Mutually unbiased bases containing a complex Hadamard matrix of Schmidt rank three

Abstract

Constructing four six-dimensional mutually unbiased bases (MUBs) is an open problem in quantum physics and measurement. We investigate the existence of four MUBs including the identity, and a complex Hadamard matrix (CHM) of Schmidt rank three. The CHM is equivalent to a controlled unitary operation on the qubit-qutrit system via local unitary transformation I₂ ⊗ V and I₂ ⊗ W. We show that V and W have no zero entry, and apply it to exclude constructed examples as members of MUBs. We further show that the maximum of entangling power of controlled unitary operation is log ₂ 3 ebits. We derive the condition under which the maximum is achieved, and construct concrete examples. Our results describe the phenomenon that if a CHM of Schmidt rank three belongs to an MUB then its entangling power may not reach the maximum.

1. Introduction

The existence of four six-dimensional mutually unbiased bases (MUBs) is one of the main open problems in quantum mechanics and information [1–3]. The problem is equivalent to showing the existence of identity matrix, and three 6 × 6 complex Hadamard matrices (CHMs) satisfying certain constraint. The n × n CHM H_n = [u_ij]_i,j=1,…,n is a matrix with orthogonal row vectors and entries of modulus one. That is, $H_n{H}_n^{†}=n{I}_n,|u_{ij}|=1$. In this paper, we let CHM be a unitary matrix with entries of identical modulus for convenience. By regarding the column vectors of each CHM as an orthonormal basis in the six-dimensional Hilbert space ${\mathbb{C}}^6$, the constraint says that every two vectors from different bases has inner product of modulus $1/\sqrt{6}$. We shall denote an MUB trio as the set of three CHMs with above constraint, though it is widely believed that the set does not exist [4–13]. MUBs play a key role in quantum tomography, key distribution, error correction, uncertainty relation and more quantum correlations. They are of pivotal importance in quantum mechanics as for the discrete Wigner function [14,15] and the solution of the Mean King problem [16]. One optimal application of MUBs is quantum random access code. It is commonly believed that the optimal performance of the 2^d → 1 quantum random access code is achieved when the measurements correspond to a pair of MUBs in dimension d. Furthermore, MUBs are important in quantifying the degree of quantum coherence [17]. It has been shown in [18] that one can obtain new uncertainty relations for coherence quantifiers averaged with respect to MUBs. The incompatibility of MUBs has been quantified using the noise robustness [19]. Zhu et al. [20] established a connection between maximally steerable assemblages and complete sets of MUBs, which is important in quantum estimation theory. Carollo et al. [21] define a ratio to measure the discrepancy between an inherently quantum and a quasi-classical multi-parameter estimation problem based on the mean Uhlmann curvature [22] and the Fisher information quantitatively. Recently, the MUB problem has been investigated using the extensively useful notion in quantum information, i.e. Schmidt rank [23],¹ see also figure 1. It has been shown that the CHM with Schmidt rank one or two does not belong to any MUB trio. As far as we know, the approach of studying MUBs in terms of Schmidt rank is not much understood. Chen & Yu [23] display a promising perspective on this long-standing problem. It is both physically meaningful and mathematically operational to investigate CHMs of larger Schmidt rank.

Figure 1.

The CHM M consists of four blocks C, D, F, G. They are all 3 × 3 submatrices of entries of modulus $1/\sqrt{6}$. The Schmidt rank of M is the number of linearly independent blocks in C, D, F, G. So the Schmidt rank is at most four. Recently, it has been shown that if M belongs to an MUB trio then M has Schmidt rank three or four. We show that if U_AB has Schmidt rank three then M = (I₂ ⊗ V) U_AB (I₂ ⊗ W) with 3 × 3 unitary matrices V, W containing no zero entry, though numerical tests indicate that such M may not exist. We further regard M as a bipartite unitary operation, and investigate its entangling power. U_AB is a controlled unitary operation controlled from the B side in the computational basis {|j〉}. It is equal to the entangling power of U_AB, namely the maximum entanglement E[(U_AB ⊗ I_ab)|δ〉_Aa|ϵ〉_Bb] of the bipartite state (U_AB ⊗ I_ab)|δ〉_Aa|ϵ〉_Bb over all input states |δ〉_Aa|ϵ〉_Bb with ancilla system a, b, where $|\delta {\rangle }_{Aa}\in {\mathbb{C}}^2\otimes {\mathbb{C}}^m,{|ϵ⟩}_{Bb}\in {\mathbb{C}}^3\otimes {\mathbb{C}}^n$.

The CHM M of Schmidt rank three is a bipartite controlled unitary operation on the space ${\mathbb{C}}^2\otimes {\mathbb{C}}^3$ controlled by system B. This is expressed as $M=(I\otimes V)(\sum _{k=1}^3{U}_k\otimes |k⟩⟨k|)(I\otimes W)$, where V, U_k and W are local unitary gates. The output systems A′ and B′ have the same size as that of A and B, respectively. It implies that the CHM M may be reliably implemented by experiments. Recently, Chen & Yu [23] have shown that the CHM of Schmidt rank one or two does not belong to any MUB trio. So the next step is to treat CHMs of Schmidt rank three. Such CHMs exist. For example,

$$\frac{1}{\sqrt{6}}\left[\begin{array}{cccccc}y& -y& z& 1& 1& 1\\ y{\omega }^2& -y{\omega }^2& z\omega & 1& 1& \omega \\ y\omega & -y\omega & z{\omega }^2& 1& 1& {\omega }^2\\ 1& 1& 1& x& -x& -z^{\ast }\\ 1& 1& \omega & x{\omega }^2& -x{\omega }^2& -z^{\ast }\omega \\ 1& 1& {\omega }^2& x\omega & -x\omega & -z^{\ast }{\omega }^2\end{array}\right],$$ 1.1

where x, y, z are complex numbers of modulus one, ω = e^2πi/3 and z/y ≠ −z*/x.

In this paper, we shall investigate the CHM M with Schmidt rank three. We review the preliminary results in lemmas 2.1 and 2.3. We also construct examples of Schmidt-rank-three CHMs satisfying certain linear dependence in lemma 2.2. Next, we characterize the expressions and properties of Schmidt-rank-three CHMs M in lemmas 3.1 and 3.2. In quantum physics, the bipartite unitary operation is used for implementing quantum computing and cryptography. If the operation has Schmidt rank larger than one then it is nonlocal and can create entanglement. We ask for the maximum entanglement a non-local operation can create using a product state as an input state. The maximum is called the entangling power of the nonlocal operation. The input state contains ancilla systems not affected by the operation. It has been proven in [24] that M is a controlled unitary operation on ${\mathbb{C}}^2\otimes {\mathbb{C}}^3$. So we obtain the decomposition M = (I₂ ⊗ V) U_AB (I₂ ⊗ W) with some 3 × 3 unitary matrices V, W and U_AB a controlled unitary operation controlled from the B side in the computational basis {|j〉}. Assisted by lemmas 3.1 and 3.2, we show that if V or W has a zero entry then M does not belong to any MUB trio in theorem 3.3. We show that constructed examples of M do not belong to any MUB trio. It indicates that no Schmidt-rank-three CHM belong to an MUB trio. Since M has Schmidt rank three, the maximum of entangling power of U_AB is log ₂ 3 ebits. In equation (4.5), we analytically derive the condition under which the maximum is achieved. In example 4.1, we construct a concrete U_AB by which the condition is satisfied. We also describe the lower bound of entangling power of general U_AB in figures 2–4. In particular, the lower bound of a CHM as bipartite unitary operation may not reach the maximum entangling power, if the CHM belongs to an MUB trio. Our results show the connection between the open problem on the existence of four six-dimensional MUBs, and the entangling power of bipartite unitary operations in terms of Schmidt rank.

Figure 3.

Let $({\beta }_1,{\beta }_3,d_1,d_2,d_3)=({\beta }_1,0,\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$ and β₁ = 0, π/6, π/4, π/2, respectively. The curves show that $-\sum _{j=1}^3{\lambda }_j{\log }_2{\lambda }_j$ increases monotonically with x ∈ [ − π/6, 0]. The curves β₁ = 0 and β₁ = π/2 coincide. The curve β₁ = 0 is above the curve π/6, and the curve β₁ = 0 is above the curve β₁ = π/6.

Figure 2.

Let $({\beta }_1,{\beta }_3,d_1,d_2,d_3)=(\pi ,{\beta }_3,\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$ and β₃ = 0, π/6, π/4, π/3, π/2, π, 3π/2, respectively. The curves show that the function $-\sum _{j=1}^3{\lambda }_j{\log }_2{\lambda }_j$ in (4.8) increases monotonically with x ∈ [ − π/6, 0]. The curves β₃ = 0 and β₃ = π coincide, and the curves β₃ = π/2 and β₃ = 3π/2 also coincide. The curve β₃ = π/6 is above the curve β₃ = π/4, and the curve β₃ = π/4 is above the curve β₃ = π/3.

Figure 4.

Let (β₁, β₃, d₁, d₂, d₃) = (0, 0, d₁, d₂, d₃) and $(d_1,d_2,d_3)=(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$, $(\frac{1}{2},\frac{1}{2},\frac{\sqrt{2}}{2})$, $(\frac{1}{\sqrt{5}},\frac{1}{\sqrt{5}},\frac{\sqrt{3}}{\sqrt{5}})$, $(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}})$, $(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{6}},\frac{1}{\sqrt{2}})$, $(\frac{1}{\sqrt{7}},\frac{\sqrt{2}}{\sqrt{7}},\frac{2}{\sqrt{7}})$, $(\frac{1}{2\sqrt{2}},\frac{1}{2},\frac{\sqrt{5}}{2\sqrt{2}})$, respectively. The curves show that $-\sum _{j=1}^3{\lambda }_j{\log }_2{\lambda }_j$ increases monotonically with x ∈ [ − π/6, 0]. And $-\sum _{j=1}^3{\lambda }_j{\log }_2{\lambda }_j$ has the maximum log ₂3 when x = 0 and $(d_1,d_2,d_3)=(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$.

The study of entangling power has received extensive attentions in the past decades [25–30]. The entangling power of a bipartite unitary operation is a lower bound of the entanglement required for realizing the operation under local operations and classical communications (LOCC). It is known that bipartite unitary operation with Schmidt rank at most three is a controlled unitary operation, and thus may be more easily implemented in experiments [24,29]. Hence, studying the six-dimensional CHM in terms of Schmidt rank connects the MUB problem and bipartite unitary operations. Besides, the technique employed in our results relate the MUB problem to other fundamental notions like multiqubit entangled states, unextendible product basis [31,32]. Very recently, bipartite operator of Schmidt rank three has been applied to the study of entanglement distillability of three bipartite reduced density operators from the same tripartite state [33].

The rest of this paper is structured as follows. In §2, we introduce preliminary results from linear algebra and quantum information. In §3, we characterize the properties of Schmidt-rank-three CHMs. The proofs are given in appendix A and B. In §4, we investigate the entangling power of Hadamard matrix, taken as a bipartite unitary operation on ${\mathbb{C}}^2\otimes {\mathbb{C}}^3$. We conclude in §5.

2. Preliminaries

In this section, we introduce the fundamental knowledge used throughout the paper. Let ${\mathbb{C}}^d$ be the d-dimensional Hilbert space, and $d=\prod _{j=1}^n{d}_j^{k_j}$ the prime factor decomposition such that $d_1^{k_1}<d_2^{k_2}<\cdots <d_n^{k_n}$. It has been shown that there are at most d + 1 and at least $d_1^{k_1}+1$ MUBs [2,3]. In particular, the upper bound d + 1 is achieved when n = 1, namely d is the prime power. On the other hand, if n > 1 say d = 6 then constructing MUBs becomes a hard problem. The traditional way of studying the existence of MUBs employs Pauli groups, while we will do it using the Schmidt rank of CHMs. Recall that the monomial unitary matrix is a unitary matrix with exactly one non-zero element in each row and column. We say that two mn × mn matrices A, B are equivalent if there exists a monomial unitary matrix P ⊗ Q and R ⊗ S with P, R on ${\mathbb{C}}^m$ and Q, S on ${\mathbb{C}}^n$, such that (P ⊗ Q)A(R ⊗ S) = B. If A is a CHM then one can show that B is also a CHM, and has the same Schmidt rank as that of A. The following result is from lemma 13 of [23]. It characterizes the expressions of order-six CHM of Schmidt rank up to three.

Lemma 2.1. —

Any Schmidt-rank-three order-six CHM can be written as

$$\begin{array}{rl}& {\mathbb{H}}_3:=(I_2\otimes V)\cdot \\ & \times \left(\begin{array}{cccccc}\cos {\alpha }_1& 0& 0& {\mathrm{e}}^{\mathrm{i}{\gamma }_1}\sin {\alpha }_1& 0& 0\\ 0& \cos {\alpha }_2& 0& 0& {\mathrm{e}}^{\mathrm{i}{\gamma }_2}\sin {\alpha }_2& 0\\ 0& 0& \cos {\alpha }_3& 0& 0& {\mathrm{e}}^{\mathrm{i}{\gamma }_3}\sin {\alpha }_3\\ {\mathrm{e}}^{\mathrm{i}{\beta }_1}\sin {\alpha }_1& 0& 0& -{\mathrm{e}}^{\mathrm{i}({\beta }_1+{\gamma }_1)}\cos {\alpha }_1& 0& 0\\ 0& {\mathrm{e}}^{\mathrm{i}{\beta }_2}\sin {\alpha }_2& 0& 0& -{\mathrm{e}}^{\mathrm{i}({\beta }_2+{\gamma }_2)}\cos {\alpha }_2& 0\\ 0& 0& {\mathrm{e}}^{\mathrm{i}{\beta }_3}\sin {\alpha }_3& 0& 0& -{\mathrm{e}}^{\mathrm{i}({\beta }_3+{\gamma }_3)}\cos {\alpha }_3\end{array}\right)\\ & \cdot (I_2\otimes W),\end{array}$$ 2.1

where V and W are order-three unitary matrices, the first column vector of W have all non-negative and real elements. Further the matrix

$${\mathbb{H}}_4:=\left[\begin{array}{cccc}\cos {\alpha }_1& {\text{e}}^{\text{i}{\beta }_1}\sin {\alpha }_1& {\text{e}}^{\text{i}{\gamma }_1}\sin {\alpha }_1& -{\text{e}}^{\text{i}({\beta }_1+{\gamma }_1)}\cos {\alpha }_1\\ \cos {\alpha }_2& {\text{e}}^{\text{i}{\beta }_2}\sin {\alpha }_2& {\text{e}}^{\text{i}{\gamma }_2}\sin {\alpha }_2& -{\text{e}}^{\text{i}({\beta }_2+{\gamma }_2)}\cos {\alpha }_2\\ \cos {\alpha }_3& {\text{e}}^{\text{i}{\beta }_3}\sin {\alpha }_3& {\text{e}}^{\text{i}{\gamma }_3}\sin {\alpha }_3& -{\text{e}}^{\text{i}({\beta }_3+{\gamma }_3)}\cos {\alpha }_3\end{array}\right]$$ 2.8

has rank three by the parameters α₁, α₂, α₃ ∈ [0, π/2], β₁, β₂, β₃, γ₁, γ₂, γ₃ ∈ [0, 2π). Hence

(i.a) If ${\mathbb{H}}_3$ is a member of some MUB trio, then α₁, α₂, α₃ ∈ (0, π/2) and two of them are not equal.

By applying lemma 2.1, we construct more examples of Schmidt-rank-three CHMs satisfying certain linear dependence.

Lemma 2.2. —

Let M be a Schmidt-rank-three order-six CHM whose four order-three submatrices are A, B, C, D. Then M exists when one of the following two conditions is satisfied.

• (i)Any three of A, B, C, D are linearly independent.
• (ii)A, B, C are linearly dependent and any two of A, B, C are linearly independent.

Proof. —

• (i)
  We choose α_i = π/4,

  $$V=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}1& 1& 1\\ 1& \omega & {\omega }^2\\ 1& {\omega }^2& \omega \end{array}\right],$$

  and W = I₃ in (2.1), where ω = e^2πi/3. Then ${\mathbb{H}}_3$ is an order-six CHM. The assertion is equivalent to finding β_i and γ_i such that any order-three submatrix of ${\mathbb{H}}_4$ in (2.8) has rank three. An example is β₁ = 0, β₂ = π/6, β₃ = π/3, γ₁ = 0, γ₂ = π/3, γ₃ = 2π/3.
• (ii)We need find α_i, β_i, γ_i such that the leftmost order-three submatrix M of ${\mathbb{H}}_4$ has rank two, and any 3 × 2 submatrix of M also has rank two. An example is α₁ = α₂ = α₃ = 1, β₁ = β₂, γ₁ = γ₂ and β₃ − β₁ ≠ γ₃ − γ₁. ▪

Next, we review the following observation from [23, Lemma 11]. It explains the necessary condition by which a 6 × 6 CHM is a member of some MUB trio. It will be used frequently in the proofs for the claims in the next section. We shall refer to a subunitary matrix as a matrix proportional to a unitary matrix.

Lemma 2.3. —

Any MUB trio contains neither of the order-six CHMs Y₂ and Y₃, where

• 1.Y₂ contains a submatrix of size 3 × 2 and rank one.
• 2.Y₃ contains an order-three submatrix whose one column vector is orthogonal to the other two column vectors.

For example, one can show that the CHM in (1.1) does not belong to any MUB trio by lemma 2.3.

3. Complex Hadamard matrices of Schmidt rank three: mutually

unbiased base

In this section, we characterize CHMs of Schmidt rank three. In lemma 3.1, we investigate the cases when V in (3.1) is a unitary matrix having exactly six and four zero entries, respectively. In lemma 3.2, we investigate the cases when V in (3.1) is a unitary matrix having exactly one zero entry. In theorem 3.3, we present the main result of this section, namely the exclusion of CHM ${\mathbb{H}}_3$ with V, W containing at least one zero entry. We construct examples of M and show that they do not belong to any MUB trio. It indicates that no Schmidt-rank-three CHM belong to an MUB trio.

Lemma 3.1. —

Let ${\mathbb{H}}_3$ be the Schmidt-rank-three order-six CHM in (2.1). We shall use the matrices V, W and parameters α_i, β_i, γ_i in (2.1).

• (i)
  If V is a monomial unitary matrix, then α₁ = α₂ = α₃ = π/4, β₁, β₂, β₃, γ₁, γ₂, γ₃ ∈ [0, 2π), such that the matrix (2.2) has rank three. Further W = W₁D₁ where

  $$W_1=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}1& 1& 1\\ 1& \omega & {\omega }^2\\ 1& {\omega }^2& \omega \end{array}\right]\text{or}\hspace{0.17em}\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}1& 1& 1\\ 1& {\omega }^2& \omega \\ 1& \omega & {\omega }^2\end{array}\right],$$

  $D_1=\text{diag}(1,{\text{e}}^{\text{i}{\delta }_2},{\text{e}}^{\text{i}{\delta }_3}),$ and ω = e^2πi/3, δ₂, δ₃ ∈ [0, 2π).
• (ii)
  If V is a unitary matrix with exactly four zero entries, then V is equivalent to $1\oplus \frac{1}{\sqrt{2}}\left[\begin{array}{cc}1& {\text{e}}^{\text{i}\theta }\\ 1& -{\text{e}}^{\text{i}\theta }\end{array}\right]$ where θ = ±π/2 when α₂ ≠ 0, π/2. Next, α₁ = π/4, α₂ + α₃ = π/2, ${\alpha }_2\in [0,\pi /4)\cup (\pi /4,\pi /2],$ β₁, β₂, β₃, γ₁, γ₂, γ₃ ∈ [0, 2π), such that the matrix (2.2) has rank three. Further W = W₂D₂, where

  $$W_2=\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}1& 1& 1\\ 1& \omega & {\omega }^2\\ 1& {\omega }^2& \omega \end{array}\right]\text{or}\hspace{0.17em}\frac{1}{\sqrt{3}}\left[\begin{array}{ccc}1& 1& 1\\ 1& {\omega }^2& \omega \\ 1& \omega & {\omega }^2\end{array}\right],$$

  $D_2=\mathrm{diag}(1,{\mathrm{e}}^{\mathrm{i}{\delta }_2},{\mathrm{e}}^{\mathrm{i}{\delta }_3})$ and ω = e^2πi/3, δ₂, δ₃ ∈ [0, 2π).

One can prove this lemma easily by computing ${\mathbb{H}}_3=[h_{jk}],h_{jk}=\frac{1}{\sqrt{6}}$. Next, we investigate the more complex case, namely ${\mathbb{H}}_3$ when V in (2.1) has exactly one zero entry.

Lemma 3.2. —

Let ${\mathbb{H}}_3$ be the Schmidt-rank-three order-six CHM in (2.1). If V in (2.1) is a unitary matrix with exactly one zero entry. Then

• (i)V is equivalent to the product of two unitary matrices $p_1\oplus \frac{1}{\sqrt{2}}\left[\begin{array}{cc}1& {\text{e}}^{\text{i}\theta }\\ 1& -{\text{e}}^{\text{i}\theta }\end{array}\right]$ and $\left[\begin{array}{cc}{g}_{21}& g_{23}\\ g_{22}& g_{24}\end{array}\right]\oplus p_2,$ where |g₂₁|² = |g₂₄|² = cos 2α₂/(cos 2α₂ − cos 2α₁), p₁ and p₂ have modulus one;
• (ii)${\cos }^2{\alpha }_1+{\cos }^2{\alpha }_2+{\cos }^2{\alpha }_3=\frac{3}{2},$ β₁, β₂, β₃, γ₁, γ₂, γ₃ ∈ [0, 2π), such that the matrix (2.2) has rank three;
• (iii)
  W = W₃D₃, where

  $$W_3=\left[\begin{array}{ccc}{d}_1& e_1& f_1\\ d_2& e_2& f_2\\ d_3& e_3& f_3\end{array}\right],$$

  d₁, d₂ and d₃ are non-negative and real numbers, e_i, f_i are complex, and $D_3=\mathrm{diag}(1,{\mathrm{e}}^{\mathrm{i}{\delta }_2},{\mathrm{e}}^{\mathrm{i}{\delta }_3})$ is a diagonal unitary matrix, δ₂, δ₃ ∈ [0, 2π).

The proof of this lemma is given in appendix A. Now we present the main result of this section.

Theorem 3.3. —

Let ${\mathbb{H}}_3$ be the Schmidt-rank-three order-six CHM in (2.1).

• (i)If V is a monomial unitary matrix then ${\mathbb{H}}_3$ is not a member of any MUB trio.
• (ii)If V is a unitary matrix with exactly four zero entries, then ${\mathbb{H}}_3$ is not a member of any MUB trio.
• (iii)If V is a unitary matrix with exactly one zero entry then ${\mathbb{H}}_3$ is not a member of any MUB trio.

Proof. —

• (i)Since V is a monomial unitary matrix, lemma 3.1 (i) shows that ${\mathbb{H}}_3$ has a 3 × 3 subunitary matrix. An example is the upper-left 3 × 3 submatrix of ${\mathbb{H}}_3$. This is the matrix Y₃ in lemma 2.3. So assertion (i) holds.
• (ii)Since $V=p\oplus G$ is a unitary matrix, lemma 3.1 (ii) shows that ${\mathbb{H}}_3$ has a 2 × 3 matrix of rank one. An example is the submatrix in the first and fourth rows, and the first three columns of ${\mathbb{H}}_3$. This is the matrix Y₂ in lemma 2.3. So assertion (ii) holds.
• (iii)The proof is based on lemma 3.2 and given in appendix B. ▪

By theorem 3.3, we conclude that the Schmidt-rank-three CHM ${\mathbb{H}}_3$ in (2.1) in an MUB trio satisfies that the 3 × 3 unitary matrix V in (2.1) has no zero entry. Since ${\mathbb{H}}_3^{†}$ also belongs to an MUB trio, theorem 3.3 shows that the 3 × 3 unitary matrix W in (2.1) has no zero entry too. This fact shows that the CHM in the proof of lemma 2.2 (i) is excluded as a member of an MUB trio. Furthermore, the CHMs in (1.1) and lemma 2.2 (ii) are excluded by lemma 2.3 and its full version in [23]. The above facts and theorem 3.3 indicate that the MUB trio do not contain any CHM of Schmidt rank three.

On the other hand, the idea of constructing CHMs M using the four blocks in figure 1 has been introduced by studying a four-parameter family of CHMs in [34]. It firstly determines the block C of M, then finds out finitely many blocks D and F, and finally determines whether the block G exists. It is possible that we may choose suitable D, F, G such that M has Schmidt rank three. Furthermore, Szöllősi [34] constructs a four-parameter family of CHMs, and conjectures that it may be the full characterization of all CHMs. So the construction assisted by computer bring about all CHMs of Schmidt rank three, especially ${\mathbb{H}}_3$ with V, W having no zero entries.

4. Complex Hadamard matrices of Schmidt rank three: entangling power

In this section, we regard the six-dimensional CHM ${\mathbb{H}}_3$ in (2.1) as a bipartite unitary operation on ${\mathbb{C}}^2\otimes {\mathbb{C}}^3$. It is known that ${\mathbb{H}}_3$ is a controlled unitary operation controlled from system B [24,35]. We evaluate the entangling power of ${\mathbb{H}}_3$. Since the entangling power is invariant under local unitary transformation, we obtain that the entangling power of ${\mathbb{H}}_3$ is the same as that of

$$U_{AB}=\sum _{k=1}^3{U}_k\otimes |k⟩⟨k|,$$ 4.1

where $U_k=\left[\begin{array}{cc}\cos {\alpha }_k& {\text{e}}^{\text{i}{\gamma }_k}\sin {\alpha }_k\\ {\text{e}}^{\text{i}{\beta }_k}\sin {\alpha }_k& -{\text{e}}^{\text{i}({\beta }_k+{\gamma }_k)}\cos {\alpha }_k\end{array}\right]$, and the real parameters α_j, β_j, γ_j satisfy equation (2.8) and cos ²α₁ + cos ²α₂ + cos ²α₃ = 3/2. This equation is from the fact that ${\mathbb{H}}_3$ has entries of modulus $1/\sqrt{6}$.

Suppose U_AB acts on the input state, which is a bipartite product state ${|\delta ⟩}_{Aa}\otimes {|ϵ⟩}_{Bb}\in ({\mathbb{C}}^2\otimes {\mathbb{C}}^m)\otimes ({\mathbb{C}}^3\otimes {\mathbb{C}}^n)$, and a, b are the ancilla systems. Up to local unitary transformation on system a, b we may assume that

$${|\delta ⟩}_{Aa}={(c_1|1,1⟩+c_2|1,2⟩+c_3|2,2⟩)}_{Aa}$$ 4.2

and

$${|ϵ⟩}_{Bb}={(d_1|1,x_1⟩+d_2|2,x_2⟩+d_3|3,x_3⟩)}_{Bb},$$ 4.3

where |x₁〉, |x₂〉, |x₃〉 are unit vectors on ancilla system b, c₁, c₂, d₁, d₂, d₃ ≥ 0, $c_1^2+c_2^2+|c_3{|}^2=d_1^2+d_2^2+d_3^2=1.$ The output state is the bipartite entangled state

$${|\psi ⟩}_{Aa:Bb}=U_{AB}({|\delta ⟩}_{Aa}\otimes {|ϵ⟩}_{Bb}).$$ 4.4

By definition, the entangling power of U_AB is the maximum entanglement of state |ψ〉 over all |δ〉 ⊗ |ϵ〉. It follows from (4.1) that |ψ〉 has Schmidt rank at most three. So it has the maximum entanglement log ₂ 3 ebits. Using (4.1)–(4.4), the maximum entanglement is achievable if and only if there exists |δ〉_Aa such that the three states (U_j)_A|δ〉_Aa, j = 1, 2, 3 are pairwise orthogonal. That is,

$$⟨\delta {|}_{Aa}{(U_2)}_A^{†}{(U_1)}_A{|\delta ⟩}_{Aa}=⟨\delta {|}_{Aa}{(U_3)}_A^{†}{(U_2)}_A{|\delta ⟩}_{Aa}=⟨\delta {|}_{Aa}{(U_1)}_A^{†}{(U_3)}_A{|\delta ⟩}_{Aa}=0.$$ 4.5

Using (4.1)–(4.5), we can determine whether U_AB has the maximum entanglement. A concrete example reaching the maximum is constructed as follows.

Example 4.1. —

Let the input state be ${|\delta ⟩}_{Aa}=(|11⟩+|22⟩)/\sqrt{2}$ by choosing $c_1=c_3=1/\sqrt{2}$, c₂ = 0 in (4.2). Then (4.5) is equivalent to the statement that ${\mathbb{H}}_4{\mathbb{H}}_4^{†}$ is diagonal, where ${\mathbb{H}}_4$ is from (2.8). Let U_AB in (4.1) satisfy β₁ + γ₁ = β₂ + γ₂ = β₃ + γ₃ + π, cos ²α₁ + cos ²α₂ = 1/2. By choosing α₁ = α₂ = π/3 and $\cos ({\beta }_1-{\beta }_2)=-\frac{1}{3}$, we have

$$\begin{array}{rl}& ({\alpha }_1,{\alpha }_2,{\alpha }_3,{\beta }_1,{\beta }_2,{\beta }_3,{\gamma }_1,{\gamma }_2,{\gamma }_3)\\ & =(\frac{\pi }{3},\frac{\pi }{3},0,{\beta }_1,{\beta }_1-\arccos (-\frac{1}{3}),{\beta }_3,{\gamma }_1,{\gamma }_1+\arccos (-\frac{1}{3}),{\beta }_1+{\gamma }_1-{\beta }_3-\pi ).\end{array}$$ 4.6

By choosing good β₁ and γ₁, we can obtain that ${\mathbb{H}}_4$ in (2.8) has rank three. For example, β₁ = γ₁ = 0 or β₁ = γ₁ = π/2. To conclude, we have shown that U_AB corresponds to the six-dimensional CHM ${\mathbb{H}}_3$ in (2.1). Furthermore, U_AB is a Schmidt-rank-three bipartite unitary operation of maximum entangling power log ₂ 3 ebits. Nevertheless, lemma 2.1 (i.a) and theorem 3.3 show that such an ${\mathbb{H}}_3$ is not a member of any MUB trio.

In the remaining of this section, we investigate the entangling power of |ψ〉. Using (4.4), one can obtain that the reduced density operator of system Aa is the two-qubit state

$${\rho }_{Aa}=d_1^2{(U_1)}_A|\delta ⟩⟨\delta {|}_{Aa}{(U_1)}_A^{†}+d_2^2{(U_2)}_A|\delta ⟩⟨\delta {|}_{Aa}{(U_2)}_A^{†}+d_3^2{(U_3)}_A|\delta ⟩⟨\delta {|}_{Aa}{(U_3)}_A^{†}.$$ 4.7

Evidently, ρ_Aa has a zero eigenvalue. Suppose the three remaining eigenvalues are λ₁, λ₂ and λ₃. Then the entanglement of |ψ〉 is $S({\rho }_{Aa})=-\sum _{j=1}^3{\lambda }_j{\log }_2{\lambda }_j$, where S(ρ) is the von Neumann entropy of a quantum state ρ. So the entangling power of U_AB is

$$\underset{\genfrac{}{}{0ex}{}{c_1,c_2,d_1,d_2,d_3\ge 0,}{c_1^2+c_2^2+|c_3{|}^2=d_1^2+d_2^2+d_3^2=1}}{max}-\sum _{j=1}^3{\lambda }_j{\log }_2{\lambda }_j.$$ 4.8

We shall investigate its lower bound. We still use the parameters in example 4.1, except that we replace α₁ = α₂ = π/3 by π/3 + x with x ∈ [ − π/6, 0]. Correspondingly, we replace U_AB in example 4.1 by U_AB(x). Note that the function $-\sum _{j=1}^3{\lambda }_j{\log }_2{\lambda }_j$ is a lower bound of the entangling power of U_AB in (4.8). By using (4.7), we describe how the function changes with β₁, β₃ and d₁, d₂, d₃ in figures 2–4. They imply that $-\sum _{j=1}^3{\lambda }_j{\log }_2{\lambda }_j$ has the maximum log ₂ 3 only if x = 0. Note that U_AB(0) is exactly the bipartite unitary operation in example 4.1. So U_AB(0) reaches the maximum entangling power and does not belong to any MUB trio. The above pictures shows that the lower bound of entangling power of U_AB(x) is smaller than that of U_AB(0) as x < 0. At the same time, we have not excluded the possibility that U_AB(x) may belong to some MUB trio when x < 0. It implies that the entangling power of a CHM may not reach the maximum, if it belongs to some MUB trio. It shows the connection between the existence of four MUBs and entangling power.

5. Conclusion

The existence of four six-dimensional MUBs consisting of an identity matrix and three CHMs has been a fundamental problem for decades. We have excluded a subset of CHMs of Schmidt rank three from the four MUBs, and apply it to exclude examples constructed in this paper. It imposes a strict constraint on the existence of four six-dimensional MUBs, and this is supported by numerical tests. We also have constructed the condition by which the entangling power of CHM as a bipartite controlled unitary operation is achieved. Our results indicate the conjecture that if a CHM of Schmidt rank three belongs to an MUB then its entangling power may not reach the maximum. The next target is to prove this conjecture, and analytically exclude any CHM of Schmidt rank three as a member of four MUBs including the identity matrix.

There are still some open problems. In lemma 2.2, we construct some examples of Schmidt-rank-three CHMs satisfying certain linear dependence. Is it possible to parametrize matrix (2.8) with appropriate α_j, β_j and γ_j (and V, W) such that ${\mathbb{H}}_3$ is a Schmidt-rank-three CHMs instead of some examples? In §3, we characterize CHMs of Schmidt rank three. We investigate the cases when V in (3.1) is a unitary matrix having exactly six, four and one zero entry, respectively. However, how to characterize Schmidt-rank-three order-six CHM ${\mathbb{H}}_3$ when V in (3.1) is a unitary matrix having no zero entry? Can we analytically exclude it as a member of four MUBs including the identity matrix? In §4, we investigate the entangling power of six-dimensional Hadamard matrix, taken as a bipartite unitary operation on ${\mathbb{C}}^2\otimes {\mathbb{C}}^3$. How about other tensor-like dimensions like four or eight?

Appendix A. Proof of lemma 3.2

Proof. —

From lemma 2.1 (i.a), we obtain that if ${\mathbb{H}}_3$ is a member of some MUB trio, then α₁, α₂, α₃ ∈ (0, π/2) and two of them are not equal.

(i) If V is a unitary matrix with exactly one zero entry, then there exist permutation matrices P₄, P₅ such that

$$V^{\mathrm{\prime }\mathrm{\prime }}=P_{4}V{P}_5=\left[\begin{array}{ccc}{v}_{11}& v_{12}& 0\\ v_{21}& v_{22}& v_{23}\\ v_{31}& v_{32}& v_{33}\end{array}\right],$$

where v_jk ≠ 0. Since I₂ ⊗ P₅ only changes its right matrix by row permutations, (1) is equivalent

$$\begin{array}{rl}{\mathbb{H}}_3& =(I_2\otimes V^{\mathrm{\prime }\mathrm{\prime }})\\ & \cdot \left(\begin{array}{cccccc}\cos {\alpha }_1& 0& 0& {\mathrm{e}}^{\mathrm{i}{\gamma }_1}\sin {\alpha }_1& 0& 0\\ 0& \cos {\alpha }_2& 0& 0& {\mathrm{e}}^{\mathrm{i}{\gamma }_2}\sin {\alpha }_2& 0\\ 0& 0& \cos {\alpha }_3& 0& 0& {\mathrm{e}}^{\mathrm{i}{\gamma }_3}\sin {\alpha }_3\\ {\mathrm{e}}^{\mathrm{i}{\beta }_1}\sin {\alpha }_1& 0& 0& -{\mathrm{e}}^{\mathrm{i}({\beta }_1+{\gamma }_1)}\cos {\alpha }_1& 0& 0\\ 0& {\mathrm{e}}^{\mathrm{i}{\beta }_2}\sin {\alpha }_2& 0& 0& -{\mathrm{e}}^{\mathrm{i}({\beta }_2+{\gamma }_2)}\cos {\alpha }_2& 0\\ 0& 0& {\mathrm{e}}^{\mathrm{i}{\beta }_3}\sin {\alpha }_3& 0& 0& -{\mathrm{e}}^{\mathrm{i}({\beta }_3+{\gamma }_3)}\cos {\alpha }_3\end{array}\right)\\ & \cdot (I_2\otimes W),\end{array}$$ A 1

up to the switching of α_j, β_j and γ_j and entries of W. Suppose W = W₃D₃, where

$$W_3=\left[\begin{array}{ccc}{d}_1& e_1& f_1\\ d_2& e_2& f_2\\ d_3& e_3& f_3\end{array}\right],$$

d₁, d₂ and d₃ are non-negative and real numbers, and $D_3=\mathrm{diag}(1,{\mathrm{e}}^{\mathrm{i}{\delta }_2},{\mathrm{e}}^{\mathrm{i}{\delta }_3})$ is a diagonal unitary matrix, δ₂, δ₃ ∈ [0, 2π).

Since V^′′ can be expressed as the product of two unitary matrices $p_1\oplus G_1$ and $G_2\oplus p_2$, where G₁ and G₂ are non-monomial unitary matrices, p₁, p₂ are modulus one. Namely,

$$\begin{array}{rl}& V^{\mathrm{\prime }\mathrm{\prime }}=(p_1\oplus G_1)\cdot (G_2\oplus p_2)=\left[\begin{array}{ccc}{p}_1& 0& 0\\ 0& g_{11}& g_{13}\\ 0& g_{12}& g_{14}\end{array}\right]\cdot \left[\begin{array}{ccc}{g}_{21}& g_{23}& 0\\ g_{22}& g_{24}& 0\\ 0& 0& p_2\end{array}\right]\\ & =\left[\begin{array}{ccc}{p}_1{g}_{21}& p_1{g}_{23}& 0\\ g_{11}{g}_{22}& g_{11}{g}_{24}& p_2{g}_{13}\\ g_{12}{g}_{22}& g_{12}{g}_{24}& p_2{g}_{14}\end{array}\right]=\left[\begin{array}{ccc}{v}_{11}& v_{12}& 0\\ v_{21}& v_{22}& v_{23}\\ v_{31}& v_{32}& v_{33}\end{array}\right].\end{array}$$ A 2

By using $W_3{W}_3^{†}=W_3^{†}{W}_3=I$, we know

$$d_i{d}_j+e_i{e}_j^{\ast }+f_i{f}_j^{\ast }=\{\begin{array}{ll}1& i=j,\\ 0& i\ne j.\end{array}$$ A 3

In (A 1), let ${\mathbb{H}}_3=[h_{jk}]$, and $|h_{jk}|=1/\sqrt{6}$. Then we can obtain

$$\begin{array}{rl}& |d_1{v}_{11}\cos {\alpha }_1+d_2{v}_{12}\cos {\alpha }_2|=|e_1{v}_{11}\cos {\alpha }_1+e_2{v}_{12}\cos {\alpha }_2|\\ & =|\hspace{0.17em}{f}_1{v}_{11}\cos {\alpha }_1+f_2{v}_{12}\cos {\alpha }_2|=\frac{1}{\sqrt{6}},\end{array}$$ A 4

$$\begin{array}{rl}& |d_1{v}_{11}{\text{e}}^{\text{i}{\beta }_1}\sin {\alpha }_1+d_2{v}_{12}{\text{e}}^{\text{i}{\beta }_2}\sin {\alpha }_2|=|e_1{v}_{11}{\text{e}}^{\text{i}{\beta }_1}\sin {\alpha }_1+e_2{v}_{12}{\text{e}}^{\text{i}{\beta }_2}\sin {\alpha }_2|\\ & =|\hspace{0.17em}{f}_1{v}_{11}{\text{e}}^{\text{i}{\beta }_1}\sin {\alpha }_1+f_2{v}_{12}{\text{e}}^{\text{i}{\beta }_2}\sin {\alpha }_2|=\frac{1}{\sqrt{6}},\end{array}$$ A 5

$$\begin{array}{rl}& |d_1{v}_{11}{\text{e}}^{\text{i}{\gamma }_1}\sin {\alpha }_1+d_2{v}_{12}{\text{e}}^{\text{i}{\gamma }_2}\sin {\alpha }_2|=|e_1{v}_{11}{\text{e}}^{\text{i}{\gamma }_1}\sin {\alpha }_1+e_2{v}_{12}{\text{e}}^{\text{i}{\gamma }_2}\sin {\alpha }_2|\\ & =|\hspace{0.17em}{f}_1{v}_{11}{\text{e}}^{\text{i}{\gamma }_1}\sin {\alpha }_1+f_2{v}_{12}{\text{e}}^{\text{i}{\gamma }_2}\sin {\alpha }_2|=\frac{1}{\sqrt{6}},\end{array}$$ A 6

$$\begin{array}{rl}& |d_1{v}_{11}{\text{e}}^{\text{i}({\beta }_1+{\gamma }_1)}\cos {\alpha }_1+d_2{v}_{12}{\text{e}}^{\text{i}({\beta }_2+{\gamma }_2)}\cos {\alpha }_2|=|e_1{v}_{11}{\text{e}}^{\text{i}({\beta }_1+{\gamma }_1)}\cos {\alpha }_1+e_2{v}_{12}{\text{e}}^{\text{i}({\beta }_2+{\gamma }_2)}\cos {\alpha }_2|\\ & =|\hspace{0.17em}{f}_1{v}_{11}{\text{e}}^{\text{i}({\beta }_1+{\gamma }_1)}\cos {\alpha }_1+f_2{v}_{12}{\text{e}}^{\text{i}({\beta }_2+{\gamma }_2)}\cos {\alpha }_2|=\frac{1}{\sqrt{6}},\end{array}$$ A 7

$$\begin{array}{rl}& |d_1{v}_{21}\cos {\alpha }_1+d_2{v}_{22}\cos {\alpha }_2+d_3{v}_{23}\cos {\alpha }_3|\\ & =|d_1{v}_{31}\cos {\alpha }_1+d_2{v}_{32}\cos {\alpha }_2+d_3{v}_{33}\cos {\alpha }_3|=\frac{1}{\sqrt{6}},\end{array}$$ A 8

$$\begin{array}{rl}& |e_1{v}_{21}\cos {\alpha }_1+e_2{v}_{22}\cos {\alpha }_2+e_3{v}_{23}\cos {\alpha }_3|\\ & =|e_1{v}_{31}\cos {\alpha }_1+e_2{v}_{32}\cos {\alpha }_2+e_3{v}_{33}\cos {\alpha }_3|=\frac{1}{\sqrt{6}}\end{array}$$ A 9

$$\begin{array}{rl}\mathrm{and}& |\hspace{0.17em}{f}_1{v}_{21}\cos {\alpha }_1+f_2{v}_{22}\cos {\alpha }_2+f_3{v}_{23}\cos {\alpha }_3|\\ & =|\hspace{0.17em}{f}_1{v}_{31}\cos {\alpha }_1+f_2{v}_{32}\cos {\alpha }_2+f_3{v}_{33}\cos {\alpha }_3|=\frac{1}{\sqrt{6}}.\end{array}$$ A 10

Using (A 2) and (A 4), we obtain

$$|v_{11}{|}^2{\cos }^2{\alpha }_1+|v_{12}{|}^2{\cos }^2{\alpha }_2=\frac{1}{2}.$$ A 11

From (A 2), (A 3)–(B 4) and (B 8)–(B 10), we have

$$\begin{array}{rl}& |v_{21}{|}^2{\cos }^2{\alpha }_1+|v_{22}{|}^2{\cos }^2{\alpha }_2+|v_{23}{|}^2{\cos }^2{\alpha }_3=\frac{1}{2},\end{array}$$ A 12

$$\begin{array}{rl}& |v_{31}{|}^2{\cos }^2{\alpha }_1+|v_{32}{|}^2{\cos }^2{\alpha }_2+|v_{33}{|}^2{\cos }^2{\alpha }_3=\frac{1}{2},\end{array}$$ A 13

$$\begin{array}{rl}& d_1^2{\cos }^2{\alpha }_1+d_2^2{\cos }^2{\alpha }_2+d_3^2{\cos }^2{\alpha }_3=\frac{1}{2},\end{array}$$ A 14

$$\begin{array}{rl}& |e_1{|}^2{\cos }^2{\alpha }_1+|e_2{|}^2{\cos }^2{\alpha }_2+|e_3{|}^2{\cos }^2{\alpha }_3=\frac{1}{2}\end{array}$$ A 15

$$\begin{array}{rl}\text{and}& |\hspace{0.17em}{f}_1{|}^2{\cos }^2{\alpha }_1+|\hspace{0.17em}{f}_2{|}^2{\cos }^2{\alpha }_2+|\hspace{0.17em}{f}_3{|}^2{\cos }^2{\alpha }_3=\frac{1}{2}.\end{array}$$ A 16

Equations (A 3) and (A 14)–(A 16) imply that

$${\cos }^2{\alpha }_1+{\cos }^2{\alpha }_2+{\cos }^2{\alpha }_3=\frac{3}{2}.$$ A 17

Recall that the first and second column vectors of (A 1) are orthogonal, the first and third column vectors of (A 1) are orthogonal, the second and third column vectors of (A 1) are orthogonal. We have

$$d_1{e}_1+d_2{e}_2+d_3{e}_3=0,d_1{f}_1+d_2{f}_2+d_3{f}_3=0,e_1^{\ast }{f}_1+e_2^{\ast }{f}_2+e_3^{\ast }{f}_3=0.$$ A 18

By simplifying (A 2) and (A 4)–(A 6), we can obtain

$$\begin{array}{rl}& \sin {\alpha }_1\sin {\alpha }_2(g_{21}^{\ast }{g}_{23}({\text{e}}^{\text{i}({\beta }_2-{\beta }_1)}-{\text{e}}^{\text{i}({\gamma }_2-{\gamma }_1)})+g_{21}{g}_{23}^{\ast }{({\text{e}}^{\text{i}({\beta }_2-{\beta }_1)}-{\text{e}}^{\text{i}({\gamma }_2-{\gamma }_1)})}^{\ast })=0,\end{array}$$ A 19

$$\begin{array}{rl}& \sin {\alpha }_1\sin {\alpha }_2(e_1^{\ast }{e}_2{g}_{21}^{\ast }{g}_{23}({\text{e}}^{\text{i}({\beta }_2-{\beta }_1)}-{\text{e}}^{\text{i}({\gamma }_2-{\gamma }_1)})+e_1{e}_2^{\ast }{g}_{21}{g}_{23}^{\ast }{({\text{e}}^{\text{i}({\beta }_2-{\beta }_1)}-{\text{e}}^{\text{i}({\gamma }_2-{\gamma }_1)})}^{\ast })=0\end{array}$$ A 20

$$\begin{array}{rl}\text{and}& \sin {\alpha }_1\sin {\alpha }_2(f_1^{\ast }{f}_2{g}_{21}^{\ast }{g}_{23}({\text{e}}^{\text{i}({\beta }_2-{\beta }_1)}-{\text{e}}^{\text{i}({\gamma }_2-{\gamma }_1)})+f_1{f}_2^{\ast }{g}_{21}{g}_{23}^{\ast }{({\text{e}}^{\text{i}({\beta }_2-{\beta }_1)}-{\text{e}}^{\text{i}({\gamma }_2-{\gamma }_1)})}^{\ast })=0.\end{array}$$ A 21

Using (A 4) and (A 7), we have

$$\begin{array}{rl}& \cos {\alpha }_1\cos {\alpha }_2(g_{21}^{\ast }{g}_{23}({\text{e}}^{\text{i}({\beta }_2+{\gamma }_2-{\beta }_1-{\gamma }_1)}-1)+g_{21}{g}_{23}^{\ast }{({\text{e}}^{\text{i}({\beta }_2+{\gamma }_2-{\beta }_1-{\gamma }_1)}-1)}^{\ast })=0,\end{array}$$ A 22

$$\begin{array}{rl}& \cos {\alpha }_1\cos {\alpha }_2(e_1^{\ast }{e}_2{g}_{21}^{\ast }{g}_{23}({\text{e}}^{\text{i}({\beta }_2+{\gamma }_2-{\beta }_1-{\gamma }_1)}-1)+e_1{e}_2^{\ast }{g}_{21}{g}_{23}^{\ast }{({\text{e}}^{\text{i}({\beta }_2+{\gamma }_2-{\beta }_1-{\gamma }_1)}-1)}^{\ast })=0\end{array}$$ A 23

$$\begin{array}{rl}\text{and}& \cos {\alpha }_1\cos {\alpha }_2(f_1^{\ast }{f}_2{g}_{21}^{\ast }{g}_{23}({\text{e}}^{\text{i}({\beta }_2+{\gamma }_2-{\beta }_1-{\gamma }_1)}-1)+f_1{f}_2^{\ast }{g}_{21}{g}_{23}^{\ast }{({\text{e}}^{\text{i}({\beta }_2+{\gamma }_2-{\beta }_1-{\gamma }_1)}-1)}^{\ast })=0.\end{array}$$ A 24

From (A 2) and (A 11), we have

$$|g_{21}{|}^2=|g_{24}{|}^2=\frac{\cos 2{\alpha }_2}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}\text{and}|g_{22}{|}^2=|g_{23}{|}^2=\frac{-\text{cos}\hspace{0.17em}2{\alpha }_1}{\cos 2{\alpha }_2-\cos 2{\alpha }_1},$$ A 25

where α₁ ∈ (0, π/4), α₂ ∈ (π/4, π/2) or α₁ ∈ (π/4, π/2), α₂ ∈ (0, π/4). So $G_2=\left[\begin{array}{cc}{g}_{21}& g_{23}\\ g_{22}& g_{24}\end{array}\right]$, where |g₂₁|² = |g₂₄|² = cos 2α₂/(cos 2α₂ − cos 2α₁) and $|g_{22}{|}^2=|g_{23}{|}^2=-\text{cos}\hspace{0.17em}2{\alpha }_1/(\cos 2{\alpha }_2-\cos 2{\alpha }_1)$. And there exist a diagonal unitary matrix $\text{diag}({\text{e}}^{\text{i}{\vartheta }_1},{\text{e}}^{\text{i}{\vartheta }_2},1)$ such that

$$\text{diag}({\text{e}}^{\text{i}{\vartheta }_1},{\text{e}}^{\text{i}{\vartheta }_2},1)(G_2\oplus 1)=\left[\begin{array}{ccc}{\text{e}}^{\text{i}{\vartheta }_1}{g}_{21}& {\text{e}}^{\text{i}{\vartheta }_1}{g}_{23}& 0\\ {\text{e}}^{\text{i}{\vartheta }_2}{g}_{22}& {\text{e}}^{\text{i}{\vartheta }_2}{g}_{24}& 0\\ 0& 0& 1\end{array}\right],$$

where ${\text{e}}^{\text{i}{\vartheta }_1}{g}_{21}$ and ${\text{e}}^{\text{i}{\vartheta }_2}{g}_{22}$ are real numbers, ${\vartheta }_1$, ${\vartheta }_2\in [0,2\pi )$. Hence $G_2=\left[\begin{array}{cc}{g}_{21}& g_{23}\\ g_{22}& g_{24}\end{array}\right]$ is equivalent to

$$\left[\begin{array}{cc}\sqrt{\frac{\cos 2{\alpha }_2}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}& {\text{e}}^{\text{i}{\vartheta }_1}{g}_{23}\\ \sqrt{\frac{-\text{cos}\hspace{0.17em}2{\alpha }_1}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}& {\text{e}}^{\text{i}{\vartheta }_2}{g}_{24}\end{array}\right],$$

where $|{\text{e}}^{\text{i}{\vartheta }_1}{g}_{23}|=\sqrt{-\text{cos}\hspace{0.17em}2{\alpha }_1/(\cos 2{\alpha }_2-\cos 2{\alpha }_1)}$, $|{\text{e}}^{\text{i}{\vartheta }_2}{g}_{24}|=\sqrt{\cos 2{\alpha }_2/(\cos 2{\alpha }_2-\cos 2{\alpha }_1)}$.

Simplifying (A 12)–(A 13) and (A 25), we can obtain $|g_{11}|=|g_{12}|=|g_{13}|=|g_{14}|=1/\sqrt{2}$. So G₁ is equivalent to $\frac{1}{\sqrt{2}}\left[\begin{array}{cc}1& {\text{e}}^{\text{i}\theta }\\ 1& -{\text{e}}^{\text{i}\theta }\end{array}\right]$ where θ ∈ [0, 2π]. ▪

Appendix B. Proof of theorem 3.3 (iii)

Proof. —

In lemma 3.2, we characterize ${\mathbb{H}}_3$. Here, we prove that if V is a unitary matrix with exactly one zero entry then ${\mathbb{H}}_3$ is not a member of any MUB trio.

From (A 19)–(A 21) to (A 22)–(A 24), we know sinα₁sinα₂ = 0 or cosα₁cosα₂ = 0 is a contradiction with the condition α₁, α₂, α₃ ∈ (0, π/2) and two of them are not equal. So we have

$$\begin{array}{rl}& g_{21}^{\ast }{g}_{23}({\text{e}}^{\text{i}({\beta }_2-{\beta }_1)}-{\text{e}}^{\text{i}({\gamma }_2-{\gamma }_1)})+g_{21}{g}_{23}^{\ast }{({\text{e}}^{\text{i}({\beta }_2-{\beta }_1)}-{\text{e}}^{\text{i}({\gamma }_2-{\gamma }_1)})}^{\ast }=0,\end{array}$$ B 1

$$\begin{array}{rl}& e_1^{\ast }{e}_2{g}_{21}^{\ast }{g}_{23}({\text{e}}^{\text{i}({\beta }_2-{\beta }_1)}-{\text{e}}^{\text{i}({\gamma }_2-{\gamma }_1)})+e_1{e}_2^{\ast }{g}_{21}{g}_{23}^{\ast }{({\text{e}}^{\text{i}({\beta }_2-{\beta }_1)}-{\text{e}}^{\text{i}({\gamma }_2-{\gamma }_1)})}^{\ast }=0,\end{array}$$ B 2

$$\begin{array}{rl}& f_1^{\ast }{f}_2{g}_{21}^{\ast }{g}_{23}({\text{e}}^{\text{i}({\beta }_2+{\gamma }_2-{\beta }_1-{\gamma }_1)}-1)+f_1{f}_2^{\ast }{g}_{21}{g}_{23}^{\ast }{({\text{e}}^{\text{i}({\beta }_2+{\gamma }_2-{\beta }_1-{\gamma }_1)}-1)}^{\ast }=0,\end{array}$$ B 3

$$\begin{array}{rl}& g_{21}^{\ast }{g}_{23}({\text{e}}^{\text{i}({\beta }_2+{\gamma }_2-{\beta }_1-{\gamma }_1)}-1)+g_{21}{g}_{23}^{\ast }{({\text{e}}^{\text{i}({\beta }_2+{\gamma }_2-{\beta }_1-{\gamma }_1)}-1)}^{\ast }=0,\end{array}$$ B 4

$$\begin{array}{rl}& e_1^{\ast }{e}_2{g}_{21}^{\ast }{g}_{23}({\text{e}}^{\text{i}({\beta }_2+{\gamma }_2-{\beta }_1-{\gamma }_1)}-1)+e_1{e}_2^{\ast }{g}_{21}{g}_{23}^{\ast }{({\text{e}}^{\text{i}({\beta }_2+{\gamma }_2-{\beta }_1-{\gamma }_1)}-1)}^{\ast }=0\end{array}$$ B 5

$$\begin{array}{rl}\text{and}& f_1^{\ast }{f}_2{g}_{21}^{\ast }{g}_{23}({\text{e}}^{\text{i}({\beta }_2+{\gamma }_2-{\beta }_1-{\gamma }_1)}-1)+f_1{f}_2^{\ast }{g}_{21}{g}_{23}^{\ast }{({\text{e}}^{\text{i}({\beta }_2+{\gamma }_2-{\beta }_1-{\gamma }_1)}-1)}^{\ast }=0.\end{array}$$ B 6

They imply the following two cases (i) and (ii).

(i) If $g_{21}^{\ast }{g}_{23}({\text{e}}^{\text{i}({\beta }_2-{\beta }_1)}-{\text{e}}^{\text{i}({\gamma }_2-{\gamma }_1)})$, $e_1^{\ast }{e}_2{g}_{21}^{\ast }{g}_{23}({\text{e}}^{\text{i}({\beta }_2-{\beta }_1)}-{\text{e}}^{\text{i}({\gamma }_2-{\gamma }_1)})$, $f_1^{\ast }{f}_2{g}_{21}^{\ast }{g}_{23}({\text{e}}^{\text{i}({\beta }_2-{\beta }_1)}-{\text{e}}^{\text{i}({\gamma }_2-{\gamma }_1)})$, $g_{21}^{\ast }{g}_{23}({\text{e}}^{\text{i}({\beta }_2+{\gamma }_2-{\beta }_1-{\gamma }_1)}-1)$, $e_1^{\ast }{e}_2{g}_{21}^{\ast }{g}_{23}({\text{e}}^{\text{i}({\beta }_2+{\gamma }_2-{\beta }_1-{\gamma }_1)}-1)$ and $f_1^{\ast }{f}_2{g}_{21}^{\ast }{g}_{23}({\text{e}}^{\text{i}({\beta }_2+{\gamma }_2-{\beta }_1-{\gamma }_1)}-1)$ are all pure imaginaries. Then e*₁ e₂ and f*₁ f₂ are real numbers. Evidently, there exists a diagonal unitary $D_3^{\prime }=\mathrm{diag}(1,{\mathrm{e}}^{\mathrm{i}{\zeta }_1},{\mathrm{e}}^{\mathrm{i}{\zeta }_2})$ such that

$$W_3{D}_3^{\prime }=\left[\begin{array}{ccc}{d}_1& {\text{e}}^{\text{i}{\zeta }_1}{e}_1& {\text{e}}^{\text{i}{\zeta }_2}{f}_1\\ d_2& {\text{e}}^{\text{i}{\zeta }_1}{e}_2& {\text{e}}^{\text{i}{\zeta }_2}{f}_2\\ d_3& {\text{e}}^{\text{i}{\zeta }_1}{e}_3& {\text{e}}^{\text{i}{\zeta }_2}{f}_3\end{array}\right]=\left[\begin{array}{ccc}{d}_1& e_1^{\prime }& f_1^{\prime }\\ d_2& e_2^{\prime }& f_2^{\prime }\\ 0& e_3^{\prime }& f_3^{\prime }\end{array}\right]$$

is a unitary where e′₁, e′₂, f′₁, f′₂ are real, e′₃, f′₃ are complex, ζ₁, ζ₂ ∈ [0, 2π]. Then from (A 19), we can obtain e₃ = f₃ = 0 or d₃ = 0.

If e₃ = f₃ = 0 then W in (A 1) is equivalent to the case $V=1\oplus G$ in lemma 3.1. Since if I₆, B₁, B₂ and B₃ are four MUBs in six-dimensional system then there exists $B_1^{†}$ such that I₆, B₁, B₂ and B₃ become $B_1^{†}$, I₆, $B_1^{†}{B}_2$ and $B_1^{†}{B}_3$, respectively. So, $B_1^{†}$, I₆, $B_1^{†}{B}_2$ and $B_1^{†}{B}_3$ are still four MUBs. Hence if e₃ = f₃ = 0 then ${\mathbb{H}}_3^{†}$ is equivalent to the case $V=1\oplus G$ in lemma 3.1.

If d₃ = 0, then from (A 14), $W_3{W}_3^{†}=W_3^{†}{W}_3=I$, d₁ and d₂ are non-negative and real numbers, one can obtain

$$d_1=\sqrt{\frac{\cos 2{\alpha }_2}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}\text{and}{d}_2=\sqrt{\frac{-\text{cos}\hspace{0.17em}2{\alpha }_1}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}.$$ B 7

Then from (A 19), we know

$$\frac{e_1}{e_2}=-\sqrt{\frac{-\text{cos}\hspace{0.17em}2{\alpha }_1}{\cos 2{\alpha }_2}}\text{and}\frac{f_1}{f_2}=-\sqrt{\frac{-\text{cos}\hspace{0.17em}2{\alpha }_1}{\cos 2{\alpha }_2}}.$$ B 8

And (A 4) becomes

$$\begin{array}{r}|d_1{g}_{21}\cos {\alpha }_1+d_2{g}_{23}\cos {\alpha }_2|=|e_1^{\prime }{g}_{21}\cos {\alpha }_1+e_2^{\prime }{g}_{23}\cos {\alpha }_2|=|\hspace{0.17em}{f}_1^{\prime }{g}_{21}\cos {\alpha }_1+f_2^{\prime }{g}_{23}\cos {\alpha }_2|=\frac{1}{\sqrt{6}}.\end{array}$$ B 9

Equations (B 7)–(B 9) imply that (e′₁, e′₂) = ±(f′₁, f′₂). Let (e′₁, e′₂) = (f′₁, f′₂), so W₃D₃′ becomes

$$\left[\begin{array}{ccc}{d}_1& e_1^{\prime }& e_1^{\prime }\\ d_2& e_2^{\prime }& e_2^{\prime }\\ 0& e_3^{\prime }& f_3^{\prime }\end{array}\right].$$

Since the second column and the third column vectors of W₃D₃′ are orthorgonal, we know e′₃ = −f′₃, $|e_3^{\prime }|=|e_3|=1/\sqrt{2}$, $e_1^{\prime 2}+e_2^{\prime 2}=1/2$. Then from (A 17) and (B 8), we have

$$|e_1^{\prime }|=\frac{1}{\sqrt{2}}\sqrt{\frac{-\text{cos}\hspace{0.17em}2{\alpha }_1}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}\text{and}|e_2^{\prime }|=\frac{1}{\sqrt{2}}\sqrt{\frac{\cos 2{\alpha }_2}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}.$$ B 10

So

$$W_3{D}_3^{\prime }=\left[\begin{array}{ccc}{d}_1& \frac{1}{\sqrt{2}}{d}_2& \frac{1}{\sqrt{2}}{d}_2\\ d_2& -\frac{1}{\sqrt{2}}{d}_1& -\frac{1}{\sqrt{2}}{d}_1\\ 0& \frac{1}{\sqrt{2}}{\text{e}}^{\text{i}{\zeta }_1}& -\frac{1}{\sqrt{2}}{\text{e}}^{\text{i}{\zeta }_1}\end{array}\right],$$

where $d_1=\sqrt{\cos 2{\alpha }_2/(\cos 2{\alpha }_2-\cos 2{\alpha }_1)}$, $d_2=\sqrt{-\text{cos}\hspace{0.17em}2{\alpha }_1/(\cos 2{\alpha }_2-\cos 2{\alpha }_1)}$, ζ₁ ∈ [0, 2π].

From (A 25), we can obtain that G₂ is equivalent to $\left[\begin{array}{cc}{d}_1& d_2{\text{e}}^{\text{i}{\vartheta }_1}\\ d_2& -d_1{\text{e}}^{\text{i}{\vartheta }_1}\end{array}\right]$, where $d_1=\sqrt{\cos 2{\alpha }_2/(\cos 2{\alpha }_2-\cos 2{\alpha }_1)}$, $d_2=\sqrt{-\text{cos}\hspace{0.17em}2{\alpha }_1/(\cos 2{\alpha }_2-\cos 2{\alpha }_1)}$, ${\vartheta }_1\in [0,2\pi )$. Hence

$$\begin{array}{rl}{V}^{″}& =(p_1\oplus G_1)\mathrm{diag}({\mathrm{e}}^{-\mathrm{i}{\vartheta }_1},{\mathrm{e}}^{-\mathrm{i}{\vartheta }_1},1)\mathrm{diag}({\mathrm{e}}^{\mathrm{i}{\vartheta }_1},{\mathrm{e}}^{\mathrm{i}{\vartheta }_1},1)(G_2\oplus p_2)\\ & =\left[\begin{array}{ccc}{d}_1& d_2& 0\\ \frac{1}{\sqrt{2}}{d}_2& -\frac{1}{\sqrt{2}}{d}_1& \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}{d}_2& -\frac{1}{\sqrt{2}}{d}_1& -\frac{1}{\sqrt{2}}\end{array}\right]\cdot \left[\begin{array}{ccc}{\mathrm{e}}^{-\mathrm{i}{\vartheta }_1}& 0& 0\\ 0& 1& 0\\ 0& 0& {\mathrm{e}}^{\mathrm{i}\theta }\end{array}\right]\end{array}$$ B 11

and

$$\begin{array}{rl}W& =(W_3{D}_3^{\prime })D_3^{\prime -1}{D}_3\\ & =\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& {\text{e}}^{\text{i}{\zeta }_1}\end{array}\right]\cdot \left[\begin{array}{ccc}{d}_1& \frac{1}{\sqrt{2}}{d}_2& \frac{1}{\sqrt{2}}{d}_2\\ d_2& -\frac{1}{\sqrt{2}}{d}_1& -\frac{1}{\sqrt{2}}{d}_1\\ 0& \frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\end{array}\right]\cdot \left[\begin{array}{ccc}1& 0& 0\\ 0& {\text{e}}^{-\text{i}{\zeta }_1}& 0\\ 0& 0& {\text{e}}^{-\text{i}{\zeta }_1}\end{array}\right]\cdot \left[\begin{array}{ccc}1& 0& 0\\ 0& {\text{e}}^{\text{i}{\delta }_1}& 0\\ 0& 0& {\text{e}}^{\text{i}{\delta }_2}\end{array}\right],\end{array}$$ B 12

where $d_1=\sqrt{\cos 2{\alpha }_2/(\cos 2{\alpha }_2-\cos 2{\alpha }_1)}$, $d_2=\sqrt{-\text{cos}\hspace{0.17em}2{\alpha }_1/(\cos 2{\alpha }_2-\cos 2{\alpha }_1)}$, ${\vartheta }_1$, θ, ζ₁, δ₁, δ₂ ∈ [0, 2π). Then (A 1) is equivalent to

$$\begin{array}{rl}{\mathbb{H}}_3& =(I_2\otimes \left(\left[\begin{array}{ccc}{d}_1& d_2& 0\\ \frac{1}{\sqrt{2}}{d}_2& -\frac{1}{\sqrt{2}}{d}_1& \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}{d}_2& -\frac{1}{\sqrt{2}}{d}_1& -\frac{1}{\sqrt{2}}\end{array}\right]\left[\begin{array}{ccc}{\mathrm{e}}^{-\mathrm{i}{\vartheta }_1}& 0& 0\\ 0& 1& 0\\ 0& 0& {\mathrm{e}}^{\mathrm{i}{\vartheta }_2}\end{array}\right]\right))\\ & \cdot \left(\begin{array}{cccccc}\cos {\alpha }_1& 0& 0& {\mathrm{e}}^{\mathrm{i}{\gamma }_1}\sin {\alpha }_1& 0& 0\\ 0& \cos {\alpha }_2& 0& 0& {\mathrm{e}}^{\mathrm{i}{\gamma }_2}\sin {\alpha }_2& 0\\ 0& 0& \cos {\alpha }_3& 0& 0& {\mathrm{e}}^{\mathrm{i}{\gamma }_3}\sin {\alpha }_3\\ {\mathrm{e}}^{\mathrm{i}{\beta }_1}\sin {\alpha }_1& 0& 0& -{\mathrm{e}}^{\mathrm{i}({\beta }_1+{\gamma }_1)}\cos {\alpha }_1& 0& 0\\ 0& {\mathrm{e}}^{\mathrm{i}{\beta }_2}\sin {\alpha }_2& 0& 0& -{\mathrm{e}}^{\mathrm{i}({\beta }_2+{\gamma }_2)}\cos {\alpha }_2& 0\\ 0& 0& {\mathrm{e}}^{\mathrm{i}{\beta }_3}\sin {\alpha }_3& 0& 0& -{\mathrm{e}}^{\mathrm{i}({\beta }_3+{\gamma }_3)}\cos {\alpha }_3\end{array}\right)\\ & \cdot (I_2\otimes \left[\begin{array}{ccc}{d}_1& \frac{1}{\sqrt{2}}{d}_2& \frac{1}{\sqrt{2}}{d}_2\\ d_2& -\frac{1}{\sqrt{2}}{d}_1& -\frac{1}{\sqrt{2}}{d}_1\\ 0& \frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\end{array}\right]),\end{array}$$ B 13

where ${\vartheta }_1$, ${\vartheta }_2\in [0,2\pi ]$ are the functions of α₁, α₂. Next, we compute ${\vartheta }_1$, ${\vartheta }_2$ such that (B 13) is a Schmidt-rank-three order-six CHM. In (B 13), $|h_{jk}|=1/\sqrt{6}$, so we have

$$\begin{array}{rl}& |d_1^2{\text{e}}^{-\text{i}{\vartheta }_1}\cos {\alpha }_1+d_2^2\cos {\alpha }_2|=\frac{1}{\sqrt{2}}|d_1{d}_2{\text{e}}^{-\text{i}{\vartheta }_1}\cos {\alpha }_1-d_1{d}_2\cos {\alpha }_2|=\frac{1}{\sqrt{6}},\end{array}$$ B 14

$$\begin{array}{rl}& \frac{1}{2}|d_2^2{\text{e}}^{-\text{i}{\vartheta }_1}\cos {\alpha }_1+d_1^2\cos {\alpha }_2+{\text{e}}^{\text{i}{\vartheta }_2}\cos {\alpha }_3|=\frac{1}{2}|d_2^2{\text{e}}^{-\text{i}{\vartheta }_1}\cos {\alpha }_1+d_1^2\cos {\alpha }_2-{\text{e}}^{\text{i}{\vartheta }_2}\cos {\alpha }_3|=\frac{1}{\sqrt{6}},\end{array}$$ B 15

$$\begin{array}{rl}& |d_1^2{\text{e}}^{\text{i}({\gamma }_1-{\vartheta }_1)}+d_2^2{\text{e}}^{\text{i}{\gamma }_2}\sin {\alpha }_2|=\frac{1}{\sqrt{2}}|d_1{d}_2{\text{e}}^{\text{i}({\gamma }_1-{\vartheta }_1)}\sin {\alpha }_1-d_1{d}_2{\text{e}}^{\text{i}{\gamma }_2}\sin {\alpha }_2|=\frac{1}{\sqrt{6}},\end{array}$$ B 16

$$\begin{array}{rl}& \frac{1}{2}|d_2^2{\text{e}}^{\text{i}({\gamma }_1-{\vartheta }_1)}+d_1^2{\text{e}}^{\text{i}{\gamma }_2}\sin {\alpha }_2+{\text{e}}^{\text{i}({\gamma }_3+{\vartheta }_2)}\sin {\alpha }_3|\\ & =\frac{1}{2}|d_2^2{\text{e}}^{\text{i}({\gamma }_1-{\vartheta }_1)}+d_1^2{\text{e}}^{\text{i}{\gamma }_2}\sin {\alpha }_2-{\text{e}}^{\text{i}({\gamma }_3+{\vartheta }_2)}\sin {\alpha }_3|=\frac{1}{\sqrt{6}},\end{array}$$ B 17

$$\begin{array}{rl}& |d_1^2{\text{e}}^{\text{i}({\beta }_1-{\vartheta }_1)}\sin {\alpha }_1+d_2^2{\text{e}}^{\text{i}{\beta }_2}\sin {\alpha }_2|\\ & =\frac{1}{\sqrt{2}}|d_1{d}_2{\text{e}}^{\text{i}({\beta }_1-{\vartheta }_1)}\sin {\alpha }_1-d_1{d}_2{\text{e}}^{\text{i}{\beta }_2}\sin {\alpha }_2|=\frac{1}{\sqrt{6}},\end{array}$$ B 18

$$\begin{array}{rl}& \frac{1}{2}|d_2^2{\text{e}}^{\text{i}({\beta }_1-{\vartheta }_1)}+d_1^2{\text{e}}^{\text{i}{\beta }_2}\sin {\alpha }_2+{\text{e}}^{\text{i}({\beta }_3+{\vartheta }_2)}\sin {\alpha }_3|\\ & =\frac{1}{2}|d_2^2{\text{e}}^{\text{i}({\beta }_1-{\vartheta }_1)}+d_1^2{\text{e}}^{\text{i}{\beta }_2}\sin {\alpha }_2-{\text{e}}^{\text{i}({\beta }_3+{\vartheta }_2)}\sin {\alpha }_3|=\frac{1}{\sqrt{6}},\end{array}$$ B 19

$$\begin{array}{rl}& |d_1^2{\text{e}}^{\text{i}({\beta }_1+{\gamma }_1-{\vartheta }_1)}+d_2^2{\text{e}}^{\text{i}({\beta }_2+{\gamma }_2)}\cos {\alpha }_2|\\ & =\frac{1}{\sqrt{2}}|d_1{d}_2{\text{e}}^{\text{i}({\beta }_1+{\gamma }_1-{\vartheta }_1)}\cos {\alpha }_1-d_1{d}_2{\text{e}}^{\text{i}({\beta }_2+{\gamma }_2)}\cos {\alpha }_2|=\frac{1}{\sqrt{6}}\end{array}$$ B 20

$$\begin{array}{rl}\mathrm{and}& \frac{1}{2}|d_2^2{\mathrm{e}}^{\mathrm{i}({\beta }_1+{\gamma }_1-{\vartheta }_1)}\cos {\alpha }_1+d_1^2{\mathrm{e}}^{\mathrm{i}({\beta }_2+{\gamma }_2)}\cos {\alpha }_2+{\mathrm{e}}^{\mathrm{i}({\beta }_3+{\gamma }_3+{\vartheta }_2)}\cos {\alpha }_3|\\ & =\frac{1}{2}|d_2^2{\mathrm{e}}^{\mathrm{i}({\beta }_1+{\gamma }_1-{\vartheta }_1)}\cos {\alpha }_1+d_1^2{\mathrm{e}}^{\mathrm{i}({\beta }_2+{\gamma }_2)}\cos {\alpha }_2-{\mathrm{e}}^{\mathrm{i}({\beta }_3+{\gamma }_3+{\vartheta }_2)}\cos {\alpha }_3|=\frac{1}{\sqrt{6}}.\end{array}$$ B 21

Simplifying (B 15) and (B 21), one can obtain

$$\cos ({\vartheta }_1+{\vartheta }_2)\cos {\alpha }_1\cos 2{\alpha }_1=\cos {\vartheta }_2\cos {\alpha }_2\cos 2{\alpha }_2$$ B 22

and

$$\begin{array}{rl}& \cos ({\vartheta }_1+{\vartheta }_2+{\beta }_3+{\gamma }_3-{\beta }_1-{\gamma }_1)\cos {\alpha }_1\cos 2{\alpha }_1\\ & =\cos ({\vartheta }_2+{\beta }_3+{\gamma }_3-{\beta }_2-{\gamma }_2)\cos {\alpha }_2\cos 2{\alpha }_2.\end{array}$$ B 23

Similarly, from (B 17) and (B 19), we have

$$\cos ({\vartheta }_1+{\vartheta }_2-{\gamma }_3-{\gamma }_1)\sin {\alpha }_1\cos 2{\alpha }_1=\cos ({\vartheta }_2+{\gamma }_3-{\gamma }_2)\sin {\alpha }_2\cos 2{\alpha }_2$$ B 24

and

$$\cos ({\vartheta }_1+{\vartheta }_2+{\beta }_3-{\beta }_1)\sin {\alpha }_1\cos 2{\alpha }_1=\cos ({\vartheta }_2+{\beta }_3-{\beta }_2)\sin {\alpha }_2\cos 2{\alpha }_2.$$ B 25

By comparing (B 14) and (B 20), (B 16) and (B 18), we obtain

$$\cos ({\vartheta }_1+{\beta }_2+{\gamma }_2-{\beta }_1-{\gamma }_1)=\cos {\vartheta }_1\text{and}\cos ({\vartheta }_1+{\gamma }_2-{\gamma }_1)=\cos ({\vartheta }_1+{\beta }_2-{\beta }_1).$$ B 26

They imply

$${\vartheta }_1+\frac{{\gamma }_2+{\beta }_2-{\gamma }_1-{\beta }_1}{2}=k_1\pi \text{or}\hspace{0.17em}\frac{{\gamma }_2+{\beta }_2-{\gamma }_1-{\beta }_1}{2}=k_2\pi$$ B 27

and

$${\vartheta }_1+\frac{{\gamma }_2+{\beta }_2-{\gamma }_1-{\beta }_1}{2}=k_1\pi \text{or}\hspace{0.17em}\frac{{\gamma }_2-{\gamma }_1-({\beta }_2-{\beta }_1)}{2}=k_3\pi ,$$ B 28

where k₁, k₂, $k_3\in \mathbb{Z}$. Thus in the following we have two subcases (i.a) and (i.b) in terms of (B 27) and (B 28).

• (i.a)
  If (γ₂ + β₂ − γ₁ − β₁)/2 = k₂π and (γ₂ − γ₁ − (β₂ − β₁))/2 = k₃π, then γ₂ − γ₁ = (k₂ + k₃)π, β₂ − β₁ = (k₂ − k₃)π, s = (k₁ − k₃)π. Then (B 22)–(B 25) can be simplified as

  $$\frac{\cos {\alpha }_2\cos 2{\alpha }_2}{\cos {\alpha }_1\cos 2{\alpha }_2}=\pm 1\text{and}\frac{\sin {\alpha }_2\cos 2{\alpha }_2}{\sin {\alpha }_1\cos 2{\alpha }_2}=\pm 1.$$ B 29

  So we have

  $$\frac{\cos {\alpha }_2\sin {\alpha }_1}{\cos {\alpha }_1\sin {\alpha }_2}=\pm 1.$$ B 30

  It implies that α₁ = α₂ or α₁ + α₂ = π. It is a contradiction with the condition α₁ ∈ (0, π/4), α₂ ∈ (π/4, π/2) or α₁ ∈ (π/4, π/2), α₂ ∈ (0, π/4).
• (i.b)
  If ${\vartheta }_1+({\gamma }_2+{\beta }_2-{\gamma }_1-{\beta }_1)/2=k_1\pi$. From (B 14) to (B 22), we have

  $$\cos {\vartheta }_1=\frac{-{(\cos 2{\alpha }_1-\cos 2{\alpha }_2)}^2+6{\cos }^{2}2{\alpha }_1{\cos }^2{\alpha }_2+6{\cos }^2{\alpha }_1{\cos }^{2}2{\alpha }_2}{12\cos {\alpha }_1\cos 2{\alpha }_1\cos {\alpha }_2\cos 2{\alpha }_2}$$ B 31

  and

  $${\cos }^2{\vartheta }_2=\frac{{(-2+\cos 2{\alpha }_1)}^2{\sec }^2{\alpha }_2+8\cos 2{\alpha }_1\sec 2{\alpha }_2(4+\cos 2{\alpha }_1(2+\sec 2{\alpha }_2))}{24(1+3\cos 2{\alpha }_1+3\cos 2{\alpha }_2)}.$$ B 32

  Note that in (B 13),

  $$(I_2\otimes \left[\begin{array}{ccc}{\text{e}}^{-\text{i}{\vartheta }_1}& 0& 0\\ 0& 1& 0\\ 0& 0& {\text{e}}^{\text{i}{\vartheta }_2}\end{array}\right])\cdot H$$

  is unitary. We have $1+{\text{e}}^{2\text{i}({\beta }_1+{\gamma }_1)}=1+{\text{e}}^{-2\text{i}({\beta }_2+{\gamma }_2)}=1+{\text{e}}^{2\text{i}({\beta }_3+{\gamma }_3)}$. It implies that β₁ + γ₁ = (m₁ + 1/2)π, β₂ + γ₂ = (m₂ + 1/2)π, β₃ + γ₃ = (m₃ + 1/2)π, where m₁, m₂, $m_3\in \mathbb{Z}$. Together with (B 22)–(B 25), one can obtain

  $$\begin{array}{rl}{\vartheta }_1& =k_1\pi +\frac{{\gamma }_1+{\beta }_1-{\gamma }_2-{\beta }_2}{2}=\frac{1}{2}(2k_1+m_1-m_2)\pi \end{array}$$ B 33

  $$\begin{array}{rl}{\vartheta }_2& =\frac{1}{2}({\beta }_2+{\gamma }_2-{\beta }_3-{\gamma }_3-2k_1\pi +k_2\pi )=\frac{1}{2}(-2k_1+k_2+m_2-m_3)\pi .\end{array}$$ B 34

Hence ${\vartheta }_1+{\vartheta }_2=(1/2)(k_2+m_1-m_3)\pi$. It means that $\cos ({\vartheta }_1+{\vartheta }_2)/\cos {\vartheta }_2=\cos {\alpha }_2\cos 2{\alpha }_2/\cos {\alpha }_1\cos 2{\alpha }_1$ is equal to −1, 0 or 1. So $(\cos ({\vartheta }_1+{\vartheta }_2),\cos {\vartheta }_2)=(1,-1),(-1,1)$ or (0, 0). In the first or second case, we have $\cos {\vartheta }_1=-1$. From the expressions of $\cos {\vartheta }_1$ and ${\cos }^2{\vartheta }_2$ in (B 31)–(B 32), one can obtain that $(\cos {\vartheta }_1,{\cos }^2{\vartheta }_2)=(-1,1)$ has no solutions with α₁ ∈ (0, π/4), α₂ ∈ (π/4, π/2) or α₁ ∈ (π/4, π/2), α₂ ∈ (0, π/4). Similarly, the third case is excluded by $(\cos {\vartheta }_1,{\cos }^2{\vartheta }_2)=(0,0)$ from the expressions in (B 31)–(B 32).

(ii) If β₂ − β₁ − (γ₂ − γ₁) = 0 and β₂ + γ₂ − β₁ − γ₁ = 0 then β₂ = β₁, γ₂ = γ₁. Then (1) is equivalent to

$$\begin{array}{rl}& (I_2\otimes \left[\begin{array}{ccc}{\mathrm{e}}^{\mathrm{i}{\vartheta }_2}\sqrt{\frac{\cos 2{\alpha }_2}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}& -{\mathrm{e}}^{\mathrm{i}{\vartheta }_2}\sqrt{\frac{-\text{cos}\hspace{0.17em}2{\alpha }_1}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}& 0\\ \frac{1}{\sqrt{2}}\sqrt{\frac{-\text{cos}\hspace{0.17em}2{\alpha }_1}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}& \sqrt{\frac{\cos 2{\alpha }_2}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}& \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\sqrt{\frac{-\text{cos}\hspace{0.17em}2{\alpha }_1}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}& \sqrt{\frac{\cos 2{\alpha }_2}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}& -\frac{1}{\sqrt{2}}\end{array}\right])\\ & \cdot \left(\begin{array}{cccccc}{\mathrm{e}}^{-\mathrm{i}{\vartheta }_2}\cos {\alpha }_1& 0& 0& {\mathrm{e}}^{\mathrm{i}({\gamma }_1-{\vartheta }_2)}\sin {\alpha }_1& 0& 0\\ 0& \cos {\alpha }_2& 0& 0& {\mathrm{e}}^{\mathrm{i}{\gamma }_1}\sin {\alpha }_2& 0\\ 0& 0& {\mathrm{e}}^{\mathrm{i}\theta }\cos {\alpha }_3& 0& 0& {\mathrm{e}}^{\mathrm{i}({\gamma }_3+\theta )}\sin {\alpha }_3\\ {\mathrm{e}}^{\mathrm{i}({\beta }_1-{\vartheta }_2)}\sin {\alpha }_1& 0& 0& -{\mathrm{e}}^{\mathrm{i}({\beta }_1+{\gamma }_1-{\vartheta }_2)}\cos {\alpha }_1& 0& 0\\ 0& {\mathrm{e}}^{\mathrm{i}{\beta }_1}\sin {\alpha }_2& 0& 0& -{\mathrm{e}}^{\mathrm{i}({\beta }_1+{\gamma }_1)}\cos {\alpha }_2& 0\\ 0& 0& {\mathrm{e}}^{\mathrm{i}({\beta }_3+\theta )}\sin {\alpha }_3& 0& 0& -{\mathrm{e}}^{\mathrm{i}({\beta }_3+{\gamma }_3+\theta )}\cos {\alpha }_3\end{array}\right)\\ & \cdot (I_2\otimes \left[\begin{array}{ccc}{d}_1& e_1& f_1\\ d_2& e_2& f_2\\ d_3& e_3& f_3\end{array}\right]).\end{array}$$ B 35

In (B 35), let $|h_{jk}|=\frac{1}{\sqrt{6}}$. One can obtain

$$\begin{array}{rl}& |-d_2{\text{e}}^{\text{i}{\vartheta }_2}\cos {\alpha }_2\sqrt{\frac{-\text{cos}\hspace{0.17em}2{\alpha }_1}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}+d_1\cos {\alpha }_1\sqrt{\frac{\cos 2{\alpha }_2}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}|=\frac{1}{\sqrt{6}},\end{array}$$ B 36

$$\begin{array}{rl}& \left|{\text{e}}^{\text{i}{\beta }_1}(d_1\sqrt{\frac{\cos 2{\alpha }_2}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}\sin {\alpha }_1-d_2{\text{e}}^{\text{i}{\vartheta }_2}\sqrt{\frac{-\text{cos}\hspace{0.17em}2{\alpha }_1}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}\sin {\alpha }_2)\right|=\frac{1}{\sqrt{6}},\end{array}$$ B 37

$$\begin{array}{rl}& |-{\text{e}}^{\text{i}{\vartheta }_2}{e}_2\cos {\alpha }_2\sqrt{\frac{-\text{cos}\hspace{0.17em}2{\alpha }_1}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}+e_1\cos {\alpha }_1\sqrt{\frac{\cos 2{\alpha }_2}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}|=\frac{1}{\sqrt{6}}\end{array}$$ B 38

$$\begin{array}{rl}\text{and}& \left|{\text{e}}^{\text{i}{\beta }_1}(e_1\sqrt{\frac{\cos 2{\alpha }_2}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}\sin {\alpha }_1-{\text{e}}^{\text{i}{\vartheta }_2}{e}_2\sqrt{\frac{-\text{cos}\hspace{0.17em}2{\alpha }_1}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}\sin {\alpha }_2)\right|=\frac{1}{\sqrt{6}}.\end{array}$$ B 39

Simplifying (B 36) and (B 37), we have

$$\begin{array}{rl}& d_1^2\frac{\cos 2{\alpha }_2}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}+d_2^2\frac{-\text{cos}\hspace{0.17em}2{\alpha }_1}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}\\ & -2d_1{d}_2\sqrt{\frac{-\text{cos}\hspace{0.17em}2{\alpha }_1}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}\sqrt{\frac{\cos 2{\alpha }_2}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}\cos ({\alpha }_1-{\alpha }_2)\cos {\vartheta }_2=\frac{1}{3}\end{array}$$ B 40

and

$$\begin{array}{rl}& d_1^2\frac{\cos 2{\alpha }_1\cos 2{\alpha }_2}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}+d_2^2\frac{-\text{cos}\hspace{0.17em}2{\alpha }_1\cos 2{\alpha }_2}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}\\ & -2d_1{d}_2\sqrt{\frac{-\text{cos}\hspace{0.17em}2{\alpha }_1}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}\sqrt{\frac{\cos 2{\alpha }_2}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}\cos ({\alpha }_1+{\alpha }_2)\cos {\vartheta }_2=0.\end{array}$$ B 41

By solving the above equations, we obtain

$$\cos {\vartheta }_2=\frac{\cos 2{\alpha }_2-\cos 2{\alpha }_1+3d_2^2\cos 2{\alpha }_1-3d_1^2\cos 2{\alpha }_2}{6d_1{d}_2\cos ({\alpha }_1-{\alpha }_2)(\cos 2{\alpha }_1-\cos 2{\alpha }_2)\sqrt{\frac{-\text{cos}\hspace{0.17em}2{\alpha }_1}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}\sqrt{\frac{\cos 2{\alpha }_2}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}}$$ B 42

and

$$\cos {\vartheta }_2=\frac{(d_2^2-d_1^2)\cos 2{\alpha }_1\cos 2{\alpha }_2}{2d_1{d}_2\cos ({\alpha }_1+{\alpha }_2)(\cos 2{\alpha }_1-\cos 2{\alpha }_2)\sqrt{\frac{-\text{cos}\hspace{0.17em}2{\alpha }_1}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}\sqrt{\frac{\cos 2{\alpha }_2}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}}}.$$ B 43

They imply that

$$d_1^2=\frac{3d_2^2\sin 2({\alpha }_1-{\alpha }_2)+(2-3d_2^2)\sin 2({\alpha }_1+{\alpha }_2)}{6\sin 2{\alpha }_1\cos 2{\alpha }_2},$$ B 44

so we know $(3d_1^2-1)\sin 2{\alpha }_1\cos 2{\alpha }_2+(3d_2^2-1)\cos 2{\alpha }_1\sin 2{\alpha }_2=0$. Note that α₁ ∈ (0, π/4), α₂ ∈ (π/4, π/2) or α₁ ∈ (π/4, π/2), α₂ ∈ (0, π/4), d₁, d₂, d₃ are non-negative and real numbers. So $d_1=d_2=d_3=\sqrt{3}/3$. Then (B 42)–(B 43) imply that $\cos {\vartheta }_2=0$. Hence ${\vartheta }_2=\pi /2$ or 3π/2.

Similarly, simplifying equations (B 37)–(B 39), we have

$$\begin{array}{rl}& |e_1{|}^2\frac{\cos 2{\alpha }_2}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}+|e_2{|}^2\frac{-\text{cos}\hspace{0.17em}2{\alpha }_1}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}\\ & +\frac{\sqrt{-\text{cos}\hspace{0.17em}2{\alpha }_1\cos 2{\alpha }_2}}{|\cos 2{\alpha }_2-\cos 2{\alpha }_1|}[{(-{\text{e}}^{\text{i}{\vartheta }_2}{e}_2)}^{\ast }{e}_1+(-{\text{e}}^{\text{i}{\vartheta }_2}{e}_2)e_1^{\ast }]\cos ({\alpha }_1-{\alpha }_2)=\frac{1}{3}\end{array}$$ B 45

and

$$\begin{array}{rl}& |e_1{|}^2\frac{\cos 2{\alpha }_1\cos 2{\alpha }_2}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}+|e_2{|}^2\frac{-\text{cos}\hspace{0.17em}2{\alpha }_1\cos 2{\alpha }_2}{\cos 2{\alpha }_2-\cos 2{\alpha }_1}\\ & +\frac{\sqrt{-\text{cos}\hspace{0.17em}2{\alpha }_1\cos 2{\alpha }_2}}{|\cos 2{\alpha }_2-\cos 2{\alpha }_1|}[{(-{\text{e}}^{\text{i}{\vartheta }_2}{e}_2)}^{\ast }{e}_1+(-{\text{e}}^{\text{i}{\vartheta }_2}{e}_2)e_1^{\ast }]\cos ({\alpha }_1+{\alpha }_2)=0.\end{array}$$ B 46

If ${\vartheta }_2=\pi /2$ then $-{\text{e}}^{\text{i}{\vartheta }_2}=-\text{i}$. Then (B 46) implies that

$$\sqrt{-\text{cos}\hspace{0.17em}2{\alpha }_1\cos 2{\alpha }_2}(|e_1{|}^2-|e_2{|}^2)\pm (e_1{e}_2^{\ast }-e_1^{\ast }{e}_2)\cos ({\alpha }_1+{\alpha }_2)i=0.$$ B 47

Since α₁ ∈ (0, π/4), α₂ ∈ (π/4, π/2) or α₁ ∈ (π/4, π/2), α₂ ∈ (0, π/4), we know |e₁| = |e₂|, e₁ e*₂ − e*₁ e₂ = 0. Applying them to (B 45), we have $|e_1|=|e_2|=|e_3|=\sqrt{3}/3$. Then suppose e₁ = a + bi. Since e₁ e*₂ − e*₁ e₂ = 0, we obtain e₂ = e₁ = a + bi. And (A 19) implies that e₃ = −2a − 2bi. It is a contradiction with $|e_1|=|e_2|=|e_3|=\sqrt{3}/3$. Hence W₃ does not exist with α₁ ∈ (0, π/4), α₂ ∈ (π/4, π/2) or α₁ ∈ (π/4, π/2), α₂ ∈ (0, π/4).

If ${\vartheta }_2=3\pi /2$, one can similarly show that W₃ does not exist with α₁ ∈ (0, π/4), α₂ ∈ (π/4, π/2) or α₁ ∈ (π/4, π/2), α₂ ∈ (0, π/4). ▪

Data accessibility

The code in this paper is provided in CODE.rar as electronic supplementary material.

Authors' contributions

L.C., M.H. and Y.S. designed and coordinated the overall study; L.C., M.H. and Y.S. wrote the manuscript; L.C. and Y.S. provided valuable suggestions for the manuscript; Y.S. and M.H. revised the manuscript; L.C. obtained funding and is responsible for this article. All authors read and approved the manuscript.

Competing interests

We declare we have no competing interest.

Funding

Authors were supported by the NNSF of China(grant no. 11871089), and the Fundamental Research Funds for the Central Universities(grant nos KG12080401 and ZG216S1902).