ t (a23 +a24 + · · ·+a2,n−1 +a2n) = 0
2r+ t (a23 +a34 + · · ·+a3,n−1 +a3n) = 0
...
2r+ t (a2,n−1 +a3,n−1 + · · ·+an−2,n−1 +an−1,n) = 0
2r+ t (a2n +a3n + · · ·+an−2,n +an−1,n) = 0
These equations can be rewritten as follows
a23 +a24 + · · ·+a2,n−1 +a2n =2rt
a23 +a34 + · · ·+a3,n−1 +a3n =2rt
...
a2,n−1 +a3,n−1 + · · ·+an−2,n−1 +an−1,n =2rt
a2n +a3n + · · ·+an−2,n +an−1,n =2rt.
(4.4)
Each of the above equations has n− 2 parameters while on the right hand sidewe have that 2r
t = 2(1− 2
n
) n2 = n− 2. Since each parameter is unimodular, all the
equations are satisfied if and only if each of the parameters is unity, i.e. ai j = 1,i = 2, . . . ,n−1, j = 3, . . . ,n, i < j. All the other 1
2 (n−2)(n−1) equations are of theform
−2rai j + t(1+∑n
l=3 aila jl)= 0, i = 2, . . . ,n−1, j = 3, . . . ,n, i 6= l 6= j, i < j.
The terms under the summations could involve the conjugates of the parameters. Allthese equations are also satisfied when all the parameters are unity. It follows that thematrix determined is precisely the smc matrix and is given as follows
7
C1,0 =
−1+ 2n
2n
2n · · · 2
n2n −1+ 2
n2n · · · 2
n2n
2n −1+ 2
n2n
......
. . ....
2n
2n
2n · · · −1+ 2
n
.
The case (n+,ν+) = (n−1,n) can be analysed in a similar way leading to thematrix
Cn−1,n =
1− 2n
2n
2n · · · 2
n2n 1− 2
n − 2n · · · − 2
n2n − 2
n 1− 2n − 2
n...
.... . .
...2n − 2
n − 2n · · · 1− 2
n
.
The two matrices are related as follows
Cn−1,n =−DC1,0D
where D is an n×n diagonal matrix given by D = diag{−1,1, . . . ,1}. This means thatfor all n≥ 2, the cases (1,0) and (n−2,n) give the same transition probabilities as theSMC.
5 Equi-transmitting matrices of order 2,3,4 and 5.This section is devoted to the discussion of nontrivial ET-matrices of orders 2,3,4 and5 for the values of r ∈ (0,1). We will only state the results for the cases (1,0) and(n−2,n) because they have been adequately covered in the preceding section.
Order 2The only case leading to r ∈ (0,1) is when the trace of the matrix is zero. This is givenby (n+,ν+) = (1,1). The corresponding representation of matrix C is
C =
(r tt −r
).
Normalisation of the rows yield the relation t =√
1− r2 where r ∈ (0,1). The corre-sponding ET-matrices C are a one-parameter family given as follows
C1,1 =
(r
√1− r2√
1− r2 −r
).
8
Observe that the matrix has zero trace and it is a one-parameter family of matrices.
Order 3When n = 3, then the only cases yielding r and t in the interval (0,1) are given by theordered pairs (1,0) and (2,3). These have been covered in the previous section. Thematrices C corresponding to these cases are respectively given as
C1,0 =13
−1 2 22 −1 22 2 −1
and C2,3 =
13
1 2 22 1 −22 −2 1
.
From equation (3.3) we see that for r to be in the interval (0,1), then the onlychoice we have is n+ = 1 and ν+ = 0. Any any other case will not yield values r and tin the interval (0,1). Thus when n = 3, then the only nontrivial ET-matrices are thosegenerated by the SMC-matrix.
Order 4The cases to consider when n = 4 in order to obtain the values of r and t in the interval(0,1) are (1,0), (3,4) and (2,2).
1. Cases (1,0) and (3,4) The corresponding matrices are equivalent to the stan-dard matching conditions matrix as discussed in the previous section. Therespective C matrices are
C1,0 =12
−1 1 1 11 −1 1 11 1 −1 11 1 1 −1
and C3,4 =
12
1 1 1 11 1 −1 −11 −1 1 −11 −1 −1 1
.
3. Case (2,2) In this case the trace is zero and r cannot be determined fromthe formula (3.2) but it is in the interval (0,1). The number of positive andnegative diagonal entries are equal and the corresponding matrix C can berepresented by
C =
r t t tt r at btt at −r ctt bt ct −r
,
where a,b,c ∈ {z ∈ C : |z|= 1}. Using the orthogonality of the rows we havethe following system of equations
2r+ t (a+b) = 0 (1)
a+ c = 0 (2)
b+ c = 0 (3)
1+bc = 0 (4)
1+ac = 0 (5)
−2rc+ t(1+ab
)= 0 (6).
9
Solving for the parameters a,b,c, it is obtained that a can be chosen arbi-trary, in which case b = a and c = −a. A relation between r and t based onthe system is obtained as r+ tℜ(a) = 0. Using this relation between r and tand that in (3.1) one obtains that t = 1√
3+ℜ(a)2 and r = −ℜ(a)√3+ℜ(a)2 . The pa-
rameter a is chosen such that ℜ(a)< 0 because r > 0. For example supposea = −eiθ , then the appropriate values of θ are in the interval
(−π
2 ,π2
). The
corresponding matrix C is given as follows
C2,2 =1√
3+cos2 θ
cosθ 1 1 11 cosθ −eiθ −e−iθ
1 −e−iθ −cosθ e−iθ
1 −eiθ eiθ −cosθ
.
Therefore the matrix C is given by
C2,2 =1√
3+ℜ(a)2
−ℜ(a) 1 1 11 −ℜ(a) a a1 a ℜ(a) −a1 a −a ℜ(a)
.
It is therefore observed that when n = 4, all nontrivial ET-matrices obtained can beput into the following two categories
1. Matrices generated by the SMC-matrix.
2. One parameter family matrices C with zero trace.
Order 5When n = 5, we establish that the only nontrivial ET-matrices are those generated bythe SMCmatrix. The cases that give rise to r, t ∈ (0,1) when n = 5 are (1,0), (4,5),(2,0), (3,5), (2,1) and (3,4). Each case is discussed in the sequel.
1. Cases (1,0) and (4,5) . The respective matrices C are
C1,0 =15
−3 2 2 2 22 −3 2 2 22 2 −3 2 22 2 2 −3 22 2 2 2 −3
and C4,5 =15
3 2 2 2 22 3 −2 −2 −22 −2 3 −2 −22 −2 −2 3 −22 −2 −2 −2 3
.
The first matrix is precisely the SMC-matrix, while the second is generatedby it in the sense of the transformation discussed in section 4.
3. Case (2,0). In this case all diagonal entries are negative and two eigenvaluesare positive. The values r and t are 1
5 and√
65 respectively. The corresponding
matrix C is represented as follows10
C =
−r t t t tt −r at bt ctt at −r dt ett bt dt −r f tt ct et f t −r
,
where a, . . . , f ∈ {z ∈ C : |z|= 1}. Since the rows of C are orthogonal, theirinner products yield the following system of equations
−2r+ t (a+b+ c) = 0 (1)−2r+ t (a+d + e) = 0 (2)−2r+ t
(b+d + f
)= 0 (3)
−2r+ t(c+ e+ f
)= 0 (4)
−2ar+ t(1+bd + ce
)= 0 (5)
−2br+ t(1+ad + c f
)= 0 (6)
−2cr+ t(1+ae+b f
)= 0 (7)
{−2dr+ t
(1+ab+ e f
)= 0 (8)
−2er+ t(1+ac+d f
)= 0 (9)
{−2 f r+ t (1+bc+de) = 0 (10).
(5.1)
Equations (2) to (4) can be written as
M1 =−~X , (5.2)
where
M =
−α d ed −α fe f −α
1 = [1,1,1]t , ~X = [a,b,c]t , α = 2rt =
√6
3
(5.3)
Equations (5) to (7) can be written as
M~X =−1. (5.4)
Substituting the value of ~X from (5.2) into (5.4) yields that
M21 = 1. (5.5)
The matrix M is Hermitian and so it possesses a complete orthonormal set ofeigenvectors. Now suppose that λ is an eigenvalue for M corresponding tothe eigenvector ψ then
M~ψ = λ~ψ, ⇐⇒ M2~ψ = λ 2~ψ. (5.6)
Comparing the second equation in (5.6) and equation (5.5) it can be seen thatλ 2 = 1 is the eigenvalue for M2 corresponding to the eigenvector 1 and itfollows that
M1 =±1. (5.7)
Comparing (5.7) and (5.2) we obtain that11
~X =±1.
Therefore either a = b = c = 1 or a = b = c = −1. The substitution of thevalues into equation (1) in (5.1) yields the contradictions that ±3 =
√6
3 . Thusthe case (2,0) does not yield an ET-matrix for r, t ∈ (0,1).
4. Case (3,5). This case is completely analogous to case (2,0) and similarcontradictions are obtained hence there is no ET-matrix.
5. Case (2,1). In this case one diagonal entry is positive and two eigenvaluesare positive.This implies that r = 1
3 and t =√
23 . The matrix associated to this
case can be written as
C =
r t t t tt −r at bt ctt at −r dt ett bt dt −r f tt ct et f t −r
,
where a, . . . , f ∈ {z ∈ C : |z|= 1}. Since the rows of C are orthogonal, theirinner products yield the following system of equations
a+b+ c = 0 (1)a+d + e = 0 (2)b+d + f = 0 (3)c+ e+ f = 0 (4)
−2ar+ t(1+bd + ce
)= 0 (5)
−2br+ t(1+ad + c f
)= 0 (6)
−2cr+ t(1+ae+b f
)= 0 (7)
{−2dr+ t
(1+ab+ e f
)= 0 (8)
−2er+ t(1+ac+d f
)= 0 (9)
{−2 f r+ t (1+bc+de) = 0 (10).
(5.8)
Equatiions (2) to (4) can be written as
A1 =−~X , (5.9)where
A =
0 d ed 0 fe f 0
and vectors 1 and ~X are as given in (5.3). Using the same notations, equations(5) to (7) can also be written as
(A−αI)~X =−1, (5.10)
where α = 2rt =√
2. Substituting the value of ~X from (5.9) into (5.10) yieldsthat
(A2−αA− I
)1 = 0. (5.11)
12
The matrix A is Hermitian and thus has a complete set of eigenvectors whichcan be chosen orthonormal. Therefore if λ is an eigenvalue for A correspond-ing to the eigenvector ψ then
A~ψ = λ~ψ, ⇔(A2−αA− I
)~ψ =
(λ 2−αλ −1
)~ψ. (5.12)
From equations (5.11) and (5.12) it follows that 1 belongs to the kernel ofA2−αA− I if and only if it is the eigenvector for A corresponding to theeigenvalue λ satisfying λ 2−αλ −1 = 0. Therefore
λ = λ± =√
2±√
62 6= 0
and
A1 = λ±1. (5.13)
Comparing the equations (5.9) and (5.13) it can be seen that
~X =−λ±1.
Therefore either a = b = c = −λ+ or a = b = c = −λ−. Substituting thesevalues into equation (1) in (5.8) one obtains that −3λ± =
√2 which are con-
tradictions. Thus the case (2,1) does not yield any nontrivial ET-matrix.
6. Case (3,4). The values of r and t are obtained as 13 and
√2
3 respectively. Onlyone of the diagonal entries is negative and so the associated matrix C can bewritten as follows
C =
r t t t tt r at bt ctt at r dt ett bt dt r f tt ct et f t −r
,
where a, . . . , f ∈ {z ∈ C : |z|= 1}. From the orthogonality of the rows of Swe obtain the following system of equations
2r+ t (a+b+ c) = 0 (1)2r+ t (a+d + e) = 0 (2)2r+ t
(b+d + f
)= 0 (3)
c+ e+ f = 0 (4)
2ar+ t(1+bd + ce
)= 0 (5)
2br+ t(1+ad + c f
)= 0 (6)
1+ae+b f = 0 (7)
{2dr+ t
(1+ab+ e f
)= 0 (8)
1+ac+d f = 0 (9)
{1+bc+de = 0 (10).
(5.14)
This case is similar to case (2,1) and can be analysed in a similar fashionwith the following modifications
13
1. From equations (1) to (3) matrix A and vector ~X are obtained respectivelyas
0 a ba 0 db d 0
and
cef
.
2. Equations (7), (9) and (10) yield the matrix equation A~X =−1.
The analysis shows that
~X =− 1λ± 1, λ± = −
√2±√
62 .
Substituting the corresponding values of the parameters into equation (4) ofthe system (5.14) yields the contradiction that −3 1
λ± = 0. Thus no nontrivialET-matix exists in this case as well.
From all the cases considered, nontrivial ET-matrices have been constructed onlyin the cases (1,0) and (4,5). As has been stated above, these cases are generated bythe SMC-matrix. No nontrivial ET-matrix can be obtained in all the other cases.
6 Equi-transmitting matrices of order 6When n = 6, the cases for which r, t ∈ (0,1) are
(1,0) , (5,6) , (2,0) , (4,6) , (2,1) , (4,5) and (3,3) . (6.1)
The cases (1,0) and (5,6) have already been covered in 4 and so here we simply statethe results. The discussion of each of the other cases begins with a matrix represen-tation in terms of some ten unimodular parameters so that C is an ET-matrix. Theparameters are then determined so as to obtain an explicit form of the matrix. By con-struction all the rows and columns are normalised. Therefore C is unitary if and onlyif the rows are orthogonal, which gives us a system of 15 equations. The system isthen reduced until either an explicit form of the matrix determined or a contradictionis obtained. In determining the parameters, the following lemma is applied to appro-priate equations1. Whenever the matrix is obtained, it may be completely determinedor it may be given in terms of some arbitrary but independent parameters.
Lemma 1. The sum of four complex numbers z1,z2,z3,z4 of equal magnitude equalszero if and only if at least one of the following cases occur:
1Details are in Appendix B14
z2 =−z1,z4 =−z3,
or
z3 =−z1,z4 =−z2,
or
z4 =−z1,z3 =−z2.
Now we consider each of the cases listed in (6.1):
1. Cases (1,0) and (5,6). These cases have been discussed in section 4 andthe respective matrices C given below are generated by the SMC-matrix (seesection 4).
C1,0 =13
−2 1 1 1 1 11 −2 1 1 1 11 1 −2 1 1 11 1 1 −2 1 11 1 1 1 −2 11 1 1 1 1 −2
and C5,6 =13
2 1 1 1 1 11 2 −1 −1 −1 −11 −1 2 −1 −1 −11 −1 −1 2 −1 −11 −1 −1 −1 2 −11 −1 −1 −1 −1 2
.
2. Case (2,0). In this case all diagonal entries are negative while two of theeigenvalues are positive. The values of r and t are therefore obtained as 1
3 and2√
23√
5respectively. The associated matrix C is represented as follows
C =
−r t t t t tt −r at bt ct dtt at −r et f t gtt bt et −r ht ltt ct f t ht −r mtt dt gt lt mt −r
,
where a,b,c,d,e, f ,g,h, l,m ∈ {z ∈ C : |z|= 1}. Since the rows are orthogo-nal, determining their inner products yield the following system of equations
−2r+ t (a+b+ c+d) = 0 (1)−2r+ t (a+ e+ f +g) = 0 (2)−2r+ t
(b+ e+h+ l
)= 0 (3)
−2r+ t(c+ f +h+m
)= 0 (4)
−2r+ t(d +g+ l +m
)= 0 (5)
−2ar+ t(1+be+ c f +dg
)= 0 (6)
−2br+ t(1+ae+ ch+dl
)= 0 (7)
−2cr+ t(1+a f +bh+dm
)= 0 (8)
−2dr+ t(1+ag+bl + cm
)= 0 (9)
−2er+ t(1+ab+ f h+gl
)= 0 (10)
−2 f r+ t(1+ac+ eh+gm
)= 0 (11)
−2gr+ t(1+ad + el + f m
)= 0 (12)
{−2hr+ t
(1+bc+ e f + lm
)= 0 (13)
−2lr+ t(1+bd + eg+hm
)= 0 (14)
{−2mr+ t
(1+ cd + f g+hl
)= 0 (15).
(6.2)15
In what follows we rewrite equations (2) to (9) as a linear system of equa-tions in a,b,c and d. Let α = 2r
t =√
5√2. Then equations (2) to (9) can be
written as follows
a = α− e− f −g(2)
b =−e+α−h− l (3)
c =− f −h+α−m(4)
d =−g− l−m+α (5)
−αa+be+ c f +dg =−1(6)
ae−αb+ ch+dl =−1(7)
a f +bh−αc+dm =−1(8)
ag+bl + cm−αd =−1(9)
(6.3)
A matrix equation corresponding to the equations (6) to (9) is given by
−α e f ge −α h lf h −α mg l m −α
abcd
=
−1−1−1−1
. (6.4)
Let the matrix equation be written as
M~X =−1, (6.5)
where M is the coefficients matrix, ~X is the vector of the parameters a,b,c,dand 1 = [1,1,1,1]t . Using these notations then equations (2) to (5) in (6.3)become
~X =−M1. (6.6)
Substituting this ~X into the matrix equation (6.5) one obtains the following
M21 = 1. (6.7)
The matrix M is Hermitian and so it is diagonalizable and has a completeorthonormal set of eigenvectors. Thus suppose λ is an eigenvalue for M cor-responding to the eigenvector ~ψ , then
M~ψ = λ~ψ, (6.8)
if and only if
M2~ψ = λ 2~ψ. (6.9)
This means that ~ψ is an eigenvector for M corresponding to the eigenvalueλ if and only if it is an eigenvector for M2 corresponding to the eigenvalueλ 2. Comparing equations (6.7) and (6.9) we see that the eigenvector for M2
corresponding to the eigenvalue λ 2 = 1 is 1. Thus we obtain that M1 = ±1.Comparing this equation with (6.5) it follows that ~X = 1, i.e. either a = b =c = d = 1 or a = b = c = d =−1. Subsituting these values in equation (1) ofthe system (6.2) one obtains that ±4 =
√5√2
which is a contradiction.It thus follows that in the case (2,0) no ET-matrix exists.
16
3. Case (4,6). This case can be analyzed in a manner similar to case (2,0)resulting in similar contradictions. Therefore no ET-matrix exists here as well.
4. Case (2,1). Only one diagonal entry is positive and two eigenvalues arepositive. It is easy to see that this yields the values of r and t to be 1
2 and√
32√
5.
The matrix C corresponding to this case has the following representation
C =
r t t t t tt −r at bt ct dtt at −r et f t gtt bt et −r ht ltt ct f t ht −r mtt dt gt lt mt −r
,
where a,b,c,d,e, f ,g,h, l,m ∈ {z ∈ C : |z|= 1}. From the orthogonality ofthe rows of C we obtain the following system of equations
a+b+ c+d = 0 (1)a+ e+ f +g = 0 (2)b+ e+h+ l = 0 (3)c+ f +h+m = 0 (4)d +g+ l +m = 0 (5)
−2ar+ t(1+be+ c f +dg
)= 0 (6)
−2br+ t(1+ae+ ch+dl
)= 0 (7)
−2cr+ t(1+a f +bh+dm
)= 0 (8)
−2dr+ t(1+ag+bl + cm
)= 0 (9)
−2er+ t(1+ab+ f h+gl
)= 0 (10)
−2 f r+ t(1+ac+ eh+gm
)= 0 (11)
−2gr+ t(1+ad + el + f m
)= 0 (12)
{−2hr+ t
(1+bc+ e f + lm
)= 0 (13)
−2lr+ t(1+bd + eg+hm
)= 0 (14)
{−2mr+ t
(1+ cd + f g+hl
)= 0 (15).
(6.10)Equations (6) to (9) can be written as follows
(A−αI)~X =−1, (6.11)where
A =
0 e f ge 0 h lf h 0 mg l m 0
and α = 2r
t = 2√
5√3.
Using the same notations as before, equations (2) and (5) can be written asfollows
A1 =−~X . (6.12)Substituting (6.12) into (6.11) one obtains that
(A2−αA− I
)1 = 0. (6.13)
Matrix A is Hermitian and so it is diagonalizable and has a complete or-thonormal set of eigenvectors. Suppose λ is an eigenvalue for A correspond-ing to the eigenvector ~ψ . Then we have that
17
A~ψ = λ~ψ, ⇔(A2−αA− I
)~ψ =
(λ 2−αλ −1
)~ψ. (6.14)
From equations (6.13) and (6.14) it follows that 1 belongs to the kernel ofA2−αA− I if and only if it is the eigenvector for A corresponding to theeigenvalue λ satisfying λ 2−αλ −1 = 0. Therefore
A1 = λ±1. (6.15)
where λ± =√
5/3± 2√
2/3. Comparing equations (6.12) and (6.15) oneobtains that
~X =−λ±1. (6.16)
Substituting the values of the parameters a,b,c,d obtained here into equation(1) of the system (6.10) one obtains the contradiction that
±4λ± = 0.
This result shows that if (n+,ν+) = (2,1), then no nontrivial ET-matrixexists.
5. Case (4,5). This case yields the same values of r and t as case (2,1) above.The corresponding matrices is given by
C =
r t t t t tt r at bt ct dtt at r et f t gtt bt et r ht ltt ct f t ht r mtt dt gt lt mt −r
,
where
a,b,c,d,e, f ,g,h, l,m ∈ {z ∈ C : |z|= 1} .
From the orthogonality of the rows of C we obtain the following system ofequations
2r+ t (a+b+ c+d) = 0 (1)2r+ t (a+ e+ f +g) = 0 (2)2r+ t
(b+ e+h+ l
)= 0 (3)
2r+ t(c+ f +h+m
)= 0 (4)
d +g+ l +m = 0 (5)
2ar+ t(1+be+ c f +dg
)= 0 (6)
2ar+ t(1+ae+ ch+dl
)= 0 (7)
2br+ t(1+a f +bh+dm
)= 0 (8)
1+ag+bl + cm = 0 (9)
2er+ t(1+ab+ f h+gl
)= 0 (10)
2 f r+ t(1+ac+ eh+gm
)= 0 (11)
1+ad + el + f m = 0 (12)
{2hr+ t
(1+bc+ e f + lm
)= 0 (13)
1+bd + eg+hm = 0 (14){1+ cd + f g+hl = 0 (15).
(6.17)18
This case is similar to case (2,1) and can be analysed in a similar fashionwith the following modifications
1. From equations (1) to (4) matrix A and vector ~X are obtained respectivelyas
0 a b ca 0 e fb e 0 hc f h 0
and
dglm
.
2. Equations (9), (12), (14) and (15) yield the matrix equation A~X =−1.
The analysis shows that
~X =− 1λ± 1, λ± =
√5±√
2√3
.
Substituting the corresponding values of the parameters into equation (5) ofthe system (6.17) yields the contradiction that −4 1
λ± = 0. Thus no nontrivialET-matix exists in this case as well.
6. Case (3,3). Here the trace of C is zero and the value of r cannot be determinedfrom the formula (3.2). Since the number of positive and negative diagonalentries are equal, the corresponding matrix C can be represented as
C =
r t t t t tt r at bt ct dtt at r et f t gtt bt et −r ht ltt ct f t ht −r mtt dt gt lt mt −r
,
where
a,b,c,d,e, f ,g,h, l,m ∈ {z ∈ C : |z|= 1} .
From the orthogonality of the rows of C we obtain the following system ofequations
2r+ t (a+b+ c+d) = 0 (1)2r+ t (a+ e+ f +g) = 0 (2)b+ e+h+ l = 0 (3)c+ f +h+m = 0 (4)d +g+ l +m = 0 (5)
2ar+ t(1+be+ c f +dg
)= 0 (6)
1+ae+ ch+dl = 0 (7)1+a f +bh+dm = 0 (8)1+ag+bl + cm = 0 (9)
1+ab+ f h+gl = 0 (10)1+ac+ eh+gm = 0 (11)1+ad + el + f m = 0 (12)
{2hr+ t
(1+bc+ e f + lm
)= 0 (13)
−2lr+ t(1+bd + eg+hm
)= 0 (14)
{2mr+ t
(1+ cd + f g+hl
)= 0 (15).
(6.18)
19
In this system there are nine equations to which Lemma 1 can be applieddirectly: equations (3), (4), (5), (7), (8), (9), (10), (11) and (12). It shouldbe noted that the application of the Lemma is done to an appropriate equa-tion in each subsequent system, starting with any of the nine equations inthe above system. For example applying the Lemma to equation (3) yieldsthree possibilities. Making substitutions from any of the possibilities yieldsan equivalent system in which equation (3) vanishes. In that subsequent sys-tem, the number of equations to which the Lemma can now be applied willbe at most eight.
The details of the computations in determining the parameters in this casecan be found in the appendix B. The results can be grouped into four cate-gories depending the kind of matrix C obtained:(a) no matrix exists,(b) unique real matrix,(c) one parameter family,(d) two parameters family.
Below we give examples from the last three categories. The cases referred tocan be found in appendix B.
Example 1 (Unique real matrix- Case 2.1.1.1). This result is obtained by mak-ing the following choices when Lemma 1 is applied to equations (3), (4) and (5) re-spectively of the system (6.18).
{h =−bl =−e
,
{c =−hm =− f
and
{l =−dm =−g.
With these substitutions, the equations (3), (4), and (5) vanish from the system andit can also be seen that c = b, e = d and g = f . Equation (7) can now be writtenas ad− 1 = 0 which implies that d = a. At this stage, other parameters that can beexpressed in terms of a are e = a and l = −a. Equations (1) and (2) of the system(6.18) now become
2r+ t(2a+b+b
)= 0 (1) 2r+ t
(2a+ f + f
)= 0 (2).
Evaluating the difference of equation (1) and the conjugate of equation (2) yieldsthat ℜ( f ) = ℜ(b) which means that either f = b or f = b. Choosing f = b and theother relations above, equations (6) and (13) of the system (6.18) now become
2ar+ t(1+2ab+b2)= 0 (6) 2br+ t
(1+b2 +2ab
)= 0 (13).
The difference of these two equations shows that b = a. With this value of b,equation (1) is now written as r+ t (a+ℜ(b)) = 0. This relation between r and t andthat in equation (3.1) show that
t = 1√5+(a+ℜ(a))2
and r = −(a+ℜ(a))√5+(a+ℜ(a))2
.
20
Since r is real, a is necessarily real implying that a = ±1. The appropriate choice ofa so that r is nonnegative is a =−1. Thus the matrix C is fully determined and real asseen below
C = 13
2 1 1 1 1 11 2 −1 −1 −1 −11 −1 2 −1 −1 −11 −1 −1 −2 1 11 −1 −1 1 −2 11 −1 −1 1 1 −2
. (6.19)
Example 2 (One parameter family - Case 2.1.1.2). Suppose in Example 1above the relation f = b is chosen instead of f = b, while the other relations areretained. Then a and b are arbitrary and equations (1), (13) and (15) of the system(6.18) can be written as follows
2r+ t(2a+b+b
)= 0 (1)
2br+ t(
1+b2 +2ab)= 0 (13)
2br+ t(
1+2ab+b2)= 0 (15).
Equation (1) then takes the form r+ t (a+ℜ(b)) = 0. Since r and t are real, a isalso real meaning that a = ±1. The resulting relation between r and t and equation(3.1) yield that
t = 1√5+α2 and r = −α√
5+α2 ,
where α =±1+ℜ(b).Since r > 0, α should be chosen negative and this is obtained when a = −1. The
sign of ℜ(b) does not determine the sign of α because |ℜ(b)| ≤ 1. In particular thevalue b = 1 is not allowed because then one obtains that r = 0 which is excluded. Thusα = −1+ℜ(b). The ET-matrix C is fully determined and is given in terms of b asseen below
C = 1√5+(1−ℜ(b))2
1−ℜ(b) 1 1 1 1 11 1−ℜ(b) −1 b b −11 −1 1−ℜ(b) −1 b b1 b −1 −1+ℜ(b) −b 11 b b −b −1+ℜ(b) −b1 −1 b 1 −b −1+ℜ(b)
.
(6.20)
Example 3 (Two parameters family - Case 2.2.3). Here the choices madewhen Lemma 1 is applied to equation (3), (4) and (5) respectively of the system (6.18)are
{h =−bl =−e
,
{f =−hm =−c
and
{g =−ml =−d
.
21
The following relations immediately emerge; e = d, f = b and g = c. With theserelations the system (6.17) reduces to the following
{2r+ t (a+b+ c+d) = 0ad−bc = 0.
(6.21)
From the first equation in (6.21) and from equation (3.1), the values of r and t in termsof a,b,c,d are
r = −α√20+α2 , t = 2√
20+α2 ,
where α = a+ b+ c+ d. Since r and t are real and nonnegative, α has to be chosenreal and negative. Remembering that the parameters are unimodular, the system (6.21)can now be written as
{a+b+ c+d = α ∈ R−a = bcd,
(6.22)
where R− = (−∞,0). These equations can be written as a linear system of a and d asfollows
{a+d = α−b− ca−bcd = 0.
(6.23)
The system yields values of a and d if and only if the determinant of the coefficientsmatrix does not vanish. Thus we require that bc 6=−1. The values of a and d are thenobtained as
a =bc(α−b−c)
1+bc , d = α−b−c1+bc . (6.24)
Given that d is unimodular, it follows that dd = 1. Substituting the value of d aboveinto this equation and simplifying one obtains that
α2−2αℜ(b+ c)+4ℑ(b)ℑ(c) = 0
from which it follows that
α = ℜ(b+ c)±√(ℜ(b+ c))2−4ℑ(b)ℑ(c). (6.25)
The value α ∈ R− is obtained in any of the following cases
1. (ℜ(b+ c))2− 4ℑ(b)ℑ(c) > 0 and ℑ(b)ℑ(c) < 0. Here there exists a uniqueα ∈ R−.
2. (ℜ(b+ c))2− 4ℑ(b)ℑ(c) > 0, ℑ(b)ℑ(c) > 0 and ℜ(b+ c) < 0. There existstwo α ∈ R−.
3. (ℜ(b+ c))2−4ℑ(b)ℑ(c) = 0, ℑ(b)ℑ(c)> 0 and ℜ(b+ c)< 0. There exists aunique α ∈ R−.
22
4. ℑ(b)ℑ(c) = 0 and ℜ(b+ c)< 0. There exists a unique α ∈ R− since α = 0 isexcluded.
The corresponding matrix C can now be written as follows
C = 1√20+α2
−α 2 2 2 2 22 −α 2a 2b 2c 2d2 2a −α −2c 2b −2d2 2b −2c α −2b 2c2 2c 2b −2b α −2c2 2d −2d 2c −2c α
. (6.26)
Here b and c are arbitrary parameters subject to bc 6= −1, α is given by the formula(6.25) and a and d by the equations (6.24).
On the other hand, suppose that bc =−1 is allowed. Then c =−b and the system(6.21) now becomes
{a+d = α−b+ba+d = 0.
This implies that α = b−b. Since α ∈R it follows that α = 0 contrary to the require-ment that α is negative.
Conclusion. If Tr(S) 6= 0, there are few possible values of r given by the formula(3.2). It appears that almost all such r do not yield an ET-matrix. Our calculationsshow that only the same r as for the standard matching conditions is possible sincesuch matrices are equivalent to standard matching conditions scattering matrices.
If Tr(S) = 0, then r cannot be determined by equation (3.2). This case occurs onlyfor matrices of even dimension and the corresponding ET-matrices exist as follows.When n = 2,4, there exixt one parameter families of matrices C. When n = 6, thereexists a unique real matrix C which is equivalent to the SMC-matrix, a one-parameterfamily and two-parameter family matrices C.
We have described all possible ET-matrices in dimensions 2 to 6. Before onlyexamples of such matrices were constructed. Our hypothesis is that in all dimensionsfor matrices having representation (2.1) with zero trace, and this occurs when the oderof the matrix is even, there exist some families of ET-matrices. In odd dimensionsthere exists only those generated by the SMC-matrix
References[1] G. Berkolaiko and P. Kuchment Introduction to quantum graphs, American Mathematical Society,
Rhode Island, 2013.[2] M. Hirobumi and S. Iwao The scattering matrix, Electronic Journal of Combinatorics, 15 (2008).[3] G. Berkolaiko, J. M. Harrison and J. H. Wilson Mathematical aspects of vacuum energy on quantum
graphs, Journal of Physics. A. Mathematical and Theoritical, 42 (2009).[4] R. Band, J. M. Harrison and C. H. Joyner Finite pseudo orbit expansions for spectral quantities of
quantum graphs, Journal of Physics. A. Mathematical and Theoritical, 45 (2012).[5] T. Ondrej and C. Taksu Quantum graph vertices with permutation-symmetric scattering probabilities,
Physics Letters. A, 375 (2011).[6] T. Ondrej, P. Exner and C. Taksu Reflectioless and equiscattering quantum graphs, IARIA, (2001).[7] T. Ondrej and C. Taksu Hermitian unitary matrices with modular permutation symmetry, arXiv, (2011)[8] T. Cheon, Reflectionless and Equiscattering Quantum Graphs and Their Applications, Int. J. System &
Measurement, 5 (2012) 34-44.23
[9] T. Cheon, Reflectionless and Equiscattering Quantum Graphs, textitInt. J. Adv. Systems and Measure-ments, 5 (2012), 18–22.
[10] M. Harmer, Hermitian symplectic geometry and extension theory, Journal of Physics. A. Mathematicaland General, 33, (2000),9193-9203.
[11] M. Harmer, Hermitian symplectic geometry and the factorization of the scattering matrix on graphs,Journal of Physics. A. Mathematical and General, 33, (2000),9015–9032.
[12] J.M. Harrison, U. Smilansky, and B. Winn, Quantum graphs where back-scattering is prohibited, Jour-nal of Physics. A. Mathematical and Theoritical, 40, (2007), 14181-14193.
[13] R.B. Holmes and V.I. Paulsen, Optimal frames for erasures, Linear Algebra Appl. 377 (2004), 3151.[14] P. Kurasov, Quantum graphs:Spectral theory and inverse problems, 2014 monograph in preparation.[15] P. Kurasov and M. Nowaczyk, Geometric properties of quantum graphs and vertex scattering matrices,
Opuscula Mathematica, 30, (2010).[16] P. Kurasov, R. Ogik and A. Rauf, On reflectionless equi-transmitting matrices, Institut Mittag-Leffler
Report No. 24 2012/2013, https://www.mittag-leffler.se/preprints/files/IML-1213s-24.pdf[17] O. Post, Spectral analysis on graph-like spaces, Springer, 2010.[18] O. Turek and T. Cheon, Quantum graph vertices with permutation-symmetric scattering probabilities,
Phys. Lett. A 375 (2011), no. 43, 37753780.[19] O. Turek and T. Cheon, Hermitian unitary matrices with modular permutation symmetry,
arXiv:1104.0408.
DEPT. OF MATHEMATICS, STOCKHOLM UNIV. 106 91 STOCKHOLM, SWEDEN
24
Part III
Appendices
55
A. Reflectioless Equi-transimtting Matrices ofOrder 6
In this appendix we give details of computations yielding the cases summarized inTable 1 of Paper 1.
C =
0 1 1 1 1 11 0 a b c d1 a 0 e f g1 b e 0 h l1 c f h 0 m1 d g l m 0
. (A.1)
The parameters a,b,c,d,e, f ,g,h, l,m ∈ C have absolute value one and are chosen sothat the rows (columns) of S are orthogonal. The orthogonality conditions yield thefollowing 15 equations in 10 unknowns:
a+b+ c+d = 0 (1)a+ e+ f +g = 0 (2)b+ e+h+ l = 0 (3)c+ f +h+m = 0 (4)d +g+ l +m = 0 (5)
1+be+ c f +dg = 0 (6)1+ae+ ch+dl = 0 (7)1+a f +bh+dm = 0 (8)1+ag+bl + cm = 0 (9)
1+ab+ f h+gl = 0 (10)1+ac+ eh+gm = 0 (11)1+ad + el + f m = 0 (12)
{1+bc+ e f + lm = 0 (13)1+bd + eg+hm = 0 (14)
{1+ cd + f g+hl = 0 (15).
(A.2)
Applying Lemma 1 in Paper I to equation (1) in system (A.2) we obtain the fol-lowing three cases.
Case 1 : b =−a, d =−c
Case 2 : c =−a, d =−b (A.3)Case 3 : d =−a, c =−b
In what follows we consider the first case by substituting into the system (A.2) thevalues of b and d and obtain
57
Case 1 : b =−a, d =−c
a+ e+ f +g = 0 (2)−a+ e+h+ l = 0 (3)c+ f +h+m = 0 (4)−c+g+ l +m = 0 (5)
1−ae+ c f − cg = 0 (6)1+ae+ ch− cl = 0 (7)1+a f −ah− cm = 0 (8)1+ag−al + cm = 0 (9)
1−aa+ f h+gl = 0 (10)1+ac+ eh+gm = 0 (11)1−ac+ el + f m = 0 (12)
{1−ac+ e f + lm = 0 (13)1+ac+ eg+hm = 0 (14)
{1− cc+ f g+hl = 0 (15).
(A.4)
Since the parameters have absolute value 1, the expressions 1−aa and 1−cc in equa-tions (10) and (15) respectively vanish. Equation (15) can then be eliminated becauseit is a multiple of equation (10). We then apply the Lemma to equation (2) of thesystem A.4 and obtain the following cases:
Case 1.1 : e =−a, g =− f
Case 1.2 : f =−a, g =−e (A.5)Case 1.3 : g =−a, f =−e
Case 1.1 : b =−a, d =−c
e =−a, g =− f
Substituting the values of e and g into the system (A.4) we obtain the following sys-tem:
−a−a+h+ l = 0 (3)c+ f +h+m = 0 (4)−c− f + l +m = 0 (5)
1+a2 + c f + c f = 0 (6)1−|a|2 + ch− cl = 0 (7)1+a f −ah− cm = 0 (8)1−a f −al + cm = 0 (9)
f h− f l = 0 (10)1+ac−ah− f m = 0 (11)1−ac−al + f m = 0 (12)
{1−ac−a f + lm = 0 (13)1+ac+a f +hm = 0 (14)
(A.6)
From equation (7) or (10) we obtain that l = h. Subsitituting the value of l into equa-tion (3) we obtain that ℜ(a) = h so that h = ±1; which in turn implies that a = ±1.Adding equations (4) and (5) and then substituting the values of h and l we obtainthat ℜ(m) = −h = ∓1 which implies that m = ∓1. Substituting the value of a into
58
equation (6) we obtain that 2+ 2c f = 0, which implies that f = −c. From these wesee that the corresponding matrix is
C1.1 =
0 1 1 1 1 11 0 ±1 ∓1 c −c1 ±1 0 ∓1 −c c1 ∓1 ∓1 0 ±1 ±11 c −c ±1 0 ∓11 −c c ±1 ∓1 0
= A4. (A.7)
Case 1.2 : b =−a, d =−c
f =−a, g =−e
Substitution of the values of f and g into the system (A.4) gives us the followingsystem:
−a+ e+h+ l = 0 (3)c−a+h+m = 0 (4)−c− e+ l +m = 0 (5)
1−ae−ac+ ce = 0 (6)1+ae+ ch− cl = 0 (7)1−|a|2−ah− cm = 0 (8)1−ae−al + cm = 0 (9)
−ah− el = 0 (10)1+ac+ eh− em = 0 (11)1−ac+ el−am = 0 (12)
{1−ac−ae+ lm = 0 (13)1+ac−|e|2 +hm = 0 (14)
(A.8)
Equation (14) can be eliminated because it is a multiple of equation (8). Applying theLemma to equation (3) in the system (A.8) we obtain the following cases:
Case 1.2.1 : e = a, l =−h
Case 1.2.2 : h = a, l =−e (A.9)Case 1.2.3 : l = a, h =−e
Case 1.2.1 : b =−a, d =−c
f =−a, g =−e
e = a, l =−h
Substituting the values of e and l into the system (A.8) we obtain the following system:
59
{c−a+h+m = 0 (4)−c−a−h+m = 0 (5)
1−|a|2−ac+ac = 0 (6)1+a2 + ch+ ch = 0 (7)−ah− cm = 0 (8)1−a2−al + cm = 0 (9)
−ah+ah = 0 (10)1+ac+ah−am = 0 (11)1−ac−ah−am = 0 (12)
{1−ac−a2−hm = 0 (13)ac+hm = 0 (14)
(A.10)
From equation (6) or equation (10) we obtain that a = a so that a = ±1. Addingequations (4) and (5) and then substituting the value of a we obtain that ℜ(m) = ±1and this implies that m = ±1. Substituting the value of a into equation (7) yields theresult that ch =−1 so that h =−c. Hence the corresponding matrix becomes
C1.2.1 =
0 1 1 1 1 11 0 ±1 ∓1 c −c1 ±1 0 ±1 ∓1 ∓11 ∓1 ±1 0 −c c1 c ∓1 −c 0 ±11 −c ∓1 c ±1 0
= A3. (A.11)
Case 1.2.2 : b =−a, d =−c
f =−a, g =−e
h = a, l =−e
Substituting the values of h and l into the system (A.8) we obtain the following system:
{c−a+a+m = 0 (4)−c− e− e+m = 0 (5)
1−ae−ac+ ce = 0 (6)1+ae+ac+ ce = 0 (7)−|a|2− cm = 0 (8)1−ae+ae+ cm = 0 (9)
−|a|2 + e2 = 0 (10)1+ac+ae− em = 0 (11)1−ac−|e|2−am = 0 (12)
{1−ac−ae− em = 0 (13)
(A.12)
From equation (4) or equation (8) we obtain that m = −c and m = −c respectively.This implies that c = c so that c = ±1 and m = ∓1. Substituting the values c and minto equation (5) ℜ(e) =∓1 so that e =∓1. The corresponding matrix becomes
60
C1.2.2 =
0 1 1 1 1 11 0 a −a ±1 ∓11 a 0 ∓1 −a ±11 −a ∓1 0 a ±11 ±1 −a a 0 ∓11 ∓1 ±1 ±1 ∓1 0
= A6. (A.13)
Case 1.2.3 : b =−a, d =−c
f =−a, g =−e
l = a, h =−e
Substituting the values of h and l into the system (A.8) we obtain the following system:
{c−a− e+m = 0 (4)−c− e+a+m = 0 (5)
1−ae−ac+ ce = 0 (6)1+ae− ce−ac = 0 (7)ae− cm = 0 (8)1−ae−|a|2 + cm = 0 (9)
−ae+ae = 0 (10)1+ac−|e|2− em = 0 (11)1−ac+ae−am = 0 (12)
{1−ac−ae+am = 0 (13)
(A.14)
Equation (11) can be discarded because it is a multiple of equation (9). From equation(10) we see that a = a so that a =±1. ae = cm in equation (9). Since a ∈R we obtainthat ae = cm. But from equation (8) we have that ae = cm. Therefore we obtain thatcm = cm. This implies that c = c so that c =±1. Substituting the values a and c intoequation (4) we obtain that m = e. Therefore the corresponding matrix becomes
C1.2.3 =
0 1 1 1 1 11 0 ±1 ∓1 ±1 ∓11 ±1 0 e ∓1 −e1 ∓1 e 0 −e ±11 ±1 ∓1 −e 0 e1 ∓1 −e ±1 e 0
= B2. (A.15)
Case 1.3 : b =−a, d =−c
g =−a, f =−e
Substitution of the values of f and g into the system (A.4) gives us the followingsystem:
61
−a+ e+h+ l = 0 (3)c− e+h+m = 0 (4)−c−a+ l +m = 0 (5)
1−ae− ce+ac = 0 (6)1+ae+ ch− cl = 0 (7)1−ae−ah− cm = 0 (8)1−|a|2−al + cm = 0 (9)
−eh−al = 0 (10)1+ac+ eh−am = 0 (11)1−ac+ el− em = 0 (12)
{1−ac−|e|2 + lm = 0 (13)1+ac−ae+hm = 0 (14)
(A.16)
Equation (13) can be eliminated because it is a multiple of equation (9). Applying theLemma to equation (3) in the system (A.16) we obtain the following cases:
Case 1.3.1 : e = a, l =−h
Case 1.3.2 : h = a, l =−e (A.17)Case 1.3.3 : l = a, h =−e
Case 1.3.1 : b =−a, d =−c
g =−a, f =−e
e = a, l =−h
Substituting the values of e and l into the system (A.16) we obtain the followingsystem:
{c−a+h+m = 0 (4)−c−a−h+m = 0 (5)
1−|a|2−ac+ac = 0 (6)1+a2 + ch+ ch = 0 (7)1−a2−ah− cm = 0 (8)ah+ cm = 0 (9)
−ah+ah = 0 (10)1+ac+ah−am = 0 (11)1−ac−ah−am = 0 (12)
{1+ac−a2 +hm = 0 (14)
(A.18)
From equation (6) or equation (10) we deduce that that a = ±1. Adding equations(4) and (5) and then substituting the value of a we obtain that ℜ(m) = ±1 and thisimplies that m = ±1. Substituting the values of a and m into equation (8) yields theresult that h =−c. Hence the corresponding matrix becomes
62
C1.3.1 =
0 1 1 1 1 11 0 ±1 ∓1 c −c1 ±1 0 ±1 ∓1 ∓11 ∓1 ±1 0 −c c1 c ∓1 −c 0 ±11 −c ∓1 c ±1 0
= A3. (A.19)
This case is the same as case 1.2.1.
Case 1.3.2 : b =−a, d =−c
g =−a, f =−e
h = a, l =−e
Substituting the values of h and l into the system (A.16) we obtain the followingsystem:
{c− e+a+m = 0 (4)−c−a− e+m = 0 (5)
1−ae− ce+ac = 0 (6)1+ae+ac+ ce = 0 (7)1−ae−|a|2− cm = 0 (8)ae+ cm = 0 (9)
−ae+ae = 0 (10)1+ac+ae−am = 0 (11)1−ac−|e|2− em = 0 (12)
{1+ac−ae+am = 0 (14)
(A.20)
Equation (12) can be discarded because it is a multiple of equation (8). From equation(10) we obtain that a = ±1. Using the fact that a ∈ R equation (8) yields ae = −cm.Combining this result and equation (9) we obtain that cm = cm which in turn impliesthat c = ±1. Substituting the values a and c into either equation (4) or equation (5)yields that m = e. The corresponding matrix becomes
C1.3.2 =
0 1 1 1 1 11 0 ±1 ∓1 ∓1 ±11 ±1 0 e −e ∓11 ∓1 e 0 ±1 −e1 ∓1 −e ±1 0 e1 ±1 ∓1 −e e 0
=C2. (A.21)
Case 1.3.3 : b =−a, d =−c
g =−a, f =−e
l = a, h =−e
63
Substituting the values of h and l into the system (A.16) we obtain the followingsystem:
{c− e− e+m = 0 (4)−c−a+a+m = 0 (5)
1−ae− ce+ac = 0 (6)1+ae− ce−ac = 0 (7)1−ae+ae− cm = 0 (8)−|a|2 + cm = 0 (9)
e2−|a|2 = 0 (10)1+ac−|e|2−am = 0 (11)1−ac+ae− em = 0 (12)
{1+ac−ae− em = 0 (14)
(A.22)
From equation (10) we see that e2 = 1 so that e = ±1. From equation (4) we obtainthat m = c while from equation (9) we obtain that m = c. These two results show thatc = c so that c =±1, since c ∈ R. Therefore the corresponding matrix becomes
C1.3.3 =
0 1 1 1 1 11 0 a −a ±1 ∓11 a 0 ±1 ∓1 −a1 −a ±1 0 ∓1 a1 ±1 ∓1 ∓1 0 ±11 ∓1 −a a ±1 0
= A5. (A.23)
Now we consider the second case in (c1) by substituting into the system (A.2) thevalues of c and d and obtain
Case 2 : c =−a, d =−b
a+ e+ f +g = 0 (2)b+ e+h+ l = 0 (3)−a+ f +h+m = 0 (4)−b+g+ l +m = 0 (5)
1+be−a f −bg = 0 (6)1+ae−ah−bl = 0 (7)1+a f +bh−bm = 0 (8)1+ag+bl−am = 0 (9)
1+ab+ f h+gl = 0 (10)1−|a|2 + eh+gm = 0 (11)1−ab+ el + f m = 0 (12)
{1−ab+ e f + lm = 0 (13)1−|b|2 + eg+hm = 0 (14)
{1+ab+ f g+hl = 0 (15).
(A.24)
Equation (14) can then be eliminated because it is a multiple of equation (11). Wethen apply the Lemma to equation (2) of the system A.24 and obtain the followingcases:
64
Case 2.1 : e =−a, g =− f
Case 2.2 : f =−a, g =−e (A.25)Case 2.3 : g =−a, f =−e
Case 2.1 : c =−a, d =−b
e =−a, g =− f
Substituting the values of e and g into the system (A.24) we obtain the followingsystem:
b−a+h+ l = 0 (3)−a+ f +h+m = 0 (4)−b− f + l +m = 0 (5)
1−ab−a f +b f = 0 (6)1−|a|2−ah−bl = 0 (7)1+a f +bh−bm = 0 (8)1−a f +bl−am = 0 (9)
1+ab+ f h− f l = 0 (10)−ah− f m = 0 (11)1−ab−al + f m = 0 (12)
{1−ab−a f + lm = 0 (13)
{1+ab−| f |2 +hl = 0 (15)
(A.26)
We eliminate equation (15) beacuse it is a multiple of equation (7). We then the applyLemma to equation (3) to obtain the following cases:
Case 2.1.1 : b = a, l =−h
Case 2.1.2 : h = a, l =−b (A.27)
Case 2.1.3 : l = a, h =−b
Case 2.1.1 : c =−a, d =−b
e =−a, g =− f
b = a, l =−h
Here we substitute the values of b and l into the system (A.26) to obtain the followingsystem:
65
{−a+ f +h+m = 0 (4)−a− f −h+m = 0 (5)
1−|a|2−a f +a f = 0 (6)−ah+ah = 0 (7)1+a f +ah−am = 0 (8)1−a f −ah−am = 0 (9)
1+a2 + f h+ f h = 0 (10)−ah− f m = 0 (11)1−a2 +ah+ f m = 0 (12)
{1−a2−a f −hm = 0 (13)
(A.28)
It is easy to see from equations (6) or (7) that a = ±1. Adding equations (4) and(5) and then substituting the value of a we obtain that ℜ(m) =±1 which implies thatm =±1. Substituting the value of a into equation (10) we obtain that f h =−1, whichimplies that h =− f . From these we see that the corresponding matrix is
C2.1.1 =
0 1 1 1 1 11 0 ±1 ±1 ∓1 ∓11 ±1 0 ∓1 f − f1 ±1 ∓1 0 − f f1 ∓1 f − f 0 ±11 ∓1 − f f ±1 0
= A2. (A.29)
Case 2.1.2 : c =−a, d =−b
e =−a, g =− f
h = a, l =−b
Here we substitute the values of h and l into the system (A.26) to obtain the followingsystem:
{−a+ f +a+m = 0 (4)−b− f −b+m = 0 (5)
1−ab−a f +b f = 0 (6)−|a|2 +b2 = 0 (7)1+a f +ab−bm = 0 (8)1−a f −|b|2−am = 0 (9)
1+ab+a f +b f = 0 (10)−|a|2− f m = 0 (11)1−ab+ab+ f m = 0 (12)
{1−ab−a f −bm = 0 (13)
(A.30)
It is easy to see from equation (7) that b = ±1. From equation (4) we obtain thatm = − f . From equation (5) we obtain that m = 2ℜ(b)+ f , which implies that m =±2+ f . It then follows that ±2+ f =− f which upon simplification yields the resultthat f =∓1. From these we see that the corresponding matrix is
66
C2.1.2 =
0 1 1 1 1 11 0 a ±1 −a ∓11 a 0 −a ∓1 ±11 ±1 −a 0 a ∓11 −a ∓1 a 0 ±11 ∓1 ±1 ∓1 ±1 0
= B6 (A.31)
Case 2.1.3 : c =−a, d =−b
e =−a, g =− f
l = a, h =−b
Here we substitute the values of h and l into the system (A.26) to obtain the followingsystem:
{−a+ f −b+m = 0 (4)−b− f +a+m = 0 (5)
1−ab−a f +b f = 0 (6)ab−ab = 0 (7)1+a f −|b|2−bm = 0 (8)1−a f +ab−am = 0 (9)
1+ab−b f −a f = 0 (10)ab− f m = 0 (11)1−ab−|a|2 + f m = 0 (12)
{1−ab−a f +am = 0 (13)
(A.32)
Equation (12) can be eliminated because it is a multiple of equation (8). Subtractingthe conjugate of equation (5) from equation (4) yields f = a. Substituting this valueof f into equation (8) yields m = b. From these we see that the corresponding matrixis
C2.1.3 =
0 1 1 1 1 11 0 ±1 b ∓1 −b1 ±1 0 ∓1 ±1 ∓11 b ∓1 0 −b ±11 ∓1 ±1 −b 0 b1 −b ∓1 ±1 b 0
= B3 (A.33)
Case 2.2 : c =−a, d =−b
f =−a, g =−e
Substitution of the values of f and g into the system (A.24) gives us the followingsystem:
67
b+ e+h+ l = 0 (3)−a−a+h+m = 0 (4)−b− e+ l +m = 0 (5)
1+be+a2 +be = 0 (6)1+ae−ah−bl = 0 (7)1−|a|2 +bh−bm = 0 (8)1−ae+bl−am = 0 (9)
1+ab−ah− el = 0 (10)eh− em = 0 (11)1−ab+ el−am = 0 (12)
{1−ab−ae+ lm = 0 (13)
{1+ab+ae+hl = 0
(A.34)
Equation (11) can be eliminated because it is a multiple of equation (8). From equation(8) we obtain that m = h. Substituting this value of m into equation (4) and thensimplifying we obtain that Re(a) = h implying that h ∈R so that h =±1. This in turnimplies that a =±1 and m =±1. Adding equations (3) and (5) and then substitutingin the values of h and m we obtain upon simplification that l = ∓1. Substituting thevalue of a into equation (6) yields the result e =−b. The resulting matrix is
Case 2.3 : c =−a, d =−b
g =−a, f =−e
Substitution of the values of f and g into the system (A.24) gives us the followingsystem:
b+ e+h+ l = 0 (3)−a− e+h+m = 0 (4)−b−a+ l +m = 0 (5)
1+be+ae+ab = 0 (6)1+ae−ah−bl = 0 (7)1−ae+bh−bm = 0 (8)1−|a|2 +bl−am = 0 (9)
1+ab− eh−al = 0 (10)eh−am = 0 (11)1−ab+ el− em = 0 (12)
{1−ab−|e|2 + lm = 0 (13)
{1+ab+ae+hl = 0 (15)
(A.36)
Equation (13) can be eliminated because it is a multiple of equation (9). Applying theLemma to equation (3) in the system (A.36) we obtain the following cases:
68
Case 2.3.1 : e =−b, l =−h
Case 2.3.2 : h =−b, l =−e (A.37)
Case 2.3.3 : l =−b, h =−e
Case 2.3.1 : c =−a, d =−b
g =−a, f =−e
e =−b, l =−h
Substituting the values of e and l into the system (A.36) we obtain the followingsystem:
{−a+b+h+m = 0 (4)−b−a−h+m = 0 (5)
1−|b|2−ab+ab = 0 (6)1−ab−ah+bh = 0 (7)1+ab+bh−bm = 0 (8)−bh−am = 0 (9)
1+ab+bh+ah = 0 (10)−bh−am = 0 (11)1−ab+bh+bm = 0 (12)
{1+ab−ab−|h|2 = 0 (15)
(A.38)
Equation (15) can be eliminated because it is a multiple of equation (6). From equation(6) we deduce that that b = ±1. Substracting the conjugate of equation (5) fromequation (4) we obtain h = −b. Substituting this value of h into equation (9) yieldsm = a. From these we see that the corresponding matrix is
C2.3.1 =
0 1 1 1 1 11 0 a ±1 −a ∓11 a 0 ∓1 ±1 −a1 ±1 ∓1 0 ∓1 ±11 −a ±1 ∓1 0 a1 ∓1 −a ±1 a 0
=C4 (A.39)
Case 2.3.2 : c =−a, d =−b
g =−a, f =−e
h =−b, l =−e
Substituting the values of h and l into the system (A.36) we obtain the followingsystem:
69
{−a− e−b+m = 0 (4)−b−a− e+m = 0 (5)
1+be+ae+ab = 0 (6)1+ae+ab+be = 0 (7)1−ae−|b|2−bm = 0 (8)−be−am = 0 (9)
1+ab+be+ae = 0 (10)−be−am = 0 (11)1−ab−|e|2− em = 0 (12)
{1+ab+ae+be = 0 (15)
(A.40)
Equation (5) can be discarded because it is a conjugate of equation (4). Equations (9)and (12) can be discarded because they are multiples of equation (8). Determining thedifference between equations (6) and (7) and then simplifying we obtain that e = ±1or b =−a. On the other hand determining the difference between equations (10) and(15) and simplifying we obtain that e = ±1 or b = −a. These two results imply thata =±1, while b =∓1. The matrix obtained is:
C2.3.2 =
0 1 1 1 1 11 0 ±1 ±1 ∓1 ∓11 ±1 0 ∓1 ±1 ∓11 ±1 ∓1 0 ∓1 ±11 ∓1 ±1 ∓1 0 ±11 ∓1 ∓1 ±1 ±1 0
= D1. (A.41)
Case 2.3.3 : c =−a, d =−b
g =−a, f =−e
l =−b, h =−e
Substituting the values of h and l into the system (A.36) we obtain the followingsystem:
{−a− e− e+m = 0 (4)−b−a−b+m = 0 (5)
1+be+ae+ab = 0 (6)1+ae+ae+b2 = 0 (7)1−ae−be−bm = 0 (8)−|b|2−am = 0 (9)
1+ab+ e2 +ab = 0 (10)−|e|2−am = 0 (11)1−ab−be− em = 0 (12)
{1+ab+ae+be = 0 (15)
(A.42)
From equation (9) or equation (11) we see that m = −a. Substituting this value of ainto equations (4) and (5) and then simplifying yields the result that a = ±1, b = ∓1and e =∓1. The resultant matrix becomes
70
C2.3.3 =
0 1 1 1 1 11 0 ±1 ∓1 ∓1 ±11 ±1 0 ∓1 ±1 ∓11 ∓1 ∓1 0 ±1 ±11 ∓1 ±1 ±1 0 ∓11 ±1 ∓1 ±1 ∓1 0
= D4. (A.43)
Case 3 : d =−a, c =−b
We substitute the values of d and c into (c1) and obtain the following system:
a+ e+ f +g = 0 (2)b+ e+h+ l = 0 (3)−b+ f +h+m = 0 (4)−a+g+ l +m = 0 (5)
1+be−b f −ag = 0 (6)1+ae−bh−al = 0 (7)1+a f +bh−am = 0 (8)1+ag+bl−bm = 0 (9)
1+ab+ f h+gl = 0 (10)1−ab+ eh+gm = 0 (11)1−|a|2 + el + f m = 0 (12)
{1−|b|2 + e f + lm = 0 (13)1−ab+ eg+hm = 0 (14)
{1+ab+ f g+hl = 0 (15).
(A.44)
Equation (13) can then be eliminated because it is a multiple of equation (12). Wethen apply the Lemma to equation (2) of the system A.44 and obtain the followingcases:
Case 3.1 : e =−a, g =− f
Case 3.2 : f =−a, g =−e (A.45)Case 3.3 : g =−a, f =−e
Case 3.1 : d =−a, c =−b
e =−a, g =− f
Substituting the values of e and g into the system (A.44) we obtain the followingsystem:
71
b−a+h+ l = 0 (3)−b+ f +h+m = 0 (4)−a− f + l +m = 0 (5)
1−ab−b f +a f = 0 (6)1−|a|2−bh−al = 0 (7)1+a f +bh−am = 0 (8)1−a f +bl−bm = 0 (9)
1+ab+ f h− f l = 0 (10)1−ab−ah− f m = 0 (11)−al + f m = 0 (12)
{1−ab+a f +hm = 0 (14)
{1+ab−| f |2 +hl = 0 (15)
(A.46)
We eliminate equation (15) beacuse it is a multiple of equation (7). We then apply theLemma to equation (3) to obtain the following cases:
Case 3.1.1 : b = a, l =−h
Case 3.1.2 : h = a, l =−b (A.47)
Case 3.1.3 : l = a, h =−b
Case 3.1.1 : d =−a, c =−b
e =−a, g =− f
b = a, l =−h
Here we substitute the values of b and l into the system (A.46) to obtain the followingsystem:
{−a+ f +h+m = 0 (4)−a− f −h+m = 0 (5)
1−|a|2−a f +a f = 0 (6)−ah+ah = 0 (7)1+a f +ah−am = 0 (8)1−a f −ah−am = 0 (9)
1+a2 + f h+ f h = 0 (10)1−a2−ah− f m = 0 (11)ah+ f m = 0 (12)
{1−a2 +a f +hm = 0 (14)
(A.48)
It is easy to see from equations (6) or (7) that a = ±1. Adding equations (4) and(5) and then substituting the value of a we obtain that ℜ(m) =±1 which implies thatm =±1. Substituting the value of a into equation (10) we obtain that f h =−1, whichimplies that h =− f . From these we see that the corresponding matrix is
72
C3.1.1 =
0 1 1 1 1 11 0 ±1 ±1 ∓1 ∓11 ±1 0 ∓1 f − f1 ±1 ∓1 0 − f f1 ∓1 f − f 0 ±11 ∓1 − f f ±1 0
= A2. (A.49)
This case is exactly the same as the case 2.1.1
Case 3.1.2 : c =−a, d =−b
e =−a, g =− f
h = a, l =−b
Here we substitute the values of h and l into the system (A.46) to obtain the followingsystem:
{−b+ f +a+m = 0 (4)−a− f −b+m = 0 (5)
1−ab−b f +a f = 0 (6)−ab+ab = 0 (7)1+a f +ab−am = 0 (8)1−a f −|b|2−bm = 0 (9)
1+ab+a f +b f = 0 (10)1−ab−|a|2− f m = 0 (11)ab+ f m = 0 (12)
{1−ab+a f +am = 0 (14)
(A.50)
It is easy to see from equation (7) that a = ±1. From equation (4) we obtain thatb = a+ f +m so that b = a+ f +m. From equation (5) we obtain that b =−a− f +m, which implies that a+ f +m = −a− f +m. Upon simplification we obtain thatℜ( f ) = −ℜ(a) = ∓1. It then follows that f = ∓1. Substituting the values of a andf into equation (9) yields the result m = b. From these we see that the correspondingmatrix is
Case 3.1.3 : d =−a, c =−b
e =−a, g =− f
l = a, h =−b
Here we substitute the values of h and l into the system (A.46) to obtain the followingsystem:
73
{−b+ f −b+m = 0 (4)−a− f +a+m = 0 (5)
1−ab−b f +a f = 0 (6)b2−|a|2 = 0 (7)1+a f −|b|2−am = 0 (8)1−a f +ab−bm = 0 (9)
1+ab−b f −a f = 0 (10)1−ab+ab− f m = 0 (11)−|a|2 + f m = 0 (12)
{1−ab+a f −bm = 0 (14)
(A.52)
From equation (7) we obtain that b = ±1. From equation (5) we obtain that m = f .Substituting the values of b and m into equation (4) f =±1 which in turn implies thatm =±1. The resultant matrix is
C3.1.3 =
0 1 1 1 1 11 0 a ±1 ∓1 −a1 a 0 −a ±1 ∓11 ±1 −a 0 ∓1 a1 ∓1 ±1 ∓1 0 ±11 −a ∓1 a ∓1 0
=C5 (A.53)
Case 3.2 : d =−a, c =−b
f =−a, g =−e
Substitution of the values of f and g into the system (A.44) gives us the followingsystem:
b+ e+h+ l = 0 (3)−b−a+h+m = 0 (4)−a− e+ l +m = 0 (5)
1+be+ab+ae = 0 (6)1+ae−bh−al = 0 (7)1−|a|2 +bh−am = 0 (8)1−ae+bl−bm = 0 (9)
1+ab−ah− el = 0 (10)1−ab+ eh− em = 0 (11)el−am = 0 (12)
{1−ab−|e|2 +hm = 0 (14)
{1+ab+ae+hl = 0 (15)
(A.54)
Equation (14) can be eliminated because it is a multiple of equation (8). Applying theLemma to equation (3) yields the following cases:
74
Case 3.2.1 : e =−b, l =−h
Case 3.2.2 : h =−b, l =−e (A.55)
Case 3.2.3 : l =−b, h =−e
Case 3.2.1 : d =−a, c =−b
f =−a, g =−e
e =−b, l =−h
Sunstituting the values of e and l into the system (A.54) gives us the following system:
{−b−a+h+m = 0 (4)−a+b−h+m = 0 (5)
1−|b|2 +ab−ab = 0 (6)1−ab−bh+ah = 0 (7)bh−am = 0 (8)1+ab−bh−bm = 0 (9)
1+ab−ah−bh = 0 (10)1−ab−bh+bm = 0 (11)bh−am = 0 (12)
{1+ab−ab+ |h|2 = 0 (15)
(A.56)
We eliminate equation (15) because it is a multiple of equation (6). From equation(6) we obtain that b = ±1. From equation (8) bh = am so that bh = am. But fromequation (12) bh = am. Therefore we see that bh = bh. Substituting the value of b intothis equation we observe that h = h so that h = ±1. Substituting the values of b andh into either equation (4) or equation (5) yields the result that m = a. The resultantmatrix is:
C3.2.1 =
0 1 1 1 1 11 0 a ±1 ∓1 −a1 a 0 ∓1 −a ±11 ±1 ∓1 0 ±1 ∓11 ∓1 −a ±1 0 a1 −a ±1 ∓1 a 0
= B4 (A.57)
Case 3.2.2 : d =−a, c =−b
f =−a, g =−e
h =−b, l =−e
Sunstituting the values of h and l into the system (A.54) gives us the following system:
75
{−b−a−b+m = 0 (4)−a− e− e+m = 0 (5)
1−be+ab+ae = 0 (6)1+ae+b2 +ae = 0 (7)−|b|2−am = 0 (8)1−ae−be−bm = 0 (9)
1+ab+ab+ e2 = 0 (10)1−ab−be− em = 0 (11)−|e|2−am = 0 (12)
{1+ab+ae+be = 0 (15)
(A.58)
From either equation (8) or equation (12) we deduce that m = −a. Substituting thisvalue of m into equation (4) we obtain that ℜ(b) = −a which implies that a ∈ R.Therefore we conclude that a =±1 and b =∓1. Similarly substituting the value of minto equation (5) yields the result that e =∓1. The resultant matrix is:
C3.2.2 =
0 1 1 1 1 11 0 ±1 ∓1 ±1 ∓11 ±1 0 ∓1 ∓1 ±11 ∓1 ∓1 0 ±1 ±11 ±1 ∓1 ±1 0 ∓11 ∓1 ±1 ±1 ∓1 0
= D6 (A.59)
Case 3.2.3 : d =−a, c =−b
f =−a, g =−e
l =−b, h =−e
Sunstituting the values of h and l into the system (A.54) gives us the following system:
{−b−a− e+m = 0 (4)−a− e−b+m = 0 (5)
1+be+ab+ae = 0 (6)1+ae+be+ab = 0 (7)−be−am = 0 (8)1−ae−|b|2−bm = 0 (9)
1+ab+ae+be = 0 (10)1−ab−|e|2− em = 0 (11)−be−am = 0 (12)
{1+ab+ae+be = 0 (15)
(A.60)
Equation (5) can be eliminated since it is the conjugate of equation (4). We eliminateequations (9), (11) and (12) because they are multiples of equation (8).Applying theLemma to equation (4) of system (A.60) yields the following cases:
76
Case 3.2.3.1 : b =−a, m = e
Case 3.2.3.2 : e =−a, m = b (A.61)
Case 3.2.3.3 : m = a, e =−b
Case 3.2.3.1 : d =−a, c =−b
f =−a, g =−e
l =−b, h =−e
b =−a, m = e
Substituting the values of b and m into the system (A.60) yields the following system:
{−a− e+a+ e = 0 (5)
1−ae−|a|2 +ae = 0 (6)1+ae−ae−|a|2 = 0 (7)−ae−ae = 0 (8)
{1−a2 +ae−ae = 0 (10)
{1−a2 +ae−ae = 0 (15)
(A.62)
From equation (8) we observe that a =±1. The corresponding matrix is:
Case 3.2.3.2 : d =−a, c =−b
f =−a, g =−e
l =−b, h =−e
e =−a, m = b
Substituting the values of e and m into the system (A.60) yields the following system:
{−a+a−b+b = 0 (5)
1−ab+ab−|a|2 = 0 (6)1−a2−ab+ab = 0 (7)ab−ab = 0 (8)
{1+ab−|a|2−ab = 0 (10)
{1+ab−a2−ab = 0 (15)
(A.64)
Equations (8), (10) and (15) can be eliminated beacuse they are mutiples of equation(6). From equation (6) we obtain that a =±1. The corresponding matrix is:
77
C3.2.3.2 =
0 1 1 1 1 11 0 ±1 b −b ∓11 ±1 0 ∓1 ∓1 ±11 b ∓1 0 ±1 −b1 −b ∓1 ±1 0 b1 ∓1 ±1 −b b 0
=C3 (A.65)
Case 3.2.3.3 : d =−a, c =−b
f =−a, g =−e
l =−b, h =−e
m = a, e =−b
Substituting the values of e and m into the system (A.60) yields the following system:
{−a+b−b+a = 0 (5)
1−b2 +ab−ab = 0 (6)1−ab−|b|2 +ab = 0 (7)b2−|a|2 = 0 (8)
{1+ab−ab−|b|2 = 0 (10)
{1+ab−ab−b
2= 0 (15)
(A.66)
From equation (6) we obtain that b =±1. The corresponding matrix is:
C3.2.3.3 =
0 1 1 1 1 11 0 a ±1 ∓1 −a1 a 0 ∓1 −a ±11 ±1 ∓1 0 ±1 ∓11 ∓1 −a ±1 0 a1 −a ±1 ∓1 a 0
= B4 (A.67)
Case 3.3 : d =−a, c =−b
g =−a, f =−e
Substitution of the values of f and g into the system (A.44) gives us the followingsystem:
78
b+ e+h+ l = 0 (3)−b− e+h+m = 0 (4)−a−a+ l +m = 0 (5)
1+be+be+a2 = 0 (6)1+ae−bh−al = 0 (7)1−ae+bh−am = 0 (8)1−|a|2 +bl−am = 0 (9)
1+ab− eh−al = 0 (10)1−ab+ eh−am = 0 (11)el− em = 0 (12)
{1−ab−ae+hm = 0 (14)
{1+ab+ae+hl = 0 (15)
(A.68)
From equation (12) we obtain that m = l. Substituting this value of m into equation(5) we obtain that ℜ(a) = l. This in turn implies that l ∈R so that l =±1 and a =±1.Substituting the value of m into equation (4) and then adding equations (3) and (4) weobtain that Re(h) = −l which implies that h = ∓1. Substituting the value of a intoequation (6) yields the result e =−b. The resultant matrix is:
C3.3 =
0 1 1 1 1 11 0 ±1 b −b ∓11 ±1 0 −b b ∓11 b −b 0 ∓1 ±11 −b b ∓1 0 ±11 ∓1 ∓1 ±1 ±1 0
=C6 (A.69)
79
B. Equi-transmitting Matrices ofOrder 6
In this appendix we show details of the computations for the case (3,3) when n = 6.The other cases for which r, t ∈ (0,1) are comprehensively covered in section 6 ofPaper II. The matrix C can be written in the form
C =
r t t t t tt r at bt ct dtt at r et f t gtt bt et −r ht ltt ct f t ht −r ltt dt gt lt mt −r
,
where a,b,c,d,e, f ,g,h, l,m ∈ {z ∈ C : |z|= 1}. Making use of the orthogonality ofthe rows of C yields the following system of 15 equations
2r+ t (a+b+ c+d) = 0 (1)2r+ t (a+ e+ f +g) = 0 (2)b+ e+h+ l = 0 (3)c+ f +h+m = 0 (4)d +g+ l +m = 0 (5)
2ar+ t(1+be+ c f +dg
)= 0 (6)
1+ae+ ch+dl = 0 (7)1+a f +bh+dm = 0 (8)1+ag+bl + cm = 0 (9)
1+ab+ f h+gl = 0 (10)1+ac+ eh+gm = 0 (11)1+ad + el + f m = 0 (12)
{−2hr+ t
(1+bc+ e f + lm
)= 0 (13)
−2lr+ t(1+bd + eg+hm
)= 0 (14)
{−2mr+ t
(1+ cd + f g+hl
)= 0 (15).
(B.1)
Equation (3) is satisfied if and only if one of the following three cases hold:
Case 1 : e =−b, l =−hCase 2 : h =−b, l =−eCase 3 : l =−b, h =−e
(1) Case 1 e =−b, l =−h.
These substitutions into (B.1) yield the following system of equations
81
2r+ t (a+b+ c+d) = 0 (1)2r+ t (a−b+ f +g) = 0 (2)c+ f +h+m = 0 (4)d +g−h+m = 0 (5)
2ar+ t(c f +dg
)= 0 (6)
1−ab+ ch−dh = 0 (7)1+a f +bh+dm = 0 (8)1+ag−bh+ cm = 0 (9)
1+ab+ f h−gh = 0 (10)1+ac−bh+gm = 0 (11)1+ad +bh+ f m = 0 (12)
−2hr+ t(1+bc−b f −hm
)= 0 (13)
2hr+ t(1+bd−bg+hm
)= 0 (14)
−2mr+ t(cd + f g
)= 0 (15).
(B.2)
Equation (4) is satisfied in the following three cases
Case 1.1 : f =−c, m =−hCase 1.2 : h =−c, m =− fCase 1.3 : m =−c, h =− f
(i) Case 1.1 e =−b, l =−h; f =−c, m =−h.Substituting the values of f and m into the equations in (B.2) yields thefollowing system of equations
2r+ t (a+b+ c+d) = 0 (1)2r+ t (a−b− c+g) = 0 (2)d +g−h−h = 0 (5)
2ar+ t(−1+dg
)= 0 (6)
1−ab+ ch−dh = 0 (7)1−ac+bh−dh = 0 (8)1+ag−bh− ch = 0 (9)
1+ab− ch−gh = 0 (10)1+ac−bh−gh = 0 (11)1+ad +bh+ ch = 0 (12)
−2hr+ t(
1+2bc+h2)= 0 (13)
2hr+ t(bd−bg
)= 0 (14)
2hr+ t(cd− cg
)= 0 (15).
(B.3)
Equation (5) is satisfied only on the following posibilities
Case 1.1.1 : g =−d, h =−hCase 1.1.2 : h = d, g = dCase 1.1.3 : h = d, g = d
(a) Case 1.1.1 e =−b, l =−h; f =−c, m =−h; g =−d, h =−h.Substituting the value of g into the system (B.3), equations (1), (2) and (6)in (B.3) can be written as follows
82
2r+ t (a+b+ c+d) = 0(1)2r+ t (a−b− c−d) = 0(2)
ar− t = 0(6).(B.4)
Substituting the value of t obtained from equation (6) into equations (1)and (2) and then evaluating the sum yields that 5 + a2 = 0 which is acontradiction.
(b) Case 1.1.2 e =−b, l =−h; f =−c, m =−h; g = d, h = d.Substituting the values of g and h into the system (B.3), equations (6), (9)and (12) in (B.3) yields the following equations
2ar+ t(−1+d
2)= 0(6)
1+ad−bd− cd = 0(9)
1+ad +bd + cd = 0(12).
(B.5)
The sum of equations (9) and (12) simplifies to d =−a. Substituting thisvalue into equation (6) yields 2r+ t (−a+a) = 0 which is a contradiction.
(c) Case 1.1.3 e =−b, l =−h; f =−c, m =−h; g = d, h = d.Substituting the values of g and h into equations (6), (9) and (12) in (B.3)yields the following equations
2ar+ t(−1+d
2)= 0(6)
1+ad−bd− cd = 0(9)
1+ad +bd + cd = 0(12).
(B.6)
The sum of equations (9) and (12) simplifies to d =−a. Substituting thisvalue into equation (6) yields 2r+ t (−a+a) = 0 which is a contradiction.
(ii) Case 1.2 e =−b, l =−h; h =−c, m =− f .Here the values of h and m are substituted into the equations in (B.2) and thefollowing system is obtained.
2r+ t (a+b+ c+d) = 0 (1)2r+ t (a−b+ f +g) = 0 (2)d +g+ c− f = 0 (5)
2ar+ t(c f +dg
)= 0 (6)
−ab+ cd = 0 (7)1+a f −bc−d f = 0 (8)1+ag+bc− c f = 0 (9)
{1+ab− c f + cg = 0 (10)1+ac+bc− f g = 0 (11)
2cr+ t(1+bc−b f − c f
)= 0 (13)
−2cr+ t(1+bd−bg+ c f
)= 0 (14)
2 f r+ t(cd + f g
)= 0 (15).
(B.7)
83
Equation (5) is satisfied only when any of the following cases is satisfied
Case 1.2.1 : d =−c, g = fCase 1.2.2 : f = c, g =−dCase 1.2.3 : g =−c, f = d
(a) Case 1.2.1 e =−b, l =−h; h =−c, m =− f ; d =−c, g = f .Substituting the values of d and g into equation (15) in (B.7) one obtains2 f r + t
(−1+ f 2
)= 0. This simplifies to 2r + t
(f − f
)= 0 which is a
contradiction.
(b) Case 1.2.2 e =−b, l =−h; h =−c, m =− f ; f = c, g =−d.Substituting the values of f and g into the equations in (B.7) yields thefollowing equations
2r+ t (a+b+ c+d) = 0(1)2r+ t (a−b+ c−d) = 0(2)
2ar+ t(c2−1
)= 0(6)
−ab+ cd = 0(7)
1+ac−bc− cd = 0(8)
1−ad +bc− c2 = 0(9)
1+ab− c2− cd = 0(10)
1+ac+bc+ cd = 0(11)2cr+ t (bc−bc) = 0(13)
−2cr+ t(1+bd +bd + c2)= 0(14)
2cr+ t(cd− cd
)= 0(15).
(B.8)
Difference of equation (1) and conjugate of equation (2) yields ℜ(d) =−ℜ(b) so that either d =−b or d =−b.
• Case 1.2.2.1 e =−b, l =−h; h =−c, m =− f ; f = c, g =−d; d =−b.
Substituting this value of d into equation (7) yields c = −a. Substitutingthese values of c and d into equation (1) we obtain 2r+ t (a−a)−0 whichis a contradiction.
• Case 1.2.2.2 e =−b, l =−h; h =−c, m =− f ; f = c, g =−d; d =−b.
Substituting this value of d into equations (8) and (11) and then evaluatingtheir sum simplifies to c = −a. Substituting these values of c and d intoequation (1) we obtain 2r+ t
(b−b
)= 0 which is a contradiction.
(c) Case 1.2.3 e =−b, l =−h; h =−c, m =− f ; g =−c, f = d.Substituting the values of g and f into equation (15) in (B.7) one obtainsthat 2r+ t (c− c) = 0 which is a contradiction.
84
(iii) Case 1.3 e =−b, l =−h; m =−c, h =− f .Substituting the values of n and h into the equations in (B.2) one obtainsthe following system of equations
2r+ t (a+b+ c+d) = 0 (1)2r+ t (a−b+ f +g) = 0 (2)d +g+ f − c = 0 (5)
2ar+ t(c f +dg
)= 0 (6)
1−ab− c f +d f = 0 (7)1+a f −b f − cd = 0 (8)ag+b f = 0 (9)
{1+ac+b f − cg = 0 (11)1+ad−b f − c f = 0 (12)
2 f r+ t(1+bc−b f − c f
)= 0 (13)
−2 f r+ t(1+bd−bg+ c f
)= 0 (14)
2cr+ t(cd + f g
)= 0 (15).
(B.9)
Equation (5) holds only when one of the following three cases is satisfied
Case 1.3.1 : d = c, g =− fCase 1.3.2 : f = c, g =−dCase 1.3.3 : g = c, f =−d
(a) Case 1.3.1 e =−b, l =−h; m =−c, h =− f ; d = c, g =− f .Substituting the values of d and g into the equations (2) and (8) in (B.9)one obtains the following equations
2r+ t (a−b) = 0(2) a f −b f = 0(8) (B.10)
From equation (8) we obtain that b = a. Substituting this into equation (2)we obtain that r = itℑ(a) which is a contradiction.
(b) Case 1.3.2 e =−b, l =−h; m =−c, h =− f ; f = c, g =−d.Substituting the values of f and g into the equations in (B.9) one obtainsthe following equations
{2r+ t (a+b+ c+d) = 0 (1)2r+ t (a−b+ c−d) = 0 (2)
2ar+ t(c2−1
)= 0 (6)
1−ab− c2 + cd = 0 (7)1+ac−bc− cd = 0 (8)−ad +bc = 0 (9)
{1+ac+bc+ cd = 0 (11)1+ad−bc− c2 = 0 (12)
2cr+ t (bc−bc) = 0 (13)−2cr+ t
(1+2bd + c2
)= 0 (14)
2cr+ t(cd− cd
)= 0 (15).
(B.11)
85
Evaluating the difference between equations (6) and (13) yields that eitherb = a or c = c. When b = a, then from equation (9) one obtains thatd = c. Substituting these values of b and d into equation (2) yields that2r+ t (a−a) = 0 which is a contradiction. On the other hand, when c = c,then from equation (15) one obtains that cr = 0 which is a contradiction.
(c) Case 1.3.3 e =−b, m =−h; n =−c, h =− f ; g = c, f =−d.Substituting the values of f and g into equation (15) in (B.9) and simpli-fying, one obtains that 2r+ t
(d−d
)= 0 which is a contradiction.
(2) Case 2 h =−b, l =−e.
Substituting the values of h and m into the equations in (B.1) one obtains thefollowing system of equations.
2r+ t (a+b+ c+d) = 0 (1)2r+ t (a+ e+ f +g) = 0 (2)c+ f −b+m = 0 (4)d +g− e+m = 0 (5)
2ar+ t(1+be+ c f +dg
)= 0 (6)
1+ae−bc−de = 0 (7)a f +dm = 0 (8)1+ag−be+ cm = 0 (9)
{1+ab−b f − eg = 0 (10)1+ac−be+gm = 0 (11)
2br+ t(1+bc+ e f − en
)= 0 (13)
2er+ t(1+bd + eg−bm
)= 0 (14)
−2mr+ t(1+ cd + f g+be
)= 0 (15).
(B.12)
According to Lemma 1 in Paper I equation (4) holds only when one of thefollowing three cases is satisfied.
Case 2.1 : c = b, m =− fCase 2.2 : f = b, m =−cCase 2.3 : m = b, f =−c
(i) Case 2.1 h =−b, l =−e; c = b, m =− f .Substituting the values of c and n into the system of equations (B.12) oneobtains the following equations
2r+ t(a+b+b+d
)= 0 (1)
2r+ t (a+ e+ f +g) = 0 (2)d +g− e− f = 0 (5)
2ar+ t(1+be+b f +dg
)= 0 (6)
ae−de = 0 (7)1+ag−be−b f = 0 (9)
{1+ab−b f − eg = 0 (10)1+ab−be− f g = 0 (11)
2br+ t(1+b2 +2e f
)= 0 (13)
2er+ t(1+bd + eg+b f
)= 0 (14)
2 f r+ t(1+bd + f g+be
)= 0 (15).
(B.13)
86
Equation (5) is satisfied only in any one of the following cases after sub-stitution of the value of d = a obtained from equation (7).
Case 2.1.1 : e = a, g = f , d = aCase 2.1.2 : f = a, g = e, d = aCase 2.1.3 : g =−a, f =−e, d = a
(a) Case 2.1.1 h =−b, m =−e; c = b, n =− f ; e = a, g = f , d = a.Substituting the values of g and d into the equations (B.13) yields thefollowing equations
2r+ t(2a+b+b
)= 0(1)
2r+ t(2a+ f + f
)= 0(2)
2ar+ t(1+ab+b f +a f
)= 0(6)
1+a f −ab−b f = 0(9)
1+ab−b f −a f = 0(10)
2br+ t(1+b2 +2a f
)= 0(13)
2ar+ t (1+ab+a f +b f ) = 0(14)
2 f r+ t(1+2ab+ f 2)= 0(15).
(B.14)
Evaluating the difference of equation (1) and cojugate of equation (2)yields ℜ( f ) = ℜ(b) which implies that f = b, or f = b.
• Case 2.1.1.1 h = −b, l = −e; c = b, m = − f ; e = a, g = f , d = a;f = b.Substituting the value of f into equations (6) and (13) one obtainsthat 2ar+t
(1+2ab+b2
)= 0 and 2br+t
(1+b2 +2ab
)= 0 respec-
tively. From the difference of these two equations we see that b = a.With this value of b equation (1) becomes r+ t (a+ℜ(a)) = 0. Therequirement that r ∈ R implies that a is real as well. Hence we havethat r =−2at. Since the rows of the matrix are normalised we obtanthat t = 1√
5+4a2and r = − 2a√
5+4a2. The appropriate choice of a so
that r > 0 is a =−1. The matrix obtained is the following.
C = 13
2 1 1 1 1 11 2 −1 −1 −1 −11 −1 2 −1 −1 −11 −1 −1 −2 1 11 −1 −1 1 −2 11 −1 −1 1 1 −2
• Case 2.1.1.2 h = −b, l = −e; c = b, m = − f ; e = a, g = f , d = a;f = b.Substituting the value of f into equations (1), (13) and (15) in (B.14)yields
87
2r+ t(2a+b+b
)= 0(1)
2br+ t(1+b2 +2ab
)= 0(13)
2br+ t(
1+2ab+b2)= 0(15).
(B.15)
Equation (1) becomes r+ t (a+ℜ(b)) = 0 implying that b is an arbi-trary parameter. The difference between equation (13) and the conju-gate of equation (15) yields that a = a. Since the rows of the matrixare normal vectors, we obtain the values of r and t as −α√
5+α2and
1√5+α2
respectively, where α = ±1+ℜ(b). To obtain a valid value
of r, a is chosen such that α < 0. The appropriate value of a then isa = −1 since |ℜ(b)| ≤ 1. All the other parameters can be given interms of b and the value of a. The corresponding matrix C is given asfollows
C = 1√5+α2
−α 1 1 1 1 11 −α −1 b b −11 −1 −α −1 b b1 b −1 α −b 11 b b −b α −b1 −1 b 1 −b α
.
(b) Case 2.1.2 h =−b, l =−e; c = b, m =− f ; f = a, g = e, d = a.Substituting the values of f , g and d into the system (B.13) one ob-tains the following system of equations
2r+ t(2a+b+b
)= 0(1)
2r+ t (2a+ e+ e) = 0(2)
2ar+ t(1+be+ab+ae
)= 0(6)
1+ae−be−ab = 0(9)
1+ab−be−ae = 0(11)
2br+ t(1+b2 +2ae
)= 0(13)
2er+ t(1+2ab+ e2)= 0(14)
2ar+ t(1+ab+ae+be
)= 0(15).
(B.16)
The difference of equation (1) and the conjugate of equation (2) yieldsℜ(e) = ℜ(b) which implies that either e = b or e = b.
• Case 2.1.2.1 h = −b, l = −e; c = b, m = − f ; f = a, g = e, d = a;e = b.
Equation (1) yields r+ t (a+ℜ(b)) = 0. From this equation and thenormalisation of the rows of the matrix yields t = 1√
5+(a+ℜ(b))2 and
r = −(a+ℜ(b))√5+(a+ℜ(b))2 . Since S Hermitian, r ∈ R. This implies that a ∈ R
so that a = ±1. The appropriate choice of a for r to be positive isa =−1. The corresponding ET-matrix is
88
C = 1√5+α2
−α 1 1 1 1 11 −α −1 b b −11 −1 −α b −1 b1 b b α −b −b1 b −1 −b α 11 −1 b −b 1 α
.
where α =−1+ℜ(b).
• Case 2.1.2.2 h = −b, l = −e; c = b, m = − f ; f = a, g = e, d = a;e = b.Substituting the value of e into the system (B.16) yields the followingsystem of equations
2r+ t(2a+b+b
)= 0(1)
2ar+ t(
1+b2+2ab
)= 0(6)
1+ab−b2−ab = 0(9)
2br+ t(1+b2 +2ab
)= 0(13)
2ar+ t(
1+2ab+b2)= 0(15).
(B.17)
Equation (1) can be written as r + t (a+ℜ(b)) = 0. Following thesame line of argument as in case 2.1.2.1, the same matrix is obtained.
(c) Case 2.1.3 h =−b, l =−e; c = b, m =− f ; g =−a, f =−e, d = a.
Substituting the values of f and g into equation (2) of the system(B.13) yields r+ t (a−a) = 0 which is a contradiction.
(ii) Case 2.2 h =−b, l =−e; f = b, m =−c.Substituting the values of f and m into equatins in (B.12) one obtains thefollowing system of equations
2r+ t (a+b+ c+d) = 0 (1)2r+ t
(a+ e+b+g
)= 0 (2)
d +g− e− c = 0 (5)
2ar+ t(1+be+bc+dg
)= 0 (6)
1+ae−bc−de = 0 (7)ab− cd = 0 (8)ag−be = 0 (9)
{1+ac−be− cg = 0 (11)2br+ t (1+bc+be+ ce) = 0 (13)
{2er+ t
(1+bd + eg+bc
)= 0 (14)
2cr+ t(1+ cd +bg+be
)= 0 (15).
(B.18)
Equation (5) holds only when any one of the following yields the followingthree cases is satisfied.
Case 2.2.1 : d = c, g = eCase 2.2.2 : e =−c, g =−dCase 2.2.3 : g = c, e = d
89
(a) Case 2.2.1 h =−b, l =−e; f = b, m =−c; d = c, g = e.Substituting the values of d and g into the equations in (B.18) oneobtains the following equations
2r+ t (a+b+ c+ c) = 0(1)
2r+ t(a+ e+b+ e
)= 0(2)
2ar+ t(1+be+bc+ ce
)= 0(6)
1+ae−bc− ce = 0(7)ab−1 = 0(8)
1+ac−be− ce = 0(11)2br+ t (1+bc+be+ ce) = 0(13)
2er+ t(1+2bc+ e2)= 0(14)
2cr+ t(1+ c2 +2be
)= 0(15).
(B.19)
Evaluating the difference between equation (1) and the conjugate ofequation (2) we obtain that ℜ(e) = ℜ(c) so that either e = c or e = c.From equation (7) we see that b = a.
• Case 2.2.1.1 h = −b, l = −e; f = b, m = −c; d = c, g = e; e = c,b = a.Substituting the value of b into equation (1) in (B.19) it can be seenthat r = −t (a+ℜ(c)). a is chosen real since r is real. This yieldsthat r = −α√
5+α2while t = 1√
5+α2, where α =−1+ℜ(c). The RET-
matrix obtained is
C = 1√5+α2
−α 1 1 1 1 11 −α −1 −1 c c1 −1 −α c −1 c1 −1 c α 1 −c1 c −1 1 α −c1 c c −c −c α
.
• Case 2.2.1.2 h = −b, l = −e; f = b, m = −c; d = c, g = e; e = c,b = a.Substituting the values of e and b into the equations in (B.19) yieldsthe following equations
2r+ t (2a+ c+ c) = 0(1)
2ar+ t(1+2ac+ c2)= 0(6)
1+ac−ac− c2 = 0(7)
2ar+ t(1+2ac+ c2)= 0(13)
2cr+ t(1+2ac+ c2)= 0(14).
(B.20)
Difference of equation (6) and the conjugate of equation (13) yieldsc = c, i.e. c = ±1. Substituting this value of c into (1) yields r +t (a+ c) = 0. Since r is real, this implies that ac ∈ R. This in turnmeans that either c = a or c = a. When a = c, then r = −2at anda ∈ R. The correct value of a is a = −1. Therefore it is to to seethat r = 2
3 and t = 13 . When c = a then r = −2tℜ(a). This implies
90
that t = 1√5+4ℜ(a)2 and r = −2ℜ(a)√
5+4ℜ(a)2 . a has to be chosen such that
ℜ(a)< 0 in order to obtain r > 0. With all these conditions, equation(1) in (B.19) is satisfied when a = a meaning that the appropriatechoice of a is a =−1. The corresponding C matrix is the same as theone obtained in case 2.1.1.1.
(b) Case 2.2.2 h =−b, l =−e; f = b, m =−c; e =−c, g =−d.Substituting the values of e and g into the equations (1), (6), (9) and(13) in (B.18) one obtains the following system of equations
2r+ t (a+b+ c+d) = 0(1)
2ar+ t(−bc+bc
)= 0(6)
−ad +bc = 0(9)2br+ t (bc−bc) = 0(13).
(B.21)
The sum of equation (6) and the conjugate of equation (13) simplifiesto b = −a. With this value of b equation (9) yields that d = −c.Substituting thevalues of b and d into equation (1) one obtains that2r+ t (c− c) = 0 which is a contradiction.
(c) Case 2.2.3 h =−b, l =−e; f = b, m =−c; g = c, e = d.Substituting the values of g and e into the equations in (B.18) oneobtains the following equations
2r+ t (a+b+ c+d) = 0(1)
2ar+ t(1+bd +bc+ cd
)= 0(6)
ad−bc = 0(7)
2br+ t(1+bc+bd + cd
)= 0(13)
2dr+ t(1+bd + cd +bc
)= 0(14)
2cr+ t(1+ cd +bc+bd
)= 0(15).
(B.22)
The result obtained in this case is covered in section ??
(iii) Case 2.3 h =−b, l =−e; m = b, f =−c.Substituting then values of n and f into the system of equations (B.12)yields the following equations
2r+ t (a+b+ c+d) = 0 (1)2r+ t (a+ e− c+g) = 0 (2)d +g− e+b = 0 (5)
2ar+ t(be+dg
)= 0 (6)
1+ae−bc−de = 0 (7)−ac+bd = 0 (8)1+ag−be+bc = 0 (9)
{1+ab+bc− eg = 0 (10)1+ac−be+bg = 0 (11)
2br+ t (1+bc− ce−be) = 0 (13)2er+ t
(bd + eg
)= 0 (14)
−2br+ t(1+ cd− cg+be
)= 0 (15).
(B.23)
Equation (5) is satisfied if and only if one of the following three casesholds
91
Case 2.3.1 : d =−b, g = eCase 2.3.2 : e = b, g =−dCase 2.3.3 : g =−b, e = d.
(a) Case 2.3.1 h =−b, l =−e; m = b, f =−c; d =−b, g = e.Substituting the value of d into equation (8) of (B.23) yields that c =−a. Substituting this value of c and the value of d into equation (1),one obtains that 2r+ t (a−a) = 0 which is a contradiction.
(b) Case 2.3.2 h =−b, l =−e; m = b, f =−c; e = b, g =−d.Substituting the values of e and g into the equations in (B.23) oneobtains the following equations
2r+ t (a+b+ c+d) = 0(1)
2r+ t(a+b− c−d
)= 0(2)
2ar+ t(
b2−1
)= 0(6)
1+ab−bc−bd = 0(7)
−ac+bd = 0(8)
1−ad−b2+bc = 0(9)
1+ab+bc+bd = 0(10)
1+ac−b2−bd = 0(11)
2br+ t(bc−bc
)= 0(13)
2br+ t(bd−bd
)= 0(14)
−2br+ t(
1+2cd +b2)= 0(15).
(B.24)
Evaluating the difference of equation (13) and the conjugate of equa-tion (14) yields bc− bc− bd + bd = 0 which when simplified yieldsthat either d =−c or b = b.
• Case 2.3.2.1 h =−b, l =−e; m = b, f =−c; e = b, g =−d; b =±1.
With this value of b, equation (6) yields that a = 0 which is a contra-diction.
• Case 2.3.2.2 h =−b, l =−e; m = b, f =−c; e = b, g =−d; d =−c.
Substituting this value of d into equations (1), (7), (8) and (10) weobtain the following
2r+ t (a+b+ c− c) = 0(1)
1+ab−bc+bc = 0(7)
−ac−bc = 0(8)
1+ab+bc−bc = 0(10).(B.25)
The sum of equations (7) and (10) yields ab+ab = −2. This meansthat ℜ(ab) = −1 which implies that ab = −1 and which in turn im-plies that b =−a. Substituting this value of b into equation (1) yieldsthat 2r+ t (c− c) = 0 which is a contradiction.
92
(c) Case 2.3.3 h =−b, l =−e; m = b, f =−c; g =−b, e = d.Substituting the values of e and g into equation (14) of (B.23) yields2dr+ t
(bd−bd
)= 0. This simplifies to 2r+ t
(b−b
)= 0 which is
a contradiction.
(3) Case 3 l =−b, h =−e.
Substituting the values of l and h into the equations in (B.1) yields the followingsystem of equations
2r+ t (a+b+ c+d) = 0 (1)2r+ t (a+ e+ f +g) = 0 (2)c+ f − e+m = 0 (4)d +g−b+m = 0 (5)
2ar+ t(1+be+ c f +dg
)= 0 (6)
1+ae− ce−bd = 0 (7)1+a f −be+dm = 0 (8)ag+ cm = 0 (9)
{1+ab− e f −bg = 0 (10)1+ad−be+ f m = 0 (12)
2er+ t(1+bc+ e f −bm
)= 0 (13)
2br+ t(1+bd + eg− em
)= 0 (14)
−2mr+ t(1+ cd + f g+be
)= 0 (15).
(B.26)
Equation (4) is satisfied if and only if any one of the following cases holds
Case 3.1 : e = c, m =− fCase 3.2 : f =−c, m = eCase 3.3 : m =−c, f = e.
(i) Case 3.1 l =−b, h =−e; e = c, m =− f .Substituting the values of e and n into the equations in (B.26) one obtainsthe following equations
2r+ t (a+b+ c+d) = 0(1)2r+ t (a+ c+ f +g) = 0(2)
d +g−b− f = 0(5)
2ar+ t(1+bc+ c f +dg
)= 0(6)
ac−bd = 0(7)
1+a f −bc−d f = 0(8)
ag− c f = 0(9)
1+ab− c f −bg = 0(10)
2cr+ t(1+bc+ c f +b f
)= 0(13)
2br+ t(1+bd + cg+ c f
)= 0(14)
2 f r+ t(1+ cd + f g+bc
)= 0(15).
(B.27)
From equation (5) we obtain that
Case 3.1.1 : d = b, g = fCase 3.1.2 : f =−b, g =−dCase 3.1.3 : g = b, f = d
93
(a) Case 3.1.1 l =−b, h =−e; e = c, m =− f ; d = b, g = f .Substituting the values of d and g into the system (B.27) yields thefollowing system of equations
2r+ t(a+b+ c+b
)= 0(1)
2r+ t(a+ c+ f + f
)= 0(2)
2ar+ t(1+bc+ c f +b f
)= 0(6)
ac−1 = 0(7)
1+a f −bc−b f = 0(8)
1+ab− c f −b f = 0(10)
2cr+ t(1+bc+ c f +b f
)= 0(13)
2br+ t(1+b2 +2c f
)= 0(14)
2 f r+ t(1+2bc+ f 2)= 0(15).
(B.28)
From equation (7) we obtain that c= a. Substituting this into equation(8) and into the conjugate of equation (10) and then evaluating thedifference, we obtain that ℜ( f ) = ℜ(b). This implies that eitherf = b or f = b.
• Case 3.1.1.1 l = −b, h = −e; e = c, m = − f ; d = b, g = f ; c = a,f = b.
Substituting these values of c and f into equations (1), (14) and (15)we obtain that
2r+ t(2a+b+b
)= 0(1)
2br+ t(1+b2 +2ab
)= 0(14)
2br+ t(1+2ab+b2)= 0(15).
The difference of equations (14) and (15) yields a = a so that a =±1.Substituting this value of a into equation (1) and simplifying, oneobtains that r+ t (±1+ℜ(b)) = 0. This equation and normalisationof the rows yields that r = −(±1+ℜ(b))√
5+(±1+ℜ(b))2 and t = 1√5+(±1+ℜ(b))2 . To
obtain a positive value of r a = −1 is chosen and the correspondingmatrix C is as follows
C = 1√5+(−1+ℜ(b))2
−α 1 1 1 1 11 −α −1 b −1 b1 −1 −α −1 b b1 b −1 α 1 −b1 −1 b 1 α −b1 b b −b −b α
,
where α =−1+ℜ(b).
• Case 3.1.1.2 l = −b, h = −e; e = c, m = − f ; d = b, g = f ; c = a,f = b.
Substituting these values of c and f into equations (1), (6) and (13)we obtain that
94
2r+ t(2a+b+b
)= 0(1)
2ar+ t(1+2ab+b2)= 0(6)
2ar+ t(1+2ab+b2)= 0(13).
The difference between equations (6) and (13) yields b = b, i.e. b =±1. Substituting this into equation (1) and simplifying, one obtainsthat r+ t (a+b) = 0. This is the same case as in case 3.1.1. above.The appropriate value of b is b =−1. The respective values of r andt are obtained as 2
3 and 13 . The corresponding matrix obtained here is
the same is in case 2.1.1.1.
(b) Case 3.1.2 l =−b, h =−e; e = c, m =− f ; f =−b, g =−d.Substituting the value of f into the equation (13) in (B.27) yieldsthe equation 2cr + t
(bc−bc
)= 0. This leads to the contradiction
2r+ t(b−b
)= 0.
(c) Case 3.1.3 l =−b, h =−e; e = c, m =− f ; g = b, f = d.Substituting the values of g and f into the system of equations (B.27)one obtains the following equations
2r+ t (a+b+ c+d) = 0(1)
2ar+ t(1+bc+ cd +bd
)= 0(6)
ac−bd = 0(7)
2cr+ t (1+bc+ cd +bd) = 0(13)
2br+ t(1+bd +bc+ cd
)= 0(14)
2dr+ t(1+ cd +bd +bc
)= 0(15).
(B.29)
The discussion of this case is covered in section 6 of Paper II.
(ii) Case 3.2 l =−b, h =−e; f =−c, m = e.The substitution of the values of f and m into the equation (B.26)yields the following system of equations
2r+ t (a+b+ c+d) = 0 (1)2r+ t (a+ e− c+g) = 0 (2)d +g−b+ e = 0 (5)
2ar+ t(be+dg
)= 0 (6)
1+ae− ce−bd = 0 (7)1−ac−be+de = 0 (8)ag+ ce = 0 (9)
{1+ab+ ce−bg = 0 (10)1+ad−be− ce = 0 (12)
2er+ t (1+bc− ce−be) = 0 (13)2br+ t
(bd + eg
)= 0 (14)
−2er+ t(1+ cd− cg+be
)= 0 (15).(B.30)
From equation (5) we obtain that
Case 3.2.1 : d = b, g =−eCase 3.2.2 : e = b, g =−dCase 3.2.3 : g = b, e =−d
95
(a) Case 3.2.1 l =−b, h =−e; f =−c, m = e; d = b, g =−e.Substituting the values of d and g into equations (2) and (7) of (B.30)yields the following equations
2r+ t (a− c) = 0(2) ae− ce = 0(7). (B.31)
From equation (7) we see that c = a. Substituting the value of c intoequation (2) and simplifying one obtains that r+ itℑ(a) = 0 which isa contradiction.
(b) Case 3.2.2 l =−b, h =−e; f =−c, m = e; e = b, g =−d.Substitution of the values of e and g into the equations (1), (2), (7)and (10) in (B.30) yields the following equations
2r+ t (a+b+ c+d) = 0(1)
2r+ t(a+b− c−d
)= 0(2)
1+ab−bc−bd = 0(7)
1+ab+bc+bd = 0(10).(B.32)
Evaluating the sum of equations (7) and (10) and then simplifying,one obtains that b=−a. The difference of equation (1) and the conju-gate of equation (2) yields that either d =−c or d =−c. Substitutingthe values of b = −a and d = −c into equation (1) yields that r = 0which is contradiction. On the other hand substituting the values ofb = −a and d = −c into equation (1) shows that 2r + t (c− c) = 0which is a contradiction.
(c) Case 3.2.3 l =−b, h =−e; f =−c, m = e; g = b, e =−d.Substitution of the values of e and g into the equation (14) in (B.30)yields the equation 2br + t
(bd−bd
)= 0 which simplifies to 2r +
t(d−d
)= 0 and as can be seen this is a contradiction.
(iii) Case 3.3 l =−b, h =−e; m =−c, f = e.Substituting the values of n and f into the equations (B.26) one obtainsthe following system of equations
2r+ t (a+b+ c+d) = 0 (1)2r+ t (a+ e+ e+g) = 0 (2)d +g−b− c = 0 (5)
2ar+ t(1+be+ ce+dg
)= 0 (6)
1+ae− ce−bd = 0 (7)1+ae−be− cd = 0 (8)ag−1 = 0 (9)
{1+ad−be− ce = 0 (12)2er+ t
(1+2bc+ e2
)= 0 (13)
{2br+ t
(1+bd + eg+ ce
)= 0 (14)
2cr+ t(1+ cd + eg+be
)= 0 (15).
(B.33)
96
From equation (9) we obtain that g= a. When this value of g is substitutedinto equation (5), we obtain the following cases
Case 3.3.1 : b = a, d = c, g = aCase 3.3.2 : c = a, d = b, g = aCase 3.3.3 : d =−a, c =−b, g = a
(a) Case 3.3.1 l =−b, h =−e; m =−c, f = e; b = a, d = c, g = a.
Substituting the values b , d, and g into equation (1) and (2) oneobtains the following equations
2r+ t (2a+ c+ c) = 0(1) 2r+ t (2a+ e+ e) = 0(2)
The difference of equation (1) and the conjugate of equation (2) yieldsthat ℜ(e) = ℜ(c) which implies that either e = c or e = c. Equation(1) simplifies to r + t (a+ℜ(c)) = 0. This together with the nor-malisation of the rows yields that r = α√
5+α2and t = 1√
5+α2where
α = a+ℜ(c). Since r ∈ R, a must be chosen real as well. Againsince r > 0, the appropriate choice of a is a = −1. When e = c thenthe corresponding matrix C is given by
C = 1√5+(−1+ℜ(c))2
−α 1 1 1 1 11 −α −1 −1 c c1 −1 −α c c −11 −1 c α −c 11 c c −c α −c1 c −1 1 −c α
.
When e = c then the corresponding matrix C is
C = 1√5+(−1+ℜ(c))2
−α 1 1 1 1 11 −α −1 −1 c c1 −1 −α c c −11 −1 c α −c 11 c c −c α −c1 c −1 1 −c α
.
(b) Case 3.3.2 l =−b, h =−e; m =−c, f = e; c = a, d = b, g = a.Substituting the values of c, d and g into the equations (B.33) yieldsthe following equations
97
2r+ t(2a+b+b
)= 0(1)
2r+ t (2a+ e+ e) = 0(2)
2ar+ t(1+be+ae+ab
)= 0(6)
1+ae−be−ab = 0(8)
1+ab−be−ae = 0(12)
2er+ t(1+2ab+ e2)= 0(13)
2br+ t(1+b2 +2ae
)= 0(14)
2ar+ t (1+ab+ae+be) = 0(15).(B.34)
Evaluating the difference of equation (8) and the conjugate of equa-tion (12) one obtains that ℜ(e) = ℜ(b) so that either e = b or e = b.It follows from equation (1) that a is necessarily real. The relationbetween r and t from equation (1) and the normalisation of the rowsyields that r = −(a+ℜ(b))√
5+(a+ℜ(b))2 and t = 1√5+(a+ℜ(b))2 . Since r is posi-
tive, the correct choice of a is a = −1. Thus when e = b the corre-sponding matrix is
C = 1√5+(−1+ℜ(b))2
−α 1 1 1 1 11 −α −1 b −1 b1 −1 −α b b −11 b b α −b −b1 −1 b −b α 11 b −1 −b 1 α
.
When e = b the corresponding matrix is
C = 1√5+(−1+ℜ(b))2
−α 1 1 1 1 11 −α −1 b −1 b1 −1 −α b b −11 b b α −b −b1 −1 b −b α 11 b −1 −b 1 α
.
Substituting value of e into the equations (1), (6), (8), (13) and (15)yields the following equations
2r+ t(2a+b+b
)= 0(1)
2ar+ t(1+b2 +2ab
)= 0(6)
1+ab−b2−ab = 0(8)
2br+ t(
1+2ab+b2)= 0(13)
2ar+ t(1+2ab+b2)= 0(15).
The difference of equation (15) and the conjugate of (6) simplifies tob = ±1. The difference of equations (1) and (13) simplifies to eithera = b =±1 or a =−b. When a = b then equation (1) can written asr±2t = 0 which yields the smc ET-matrix entries. On the other hand,when a =−b, we obtain a RET-matrix.
(c) Case 3.3.3 l =−b, h =−e; m =−c, f = e; d =−a, c =−b, g = a.Substituting the values of c and d into equation (1) of (B.33) yields2r+ t (a−a) = 0 which is a contradiction.
98