Eigenvalues of Two-Parameter Complex Hadamard Matrices of Order Six

Yanzu Huang; Lin Chen

doi:10.2478/qic-2026-0010

1. Introduction

Characterizing complex Hadamard matrices of order n is a fundamental problem in quantum physics, as it is a way to find the complete mutually unbiased bases (MUBs). In general, k MUBs in Hilbert space ℂⁿ are k orthogonal bases, the inner product of any two vectors from different bases has a modulus of $1 / \sqrt{n}$ . When there are n + 1 MUBs in ℂⁿ, then they are referred to as complete MUBs. So far, it has been proven that complete MUBs exist in ℂⁿ when n is a prime power [1]. The complete classification of CHMs gives us a new way to find the maximal MUBs in the unsolved dimensions. In particular, the dimension six is the smallest one. Paper [2] investigated that the average distance between four bases in dimension six, providing a strong evidence against the existence of four MUBs in ℂ⁶. Paper [3] showed that if complete MUBs in dimension six exist, it cannot include more than one product basis. The examination of the number of product vectors in a set of four MUBs in dimension six showed that each of the remaining three MUBs contains at most two product vectors [4]. The existence and limitations of MUB product bases in dimension six were explored in [5–7]. The skills such as Schmidt ranks, Lie algebra, semidefinite programming, entanglement theory and so on are used in [8–14].

The complete classification of 6 × 6 CHMs is also unsolved. Karlsson presented a three-parameter CHMs family in ℂ⁶ [15], and named the family “the H₂-reducible matrices”. As far as we know, most known 6 × 6 CHMs belong to the H₂-reducible matrices such as the Haagerup matrix [16], but the Tao matrix is an exception [17]. The paper [18] presented a four-parameter 6 × 6 CHM family, though the analytic form of these matrices is still unknown. Another idea is analyzing the eigenvalues of the CHM. It was shown that CHMs of order six with dephased form have eigenvalues $\sqrt{6}$ , – $\sqrt{6}$ , and for the remaining four eigenvalues, at most two of them are the same [19]. More facts investigating the classification problem have been presented in [20–25].

CHMs in dimension six are also bipartite unitary matrices from the point of view in quantum information theory. Such matrices can be applied to create quantum entanglement and to implement quantum circuits and cryptography [26]. The so-called controlled-unitary (say controlled-NOT) gate, which is a fundamental bipartite unitary operator, serves as both the primary entanglement resource and a critical component of universal quantum computation [27]. Finding the decompositions by controlled-unitary gates of bipartite unitary operators generated by CHMs will provide a simple physical realization of complex operations. Schmidt rank of bipartite unitary matrices was shown to be closely connected to the existence of this decomposition [28,29].

In this paper, we investigate CHMs with certain type of eigenvalues. In Theorem 1 we prove that CHMs with only two sorts of eigenvalues must be complex equivalent to a family we will refer to as K(d, x) in (8) with two parameters d, x. Lemma 1 gives the solutions of parameters in Theorem 1, and gives the suitable value range of the parameters. The main technical treatment in the proof of Theorem 1 is the construction in (A4), which exhausts the form of CHMs with only two sorts of eigenvalues. Next we analyze the CHMs in dephased form whose eigenvalues are $\sqrt{6}$ , – $\sqrt{6}$ , λ₁, λ₁, λ₂, λ₂. We give a complete classification of these CHMs, containing the Tao matrix in (3) and a family K′ (d, x) in (21) with parameters d, x. That is the result in Theorem 2. The key equation in the proof of Theorem 2 is shown in (A35), which gives the main restriction of this class of CHMs. In Corollary 1 we present the outcome of a special case of Theorem 2, we claim that if λ₂ = –λ₁, then such CHM must be complex equivalent to the Hermitian matrices. In Theorem 3 we show that a family of K(d, x) has Schmidt rank at most three, and this family can be carried out as controlled unitary gates.

The rest of this paper will be presented as follows. In Section 2 we introduce some facts used in this paper, such as the definition of MUBs and CHMs, the eigenvalues of some famous CHMs. In Section 3, we classify all the CHMs with the specified eigenvalue structure. In Section 4, we find the application of some CHMs in implementable unitary gate. We conclude in Section 5.

2. Preliminaries

In this section, we present the definition of CHMs and MUBs, as well as the eigenvalues of some special CHMs.

Definition 1.

Suppose ℬ₁ = {|ϕ_1j〉}_j=1,…,d,…, ℬ_n = {|ϕ_nj〉}_j=1,…,d are n orthonormal basis in ℂ^d, if

| 〈 ϕ_{j l} ∣ ϕ_{k m} 〉 | = \frac{1}{\sqrt{d}},

for any j ≠ k, j, k = 1, …, n; l, m = 1, …, d, then ℬ₁, …, ℬ_n form n MUBs.

Definition 2.

If H = [h_jk] is an n × n complex matrix satisfying

(i). |h_jk| = 1 for any j, k = 1, …, n;
(ii). H · H^† = nI,

then H is an n × n complex Hadamard matrix(CHM).

We also introduce the definition of complex equivalence. We refer to the monomial unitary matrix as a unitary matrix that every row and column have exactly one nonzero element. Then two matrices U and V are complex equivalent when U = PVQ where P, Q are monomial unitary matrices.

Next we list some famous CHMs and their eigenvalues. The Haagerup matrices [16] are

1

H_{6}^{q} = [\begin{matrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & - 1 & i & i & - i & - i \\ 1 & i & - 1 & - i & q & - q \\ 1 & i & - i & - 1 & - q & q \\ 1 & - i & - \frac{1}{q} & \frac{1}{q} & - 1 & i \\ 1 & - i & \frac{1}{q} & - \frac{1}{q} & i & - 1 \end{matrix}],

where |q| = 1. The eigenvalues of

H_{6}^{q}

are

\sqrt{6}

, –

\sqrt{6}

, –1 +

\sqrt{5} i

, –1 +

\sqrt{5} i

, –1 –

\sqrt{5} i

, –1 –

\sqrt{5} i

Second, the Diţă matrix D₀ [30] is equivalent to $H_{6}^{q}$ for q = ±i. D₀ has the same eigenvalues as $H_{6}^{q}$ .

2

D_{0} = [\begin{matrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & - 1 & i & - i & - i & i \\ 1 & i & - 1 & i & - i & - i \\ 1 & - i & i & - 1 & i & - i \\ 1 & - i & - i & i & - 1 & i \\ 1 & i & - i & - i & i & - 1 \end{matrix}] .

Third, the Tao matrix is isolated and belongs to the Buston matrices [31,32].

3

S_{6}^{(0)} = [\begin{matrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & ω & ω & ω^{2} & ω^{2} \\ 1 & ω & 1 & ω^{2} & ω^{2} & ω \\ 1 & ω & ω^{2} & 1 & ω & ω^{2} \\ 1 & ω^{2} & ω^{2} & ω & 1 & ω \\ 1 & ω^{2} & ω & ω^{2} & ω & 1 \end{matrix}],

where

ω = e^{\frac{2 π i}{3}}

. The eigenvalues of

S_{6}^{(0)}

are

\sqrt{6}, - \sqrt{6}, \frac{3 + \sqrt{15} i}{2}, \frac{3 + \sqrt{15} i}{2}, \frac{3 - \sqrt{15} i}{2}, \frac{3 - \sqrt{15} i}{2}

The Hermitian matrices, by exchanging rows and columns, can take the following form [33],

4

H (t) = [\begin{matrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & - 1 & \frac{1}{x} & - y & - \frac{1}{x} & y \\ 1 & x & - 1 & t & - t & - x \\ 1 & - \frac{1}{y} & \frac{1}{t} & - 1 & \frac{1}{y} & - \frac{1}{t} \\ 1 & - x & - \frac{1}{t} & y & 1 & \frac{1}{z} \\ 1 & \frac{1}{y} & - \frac{1}{x} & - t & z & 1 \end{matrix}],

where θ ∈ [–π, – arccos

(\frac{- 1 + \sqrt{3}}{2})

] ∪ [arccos

(\frac{- 1 + \sqrt{3}}{2})

, π], and the parameters x, y, t, z are given by

5

y = e^{i θ}, z = \frac{1 + 2 y - y^{2}}{y (- 1 + 2 y + y^{2})};

6

x = \frac{1 + 2 y + y^{2} - \sqrt{2 (1 + 2 y + 2 y^{3} + y^{4})}}{1 + 2 y - y^{2}};

7

t = \frac{1 + 2 y + y^{2} - \sqrt{2 (1 + 2 y + 2 y^{3} + y^{4})}}{- 1 + 2 y + y^{2}} .

The eigenvalues of the Hermitian matrix H(t) are $\sqrt{6}, \sqrt{6}, \sqrt{6}, - \sqrt{6}, - \sqrt{6}, - \sqrt{6}$ .

3. Result

In this section, we will present our results in this paper. We will discuss CHMs with special type of eigenvalues up to complex equivalence. Though eigenvalues are not invariant with respect to the equivalence relation, we can simplify the result by this relation and at the same time, no results will be omitted. We first consider a family of CHMs with exactly two sorts of eigenvalues. Paper [19] claimed that there is no 6 × 6 CHM with four identical eigenvalues. So the two distinct eigenvalues must be of algebraic multiplicity three. The following theorem shows the complete classification of such CHMs.

Theorem 1.

The order-six CHM with exactly two sorts of eigenvalues is complex equivalent to the following CHM.

8

K (d, x) = [\begin{matrix} d & 1 & 1 & 1 & 1 & 1 \\ 1 & - \bar{d} & x & \bar{y} & - x & - \bar{y} \\ 1 & \bar{x} & - \bar{d} & z & - z & - \bar{x} \\ 1 & y & \bar{z} & - \bar{d} & - y & - \bar{z} \\ 1 & - \bar{x} & - \bar{z} & - \bar{y} & d & - \bar{x y z} \\ 1 & - y & - x & - z & - x y z & d \end{matrix}] .

Here the parameters d, x, y, z are all of modulus one and satisfy

9

- 2 R e [d] + x + y + z + x y z = 0.

If d = e^iθ, then the eigenvalues of K(d, x) are

10

λ_{1} = - \sqrt{6 - \sin^{2} θ} + i \sin θ, λ_{2} = \sqrt{6 - \sin^{2} θ} + i \sin θ .

The proof of Theorem 1 will be present in Appendix A. The restriction in (9) can be solved by the following lemma. Note that if |d| = 1, then –2Re[d] ∈ [–2,2].

Lemma 1.

For a real number r ∈ [–2,2], suppose the complex numbers x, y, z of modulus one satisfy the following equation,

11

r + x + y + z + x y z = 0 .

We denote x = e^iϕ₁, y = e^iϕ₂, z = e^iϕ₃, then x, y, z can be solved as follows.

(i) If r = 0, then {x, y} = {1,−1} or {y, z} = {1, –1} or {x, z} = {1, –1}.
(ii) If ϕ₁ = – ϕ₂, then cos $ϕ_{1} = \frac{- r - 2}{2}$ , ϕ₃ = 0 or cos $ϕ_{1} = \frac{- r + 2}{2}$ , ϕ₃ = ±π.
(iii) If ϕ₁ – ϕ₂ = ±π, i.e. x = –y, then $z + \bar{z} = - r, {x, y} = {\bar{i} \bar{z}, - i \bar{z}}$ .
(iv) For other cases, a, b, c can be solved by
12
$y = \frac{r x^{2} + r^{2} x + r \pm \sqrt{Δ}}{2 (x^{2} - r x - 1)},$

13
$c = - \frac{r + x + y}{1 + x y},$
where
14
$Δ = (r^{2} + 4) x^{4} + 2 r^{3} x^{3} + (r^{4} - 2 r^{2} - 8) x^{2} + 2 r^{3} x + r^{2} + 4,$
and the range of ϕ₁ is
15
$\frac{- r^{3} - 4 \sqrt{2} \sqrt{r^{2} + 2}}{2 (r^{2} + 4)} \leq \cos ϕ_{1} \leq \frac{- r^{3} + 4 \sqrt{2} \sqrt{r^{2} + 2}}{2 (r^{2} + 4)} .$

The proof of Lemma 1 will be presented in Appendix B. Noting that the solutions of (9) can be expressed by parameter ϕ₁ below (11) or x. So the notation K(d, x) means a class of two-parameter CHMs, here d controls the eigenvalues, d and x control the solutions of restriction (9).

Recall that the CHM of order-six with dephased form has eigenvalues $\sqrt{6}, - \sqrt{6}$ [19]. If the remaining four eigenvalues have at most two distinct elements, then [19] claims that the remaining four eigenvalues are λ₃, λ₃, λ₄, λ₄ with λ₃ ≠ λ₄. Next we present another theorem to classify such CHMs.

Theorem 2.

Suppose that an order-six CHM H with dephased form has eigenvalues $\sqrt{6}, - \sqrt{6}, λ_{3}, λ_{3}, λ_{4}, λ_{4}$ . Then H is complex equivalent to the Tao matrix $S_{6}^{(0)}$ in (3) or the following two-parameter H2-reducible CHM

16

K^{'} (d, x) = [\begin{matrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & - 1 & x d & \bar{y} d & - x d & - \bar{y} d \\ 1 & \bar{x} d & - 1 & z d & - z d & - \bar{x} d \\ 1 & y d & \bar{z} d & - 1 & - y d & - \bar{z} d \\ 1 & - \bar{x} d & - \bar{z} d & - \bar{y} d & d^{2} & - \bar{x y z} d \\ 1 & - y d & - x d & - z d & - x y z d & d^{2} \end{matrix}],

where

17

λ_{3} = d λ_{1}, λ_{4} = d λ_{2}

with d, λ₁, λ₂ in (10).

In particular, if d = ±i, then K′(±i, x) is complex equivalent to Haagerup matrices in (1). If d = ±1, then K′(±1, x) in (16) is complex equivalent to Hermitian matrices in (4).

We present the brief version of proof of Theorem 2 below. More details can be found in Appendix C. The proof shows the connection between Theorems 1 and 2. In fact, K(d, x) and K′ (d, x) are equivalent by the equality

18

K^{'} (d, x) = d \cdot p \cdot K (d, x) \cdot p

where

19

P = [\begin{array}{l} \bar{d} & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{array}] .

Proof.

For a given H in dephased form with eigenvalues $\sqrt{6}, - \sqrt{6}, λ_{3}, λ_{3}, λ_{4}, λ_{4}$ , we choose

20

| u_{1} 〉 = \frac{1}{\sqrt{12 + 2 \sqrt{6}}} [\begin{matrix} 1 + \sqrt{6} \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \end{matrix}], | u_{2} 〉 = \frac{1}{\sqrt{12 - 2 \sqrt{6}}} [\begin{matrix} 1 - \sqrt{6} \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \end{matrix}] .

as the eigenvectors of

\sqrt{6}, - \sqrt{6}

, respectively. We have

21

H = H_{2} + λ_{3} \sum_{j = 3}^{4} | u_{j} 〉 〈 u_{j} | + λ_{4} \sum_{j = 5}^{6} | u_{j} 〉 〈 u_{j} | .

Here

22

H_{2} = \sqrt{6} | u_{1} 〉 〈 u_{1} | - \sqrt{6} | u_{2} 〉 〈 u_{2} | = \frac{1}{5} [\begin{matrix} 5 & 5 & 5 & 5 & 5 & 5 \\ 5 & - 1 & - 1 & - 1 & - 1 & - 1 \\ 5 & - 1 & - 1 & - 1 & - 1 & - 1 \\ 5 & - 1 & - 1 & - 1 & - 1 & - 1 \\ 5 & - 1 & - 1 & - 1 & - 1 & - 1 \\ 5 & - 1 & - 1 & - 1 & - 1 & - 1 \end{matrix}] .

The equation (21) can be rewritten as follows.

23

H = \sqrt{6} | u_{1} 〉 〈 u_{1} | - \sqrt{6} | u_{2} 〉 〈 u_{2} | + λ_{3} \sum_{i = 3}^{4} | u_{i} 〉 〈 u_{i} | + λ_{4} \sum_{j = 5}^{6} | u_{j} 〉 〈 u_{j} |

24

= \sqrt{6} | u_{1} 〉 〈 u_{1} | - \sqrt{6} | u_{2} 〉 〈 u_{2} | + λ_{3} (I - | u_{1} 〉 〈 u_{1} | - | u_{2} 〉 〈 u_{2} |) + (λ_{4} - λ_{3}) \sum_{j = 5}^{6} | u_{j} 〉 〈 u_{j} |

25

= \frac{1}{5} [\begin{matrix} 5 & 5 & 5 & 5 & 5 & 5 \\ 5 & - 1 + 4 λ_{3} & - 1 - λ_{3} & - 1 - λ_{3} & - 1 - λ_{3} & - 1 - λ_{3} \\ 5 & - 1 - λ_{3} & - 1 + 4 λ_{3} & - 1 - λ_{3} & - 1 - λ_{3} & - 1 - λ_{3} \\ 5 & - 1 - λ_{3} & - 1 - λ_{3} & - 1 + 4 λ_{3} & - 1 - λ_{3} & - 1 - λ_{3} \\ 5 & - 1 - λ_{3} & - 1 - λ_{3} & - 1 - λ_{3} & - 1 + 4 λ_{3} & - 1 - λ_{3} \\ 5 & - 1 - λ_{3} & - 1 - λ_{3} & - 1 - λ_{3} & - 1 - λ_{3} & - 1 + 4 λ_{3} \end{matrix}] + (λ_{4} - λ_{3}) \sum_{j = 5}^{6} | u_{j} 〉 〈 u_{j} | .

Let the vectors $| u_{j} 〉 = {[g_{1}^{j}, \dots, g_{6}^{j}]}^{T}, j = 5, 6$ , and we can deduce the expressions of all elements in H. The details will be present in Appendix C.

By discussing three distinct cases, we obtain the relations of λ₃, λ₄. Further construction shows that for given CHM in dephased form with eigenvalues $\sqrt{6}, - \sqrt{6}, λ_{3}, λ_{3}, λ_{4}, λ_{4}$ , we will deduce the Tao matrix in (3) or a CHM with two sorts of eigenvalues, that is K(d, x). By Theorem 1, we have finished the proof of Theorem 2. The details will be present in Appendix C.

Corollary 1.

Suppose H is an order-six CHM in dephased form with the first two eigenvalues $\sqrt{6}$ and – $\sqrt{6}$ .

(i) If H has the remaining four eigenvalues λ, λ, – λ, – λ, then H is the Hermitian matrix in (4).
(ii) If H has the remaining four eigenvalues of at most two distinct elements, then H has been completely studied in [19] and Theorem 2.

Proof.

(i) By using Theorem 1, we substitute λ₃ = – λ₄ into (10). We know that d = ±1, and thus ${λ_{3}, λ_{4}} = {\sqrt{6}, - \sqrt{6}}$ . Hence H is the Hermitian matrix in (4). (ii) If the four remaining eigenvalues satisfy λ₃ = λ₄ = λ₅, then by the result of Theorem 5 in paper [19] we deduce the contradiction. If λ₃ = λ₄ and λ₅ = λ₆, then Theorem 2 has done the complete classification of this case.

4. Application

In this section, we introduce an application of our results. For a bipartite Hilbert space ℋ_A ⊗ ℋ_B, a controlled unitary gate up to local unitary equivalence takes the form

26

U = \sum_{i = 1}^{d_{A}} | i 〉 {〈 i |}_{A} \otimes U_{B}^{(i)} .

where {|i〉_A} is an orthonormal basis for subsystem A and

U_{B}^{(i)}

are unitary operators acting on subsystem B. d_A is the dimension of subsystem A.

It is known that every bipartite unitary matrix can be regarded a non-local operation from the viewpoint of quantum information. For example, the so-called controlled-NOT and other types of controlled unitary gates play a key role in quantum computing and so on, as every bipartite unitary gate can be decomposed into the product of some controlled unitary gates [34].

Further, it has been proven that a bipartite unitary gate of Schmidt rank at most three is actually a controlled unitary gate [28], which is more easily implementable in experiments. Here, for a bipartite unitary matrix U ∈ ℋ_A ⊗ ℋ_B, the Schmidt rank is the smallest integer r such that

27

U = \sum_{i = 1}^{r} λ_{i} M_{i}^{(A)} \otimes N_{i}^{(B)}

with linearly independent sets {M_i} and {N_i} and all λ_i > 0.

In the following, we show that a family of CHMs K(d, x) are complex equivalent to some CHMs of Schmidt rank three. In particular, all order-six Hermitian CHMs are included in this family. Hence, such CHMs can be actually carried out as a controlled unitary gate, and they can be applied in experiments more easily. Our result help realize more non-local operations in quantum system with less coherence time and lower error rates.

Theorem 3.

For the CHM K(d, x) in (8), if x = d or $x = \bar{d}$ and ${y, z} = {i \bar{x}, - i \bar{x}}$ , then K(d, x) is complex equivalent to a CHM whose Schmidt rank is at most three.

Proof.

We consider the monomial unitary matrix

28

Q = [\begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & - \bar{y} \\ 0 & 0 & 0 & 0 & - z & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \end{matrix}],

then we have

29

M = Q^{†} \cdot K (d, x) \cdot Q = [\begin{matrix} d & 1 & 1 & 1 & - z & - \bar{y} \\ 1 & d & - \bar{x y z} & - \bar{y} & 1 & \bar{x y} \\ 1 & - x y z & d & - z & x z & 1 \\ 1 & - y & - \bar{z} & - \bar{d} & - 1 & - 1 \\ - \bar{z} & 1 & \bar{x z} & - 1 & - \bar{d} & \bar{x y z} \\ - y & x y & 1 & - 1 & x y z & - \bar{d} \end{matrix}] .

We denote $M = [\begin{array}{l} A & B \\ C & D \end{array}]$ and A, B, C, D are four 3 × 3 submatrices of M. Then we have

30

A + D = (d - \bar{d}) I_{3},

and

31

B + C = [\begin{matrix} 2 & - y - z & - \bar{y} - \bar{z} \\ - \bar{y} - \bar{z} & 2 & \bar{x} (\bar{y} + \bar{z}) \\ - y - z & x (y + z) & 2 \end{matrix}] .

If y = –z, then B + C = 2I₃, and A + D, B + C are linear dependent. It means that the CHM M has Schmidt rank at most three. Using the Lemma 1 we know that if y = –z, then x = d or $x = \bar{d}$ and ${y, z} = {i \bar{x}, - i \bar{x}}$ . Hence we have finished the proof.

For example, if d = 1, then x = 1 and K(d, x) is indeed Hermitian. We shall take y = i, z = –i, then M in (29) turns to be

32

[\begin{matrix} 1 & 1 & 1 & 1 & i & i \\ 1 & 1 & - 1 & i & 1 & - i \\ 1 & - 1 & 1 & i & - i & 1 \\ 1 & - i & - i & - 1 & - 1 & - 1 \\ - i & 1 & i & - 1 & - 1 & 1 \\ - i & i & 1 & - 1 & 1 & - 1 \end{matrix}]

and we can rewrite M as follows:

33

M = [\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}] \otimes [\begin{matrix} 1 & 1 & 1 \\ 1 & 1 & - 1 \\ 1 & - 1 & 1 \end{matrix}] + [\begin{array}{l} 0 & 1 \\ 0 & 0 \end{array}] \otimes [\begin{matrix} 1 & i & i \\ i & 1 & - i \\ i & - i & 1 \end{matrix}] + [\begin{array}{l} 0 & 0 \\ 1 & 0 \end{array}] \otimes [\begin{matrix} 1 & - i & - i \\ - i & 1 & i \\ - i & i & 1 \end{matrix}]

We take

34

Q_{1} = I_{2} \otimes \frac{1}{\sqrt{6}} [\begin{matrix} \sqrt{2} & \sqrt{3} & 1 \\ - \sqrt{2} & \sqrt{3} & - 1 \\ - \sqrt{2} & 0 & 2 \end{matrix}] .

Then

35

Q_{1}^{+} M Q_{1} = [\begin{matrix} - 1 & 1 - 2 i \\ 1 + 2 i & 1 \end{matrix}] \otimes | 0 〉 〈 0 | + [\begin{matrix} 2 & 1 + i \\ 1 - i & - 2 \end{matrix}] \otimes (| 1 〉 〈 1 | + | 2 〉 〈 2 |) .

Where |0〉 = [1,0,0]^T, |1〉 = [0,1,0]^T and |2〉 = [0,0,1]^T. So we know that up to local unitary equivalence, M is a controlled unitary gate. Thus it can be applied in experiments more easily.

5. Conclusion

In this paper, we have presented the complete classification of CHMs with two sorts of eigenvalues λ₁, λ₂, these CHMs are included in a two-parameter family K(d, x) in (8). Moreover, we have classified all CHMs with dephased form whose eigenvalues are exactly $\sqrt{6}, - \sqrt{6}, λ_{1}, λ_{1}, λ_{2}, λ_{2}$ , this class consists of K′(d, x) in (21) and Tao matrix in (3), here K′(d, x) is complex equivalent to K(d, x) for the same d, x. One interesting thing is that some known CHMs such as Diţă matrix, Haagerup matrices and Hermitian matrices, are included in the class K′(d, x) and all CHMs in this class are H₂-reducible. We have shown that the Schmidt rank of some CHMs including all Hermitian CHMs in K(d, x) is at most three up to complex equivalence. Our future research will focus on the Schmidt rank of every CHM in K(d, x) and more CHMs with three or more sorts of eigenvalues to find more connections between the classification of CHMs and their eigenvalues.

Notes

[1] Contributed by Author Contributions

Authors equally contribute to the paper. All authors have read and agreed to the published version of the manuscript.

[2] Conflicts of interest Conflicts of Interest

No conflict.

[3] Data Availability Statement

No data.

Appendices

Appendix A. The proof of Theorem 1

Proof.

We consider the spectral decomposition of a CHM H whose eigenvalues are λ₁, λ₁, λ₁, λ₂, λ₂, λ₂, that is, H = λ₁H₁ + λ₂H₂, with projection matrices H₁, H₂. It implies that $H_{j} \cdot H_{k}^{†} = 0$ for j ≠ k and $H_{j} \cdot H_{j}^{†} = H_{j}$ and $H_{j}^{†} = H_{j}$ for j = 1,2. We have $H_{1} = \sum_{j = 1}^{3} | u_{j} 〉〈 u_{j} |$ and $H_{2} = \sum_{j = 4}^{6} | u_{j} 〉〈 u_{j} |$ , such that H₁ + H₂ = I₆. If we consider the matrix $M = \frac{\sqrt{6}}{λ_{1}} H$ , then $M = \sqrt{6} H_{1} + λ H_{2}$ with $λ = \frac{\sqrt{6} λ_{2}}{λ_{1}}$ . So we obtain that $M = \sqrt{6} I_{6} + (λ - \sqrt{6}) H_{2}$ . The point of our proof is to design the projection matrix H₂ such that M is indeed a CHM for a given λ. Since all the entries of M have modulus one, we consider the diagonal elements and non-diagonal elements of M respectively. If we denote M = [h_jk] and H₂ = [g_jk], then the diagonal elements are $h_{j j} = \sqrt{6} + (λ - \sqrt{6}) g_{j j}$ and the non-diagonal elements can be expressed by $h_{j k} = (λ - \sqrt{6}) g_{j k}$ for j ≠ k. So we have

A1

| \sqrt{6} + (λ - \sqrt{6}) g_{j j} | = 1, | (λ - \sqrt{6}) g_{j k} | = 1, j, k \in {1, \dots, 6}; j \neq k .

Recall that $H_{2} = \sum_{j = 4}^{6} | u_{j} 〉〈 u_{j} |$ , we see that g_jj is real for all j ∈ {1, …, 6}. The first equation shows that

A2

[\sqrt{6} + (λ - \sqrt{6}) g_{j j}] \cdot [\sqrt{6} + (\bar{λ} - \sqrt{6}) g_{j j}] = 1,

which means

(12 - \sqrt{6} (λ + \bar{λ})) g_{j j}^{2} - (12 - \sqrt{6} (λ + \bar{λ})) g_{j j} + 5 = 0

. If we denote

m = \frac{Re [λ]}{\sqrt{6}}

, then the solution will be expressed by

g_{j j} = \frac{1}{2} \pm \sqrt{\frac{3 m + 2}{12 (m - 1)}}

. The second equation in (A1) shows that

| g_{j k} | = \frac{1}{| λ - \sqrt{6} |} = \frac{1}{\sqrt{12 - 12 m}}

. The range of m is in [–1, 1], and the above g_jj and |g_jk| are both real. So

\frac{3 m + 2}{12 (m - 1)} \geq 0

and 12 – 12m < 0, hence the range of m should be restricted in

[- 1, - \frac{2}{3}]

We then rewrite the projection matrix H₂ by $H_{2} = \frac{1}{2} I_{6} + k M (a)$ , and once we take $k = \frac{1}{\sqrt{12 - 12 m}}$ , then the non-diagonal elements of M(a) have modulus one. The diagonal elements have modulus $\sqrt{- 3 m - 2}$ . We denote $a = \sqrt{- 3 m - 2}$ , then $k = \frac{1}{\sqrt{20 + 4 a^{2}}}$ and we have already constructed the following form of Hermitian matrices M(a), here a is real, 0 ≤ a ≤ 1. For M(a) = [f_jk]_6×6, the entries of M(a) satisfies the following conditions: For j ≠ k, we have |f_jk| = 1, $f_{j k} = \bar{f_{k j}}$ , and f_jj = ±a. Since the trace of M is $3 (\sqrt{6} + λ)$ and the trace of M can also be $6 \sqrt{6} + 6 \times \frac{1}{2} (λ - \sqrt{6}) + \sum_{j = 1}^{6} g_{j j} = 3 (\sqrt{6} + λ) + \sum_{j = 1}^{6} g_{j j}$ , we must have $\sum_{j = 1}^{6} g_{j j} = 0$ so $\sum_{j = 1}^{6} f_{j j} = 0$ . We shall assume f₁₁ = f₅₅ = f₆₆ = a and f₂₂ = f₃₃ = f₄₄ = –a, then M(a) takes the following form.

A3

[\begin{matrix} a & 1 & 1 & 1 & 1 & 1 \\ 1 & - a \\ 1 & - a \\ 1 & - a \\ 1 & a \\ 1 & a \end{matrix}]

Now we recall that M is a CHM, then M · M^† = 6I₆, and

A4

M = \sqrt{6} I_{6} + (λ - \sqrt{6}) (\frac{1}{2} I_{6} + \frac{1}{\sqrt{20 + 4 a^{2}}} M (a)),

it means that M(a) · M(a)^† = M(a)² = (5 + a²)I₆.

We analyze the orthogonality of row 1 and rows 2,3,4 of M(a) in (A3), and we assume the entries of M(a) to be

A5

M (a) = [\begin{matrix} a & 1 & 1 & 1 & 1 & 1 \\ 1 & - a & h_{23} & h_{24} & h_{25} & h_{26} \\ 1 & \bar{h_{23}} & - a & h_{34} & h_{35} & h_{36} \\ 1 & \bar{h_{24}} & \bar{h_{34}} & - a & h_{45} & h_{46} \\ 1 & \bar{h_{25}} & \bar{h_{35}} & \bar{h_{45}} & a & h_{56} \\ 1 & \bar{h_{26}} & \bar{h_{36}} & \bar{h_{46}} & \bar{h_{56}} & a \end{matrix}]

The orthoganality of rows 1,2 shows that h₂₃ + h₂₄ + h₂₅ + h₂₆ = 0. By permuting rows and columns, we divide into two cases namely Case A, B, that is h₂₃ = –h₂₄ or h₂₃ = –h₂₅.

Case A.

h₂₃ = –h₂₄, we denote x₁ = h₂₃, x₂ = h₂₅, then M(a) become

A6

[\begin{matrix} a & 1 & 1 & 1 & 1 & 1 \\ 1 & - a & x_{1} & - x_{1} & x_{2} & - x_{2} \\ 1 & \bar{x_{1}} & - a & h_{34} & h_{35} & h_{36} \\ 1 & - \bar{x_{1}} & \bar{h_{34}} & - a & h_{45} & h_{46} \\ 1 & \bar{x_{2}} & - x_{1} & \bar{h_{45}} & a & h_{56} \\ 1 & - \bar{x_{2}} & - \bar{h_{34}} & \bar{h_{46}} & \bar{h_{56}} & a \end{matrix}]

Considering the orthogonality of rows 1,3, we have $\bar{x_{1}} + h_{34} + h_{35} + h_{36} = 0$ , then we discuss the following two cases by permuting columns 5,6, namely Case A.1 and A.2, that is $h_{34} = - \bar{x_{1}}$ or $h_{35} = - \bar{x_{1}}$ .

Case A.1.

$h_{34} = - \bar{x_{1}}$ , then h₃₆ = –h₃₅ and the orthogonality of rows 3,4 shows that $a (x_{1} + \bar{x_{1}}) + h_{35} (\bar{h_{45}} - \bar{h_{46}}) = 0$ . The orthogonality of rows 2,4 shows that $a (x_{1} + \bar{x_{1}}) + x_{2} (\bar{h_{45}} - \bar{h_{46}}) = 0$ . So h₃₅ = x₂ or h₄₅ = h₄₆. If h₃₅ = x₂, then the orthogonality of rows 2,3 shows $3 - a (x_{1} + \bar{x_{1}}) + x_{1}^{2} = 0$ , the equation has no solution, which is a contradiction. If h₄₅ = h₄₆, then a = 0 or x₁ = ±i. Now if a = 0, then the inner product of rows 2,3 is $1 + x_{1}^{2} + 2 x_{2} \bar{h_{35}} = 0$ . Then $x_{1}^{2} = 1$ and $x_{2} \bar{h_{35}} = - 1$ , then x₁ = ±1, h₃₅ = –x₂, M(a) becomes

A7

[\begin{matrix} 0 & 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & x_{1} & - x_{1} & x_{2} & - x_{2} \\ 1 & x_{1} & 0 & - x_{1} & - x_{2} & x_{2} \\ 1 & - x_{1} & - x_{1} & 0 & x_{1} & x_{1} \\ 1 & \bar{x_{2}} & - \bar{x_{2}} & x_{1} & 0 & h_{56} \\ 1 & - \bar{x_{2}} & \bar{x_{2}} & x_{1} & \bar{h_{56}} & 0 \end{matrix}]

The orthogonality of rows 4,5 shows h₅₆ = –x₁, then for any x₂, M(a) indeed satisfies that M(a)² = 5I₆. For x₁ = 1 weobtain M(0)₁, for x₁ = –1 we obtain M(0)₂. The expression of M in (A4) deduced by M(0)₁ or M(0)₂ will be shown in the latter case.

If x₁ = ±i, then the inner product of rows 2,3 is $- 2 a x_{1} + 2 x_{2} \bar{h_{35}} = 0$ , so a = 1. We then find that M(a) is indeed a self-adjoint CHM, which means M(a) is complex equivalent to Hermitian matrices in (4). Moreover, m = –1 and $λ = - \sqrt{6}$ . It is known that CHM with eigenvalues $\sqrt{6}, \sqrt{6}, \sqrt{6}, - \sqrt{6}, - \sqrt{6}, - \sqrt{6}$ is complex equivalent to Hermitian matrices in (4).

Case A.2.

$h_{35} = - \bar{x_{1}}$ , then h₃₆ = –h₃₄, and the row 5 turns to [1, $\bar{x_{2}}$ , –x₁, $\bar{h_{45}}$ , a, h₅₆], the orthogonality of rows 2,5 shows that $- x_{1} h_{45} - x_{2} \bar{h_{56}} = 0$ . And the orthogonality of rows 3,5 and shows that $1 + \bar{x_{1}} x_{2} + h_{34} (h_{45} - \bar{h_{56}}) = 0$ , then $(1 + \bar{x_{1}} x_{2}) (1 - h_{34} \bar{h_{56}}) = 0$ , then h₃₄ = h₅₆ or x₂ = –x₁. If h₃₄ = h₅₆, then the orthogonality of rows 1,6 shows $1 - \bar{x_{2}} + \bar{h_{46}} + a = 0$ , since we let 0 ≤ a ≤ 1, then $h_{46} = - \bar{x_{2}}$ . The orthogonality of rows 2,6 and rows 4,5 show that $1 + x_{1} \bar{x_{2}} + h_{34} (x_{2} - x_{1}) = 0$ and $1 - x_{1} \bar{x_{2}} - h_{34} (x_{1} + x_{2}) = 0$ . Then $h_{34} = \bar{x_{1}}$ , then h₄₅ = –x₂. The inner product of rows 1,5 is $2 a + \bar{x_{1}} - x_{1} = 0$ , which shows x₁ is real and a = 0. The orthogonality of rows 1,4 shows that h₄₆ = x₂, so M(a) takes the following form.

A8

[\begin{matrix} 0 & 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & x_{1} & - x_{1} & x_{2} & - x_{2} \\ 1 & x_{1} & 0 & x_{1} & - x_{1} & - x_{1} \\ 1 & - x_{1} & x_{1} & 0 & - x_{2} & x_{2} \\ 1 & \bar{x_{2}} & - x_{1} & - \bar{x_{2}} & 0 & x_{1} \\ 1 & - \bar{x_{2}} & - x_{1} & \bar{x_{2}} & x_{1} & 0 \end{matrix}]

We denote this matrix by M(0)₃. The expression of M in (A4) deduced by M(0)₃ will be shown in the latter case.

If x₂ = –x₁, then rows 2 and 4 are respectively [1, –a, x₁, –x₁, –x₁, x₁] and [1, $- \bar{x_{1}}$ , x₁, –a, –x₁, $\bar{x_{1}}$ ], the orthogonality of rows 2,4 shows that $3 + a (x_{1} + \bar{x_{1}}) + x_{1}^{2} = 0$ , which has no solution.

Case B.

h₂₃ = –h₂₅, then by permuting rows and columns, the first four rows of M(a) must be

A9

[\begin{matrix} a & 1 & 1 & 1 & 1 & 1 \\ 1 & - a & x_{1} & x_{2} & - x_{1} & - x_{2} \\ 1 & \bar{x_{1}} & - a & x_{3} & h_{35} & h_{36} \\ 1 & \bar{x_{2}} & \bar{x_{3}} & - a & h_{45} & h_{46} \end{matrix}]

Here x₁ ≠ –x₂, $x_{3} \neq - \bar{x_{1}}$ , x₃ ≠ –x₂, otherwise Case B will degenerate into Case A. By permuting rows and columns and redefining x₁, x₂, x₃, there are only two distinct cases to discuss, that is $h_{35} = - \bar{x_{1}}, h_{36} = - x_{3}, h_{45} = - \bar{x_{2}}, h_{46} = - \bar{x_{3}} or h_{35} = - x_{3}, h_{36} = - \bar{x_{1}}, h_{45} = - \bar{x_{2}}, h_{46} = - \bar{x_{3}}$ . We denote the two cases by Case B.1 and Case B.2.

Case B.1.

Now M(a) is

A10

[\begin{matrix} a & 1 & 1 & 1 & 1 & 1 \\ 1 & - a & x_{1} & x_{2} & - x_{1} & - x_{2} \\ 1 & \bar{x_{1}} & - a & x_{3} & - \bar{x_{1}} & - x_{3} \\ 1 & \bar{x_{2}} & \bar{x_{3}} & - a & - \bar{x_{2}} & - \bar{x_{3}} \\ 1 & - \bar{x_{1}} & - x_{1} & - x_{2} & a & h_{56} \\ 1 & - \bar{x_{2}} & - \bar{x_{3}} & - x_{3} & \bar{h_{56}} & a \end{matrix}]

The orthogonality of rows 3,5 shows that h₅₆ = –x₂. The orthogonality of rows 1,5 and 1,6 show that x₂ is real and either x₁ = x₃ or $x_{1} = \bar{x_{3}}$ . The orthogonality of rows 4,5 implies that $x_{1} = \bar{x_{3}}$ . By solving the equation M(a)² = (5 + a²)I₆, we obtain the only restriction, that is

A11

- 2 a x_{1} + 2 x_{1} x_{2} + x_{1}^{2} + 1 = 0.

We substitute x = x₁, $y = x_{3} = \bar{x}$ , z = x₂, the above equation turns to – 2a + 2z + x + y = 0, i.e. – 2a + x + y + z + xyz = 0. This case corresponds to the case ϕ₁ = –ϕ₂ in Lemma 1. The solutions and the expression of M in (A4) will be shown in the latter case.

Case B.2.

Now M(a) is

A12

[\begin{matrix} a & 1 & 1 & 1 & 1 & 1 \\ 1 & - a & x_{1} & x_{2} & - x_{1} & - x_{2} \\ 1 & \bar{x_{1}} & - a & x_{3} & - x_{3} & - \bar{x_{1}} \\ 1 & \bar{x_{2}} & \bar{x_{3}} & - a & - \bar{x_{2}} & - \bar{x_{3}} \\ 1 & - \bar{x_{1}} & - \bar{x_{3}} & - x_{2} & a & h_{56} \\ 1 & - \bar{x_{2}} & - x_{1} & - x_{3} & \bar{h_{56}} & a \end{matrix}]

The orthogonality of rows 3,5 shows that $h_{56} = - \bar{x_{1}} x_{2} \bar{x_{3}}$ . Computing M(a)², we denote M(a)² = [f_jk]_6×6, then f₁₁ = f₂₂ = ⋯ = f₆₆ = 5 + a², and $| f_{15} | = | 2 a - x_{1} - x_{3} - \bar{x_{2}} - x_{1} \bar{x_{2}} x_{3} |$ . To satisfy M(a)² = (5 + a²)I₆, we must have

A13

x_{1} + \bar{x_{2}} + x_{3} + x_{1} \bar{x_{2}} x_{3} = 2 a .

The equation above is the only restriction such that M(a)² = (5 + a²)I₆ in Case B.2. To solve this equation, we substitute x = x₁, $y = \bar{x_{2}}$ , z = x₃, then equation A13 becomes

A14

x + y + z + x y z = 2 a .

By Lemma 1 we can solve the above equation. In fact, this equation takes the same form as equation (9).

In fact, If we take a = 0, the equation (A13) has solutions {x₁, x₂} = {1, –1} or {x₁, x₃} = {1, –1} or {x₂, x₃} = {1, –1}. If we take {x₁, x₃} = {1, –1}, then the matrix in (A12) gives M(0)₁ and M(0)₂ in Case A.1 and M(0)₃ in Case A.2 up to complex equivalence. Moreover, if we take $y = \bar{x}$ , then by the result in Lemma 1 (ii) we can solve the case B.1.

In conclusion, M(a) must be complex equivalent to

A15

[\begin{matrix} a & 1 & 1 & 1 & 1 & 1 \\ 1 & - a & x & \bar{y} & - x & - \bar{y} \\ 1 & \bar{x} & - a & z & - z & - \bar{x} \\ 1 & y & \bar{z} & - a & - y & - \bar{z} \\ 1 & - \bar{x} & - \bar{z} & - \bar{y} & a & - \bar{x y z} \\ 1 & - y & - x & - z & - x y z & a \end{matrix}]

Then

A16

M = \frac{λ - \sqrt{6}}{\sqrt{20 + 4 a^{2}}} [\begin{matrix} d & 1 & 1 & 1 & 1 & 1 \\ 1 & - \bar{d} & x & \bar{y} & - x & - \bar{y} \\ 1 & \bar{x} & - \bar{d} & z & - z & - \bar{x} \\ 1 & y & \bar{z} & - \bar{d} & - y & - \bar{z} \\ 1 & - \bar{x} & - \bar{z} & - \bar{y} & d & - \bar{x y z} \\ 1 & - y & - x & - z & - x y z & d \end{matrix}]

Here $d = \frac{\sqrt{20 + 4 a^{2}}}{λ - \sqrt{6}} \cdot [\sqrt{6} + (λ - \sqrt{6}) (\frac{1}{2} + \frac{a}{\sqrt{4 a^{2} + 20}})]$ . We know that $Re [λ] = \sqrt{6} m = - \frac{\sqrt{6}}{3} (2 + a^{2})$ , and we may assume $Im [λ] = \sqrt{\frac{10 - 8 a^{2} - 2 a^{4}}{3}}$ , then $d = 2 \sqrt{5 + a^{2}} (a - \sqrt{1 - a^{2}} i)$ . The eigenvalues of M are $\sqrt{6}, \sqrt{6}, \sqrt{6}, λ, λ, λ$ , λ, λ, λ. For the simplicity of the form, we denote

A17

K (d, x) = [\begin{matrix} d & 1 & 1 & 1 & 1 & 1 \\ 1 & - \bar{d} & x & \bar{y} & - x & - \bar{y} \\ 1 & \bar{x} & - \bar{d} & z & - z & - \bar{x} \\ 1 & y & \bar{z} & - \bar{d} & - y & - \bar{z} \\ 1 & - \bar{x} & - \bar{z} & - \bar{y} & d & - \bar{x y z} \\ 1 & - y & - x & - z & - x y z & d \end{matrix}]

In conclusion, CHMs with two sorts of eigenvalues must be complex equivalent to two-parameter matrices K(d, x). The meaning of parameter d, x will be explained in detail in Lemma 1. The trace of K(d, x) is 6Im[d], we denote the two sorts of eigenvalues by

A18

λ_{1} = \frac{\sqrt{20 + 4 a^{2}}}{λ - \sqrt{6}} \cdot \sqrt{6}, λ_{2} = \frac{\sqrt{20 + 4 a^{2}}}{λ - \sqrt{6}} \cdot λ,

and rewrite d = e^iθ, then by computation we know that

A19

λ_{1} = \sqrt{6 - \sin^{2} θ} + i \sin θ, λ_{2} = - \sqrt{6 - \sin^{2} θ} + i \sin θ .

Hence we have finished the proof of Theorem 1.

Appendix B. The Proof of Lemma 1

Proof.

If r = 0 then we have x = –y or x = –z or y = –z. If x = –y, then equation (11) turns to z –a y²z = 0. Hence y = ±1, and {x, y} = {1, –1}. For the other two cases, we can deduce {x, z} = {1, –1} or {y, z} = {1, –1} similarly.

If r ≠ 0, then r ∈ [–2, 0) ∪ (0, 2]. Now if 1 + xy = 0, then $y = - \bar{x}$ , then (11) turns to $r + x - \bar{x} = 0$ , which means x ∈ ℝ and r = 0, hence the contradiction. So we obtain that $z = - \frac{r + x + y}{1 + x y}$ .

To satisfy |z| = 1, we have

A20

(1 + \bar{x} \bar{y}) (1 + x y) = (r + x + y) (r + \bar{x} + \bar{y}) .

Then

A21

(x - \bar{x} - r) y^{2} - r (x + \bar{x} + r) y + (\bar{x} - x - r) = 0.

We obtain that

A22

y = \frac{r x^{2} + r^{2} x + r \pm \sqrt{Δ}}{2 (x^{2} - r x - 1)} .

where

A23

Δ = (r^{2} + 4) x^{4} + 2 r^{3} x^{3} + (r^{4} - 2 r^{2} - 8) x^{2} + 2 r^{3} x + r^{2} + 4.

here the square root denotes the principal value of the complex power function

z \to z^{\frac{1}{2}}

. Next we discuss the condition that keeps |y| = 1.

We assume x = e^iϕ1, y = e^iϕ2, z = e^iϕ3, then

A24

{\begin{array}{l} \cos ϕ_{1} + \cos ϕ_{2} + \cos ϕ_{3} + \cos (ϕ_{1} + ϕ_{2} + ϕ_{3}) = - r; \\ \sin ϕ_{1} + \sin ϕ_{2} + \sin ϕ_{3} + \sin (ϕ_{1} + ϕ_{2} + ϕ_{3}) = 0. \end{array}

If ϕ₁ = –ϕ₂, then (A24) becomes

A25

{\begin{array}{l} 2 \cos ϕ_{1} + 2 \cos ϕ_{3} = - r; \\ 2 \sin ϕ_{3} = 0. \end{array}

Once λ₁, λ₂ are given, then r is also given, we can solve x, y, z due to (A25). That is, $\cos ϕ_{1} = \frac{- r - 2}{2}, ϕ_{3} = 0$ or $\cos ϕ_{1} = \frac{- r + 2}{2}$ , ϕ₃ = ±π.

If ϕ₁ + ϕ₂ = π, then $x = - \bar{y}$ , and (A24) becomes

A26

{\begin{array}{l} 0 = - r; \\ 2 \sin ϕ_{1} = 0. \end{array}

Hence r = 0, which is a contradiction.

If ϕ₁ – ϕ₂ = ±π, then x = –y, (11) becomes r + z – y²z = 0, then y = ±1 or y²z² = –1. y = ±1 violates the restriction; y²z² = –1 means $z + \bar{z} = r, {x, y} = {i \bar{z}, - i \bar{z}}$ .

For other cases, we have

A27

\begin{array}{l} \sin ϕ_{3} & = & \frac{r + \cos ϕ_{1} + \cos ϕ_{2} + (1 + \cos (ϕ_{1} + ϕ_{2})) \cos ϕ_{3}}{\sin (ϕ_{1} + ϕ_{2})} \\ = & - \frac{\sin ϕ_{1} + \sin ϕ_{2} + \sin (ϕ_{1} + ϕ_{2}) \cos ϕ_{3}}{1 + \cos (ϕ_{1} + ϕ_{2})} . \end{array}

We square both sides of equations of (A24). Their sum is

A28

\cos ϕ_{3} = \frac{\frac{r^{2} - 4}{4} - \cos ϕ_{1} \cos ϕ_{2}}{\cos ϕ_{1} + \cos ϕ_{2}} .

Substituting the value of cos ϕ₃ in (A28) to (A27), we have

A29

\frac{1}{2} \cos \frac{ϕ_{1} + ϕ_{2}}{2} \sec \frac{ϕ_{1} - ϕ_{2}}{2} (r^{2} + 2 r (\cos ϕ_{1} + \cos ϕ_{2}) + 4 \sin ϕ_{1} \sin ϕ_{2}) = 0.

Then we have

A30

- r^{2} - 2 r \cos ϕ_{1} = 2 r \cos ϕ_{2} + 4 \sin ϕ_{1} \sin ϕ_{2} .

The right side of (A30) can be written as $\sqrt{4 r^{2} + 16 \sin^{2} ϕ_{1}} \sin (ϕ_{2} + ψ)$ , where $\tan ψ = \frac{r}{2 \sin ϕ_{1}}$ . For the case that sin ϕ₁ = 0, we stipulate that $ψ = \frac{π}{2}$ . Since sin( ϕ₂ + ψ) ∈ [–1, 1], we deduce the value condition of ϕ₁ below.

A31

- 1 \leq - \frac{r^{2} + 2 r \cos ϕ_{1}}{\sqrt{4 r^{2} + 16 \sin^{2} ϕ_{1}}} \leq 1.

So we have

A32

\frac{- r^{3} - 4 \sqrt{2} \sqrt{r^{2} + 2}}{2 (r^{2} + 4)} \leq \cos ϕ_{1} \leq \frac{- r^{3} + 4 \sqrt{2} \sqrt{r^{2} + 2}}{2 (r^{2} + 4)} .

This equation ensures that |y| = 1 and determines the range of ϕ₁. Hence we have finished the proof.

Appendix C. The Proof of Theorem 2

Proof.

We assume |u₅〉 = [0, g₂, g₃e^is₃, g₄e^is₄, g₅e^is₅, g₆e^is₆]^T, |u₆〉 = [0, h₂, h₃e^it₃, h₄e^it₄, h₅e^it₅, h₆e^it₆]^T and denote s₂ = t₂ = 0 and $λ_{3} = \sqrt{6} e^{i θ_{1}}, λ_{4} = \sqrt{6} e^{i θ_{2}}$ . We have

A33

h_{j k} = - \frac{1}{5} - \frac{1}{5} λ_{3} + (λ_{4} - λ_{3}) (g_{j} g_{k} e^{i (s_{j} - s_{k})} + h_{j} h_{k} e^{i (t_{j} - t_{k})}), j, k \geq 2, j \neq k,

A34

h_{j j} = - \frac{1}{5} + \frac{4}{5} λ_{3} + (λ_{4} - λ_{3}) (g_{j}^{2} + h_{j}^{2}), j \geq 2,

A35

\begin{array}{l} ‖ {h_{j k} |}^{2} - {| h_{k j} |}^{2} = - \frac{8}{5} \sqrt{6} (\sqrt{6} \cos \frac{θ_{1} - θ_{2}}{2} + \cos \frac{θ_{1} + θ_{2}}{2}) \cdot \sin \frac{θ_{1} - θ_{2}}{2} . \\ g_{j} g_{k} \sin (s_{j} - s_{k}) + h_{j} h_{k} \sin (t_{j} - t_{k}) = 0 . \end{array}

To satisfy the condition in (A35), we divide our discussion into three cases, namely Case A, B and C.

Case A.

If sin $\sin \frac{θ_{1} - θ_{2}}{2} = 0$ , then θ₁ = θ₂, and λ₃ = λ₄. From the first sentence below (19), we see that K′(d, x) has four identical eigenvalues. However the CHM with four identical eigenvalues does not exist due to Theorem 6 in [19].

Case B.

If g_jg_k sin(s_j – s_k) + h_jh_k sin(t_j – t_k) = 0, then for any j, k ≥ 2, f_jk := g_jg_ke^{i(s_j – s_k)} + h_jh_ke^{i(t_j – t_k)} is real. Now $h_{j k} = - \frac{1}{5} - \frac{1}{5} λ_{3} + f_{j k} (λ_{4} - λ_{3})$ for j ≠ k, so we have h_jk = h_kj and H is symmetric. Note that |h_jk| = 1, if λ₃, λ₄ are given, then $- \frac{1}{5} - \frac{1}{5} λ_{3}$ and λ₄ – λ₃ are both given. Using geometry properties, we see that f_jk takes only two distinct values. It means for all j, k ≥ 2 and j ≠ k, h_jk can only take two distinct values. We denote them by a₁, a₂. For hjj, j ≥ 2, by a similar discussion, we obtain that h_jj can take at most two distinct values b₁, b₂. In particular if h_jj can only take one value, then we take b₁ = b₂.

We consider the orthogonality of rows 1,2, that is, $1 + \sum_{k = 2}^{6} h_{2 k} = 0$ . If three of h₂₃, …, h₂₆ are the same, we shall assume 1 + h₂₂ + 3h₂₃ + h₂₆ = 0. Then h₂₂ = h₂₆ = 1 and h₂₃ = – 1, so λ₃, λ₄ are both real, which means for any j, k ≥ 2, h_jk is real, too. So H contains only real elements, i.e. {1, –1}. Such a CHM is known to be non-existent. So two of h₂₃, …, h₂₆ are the same, then 1 + h₂₂ + 2a₁ + 2a₂ = 0. By simple vector theory we know that the vector a₁ + a₂ is orthogonal to λ₄ – λ₃.

Then 1 + h₂₂ must be orthogonal to λ₄ – λ₃. If b₁ = b₂ then h₂₂ is orthogonal to λ₄ – λ₃, so b₁ = b₂ = 1. If b₁ ≠ b₂, we shall assume h₂₂ = b₁. Since b₁ + b₂ is orthogonal to λ₄ – λ₃, then we have b₂ = 1. So in both cases, we have b₂ = 1.

Since h_jk = h_kj, if b₁ appears no less than three times in h₂₂, …, h₆₆, then we select three rows that contain b₁, and by permuting columns, the three rows should be

[\begin{array}{l} 1 & b_{1} & a_{1} & a_{1} & a_{2} & a_{2} \\ 1 & a_{1} & b_{1} & x_{1} & x_{2} & x_{3} \\ 1 & a_{1} & x_{1} & b_{1} & x_{4} & x_{5} \end{array}]

Here {x₁, x₂, x₃} = {x₁, x₄, x₅} = {a₁, a₂, a₂}. If x₁ = a₁, then the orthogonality of rows 2, j never holds. Then x₁ = a₂, then x₄ = x₃, x₅ = x₂, the inner products of the three rows shows that $2 + b_{1} \bar{a_{1}} + a_{1} \bar{b_{1}} + a_{1} \bar{a_{2}} + a_{2} \bar{a_{1}} = 0$ and $2 + b_{1} \bar{a_{2}} + a_{1} \bar{b_{2}} + a_{1} \bar{a_{2}} + a_{2} \bar{a_{1}} = 0$ . Then we know that $- b_{1}^{2} {| a_{2} - a_{1} |}^{2} = {(a_{1} - a_{2})}^{2}$ . Then b₁ is orthogonal to a₁ – a₂. By simple goemetry property we know that b₁ – 1 isparallelto a₁ – a₂, then b₁ = 1 or a₁ = a₂. The latter case violates the orthogonal of the three rows, then b₁ = 1. And by solving $2 + a_{2} + \bar{a_{2}} + a_{1} \bar{a_{2}} + a_{2} \bar{a_{1}} = 0$ , we know that {a₁, a₂} = {ω, ω²} with $ω = e^{\frac{2 π i}{3}}$ . Other rows contains the same elements, so H is complex equivalent to Tao matrix $S_{6}^{(0)}$ . If b₂ appears no less than three times, we can also solve that {a₁, a₂} = {ω, ω²}, and we then deduce that b₁ = 1.

Case C.

If $\sqrt{6} \cos \frac{θ_{1} - θ_{2}}{2} + \cos \frac{θ_{1} + θ_{2}}{2} = 0$ , then $\tan \frac{θ_{1}}{2} \tan \frac{θ_{2}}{2} = - \frac{7 + 2 \sqrt{6}}{5}$ . Since λ₃ ≠ λ₄, we denote $a^{'} = - \frac{1 + λ_{3}}{5 (λ_{4} - λ_{3})}, b^{'} = g_{j} g_{k} e^{i (s_{j} - s_{k})} + h_{j} h_{k} e^{i (t_{j} - t_{k})}$ , then |h_jk| = |h_kj| = 1 shows that $| a^{'} + b^{'} | = | a^{'} + {\bar{b}}^{'} |$ , where ${\bar{b}}^{'}$ is the conjugate of b′. Since a′ is a given complex number, the equality shows that either a′ is real or b′ is always real whatever j, k are. The latter case was discussed in Case B, it is just the case that g₁g₂ sin(s₁ – s₂) + h₁h₂ sin(t₁ – t₂) = 0.

Now a′ is real, so there exists a real number k such that a′ = k, then $- \frac{1}{5} - \frac{1}{5} λ_{3} = k (λ_{4} - λ_{3})$ . So

A36

λ_{4} = - \frac{1}{5 k} + (1 - \frac{1}{5 k}) λ_{3} .

Compute the real and imaginary part of two sides, we obtain the following system of equations.

A37

{\begin{array}{l} \sqrt{6} \cos θ_{2} = - \frac{1}{5 k} + \sqrt{6} (1 - \frac{1}{5 k}) \cos θ_{1}; \\ \sin θ_{2} = (1 - \frac{1}{5 k}) \sin θ_{1} . \end{array}

We compute

A38

| λ_{4} - λ_{3} | = \sqrt{(λ_{4} - λ_{3}) (\bar{λ_{4}} - \bar{λ_{3}})} = \frac{1}{\sqrt{k - 5 k^{2}}} .

We claim that the following equality holds:

A39

1 + λ_{3} + λ_{4} = \frac{λ_{4} - λ_{3}}{\bar{λ_{4}} - \bar{λ_{3}}} .

From (A36), the equality turns to be

A40

(1 - \frac{1}{5 k} + (2 - \frac{1}{5 k}) λ_{3}) (1 + {\bar{λ}}_{3}) = 1 + λ_{3} .

Using $| λ_{3} | = \sqrt{6}$ , it turns to be

A41

(1 - \frac{1}{5 k}) λ_{3}^{2} + (12 - \frac{7}{5 k}) λ_{3} + 6 (1 - \frac{1}{5 k}) = 0.

From (A36) and ${| λ_{4} |}^{2} = λ_{4} \cdot {\bar{λ}}_{4} = 6$ we deduce the same equation as (A41). So we have proven the equality (A39).

Now we denote

A42

d = \frac{λ_{4} - λ_{3}}{| λ_{4} - λ_{3} |} .

So from (A39) we know that

A43

d^{2} = \frac{{(λ_{4} - λ_{3})}^{2}}{(λ_{4} - λ_{3}) ({\bar{λ}}_{4} - {\bar{λ}}_{3})} = λ_{3} + λ_{4} + 1.

We select

A44

| v_{1} 〉 = {[\sqrt{5 k}, - \sqrt{\frac{1 - 5 k}{5}}, - \sqrt{\frac{1 - 5 k}{5}}, - \sqrt{\frac{1 - 5 k}{5}}, - \sqrt{\frac{1 - 5 k}{5}}, - \sqrt{\frac{1 - 5 k}{5}}]}^{T},

A45

| v_{2} 〉 = {[\sqrt{1 - 5 k}, \sqrt{k}, \sqrt{k}, \sqrt{k}, \sqrt{k}, \sqrt{k}]}^{T} .

If we denote λ₄₃ = λ₄ – λ₃, then

A46

λ_{3} | v_{1} 〉 〈 v_{1} | + λ_{4} | v_{2} 〉 〈 v_{2} | = [\begin{matrix} A_{1} & A_{2} & A_{2} & A_{2} & A_{2} & A_{2} \\ A_{2} & A_{3} & A_{3} & A_{3} & A_{3} & A_{3} \\ A_{2} & A_{3} & A_{3} & A_{3} & A_{3} & A_{3} \\ A_{2} & A_{3} & A_{3} & A_{3} & A_{3} & A_{3} \\ A_{2} & A_{3} & A_{3} & A_{3} & A_{3} & A_{3} \\ A_{2} & A_{3} & A_{3} & A_{3} & A_{3} & A_{3} \end{matrix}]

with A₁ – λ₄ – 5kλ₄₃,

A_{2} = \sqrt{k - 5 k^{2}} λ_{43}, A_{3} = \frac{1}{5} λ_{3} + k λ_{43}

Substituting the expression of λ₃, λ₄ discussed in (A36), (A38), (A42), (A43), we have

A47

A_{1} = λ_{4} - 5 k λ_{43} = (1 - 5 k) λ_{43} + λ_{3} = 1 + λ_{4} + λ_{3} = d^{2},

A48

A_{2} = \sqrt{k - 5 k^{2}} λ_{43} = \frac{λ_{4} - λ_{3}}{| λ_{4} - λ_{3} |} = d,

A49

A_{3} = \frac{1}{5} λ_{3} + k λ_{43} = \frac{1}{5} λ_{3} + k [- \frac{1}{5 k} + (1 - \frac{1}{5 k}) λ_{3} - λ_{3}] = - \frac{1}{5} .

And

A50

〈 v_{1} ∣ v_{2} 〉 = \sqrt{5 k (1 - 5 k)} - 5 \sqrt{\frac{k (1 - 5 k)}{5}} = 0.

So |v₁〉, |v₂〉 are orthogonal. From (A46), (A47), (A48), (A49) we know that

A51

λ_{3} | v_{1} 〉 〈 v_{1} | + λ_{4} | v_{2} 〉 〈 v_{2} | = \frac{1}{5} [\begin{matrix} 5 d^{2} & 5 d & 5 d & 5 d & 5 d & 5 d \\ 5 d & - 1 & - 1 & - 1 & - 1 & - 1 \\ 5 d & - 1 & - 1 & - 1 & - 1 & - 1 \\ 5 d & - 1 & - 1 & - 1 & - 1 & - 1 \\ 5 d & - 1 & - 1 & - 1 & - 1 & - 1 \\ 5 d & - 1 & - 1 & - 1 & - 1 & - 1 \end{matrix}] .

For the vector $| u_{j} 〉 = {[g_{1}^{j}, \dots, g_{6}^{j}]}^{T}, j = 3, 4, 5, 6$ , we have 〈u₁|u_j〉 = 〈u₂|u_j〉 = 0. Then 〈u₁ – u₂|u_j〉 = 0, from (20) we know that $g_{1}^{j} = 0$ for j = 3,4,5,6. So once we take the matrix P in (19) and set the parameter d to be in (A42), we have P^† · 〈u_j|u_j〉 · P^† = 〈u_j|u_j〉 for j = 3, 4, 5, 6. And

A52

P^{†} \cdot H \cdot P^{†} = P^{†} H_{2} P^{†} + λ_{3} \sum_{j = 3}^{4} | u_{j} 〉 〈 u_{j} | + λ_{4} \sum_{j = 5}^{6} | u_{j} 〉 〈 u_{j} |,

here H₂ was defined in (22) and

A53

P^{†} H_{2} P^{†} = \frac{1}{5} [\begin{matrix} 5 d^{2} & 5 d & 5 d & 5 d & 5 d & 5 d \\ 5 d & - 1 & - 1 & - 1 & - 1 & - 1 \\ 5 d & - 1 & - 1 & - 1 & - 1 & - 1 \\ 5 d & - 1 & - 1 & - 1 & - 1 & - 1 \\ 5 d & - 1 & - 1 & - 1 & - 1 & - 1 \\ 5 d & - 1 & - 1 & - 1 & - 1 & - 1 \end{matrix}] .

Then we find that

A54

λ_{3} | v_{1} 〉 〈 v_{1} | + λ_{4} | v_{2} 〉 〈 v_{2} | = P^{†} H_{2} P^{†} .

We shall assume |u_k〉 = [0, x₁, x₂, x₃, x₄, x₅]^T for any k = 3,4,5,6, from (20) and the orthogonality of |u₁〉 and |u_k〉, we know that

A55

x_{1} + x_{2} + x_{3} + x_{4} + x_{5} = 0 .

From (A44), (A45) we know that 〈v_j|u_k〉 = 0 for any j = 1,2. So we obtain a orthonormal basis {|v₁〉, |v₂〉, |u₃〉, |u₄〉, |u₅〉, |u₆〉} and from (A52), (A54) we know that

A56

P^{†} H P^{†} = λ_{3} | v_{1} 〉 〈 v_{1} | + λ_{4} | v_{2} 〉 〈 v_{2} | + λ_{3} \sum_{j = 3}^{4} | u_{j} 〉 〈 u_{j} | + λ_{4} \sum_{j = 5}^{6} | u_{j} 〉 〈 u_{j} | .

So P^†HP^† exactly has eigenvalues λ₃, λ₃, λ₃, λ₄, λ₄, λ₄. Hence given a CHM H in dephased form with eigenvalues $\sqrt{6}, - \sqrt{6}, λ_{3}, λ_{3}, λ_{4}, λ_{4}$ , excluding the Tao matrix (which has been discussed in Case B) in (3), we will deduce a CHM with two sorts of eigenvalues, the latter one must be complex equivalent to K(d, x) due to Theorem 1. So H must be complex equivalent to P · K(d, x) · P with P in (19). We take K′(d, x) = d · P · K(d, x) · P, now K′(d, x) here is in dephased form. And if we assume the eigenvalues of K′(d, x) are $\sqrt{6}, - \sqrt{6}, λ_{3}, λ_{3}, λ_{4}, λ_{4}$ , we obtain that {λ₃, λ₄} = {dλ₁, dλ₂} with λ₁, λ₂ in (10). Hence we deduce the expression of K′(d, x) in (21).

Now if d = ±i, then by the result of Lemma 1 (i), we obtain that {x, y} or {y, z} or {x, z} is equal to {1, –1}. It is easy to verify that in all these cases, K′(d, x) must be complex equivalent to Haagerup matrices in (1).

If d = ±1, then from (10) we know that $λ_{1} = \sqrt{6}, λ_{2} = - \sqrt{6}$ . So ${λ_{3}, λ_{4}} = {\sqrt{6}, - \sqrt{6}}$ , then K′(d, x) is complex equivalent to Hermitian matrices in (4). Hence we have finished the proof of Theorem 2.

Eigenvalues of Two-Parameter Complex Hadamard Matrices of Order Six

Full Article

1. Introduction

2. Preliminaries

Definition 1.

Definition 2.

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3

4

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6

7

3. Result

Theorem 1.

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10

Lemma 1.

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Theorem 2.

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Proof.

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Corollary 1.

Proof.

4. Application

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27

Theorem 3.

Proof.

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35

5. Conclusion

Notes

Appendices

Appendix A. The proof of Theorem 1

Proof.

A1

A2

A3

A4

A5

Case A.

A6

Case A.1.

A7

Case A.2.

A8

Case B.

A9

Case B.1.

A10

A11

Case B.2.

A12

A13

A14

A15

A16

A17

A18