Novel methods to construct nonlocal sets of orthogonal product states in an arbitrary bipartite high-dimensional system

Abstract

Nonlocal sets of orthogonal product states (OPSs) are widely used in quantum protocols owing to their good property. In [Phys. Rev. A 101, 062329 (2020)], the authors constructed some unextendible product bases in ${\mathbb {C}}^{m} \otimes {\mathbb {C}}^{n}$ quantum system for $n\ge m\ge 3$. We find that a subset of their unextendible product basis (UPB) cannot be perfectly distinguished by local operations and classical communication (LOCC). We give a proof for the nonlocality of the subset with Vandermonde determinant and Kramer’s rule. Meanwhile, we give a novel method to construct a nonlocal set with only $2(m+n)-4$ OPSs in ${\mathbb {C}}^{m} \otimes {\mathbb {C}}^{n}$ quantum system for $m\ge 3$ and $n\ge 3$. By comparing the number of OPSs in our nonlocal set with that of the existing results, we know that $2(m+n)-4$ is the minimum number of OPSs to construct a nonlocal and completable set in ${\mathbb {C}}^{m} \otimes {\mathbb {C}}^{n}$ quantum system so far. This means that we give the minimum number of elements to construct a completable and nonlocal set in an arbitrary given space. Our work is of great help to understand the structure and classification of locally indistinguishable OPSs in an arbitrary bipartite high-dimensional system.

Appendices

Appendix A: The process to solve the equations involved in the proof of Theorem 1.

Because $(M_{k}\otimes I_{4\times 4})|\phi _{1}\rangle$ is orthogonal to $(M_{k}\otimes I_{4\times 4})|\phi _{10}\rangle$, $(M_{k}\otimes I_{4\times 4})|\phi _{11}\rangle$ and $(M_{k}\otimes I_{4\times 4})|\phi _{12}\rangle$, i.e.,

$$\begin{aligned} \left\{ \begin{aligned} \langle \phi _{1}|(M_{k}^{\dag }M_{k}\otimes I_{4\times 4})|\phi _{10}\rangle =0\\ \langle \phi _{1}|(M_{k}^{\dag }M_{k}\otimes I_{4\times 4})|\phi _{11}\rangle =0\\ \langle \phi _{1}|(M_{k}^{\dag }M_{k}\otimes I_{4\times 4})|\phi _{12}\rangle =0 \end{aligned} \right. \end{aligned}$$

and

$$\begin{aligned} \left\{ \begin{aligned} \langle \phi _{10}|(M_{k}^{\dag }M_{k}\otimes I_{4\times 4})|\phi _{1}\rangle =0\\ \langle \phi _{11}|(M_{k}^{\dag }M_{k}\otimes I_{4\times 4})|\phi _{1}\rangle =0\\ \langle \phi _{12}|(M_{k}^{\dag }M_{k}\otimes I_{4\times 4})|\phi _{1}\rangle =0 \end{aligned} \right. , \end{aligned}$$

we get two systems of linear equations

$$\begin{aligned} \left\{ \begin{aligned} a_{01}^{k}+a_{02}^{k}+a_{03}^{k}=0\qquad \quad \,\,\\ a_{01}^{k}+\omega a_{02}^{k}+\omega ^{2}a_{03}^{k}=0\quad \,\,\,\,\\ a_{01}^{k}+\omega ^{2}a_{02}^{k}+(\omega ^{2})^{2}a_{03}^{k}=0 \end{aligned} \right. \end{aligned}$$

(19)

and

$$\begin{aligned} \left\{ \begin{aligned} a_{10}^{k}+a_{20}^{k}+a_{30}^{k}=0\qquad \quad \,\,\\ a_{10}^{k}+{\overline{\omega }} a_{20}^{k}+{\overline{\omega }}^{2}a_{30}^{k}=0\quad \,\,\,\,\\ a_{10}^{k}+{\overline{\omega }}^{2}a_{20}^{k}+({\overline{\omega }}^{2})^{2}a_{30}^{k}=0 \end{aligned} \right. , \end{aligned}$$

(20)

where ${\overline{\omega }}$ is the conjugate complex number of $\omega$.

By Lemmas 1–3, we get the unique solution of Eq. (19),

$$\begin{aligned} \begin{aligned} a_{01}^{k}=a_{02}^{k}=a_{03}^{k}=0\\ \end{aligned} \end{aligned}$$

(21)

and the unique solution of Eq. (20),

$$\begin{aligned} \begin{aligned} a_{10}^{k}=a_{20}^{k}=a_{30}^{k}=0. \end{aligned} \end{aligned}$$

(22)

Similarly, we can get two systems of linear equations

$$\begin{aligned} \left\{ \begin{aligned} a_{30}^{k}+a_{31}^{k}+a_{32}^{k}=0\qquad \quad \,\,\\ a_{30}^{k}+\omega a_{31}^{k}+\omega ^{2}a_{32}^{k}=0\quad \,\,\,\,\\ a_{30}^{k}+\omega ^{2}a_{31}^{k}+(\omega ^{2})^{2}a_{32}^{k}=0 \end{aligned} \right. \end{aligned}$$

(23)

and

$$\begin{aligned} \left\{ \begin{aligned} a_{03}^{k}+a_{13}^{k}+a_{23}^{k}=0\qquad \quad \,\,\\ a_{03}^{k}+{\overline{\omega }} a_{13}^{k}+{\overline{\omega }}^{2}a_{23}^{k}=0\quad \,\,\,\,\\ a_{03}^{k}+{\overline{\omega }}^{2}a_{13}^{k}+({\overline{\omega }}^{2})^{2}a_{23}^{k}=0 \end{aligned} \right. \end{aligned}$$

(24)

since $(M_{k}\otimes I_{4\times 4})|\phi _{7}\rangle$ is orthogonal to $(M_{k}\otimes I_{4\times 4})|\phi _{4}\rangle$, $(M_{k}\otimes I_{4\times 4})|\phi _{5}\rangle$ and $(M_{k}\otimes I_{4\times 4})|\phi _{6}\rangle$. By Lemmas 1–3, we get the unique solution of Eq. (23),

$$\begin{aligned} \begin{aligned} a_{30}^{k}=a_{31}^{k}=a_{32}^{k}=0 \end{aligned} \end{aligned}$$

(25)

and the unique solution of Eq. (24),

$$\begin{aligned} \begin{aligned} a_{03}^{k}=a_{13}^{k}=a_{23}^{k}=0. \end{aligned} \end{aligned}$$

(26)

Because these three states $(M_{k}\otimes I_{4\times 4})|\phi _{4}\rangle$, $(M_{k}\otimes I_{4\times 4})|\phi _{5}\rangle$ and $(M_{k}\otimes I_{4\times 4})|\phi _{6}\rangle$ are mutually orthogonal, we have

$$\begin{aligned}&\left\{ \begin{aligned} \langle \phi _{4}|(M_{k}^{\dag }M_{k}\otimes I_{4\times 4})|\phi _{5}\rangle =0\\ \langle \phi _{4}|(M_{k}^{\dag }M_{k}\otimes I_{4\times 4})|\phi _{6}\rangle =0 \end{aligned} \right. , \\&\left\{ \begin{aligned} \langle \phi _{5}|(M_{k}^{\dag }M_{k}\otimes I_{4\times 4})|\phi _{4}\rangle =0\\ \langle \phi _{5}|(M_{k}^{\dag }M_{k}\otimes I_{4\times 4})|\phi _{6}\rangle =0 \end{aligned} \right. \end{aligned}$$

and

$$\begin{aligned} \left\{ \begin{aligned} \langle \phi _{6}|(M_{k}^{\dag }M_{k}\otimes I_{4\times 4})|\phi _{4}\rangle =0\\ \langle \phi _{6}|(M_{k}^{\dag }M_{k}\otimes I_{4\times 4})|\phi _{5}\rangle =0 \end{aligned} \right. . \end{aligned}$$

That is,

$$\begin{aligned}&\left\{ \begin{aligned} \sum _{p=1}^{2}\left( \omega ^{p}\sum _{j=0}^{2}a_{jp}^{k}\right) =-\sum _{j=0}^{2}a_{j0}^{k}\quad \,\,\\ \sum _{p=1}^{2}\left[ (\omega ^{2})^{p}\sum _{j=0}^{2}a_{jp}^{k}\right] =-\sum _{j=0}^{2}a_{j0}^{k} \end{aligned} \right. , \end{aligned}$$

(27)

$$\begin{aligned}&\left\{ \begin{aligned} \sum _{p=1}^{2}\sum _{j=0}^{2}{\overline{\omega }}^{j}a_{jp}^{k} =-\sum _{j=0}^{2}{\overline{\omega }}^{j}a_{j0}^{k}\qquad \quad \\ \sum _{p=1}^{2}\left[ (\omega ^{2})^{p}\sum _{j=0}^{2}{\overline{\omega }}^{j}a_{jp}^{k}\right] =-\sum _{j=0}^{2}{\overline{\omega }}^{j}a_{j0}^{k}\\ \end{aligned} \right. , \end{aligned}$$

(28)

and

$$\begin{aligned} \left\{ \begin{aligned} \sum _{p=1}^{2}\sum _{j=0}^{2}({\overline{\omega }}^{2})^{j}a_{jp}^{k} =-\sum _{j=0}^{2}({\overline{\omega }}^{2})^{j}a_{j0}^{k}\quad \,\,\,\\ \sum _{p=1}^{2}\left[ \omega ^{p}\sum _{j=0}^{2}({\overline{\omega }}^{2})^{j}a_{jp}^{k}\right] =-\sum _{j=0}^{2}({\overline{\omega }}^{2})^{j}a_{j0}^{k}\\ \end{aligned} \right. , \end{aligned}$$

(29)

where ${\overline{\omega }}$ is the conjugate complex number of $\omega$.

For simplicity, we denote the coefficient determinants of Eqs. (27), (28) and (29) as $D_{1}$, $D_{2}$ and $D_{3}$, respectively. Since

$$\begin{aligned}&D_{1}= \left| \begin{array}{cc} \omega &{}\omega ^{2} \\ \omega ^{2} &{}(\omega ^{2})^{2} \\ \end{array} \right| =\omega ^{3}\left| \begin{array}{cc} 1 &{}\omega \\ 1 &{}\omega ^{2} \\ \end{array} \right| =\omega ^{2}-\omega \ne 0,\\ \\&D_{2}= \left| \begin{array}{cc} 1 &{}1 \\ \omega ^{2} &{}(\omega ^{2})^{2} \\ \end{array} \right| =(\omega ^{2})^{2}-\omega ^{2} \ne 0, \qquad \qquad \\&D_{3}= \left| \begin{array}{cc} 1 &{}1 \\ \omega &{}\omega ^{2} \\ \end{array} \right| =\omega ^{2}-\omega \ne 0, \end{aligned}$$

we obtain the unique solutions of Eqs. (27), (28) and (29), respectively, i.e.,

$$\begin{aligned}&\left\{ \begin{aligned} \sum _{j=0}^{2}a_{j1}^{k}=\sum _{j=0}^{2}a_{j0}^{k}\\ \sum _{j=0}^{2}a_{j2}^{k}=\sum _{j=0}^{2}a_{j0}^{k} \end{aligned} \right. , \end{aligned}$$

(30)

$$\begin{aligned}&\left\{ \begin{aligned} \sum _{j=0}^{2}{\overline{\omega }}^{j}a_{j1}^{k}= {\overline{\omega }}\sum _{j=0}^{2}{\overline{\omega }}^{j}a_{j0}^{k}\\ \sum _{j=0}^{2}{\overline{\omega }}^{j}a_{j2}^{k}= {\overline{\omega }}^{2}\sum _{j=0}^{2}{\overline{\omega }}^{j}a_{j0}^{k}\\ \end{aligned} \right. ,\quad \end{aligned}$$

(31)

and

$$\begin{aligned} \left\{ \begin{aligned} \sum _{j=0}^{2}({\overline{\omega }}^{2})^{j}a_{j1}^{k}={\overline{\omega }}^{2} \sum _{j=0}^{2}({\overline{\omega }}^{2})^{j}a_{j0}^{k}\quad \\ \sum _{j=0}^{2}({\overline{\omega }}^{2})^{j}a_{j2}^{k}=({\overline{\omega }}^{2})^{2} \sum _{j=0}^{2}({\overline{\omega }}^{2})^{j}a_{j0}^{k}\\ \end{aligned} \right. . \end{aligned}$$

(32)

By Eqs. (22), (30), (31) and (32), we get

$$\begin{aligned} \left\{ \begin{aligned} \sum _{j=0}^{2}a_{j1}^{k}=a_{00}^{k}\qquad \quad \\ \sum _{j=0}^{2}{\overline{\omega }}^{j}a_{j1}^{k}={\overline{\omega }} a_{00}^{k}\quad \,\,\\ \sum _{j=0}^{2}({\overline{\omega }}^{2})^{j}a_{j1}^{k}={\overline{\omega }}^{2}a_{00}^{k} \end{aligned} \right. \end{aligned}$$

(33)

and

$$\begin{aligned} \left\{ \begin{aligned} \sum _{j=0}^{2}a_{j2}^{k}=a_{00}^{k}\qquad \qquad \\ \sum _{j=0}^{2}{\overline{\omega }}^{j}a_{j2}^{k}={\overline{\omega }}^{2} a_{00}^{k}\qquad \\ \sum _{j=0}^{2}({\overline{\omega }}^{2})^{j}a_{j2}^{k}=({\overline{\omega }}^{2})^{2}a_{00}^{k} \end{aligned} \right. . \end{aligned}$$

(34)

By Lemmas 1–3, we easily get the unique solutions of Eqs. (33) and (34), respectively, i.e.,

$$\begin{aligned} \left\{ \begin{aligned} a_{11}^{k}=a_{00}^{k}\qquad \\ a_{01}^{k}=a_{21}^{k}=0 \end{aligned} \right. \end{aligned}$$

(35)

and

$$\begin{aligned} \left\{ \begin{aligned} a_{22}^{k}=a_{00}^{k}\quad \,\,\,\,\,\\ a_{02}^{k}=a_{12}^{k}=0 \end{aligned} \right. . \end{aligned}$$

(36)

Similarly, because the three product states $\{(M_{k}\otimes I_{4\times 4})|\phi _{10}\rangle$, $(M_{k}\otimes I_{4\times 4})|\phi _{11}\rangle$ and $(M_{k}\otimes I_{4\times 4})|\phi _{12}\rangle \}$ are mutually orthogonal, we have

$$\begin{aligned}&\left\{ \begin{aligned} \sum _{p=0}^{1}\left[ \omega ^{p}\sum _{j=1}^{3}a_{j(p+1)}^{k}\right] =-\omega ^{2}\sum _{j=1}^{3}a_{j3}^{k}\qquad \,\,\\ \sum _{p=0}^{1}\left[ (\omega ^{2})^{p}\sum _{j=1}^{3}a_{j(p+1)}^{k}\right] =-(\omega ^{2})^{2}\sum _{j=1}^{3}a_{j3}^{k}\\ \end{aligned} \right. , \end{aligned}$$

(37)

$$\begin{aligned}&\left\{ \begin{aligned} \sum _{p=0}^{1}\sum _{j=1}^{3}{\overline{\omega }}^{j-1}a_{j(p+1)}^{k} =-\sum _{j=1}^{3}{\overline{\omega }}^{j-1}a_{j3}^{k}\qquad \qquad \quad \,\\ \sum _{p=0}^{1}\left[ (\omega ^{2})^{p}\sum _{j=1}^{3}{\overline{\omega }}^{j-1}a_{j(p+1)}^{k}\right] =-(\omega ^{2})^{2}\sum _{j=1}^{3}{\overline{\omega }}^{j-1}a_{j3}^{k}\\ \end{aligned} \right. \end{aligned}$$

(38)

and

$$\begin{aligned} \left\{ \begin{aligned} \sum _{p=0}^{1}\sum _{j=1}^{3}({\overline{\omega }}^{2})^{j-1}a_{j(p+1)}^{k} =-\sum _{j=1}^{3}({\overline{\omega }}^{2})^{j-1}a_{j3}^{k}\qquad \,\,\,\,\\ \sum _{p=0}^{1}\left[ \omega ^{p}\sum _{j=1}^{3}({\overline{\omega }}^{2})^{j-1}a_{j(p+1)}^{k}\right] =-\omega ^{2}\sum _{j=1}^{3}({\overline{\omega }}^{2})^{j-1}a_{j3}^{k}\\ \end{aligned} \right. . \end{aligned}$$

(39)

For simplicity, we denote the coefficient determinants of Eqs. (37), (38) and (39) as $D_{4}$, $D_{5}$ and $D_{6}$, respectively. Since

$$\begin{aligned} \begin{aligned} D_{4}&=\left| \begin{array}{cc} 1 &{}\omega \\ 1 &{}\omega ^{2} \\ \end{array} \right| \ne 0, \\ D_{5}&=\left| \begin{array}{cc} 1 &{}1 \\ 1 &{}\omega ^{2} \\ \end{array} \right| \ne 0, \\ D_{6}&=\left| \begin{array}{cc} 1 &{}1 \\ 1 &{}\omega \\ \end{array} \right| \ne 0, \end{aligned} \end{aligned}$$

we can get the unique solutions of Eqs. (37), (38) and (39), respectively, i.e.,

$$\begin{aligned}&\left\{ \begin{aligned} \sum _{j=1}^{3}a_{j1}^{k}=\sum _{j=1}^{3}a_{j3}^{k}\\ \sum _{j=1}^{3}a_{j2}^{k}=\sum _{j=1}^{3}a_{j3}^{k} \end{aligned} \right. , \end{aligned}$$

(40)

$$\begin{aligned}&\left\{ \begin{aligned} \sum _{j=1}^{3}{\overline{\omega }}^{j-1}a_{j1}^{k}= \omega ^{2}\sum _{j=1}^{3}{\overline{\omega }}^{j-1}a_{j3}^{k}\\ \sum _{j=1}^{3}{\overline{\omega }}^{j-1}a_{j2}^{k}= \omega \sum _{j=1}^{3}{\overline{\omega }}^{j-1}a_{j3}^{k}\,\,\\ \end{aligned} \right. , \end{aligned}$$

(41)

and

$$\begin{aligned} \left\{ \begin{aligned} \sum _{j=1}^{3}({\overline{\omega }}^{2})^{j-1}a_{j1}^{k}=(\omega ^{2})^{2} \sum _{j=1}^{3}({\overline{\omega }}^{2})^{j-1}a_{j3}^{k}\\ \sum _{j=1}^{3}({\overline{\omega }}^{2})^{j-1}a_{j2}^{k}=\omega ^{2} \sum _{j=1}^{3}({\overline{\omega }}^{2})^{j-1}a_{j3}^{k}\quad \\ \end{aligned} \right. . \end{aligned}$$

(42)

By Eqs. (26), (40), (41) and (42), we can get

$$\begin{aligned} \left\{ \begin{aligned} \sum _{j=1}^{3}a_{j1}^{k}=a_{33}^{k}\qquad \quad \\ \sum _{j=1}^{3}{\overline{\omega }}^{j-1}a_{j1}^{k}= a_{33}^{k}\quad \\ \sum _{j=1}^{3}({\overline{\omega }}^{2})^{j-1}a_{j1}^{k}=a_{33}^{k} \end{aligned} \right. \end{aligned}$$

(43)

and

$$\begin{aligned} \left\{ \begin{aligned} \sum _{j=1}^{3}a_{j2}^{k}=a_{33}^{k}\qquad \quad \quad \\ \sum _{j=1}^{3}{\overline{\omega }}^{j-1}a_{j2}^{k}={\overline{\omega }} a_{33}^{k}\quad \\ \sum _{j=1}^{3}({\overline{\omega }}^{2})^{j-1}a_{j2}^{k}={\overline{\omega }}^{2}a_{33}^{k} \end{aligned} \right. . \end{aligned}$$

(44)

By Lemmas 1–3, we can immediately get the unique solutions of Eqs. (43) and (44), i.e.,

$$\begin{aligned} \left\{ \begin{aligned} a_{11}^{k}=a_{33}^{k}\quad \quad \\ a_{21}^{k}=a_{31}^{k}=0 \end{aligned} \right. \end{aligned}$$

(45)

and

$$\begin{aligned} \left\{ \begin{aligned} a_{22}^{k}=a_{33}^{k}\quad \quad \\ a_{12}^{k}=a_{32}^{k}=0 \end{aligned} \right. . \end{aligned}$$

(46)

Appendix B: Proof of Theorem 2

Proof

Suppose that Alice firstly performs a POVM measurement with a set of general $m\times m$ POVM elements $\{M_{k}^{\dag }M_{k}: k=1,\,2,\,\cdots ,\,l\}$, where

$$\begin{aligned} \begin{aligned} M_{k}^{\dag }M_{k}= \left[ \begin{array}{ccccc} a_{00}^{k} &{}a_{01}^{k} &{}\cdots &{}a_{0(m-1)}^{k}\\ a_{10}^{k} &{}a_{11}^{k} &{}\cdots &{}a_{1(m-1)}^{k}\\ \vdots &{}\vdots &{}\ddots &{}\vdots \\ a_{(m-1)0}^{k} &{}a_{(m-1)1}^{k} &{}\cdots &{}a_{(m-1)(m-1)}^{k}\\ \end{array} \right] \end{aligned} \end{aligned}$$

in the basis $\{|0\rangle ,$ $|1\rangle ,$ $\cdots ,$ $|(m-1)\rangle \}$. The post-measurement states $\{(M_{k}\otimes I_{n\times n})|\phi _{j}\rangle :$ $j=1,$ 2, $\cdots$, $2(m+n)-4\}$ should be mutually orthogonal to make further discrimination possible.

Because $(M_{k}\otimes I_{n\times n})|\phi _{1}\rangle$ should be orthogonal to $(M_{k}\otimes I_{n\times n})|\phi _{\eta +2n+m-2}\rangle$ on Alice’s side for $\eta =0,\,1,\,\cdots ,\,m-2$, we get

$$\begin{aligned} \left\{ \begin{aligned} \sum _{j=1}^{m-1}a_{0j}^{k}=0\qquad \qquad \quad \\ \sum _{j=1}^{m-1}(\omega _{2})^{j-1} a_{0j}^{k}=0\qquad \\ \sum _{j=1}^{m-1}[(\omega _{2})^{2}]^{j-1} a_{0j}^{k}=0\quad \\ \vdots \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\ \sum _{j=1}^{m-1}[(\omega _{2})^{m-2}]^{j-1} a_{0j}^{k}=0\\ \end{aligned} \right. \end{aligned}$$

(47)

and

$$\begin{aligned} \left\{ \begin{aligned} \sum _{j=1}^{m-1}a_{j0}^{k}=0\qquad \qquad \,\,\,\\ \sum _{j=1}^{m-1}\overline{\omega _{2}}^{j-1}a_{j0}^{k}=0\qquad \\ \sum _{j=1}^{m-1}(\overline{\omega _{2}}^{2})^{j-1}a_{j0}^{k}=0\quad \\ \vdots \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\ \sum _{j=1}^{m-1}(\overline{\omega _{2}}^{m-2})^{j-1}a_{j0}^{k}=0\\ \end{aligned} \right. , \end{aligned}$$

(48)

where $\overline{\omega _{2}}$ is the conjugate complex number of $\omega _{2}$. The coefficient determinants the two systems of Eqs. (47) and (48) are not equal to zero by Lemma 1 and Lemma 3, i.e.,

$$\begin{aligned} \begin{aligned} \left| \begin{array}{ccccc} 1 &{}1 &{}1 &{}\cdots &{}1\\ 1 &{}\omega _{2} &{}(\omega _{2})^{2} &{}\cdots &{}(\omega _{2})^{m-2}\\ 1 &{}(\omega _{2})^{2} &{}[(\omega _{2})^{2}]^{2} &{}\cdots &{}[(\omega _{2})^{2}]^{m-2}\\ \vdots &{}\vdots &{}\vdots &{}\ddots &{}\vdots \\ 1 &{}(\omega _{2})^{m-2}&{}[(\omega _{2})^{m-2}]^{2} &{}\cdots &{}[(\omega _{2})^{m-2}]^{m-2} \end{array} \right| \ne 0, \end{aligned} \\ \begin{aligned} \left| \begin{array}{ccccc} 1 &{}1 &{}1 &{}\cdots &{}1\\ 1 &{}\overline{\omega _{2}} &{}\overline{\omega _{2}}^{2} &{}\cdots &{}\overline{\omega _{2}}^{m-2}\\ 1 &{}\overline{\omega _{2}}^{2} &{}(\overline{\omega _{2}}^{2})^{2} &{}\cdots &{}(\overline{\omega _{2}}^{2})^{m-2}\\ \vdots &{}\vdots &{}\vdots &{}\ddots &{}\vdots \\ 1 &{}\overline{\omega _{2}}^{m-2} &{}(\overline{\omega _{2}}^{m-2})^{2} &{}\cdots &{}(\overline{\omega _{2}}^{m-2})^{m-2} \end{array} \right| \ne 0, \end{aligned} \end{aligned}$$

we have the unique solutions of Eqs. (47) and (48), respectively, i.e.,

$$\begin{aligned} \begin{aligned} a_{01}^{k}=a_{02}^{k}=\cdots =a_{0(m-1)}^{k}=0 \end{aligned} \end{aligned}$$

(49)

and

$$\begin{aligned} \begin{aligned} a_{10}^{k}=a_{20}^{k}=\cdots =a_{(m-1)0}^{k}=0 \end{aligned} \end{aligned}$$

(50)

by Lemma 2. Similarly, because $(M_{k}\otimes I_{n\times n})|\phi _{n+m-1}\rangle$ should be orthogonal to $(M_{k}\otimes I_{n\times n})|\phi _{\eta +n}\rangle$ on Alice’s side for $\eta =0$, 1, $\cdots$, $m-2$, we have

$$\begin{aligned} \left\{ \begin{aligned} \sum _{j=0}^{m-2}a_{(m-1)j}^{k}=0\qquad \qquad \,\\ \sum _{j=0}^{m-2}(\omega _{2})^{j}a_{(m-1)j}^{k}=0\qquad \\ \sum _{j=0}^{m-2}[(\omega _{2})^{2}]^{j}a_{(m-1)j}^{k}=0\quad \\ \vdots \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\ \sum _{j=0}^{m-2}[(\omega _{2})^{m-2}]^{j}a_{(m-1)j}^{k}=0 \end{aligned} \right. \end{aligned}$$

(51)

and

$$\begin{aligned} \left\{ \begin{aligned} \sum _{j=0}^{m-2}a_{j(m-1)}^{k}=0\qquad \quad \,\,\\ \sum _{j=0}^{m-2}\overline{\omega _{2}}^{j}a_{j(m-1)}^{k}=0\qquad \\ \sum _{j=0}^{m-2}(\overline{\omega _{2}}^{2})^{j}a_{j(m-1)}^{k}=0\quad \\ \vdots \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\ \sum _{j=0}^{m-2}(\overline{\omega _{2}}^{m-2})^{j}a_{j(m-1)}^{k}=0\\ \end{aligned} \right. . \end{aligned}$$

(52)

By Lemmas 1–3, we get the unique solutions of Eqs. (51) and (52), respectively, i.e.,

$$\begin{aligned} a_{(m-1)0}^{k}= & {} a_{(m-1)1}^{k}=\cdots =a_{(m-1)(m-2)}^{k}=0, \end{aligned}$$

(53)

$$\begin{aligned} a_{0(m-1)}^{k}= & {} a_{1(m-1)}^{k}=\cdots =a_{(m-2)(m-1)}^{k}=0. \end{aligned}$$

(54)

Because the states in $\{(M_{k}\otimes I_{n\times n})|\phi _{\eta +2n+m-2}\rangle :$ $\eta =0,$ 1,  $\cdots$, $m-2\}$ should be mutually orthogonal on Alice’s side, we get the following $m-1$ systems of equations

$$\begin{aligned}&\left\{ \begin{aligned} \sum _{p=0}^{m-3}\left[ (\omega _{2})^{p}\sum _{j=1}^{m-1}a_{j(p+1)}^{k}\right] =-(\omega _{2})^{m-2}\sum _{j=1}^{m-1}a_{j(m-1)}^{k}\qquad \qquad \,\,\\ \sum _{p=0}^{m-3}\{[(\omega _{2})^{2}]^{p}\sum _{j=1}^{m-1}a_{j(p+1)}^{k}\} =-[(\omega _{2})^{2}]^{m-2}\sum _{j=1}^{m-1}a_{j(m-1)}^{k}\qquad \\ \vdots \qquad \qquad \qquad \qquad \qquad \qquad \quad \,\,\,\\ \sum _{p=0}^{m-3}\{[(\omega _{2})^{m-2}]^{p}\sum _{j=1}^{m-1}a_{j(p+1)}^{k}\} =-[(\omega _{2})^{m-2}]^{m-2}\sum _{j=1}^{m-1}a_{j(m-1)}^{k} \end{aligned} \right. , \\&\left\{ \begin{aligned} \sum _{p=0}^{m-3}\left( \sum _{j=1}^{m-1}\overline{\omega _{2}}^{j-1}a_{j(p+1)}^{k}\right) =-\sum _{j=1}^{m-1}\overline{\omega _{2}}^{j-1}a_{j(m-1)}^{k}\qquad \qquad \qquad \qquad \qquad \,\,\\ \sum _{p=0}^{m-3}\{[(\omega _{2})^{2}]^{p}\sum _{j=1}^{m-1}\overline{\omega _{2}}^{j-1}a_{j(p+1)}^{k}\} =-[(\omega _{2})^{2}]^{m-2}\sum _{j=1}^{m-1}\overline{\omega _{2}}^{j-1}a_{j(m-1)}^{k}\qquad \\ \sum _{p=0}^{m-3}\{[(\omega _{2})^{3}]^{p}\sum _{j=1}^{m-1}\overline{\omega _{2}}^{j-1}a_{j(p+1)}^{k}\} =-[(\omega _{2})^{3}]^{m-2}\sum _{j=1}^{m-1}\overline{\omega _{2}}^{j-1}a_{j(m-1)}^{k}\qquad \\ \vdots \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \\ \sum _{p=0}^{m-3}\{[(\omega _{2})^{m-2}]^{p}\sum _{j=1}^{m-1}{\overline{\omega }}^{j-1}a_{j(p+1)}^{k}\} =-[(\omega _{2})^{m-2}]^{m-2}\sum _{j=1}^{m-1}\overline{\omega _{2}}^{j-1}a_{j(m-1)}^{k} \end{aligned} \right. , \\&\left\{ {\begin{aligned} \sum _{p=0}^{m-3}\left[ \sum _{j=1}^{m-1}(\overline{\omega _{2}}^{2})^{j-1}a_{j(p+1)}^{k}\right] =-\sum _{j=1}^{m-1}(\overline{\omega _{2}}^{2})^{j-1}a_{j(m-1)}^{k}\qquad \qquad \qquad \qquad \qquad \,\,\,\,\,\\ \sum _{p=0}^{m-3}\left[ (\omega _{2})^{p}\sum _{j=1}^{m-1}(\overline{\omega _{2}}^{2})^{j-1}a_{j(p+1)}^{k}\right] =-(\omega _{2})^{m-2}\sum _{j=1}^{m-1}(\overline{\omega _{2}}^{2})^{j-1}a_{j(m-1)}^{k}\qquad \qquad \,\,\,\\ \sum _{p=0}^{m-3}\{[(\omega _{2})^{3}]^{p}\sum _{j=1}^{m-1}(\overline{\omega _{2}}^{2})^{j-1}a_{j(p+1)}^{k}\} =-[(\omega _{2})^{3}]^{m-2}\sum _{j=1}^{m-1}(\overline{\omega _{2}}^{2})^{j-1}a_{j(m-1)}^{k}\qquad \,\,\\ \vdots \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \\ \sum _{p=0}^{m-3}\{[(\omega _{2})^{m-2}]^{p}\sum _{j=1}^{m-1}(\overline{\omega _{2}}^{2})^{j-1}a_{j(p+1)}^{k}\} =-[(\omega _{2})^{m-2}]^{m-2}\sum _{j=1}^{m-1}(\overline{\omega _{2}}^{2})^{j-1}a_{j(m-1)}^{k}\\ \end{aligned}} \right. , \\&\vdots \quad \\&\left\{ {\begin{aligned} \sum _{p=0}^{m-3}\left[ \sum _{j=1}^{m-1}(\overline{\omega _{2}}^{m-2})^{j-1}a_{j(p+1)}^{k}\right] =-\sum _{j=1}^{m-1}(\overline{\omega _{2}}^{d-2})^{j-1}a_{j(m-1)}^{k}\qquad \qquad \qquad \qquad \qquad \,\,\,\,\,\,\\ \sum _{p=0}^{m-3}\left[ (\omega _{2})^{p}\sum _{j=1}^{m-1}(\overline{\omega _{2}}^{m-2})^{j-1}a_{j(p+1)}^{k}\right] =-(\omega _{2})^{m-2}\sum _{j=1}^{m-1}(\overline{\omega _{2}}^{m-2})^{j-1}a_{j(m-1)}^{k}\qquad \qquad \,\,\,\\ \sum _{p=0}^{m-3}\{[(\omega _{2})^{2}]^{p}\sum _{j=1}^{m-1}(\overline{\omega _{2}}^{m-2})^{j-1}a_{j(p+1)}^{k}\} =-[(\omega _{2})^{2}]^{m-2}\sum _{j=1}^{m-1}(\overline{\omega _{2}}^{m-2})^{j-1}a_{j(m-1)}^{k}\qquad \,\,\\ \vdots \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \\ \sum _{p=0}^{m-3}\{[(\omega _{2})^{m-3}]^{p}\sum _{j=1}^{m-1}(\overline{\omega _{2}}^{m-2})^{j-1}a_{j(p+1)}^{k}\} =-[(\omega _{2})^{d-3}]^{m-2}\sum _{j=1}^{m-1}(\overline{\omega _{2}}^{m-2})^{j-1}a_{j(m-1)}^{k}\\ \end{aligned}} \right. , \end{aligned}$$

By Lemmas 1–3 and Eq. (54), we can get the unique solutions of the above $m-1$ systems of equations, i.e., 

$$\begin{aligned} \left\{ \begin{aligned} \sum _{j=1}^{m-1}a_{jq}^{k}=a_{(m-1)(m-1)}^{k}\qquad \qquad \qquad \qquad \quad \,\,\,\,\\ \sum _{j=1}^{m-1}\overline{\omega _{2}}^{j-1}a_{jq}^{k}=\overline{\omega _{2}}^{q-1}a_{(m-1)(m-1)}^{k} \qquad \qquad \,\,\\ \sum _{j=1}^{m-1}(\overline{\omega _{2}}^{2})^{j-1}a_{jq}^{k}=(\overline{\omega _{2}}^{2})^{q-1} a_{(m-1)(m-1)}^{k}\qquad \\ \vdots \qquad \qquad \qquad \\ \sum _{j=1}^{m-1}(\overline{\omega _{2}}^{m-2})^{j-1}a_{jq}^{k}=(\overline{\omega _{2}}^{m-2})^{q-1} a_{(m-1)(m-1)}^{k} \end{aligned} \right. , \end{aligned}$$

(55)

where $q=1,\,2,\,\cdots ,\,m-2.$

By Lemmas 1-3 and Eq. (55), we have

$$\begin{aligned} \left\{ \begin{aligned} a_{jq}^{k}=0\qquad \qquad \quad \,\,\\ a_{qq}^{k}=a_{(m-1)(m-1)}^{k}\\ \end{aligned} \right. \end{aligned}$$

(56)

for $q=1,2,\cdots ,m-2;$ $j=1,2,\cdots ,m-1$ and $j\ne q.$

Similarly, since the states in the set $\{(M_{k}\otimes I_{n\times n})|\phi _{\eta +n}\rangle :$ $\eta =0,$ 1,  $\cdots$, $m-2\}$ should be orthogonal on Alice’s side, we have

$$\begin{aligned} \left\{ \begin{aligned} a_{jq}^{k}=0\,\,\,\,\,\,\,\\ a_{qq}^{k}=a_{00}^{k}\\ \end{aligned} \right. \end{aligned}$$

(57)

for $q=1,\,2,\,\cdots ,\,m-2;$ $j=0,1,\,2,\,\cdots ,\,m-2$ and $j\ne q.$

By Eqs. (49), (50), (53), (54), (56) and (57), we have

$$\begin{aligned} \begin{aligned} M_{k}^{\dag }M_{k}= \left[ \begin{array}{cccc} a_{(m-1)(m-1)}^{k} &{}0 &{}\cdots &{}0\\ 0 &{}a_{(m-1)(m-1)}^{k} &{}\cdots &{}0\\ 0 &{}0 &{}\ddots &{}0\\ 0 &{}0 &{}\cdots &{}a_{(m-1)(m-1)}^{k}\\ \end{array} \right] \end{aligned} \end{aligned}$$

for $k=1,2,\cdots ,l.$ This means that all the POVM elements are proportional to identity matrix. That is, Alice cannot start with a nontrivial measurement to keep the post-measurement states orthogonal.

In fact, Bob will face a similar case as Alice does since the set of these $2(m+n)-4$ states has a symmetrical structure. Therefore, these $2(m+n)-4$ states cannot be exactly distinguished by using only LOCC. This completes the proof. $\square$

Appendix C: Proof of Theorem 3

To discriminate these states, someone needs to perform a measurement of preserving orthogonality. That is, the states that are orthogonal only on Alice’s (Bob’s) side are still orthogonal on this side after measurement. Thus we need to show that Alice or Bob only can perform a nontrivial measurement no matter who goes first. Without loss of generality, suppose that Alice performs a general measurement with POVM element

$$\begin{aligned} \begin{aligned} M_{k}^{\dag }M_{k}= \left[ \begin{array}{cccc} a_{00}^{k} &{}a_{01}^{k} &{}\cdots &{}a_{0,m-1}^{k}\\ a_{10}^{k} &{}a_{11}^{k} &{}\cdots &{}a_{1,m-1}^{k}\\ \vdots &{}\vdots &{}\ddots &{}\vdots \\ a_{m-1,0}^{k} &{}a_{m-1,1}^{k} &{}\cdots &{}a_{m-1,m-1}^{k} \\ \end{array} \right] \end{aligned} \end{aligned}$$

in the basis $\{|0\rangle ,\,|1\rangle ,\,\cdots ,\,|(m-1)\rangle \}.$

Because $(M_{k}\otimes I_{n\times n} )|\phi _{1}\rangle$ is orthogonal to each state of the set $\{(M_{k}\otimes I_{n\times n})|\phi _{2n+m-2+2\sigma }\rangle ,\, (M_{k}\otimes I_{n\times n})|\phi _{2n+m-1+2\sigma }\rangle \}$, we have

$$\begin{aligned} \left\{ \begin{aligned} \langle \phi _{1}|M_{k}^{\dag }M_{k}\otimes I_{n\times n}|\phi _{2n+m-2+2\sigma }\rangle =0\\ \langle \phi _{1}|M_{k}^{\dag }M_{k}\otimes I_{n\times n}|\phi _{2n+m-1+2\sigma }\rangle =0\\ \langle \phi _{2n+m-2+2\sigma }|M_{k}^{\dag }M_{k}\otimes I_{n\times n}|\phi _{1}\rangle =0\\ \langle \phi _{2n+m-1+2\sigma }|M_{k}^{\dag }M_{k}\otimes I_{n\times n}|\phi _{1}\rangle =0\\ \end{aligned} \right. , \end{aligned}$$

where $\sigma =0,$ 1,  $\cdots ,$ $d_{1}-1$. Thus we have

$$\begin{aligned} \left\{ \begin{aligned} a_{01}^{k}=a_{02}^{k}=a_{03}^{k}=\cdots =a_{0,m-1}^{k}=0\\ a_{10}^{k}=a_{20}^{k}=a_{30}^{k}=\cdots =a_{m-1,0}^{k}=0\\ \end{aligned} \right. . \end{aligned}$$

(58)

Similarly, we have

$$\begin{aligned} \left\{ \begin{aligned} a_{0,m-1}^{k}=a_{1,m-1}^{k}=\cdots =a_{m-2,m-1}^{k}=0\\ a_{m-1,0}^{k}=a_{m-1,1}^{k}=\cdots =a_{m-1,m-2}^{k}=0\\ \end{aligned} \right. \end{aligned}$$

(59)

by the orthogonality of $(M_{k}\otimes I_{n\times n})|\phi _{2n+m-3}\rangle$ and each state of the set $\{(M_{k}\otimes I_{n\times n})|\phi _{n+2\sigma }\rangle ,\, (M_{k}\otimes I_{n\times n})|\phi _{n+1+2\sigma }\rangle \}$.

Due to the orthogonality of $(M_{k}\otimes I_{n\times n})|\phi _{2n+m-2+2\sigma }\rangle$ and $(M_{k}\otimes I_{n\times n})|\phi _{2n+m-1+2\sigma }\rangle$, we have

$$\begin{aligned} \left\{ \begin{aligned} \sum _{i=2\sigma +1}^{2\sigma +2}(-1)^{i+1}a_{2\sigma +1,i}^{k}+ \sum _{i=2\sigma +1}^{2\sigma +2}(-1)^{i+1}a_{2\sigma +2,i}^{k}=0\\ \sum _{i=2\sigma +1}^{2\sigma +2}a_{2\sigma +1,i}^{k}- \sum _{i=2\sigma +1}^{2\sigma +2}a_{2\sigma +2,i}^{k}=0\qquad \qquad \qquad \,\,\\ \end{aligned} \right. . \end{aligned}$$

Therefore, we obtain

$$\begin{aligned} a_{2\sigma +1,2\sigma +1}^{k}=a_{2\sigma +2,2\sigma +2}^{k} \end{aligned}$$

for $\sigma =0,$ 1,  2,  $\cdots ,$ $d_{1}-1$. Thus we have

$$\begin{aligned} a_{11}^{k}=a_{22}^{k}=\cdots =a_{m-1,m-1}^{k}. \end{aligned}$$

(60)

Similarly, due to the orthogonality of $(M_{k}\otimes I_{n\times n})|\phi _{n+2\sigma } \rangle$ and $(M_{k}\otimes I_{n\times n})|\phi _{n+1+2\sigma }\rangle$, we have

$$\begin{aligned} \left\{ \begin{aligned} a_{2\sigma ,2\sigma }^{k}-a_{2\sigma ,2\sigma +1}^{k}+a_{2\sigma +1,2\sigma }^{k} -a_{2\sigma +1,2\sigma +1}^{k}=0\\ a_{2\sigma ,2\sigma }^{k}+a_{2\sigma ,2\sigma +1}^{k}-a_{2\sigma +1,2\sigma }^{k}- a_{2\sigma +1,2\sigma +1}^{k}=0\\ \end{aligned} \right. . \end{aligned}$$

So

$$\begin{aligned} a_{2\sigma ,2\sigma }^{k}=a_{2\sigma +1,2\sigma +1}^{k}\\ \end{aligned}$$

where $\sigma =0,$ 1,  2,  $\cdots ,$ $d_{1}-1$. Thus we can get

$$\begin{aligned} a_{00}^{k}=a_{11}^{k}=\cdots =a_{m-2,m-2}^{k}. \end{aligned}$$

(61)

Because $(M_{k}\otimes I_{n\times n})|\phi _{2n+m-2+2\mu }\rangle$ and $(M_{k}\otimes$ $I_{n\times n})|\phi _{2n+m-1+2\mu }\rangle$ are orthogonal to $(M_{k}\otimes I_{n\times n})|\phi _{2n+m-2+2\lambda }\rangle$ and $(M_{k}\otimes I_{n\times n})|\phi _{2n+m-1+2\lambda }\rangle$, where $\mu ,\,\lambda =0,$ 1,  2,  $\cdots ,$ $d_{1}-1$ and $\mu \ne \lambda$, i.e.,

$$\begin{aligned} \left\{ \begin{aligned} \langle \phi _{2n+m-2+2\mu }|M_{k}^{\dag }M_{k}\otimes I_{n\times n}|\phi _{2n+m-2+2\lambda } \rangle =0\\ \langle \phi _{2n+m-2+2\mu }|M_{k}^{\dag }M_{k}\otimes I_{n\times n}|\phi _{2n+m-1+2\lambda } \rangle =0\\ \langle \phi _{2n+m-1+2\mu }|M_{k}^{\dag }M_{k}\otimes I_{n\times n}|\phi _{2n+m-2+2\lambda } \rangle =0\\ \langle \phi _{2n+m-1+2\mu }|M_{k}^{\dag }M_{k}\otimes I_{n\times n}|\phi _{2n+m-1+2\lambda } \rangle =0 \end{aligned} \right. , \end{aligned}$$

so we have

$$\begin{aligned} \left\{ \begin{aligned} \sum _{i=2\lambda +1}^{2\lambda +2}a_{2\mu +1,i}^{k}+ \sum _{i=2\lambda +1}^{2\lambda +2}a_{2\mu +2,i}^{k}=0\qquad \qquad \qquad \,\,\\ \sum _{i=2\lambda +1}^{2\lambda +2}(-1)^{i+1}a_{2\mu +1,i}^{k}+ \sum _{i=2\lambda +1}^{2\lambda +2}(-1)^{i+1}a_{2\mu +2,i}^{k}=0 \\ \sum _{i=2\lambda +1}^{2\lambda +2}a_{2\mu +1,i}^{k}- \sum _{i=2\lambda +1}^{2\lambda +2}a_{2\mu +2,i}^{k}=0\qquad \qquad \qquad \,\,\\ \sum _{i=2\lambda +1}^{2\lambda +2}(-1)^{i+1}a_{2\mu +1,i}^{k}+ \sum _{i=2\lambda +1}^{2\lambda +2}(-1)^{i}a_{2\mu +2,i}^{k}=0\,\,\,\, \\ \end{aligned} \right. . \end{aligned}$$

Thus we get

$$\begin{aligned} \begin{aligned} a_{2\mu +1,2\lambda +1}^{k}=a_{2\mu +1,2\lambda +2}^{k}=0,\\ a_{2\mu +2,2\lambda +1}^{k}=a_{2\mu +2,2\lambda +2}^{k}=0, \end{aligned} \end{aligned}$$

(62)

where $\mu ,\,\lambda =0,$ 1,  2,  $\cdots ,$ $d_{1}-1$ and $\mu \ne \lambda$. Similarly, because $(M_{k}\otimes I_{n\times n})|\phi _{n+2\mu }\rangle$ and $(M_{k}\otimes I_{n\times n})|\phi _{n+1+2\mu }\rangle$ are orthogonal to $(M_{k}\otimes I)|\phi _{n+2\lambda }\rangle$ and $(M_{k}\otimes I)|\phi _{n+1+2\lambda }\rangle$, we get

$$\begin{aligned} \begin{aligned} a_{2\mu ,2\lambda }^{k}=a_{2\mu ,2\lambda +1}^{k}=0,\\ a_{2\mu +1,2\lambda }^{k}=a_{2\mu +1,2\lambda +1}^{k}=0. \end{aligned} \end{aligned}$$

(63)

where $\mu ,\,\lambda =0,$ 1,  2,  $\cdots ,$ $d_{1}-1$ and $\mu \ne \lambda$.

By Eqs. (58), (59), (60), (61), (62) and (63), we have

$$\begin{aligned} \begin{aligned} M_{k}^{\dag }M_{k}= \left[ \begin{array}{cccc} a_{00}^{k} &{}0 &{}\cdots &{}0\\ 0 &{}a_{00}^{k} &{}\cdots &{}0\\ \vdots &{}\vdots &{}\ddots &{}\vdots \\ 0 &{}0 &{}\cdots &{}a_{00}^{k} \\ \end{array} \right] \end{aligned} \end{aligned}$$

for $i=1,2,\cdots ,l.$ This means that all the POVM elements of Alice are proportional to identity operator. Therefore, Alice only can perform a trivial measurement to preserve the orthogonality of the post-measurement states. So does Bob by the symmetry of the set of these $2(m+n)-4$ product states. Therefore, these states cannot be reliably discriminated by LOCC no matter who goes first. This completes the proof.

Appendix D: Proof of Theorem 4

Proof

We prove whether Alice or Bob cannot perform a nontrivial POVM measurement to preserve the orthogonality of the OPSs that are orthogonal only on one side. Without loss of generality, Suppose that Alice firstly performs a general POVM measurement with a set of POVM elements

$$\begin{aligned} \begin{aligned} M_{k}^{\dag }M_{k}= \left[ \begin{array}{cccc} a_{00}^{k} &{}a_{01}^{k} &{}\cdots &{}a_{0(m-1)}^{k}\\ a_{10}^{k} &{}a_{11}^{k} &{}\cdots &{}a_{1(m-1)}^{k}\\ \vdots &{}\vdots &{}\ddots &{}\vdots \\ a_{(m-1)0}^{k} &{}a_{(m-1)1}^{k} &{}\cdots &{}a_{(m-1)(m-1)}^{k}\\ \end{array} \right] \end{aligned} \end{aligned}$$

in the basis $\{|0\rangle ,\,|1\rangle ,\,|2\rangle ,\,\cdots ,\, |(m-1)\rangle \}$.

Because $(M_{k}\otimes I_{n\times n})|\phi _{1}\rangle$ is orthogonal to $(M_{k}\otimes I_{n\times n})|\phi _{2n+m-2}\rangle$, $(M_{k}\otimes I_{n\times n})|\phi _{2n+m-1}\rangle$ and $(M_{k}\otimes I_{n\times n})|\phi _{2n+m}\rangle$, i.e.,

$$\begin{aligned} \left\{ \begin{aligned} \langle \phi _{1}|M^{\dag }_{k}M_{k}\otimes I_{n\times n}|\phi _{2n+m-2}\rangle =0\\ \langle \phi _{1}|M^{\dag }_{k}M_{k}\otimes I_{n\times n}|\phi _{2n+m-1}\rangle =0\\ \langle \phi _{1}|M^{\dag }_{k}M_{k}\otimes I_{n\times n}|\phi _{2n+m}\rangle =0\quad \\ \end{aligned} \right. \end{aligned}$$

and

$$\begin{aligned} \left\{ \begin{aligned} \langle \phi _{2n+m-2}|M^{\dag }_{k}M_{k}\otimes I_{n\times n}|\phi _{1}\rangle =0\\ \langle \phi _{2n+m-1}|M^{\dag }_{k}M_{k}\otimes I_{n\times n}|\phi _{1}\rangle =0\\ \langle \phi _{2n+m}|M^{\dag }_{k}M_{k}\otimes I_{n\times n}|\phi _{1}\rangle =0\quad \\ \end{aligned} \right. , \end{aligned}$$

we have

$$\begin{aligned} \left\{ \begin{aligned} a_{01}^{k}+a_{02}^{k}+a_{03}^{k}=0\qquad \quad \,\,\\ a_{01}^{k}+\omega a_{02}^{k}+\omega ^{2}a_{03}^{k}=0\quad \,\,\,\,\\ a_{01}^{k}+\omega ^{2} a_{02}^{k}+(\omega ^{2})^{2}a_{03}^{k}=0\\ \end{aligned} \right. \end{aligned}$$

(64)

and

$$\begin{aligned} \left\{ \begin{aligned} a_{10}^{k}+a_{20}^{k}+a_{30}^{k}=0\qquad \quad \,\,\\ a_{10}^{k}+{\overline{\omega }} a_{20}^{k}+{\overline{\omega }}^{2}a_{30}^{k}=0\quad \,\,\,\,\\ a_{10}^{k}+{\overline{\omega }}^{2}a_{20}^{k}+({\overline{\omega }}^{2})^{2}a_{30}^{k}=0\\ \end{aligned} \right. , \end{aligned}$$

(65)

where ${\overline{\omega }}$ is the conjugate complex number of $\omega$. By Lemmas 1–3, Eqs. (64) and (65), we have

$$\begin{aligned} a_{01}^{k}=a_{02}^{k}=a_{03}^{k}=a_{10}^{k}=a_{20}^{k}=a_{30}^{k}=0. \end{aligned}$$

(66)

Because $(M_{k}\otimes I_{n\times n})|\phi _{1}\rangle$ is orthogonal to $(M_{k}\otimes I_{n\times n})|\phi _{2n+m+2\sigma +1}\rangle$ and $(M_{k}\otimes I_{n\times n})|\phi _{2n+m+2\sigma +2}\rangle$ for $\sigma =0,\,\, 1,\,\, 2,\,\, \cdots ,\,\, d_{1}-3$, we have

$$\begin{aligned} \left\{ \begin{aligned} a_{0,2\sigma +4}^{k}+a_{0,2\sigma +5}^{k}=0\\ a_{0,2\sigma +4}^{k}-a_{0,2\sigma +5}^{k}=0\\ \end{aligned} \right. \end{aligned}$$

(67)

and

$$\begin{aligned} \left\{ \begin{aligned} a_{2\sigma +4,0}^{k}+a_{2\sigma +5,0}^{k}=0\\ a_{2\sigma +4,0}^{k}-a_{2\sigma +5,0}^{k}=0\\ \end{aligned} \right. . \end{aligned}$$

(68)

By Eqs. (67) and (68), we have

$$\begin{aligned} \begin{aligned} a_{0,2\sigma +4}^{k}=a_{0,2\sigma +5}^{k}=a_{2\sigma +4,0}^{k}=a_{2\sigma +5,0}^{k}=0\\ \end{aligned} \end{aligned}$$

(69)

for $\sigma =0,\,\, 1,\,\, 2,\,\, \cdots ,\,\, d_{1}-3$.

Because any two states of $\{(M_{k}\otimes I_{n\times n})|\phi _{2n+m-2}\rangle$, $(M_{k}\otimes I_{n\times n})|\phi _{2n+m-1}\rangle$ and $(M_{k}\otimes I_{n\times n})|\phi _{2n+m}\rangle \}$, i.e.,

$$\begin{aligned}&\left\{ \begin{aligned} \langle \phi _{2n+m-2}|M_{k}^{\dag }M_{k}\otimes I_{n\times n}|\phi _{2n+m-1}\rangle =0\\ \langle \phi _{2n+m-2}|M_{k}^{\dag }M_{k}\otimes I_{n\times n}|\phi _{2n+m}\rangle =0\quad \\ \end{aligned} \right. , \\&\left\{ \begin{aligned} \langle \phi _{2n+m-1}|M_{k}^{\dag }M_{k}\otimes I_{n\times n}|\phi _{2n+m-2}\rangle =0\\ \langle \phi _{2n+m-1}|M_{k}^{\dag }M_{k}\otimes I_{n\times n}|\phi _{2n+m}\rangle =0\quad \\ \end{aligned} \right. , \\&\left\{ \begin{aligned} \langle \phi _{2n+m}|M_{k}^{\dag }M_{k}\otimes I_{n\times n}|\phi _{2n+m-2}\rangle =0\\ \langle \phi _{2n+m}|M_{k}^{\dag }M_{k}\otimes I_{n\times n}|\phi _{2n+m-1}\rangle =0\\ \end{aligned} \right. , \end{aligned}$$

we have

$$\begin{aligned}&\left\{ \begin{aligned} \sum _{j=1}^{3}a_{j1}+\omega \sum _{j=1}^{3}a_{j2}=-\omega ^{2}\sum _{j=1}^{3}a_{j3}\quad \,\,\,\\ \sum _{j=1}^{3}a_{j1}+\omega ^{2}\sum _{j=1}^{3}a_{j2}=-(\omega ^{2})^{2}\sum _{j=1}^{3}a_{j3}\\ \end{aligned} \right. , \end{aligned}$$

(70)

$$\begin{aligned}&\left\{ \begin{aligned} \sum _{i=1}^{2}\sum _{j=1}^{3}{\overline{\omega }}^{j-1}a_{ji} =-\sum _{j=1}^{3}{\overline{\omega }}^{j-1}a_{j3}\qquad \qquad \,\,\\ \sum _{i=1}^{2}[(\omega ^{2})^{i-1}\sum _{j=1}^{3}{\overline{\omega }}^{j-1}a_{ji}] =-\omega ^{4}\sum _{j=1}^{3}{\overline{\omega }}^{j-1}a_{j3}\\ \end{aligned} \right. , \end{aligned}$$

(71)

$$\begin{aligned}&\left\{ \begin{aligned} \sum _{i=1}^{2}\sum _{j=1}^{3}({\overline{\omega }}^{2})^{j-1}a_{ji} =-\sum _{j=1}^{3}({\overline{\omega }}^{2})^{j-1}a_{j3}\qquad \quad \\ \sum _{i=1}^{2}[\omega ^{i-1}\sum _{j=1}^{3}({\overline{\omega }}^{2})^{j-1}a_{ji}] =-\omega ^{2}\sum _{j=1}^{3}({\overline{\omega }}^{2})^{j-1}a_{j3}\\ \end{aligned} \right. . \end{aligned}$$

(72)

By Eqs. (70), (71) and (72), we have

$$\begin{aligned}&\left\{ \begin{aligned} \sum _{j=1}^{3}a_{j1}=\sum _{j=1}^{3}a_{j3}\\ \sum _{j=1}^{3}a_{j2}=\sum _{j=1}^{3}a_{j3}\\ \end{aligned} \right. , \end{aligned}$$

(73)

$$\begin{aligned}&\left\{ \begin{aligned} \sum _{j=1}^{3}{\overline{\omega }}^{j-1}a_{j1}= \omega ^{2}\sum _{j=1}^{3}{\overline{\omega }}^{j-1}a_{j3}\\ \sum _{j=1}^{3}{\overline{\omega }}^{j-1} a_{j2}=\omega \sum _{j=1}^{3}{\overline{\omega }}^{j-1}a_{j3}\\ \end{aligned} \right. , \end{aligned}$$

(74)

and

$$\begin{aligned} \left\{ \begin{aligned} \sum _{j=1}^{3}({\overline{\omega }}^{2})^{j-1}a_{j1} =\omega \sum _{j=1}^{3}({\overline{\omega }}^{2})^{j-1}a_{j3}\\ \sum _{j=1}^{3}({\overline{\omega }}^{2})^{j-1}a_{j2} =\omega ^{2}\sum _{j=1}^{3}({\overline{\omega }}^{2})^{j-1}a_{j3}\\ \end{aligned} \right. , \end{aligned}$$

(75)

respectively. By Eqs. (73), (74) and (75), we have

$$\begin{aligned} \left\{ \begin{aligned} a_{11}^{k}=a_{33}^{k}\\ a_{21}^{k}=a_{13}^{k}\\ a_{31}^{k}=a_{23}^{k}\\ \end{aligned} \right. , \end{aligned}$$

(76)

$$\begin{aligned} \left\{ \begin{aligned} a_{22}^{k}=a_{33}^{k}\\ a_{12}^{k}=a_{23}^{k}\\ a_{32}^{k}=a_{13}^{k}\\ \end{aligned} \right. . \end{aligned}$$

(77)

Similarly, because any two states of $\{(M_{k}\otimes I_{n\times n})|\phi _{n}\rangle$, $(M_{k}\otimes I_{n\times n})|\phi _{n+1}\rangle$ and $(M_{k}\otimes I_{n\times n})|\phi _{n+2}\rangle \}$, we have

$$\begin{aligned}&\left\{ \begin{aligned} a_{00}^{k}=a_{22}^{k}\\ a_{10}^{k}=a_{02}^{k}\\ a_{20}^{k}=a_{12}^{k}\\ \end{aligned} \right. , \end{aligned}$$

(78)

$$\begin{aligned}&\left\{ \begin{aligned} a_{11}^{k}=a_{22}^{k}\\ a_{01}^{k}=a_{12}^{k}\\ a_{21}^{k}=a_{02}^{k}\\ \end{aligned} \right. . \end{aligned}$$

(79)

By Eqs. (66), (76), (77), (78) and (79), we have

$$\begin{aligned} \left\{ \begin{aligned} a_{00}^{k}=a_{11}^{k}=a_{22}^{k}=a_{33}^{k}\qquad \qquad \qquad \quad \,\\ a_{12}^{k}=a_{13}^{k}=a_{21}^{k}=a_{23}^{k}=a_{31}^{k}=a_{32}^{k}=0\\ \end{aligned} \right. . \end{aligned}$$

(80)

Because $(M_{k}\otimes I_{n\times n})|\phi _{2n+m+2\sigma +1}\rangle$ and $(M_{k}\otimes I_{n\times n})$ $|\phi _{2n+m+2\sigma +2}\rangle$ are mutually orthogonal only on Alice’s side, i.e.,

$$\begin{aligned} \left\{ \begin{aligned} \langle \phi _{2n+m+2\sigma +1}|M_{k}^{\dag }M_{k}\otimes I_{n\times n}|\phi _{2n+m+2\sigma +2}\rangle =0\\ \langle \phi _{2n+m+2\sigma +2}|M_{k}^{\dag }M_{k}\otimes I_{n\times n}|\phi _{2n+m+2\sigma +1}\rangle =0\\ \end{aligned} \right. , \end{aligned}$$

(81)

we have

$$\begin{aligned} \left\{ \begin{aligned} a^{k}_{2\sigma +4,\,2\sigma +4}=a^{k}_{2\sigma +5,\,2\sigma +5}\\ a^{k}_{2\sigma +4,\,2\sigma +5}=a^{k}_{2\sigma +5,\,2\sigma +4}\\ \end{aligned} \right. , \end{aligned}$$

(82)

for $\sigma =0,\,\, 1,\,\, 2,\,\, \cdots ,\,\, d_{1}-3$. Similarly, because $(M_{k}\otimes I_{n\times n})|\phi _{n+2\sigma +3}\rangle$ and $(M_{k}\otimes I_{n\times n})|\phi _{n+2\sigma +4}\rangle$ are mutually orthogonal only on Alice’s side, i.e., we have

$$\begin{aligned} \left\{ \begin{aligned} a^{k}_{2\sigma +3,\,2\sigma +3}=a^{k}_{2\sigma +4,\,2\sigma +4}\\ a^{k}_{2\sigma +3,\,2\sigma +4}=a^{k}_{2\sigma +4,\,2\sigma +3}\\ \end{aligned} \right. , \end{aligned}$$

(83)

for $\sigma =0,\,\, 1,\,\, 2,\,\, \cdots ,\,\, d_{1}-3$.

Because any of $\{(M_{k}\otimes I_{n\times n})|\phi _{2n+m-2}\rangle$, $(M_{k}\otimes I_{n\times n})|\phi _{2n+m-1}\rangle$, $(M_{k}\otimes I_{n\times n})|\phi _{2n+m}\rangle \}$ is orthogonal to each of $\{(M_{k}\otimes I_{n\times n})|\phi _{2n+m+2\sigma +1}\rangle$, $(M_{k}\otimes I_{n\times n})|\phi _{2n+m+2\sigma +2}\rangle \}$ only on Alice’s side, we have

$$\begin{aligned}&\langle \phi _{t}|M^{\dag }_{k}M_{k}\otimes I_{n\times n}|\phi _{j}\rangle =0, \end{aligned}$$

(84)

$$\begin{aligned}&\langle \phi _{j}|M^{\dag }_{k}M_{k}\otimes I_{n\times n}|\phi _{t}\rangle =0, \end{aligned}$$

(85)

for $t=2n+m-2,\,2n+m-1,\,2n+m$ and $j=2n+m+2\sigma +1,\,2n+m+2\sigma +2.$ By Eqs. (84 and 85), we have

$$\begin{aligned} \left\{ \begin{aligned} a^{k}_{1,\,2\sigma +4}=a^{k}_{2,\,2\sigma +4}=a^{k}_{3,\,2\sigma +4}=0\\ a^{k}_{1,\,2\sigma +5}=a^{k}_{2,\,2\sigma +5}=a^{k}_{3,\,2\sigma +5}=0\\ \end{aligned} \right. \end{aligned}$$

(86)

and

$$\begin{aligned} \left\{ \begin{aligned} a^{k}_{2\sigma +4,\,1}=a^{k}_{2\sigma +4,\,2}=a^{k}_{2\sigma +4,\,3}=0\\ a^{k}_{2\sigma +5,\,1}=a^{k}_{2\sigma +5,\,2}=a^{k}_{2\sigma +5,\,3}=0\\ \end{aligned} \right. \end{aligned}$$

(87)

respectively, for $\sigma =0,\,\, 1,\,\, 2,\,\, \cdots ,\,\, d_{1}-3$.

Because $(M_{k}\otimes I_{n\times n})|\phi _{2n+m+2\sigma +1}\rangle$ and $(M_{k}\otimes$ $I_{n\times n})|\phi _{2n+m+2\sigma +2}\rangle$ are orthogonal to $(M_{k}\otimes$ $I_{n\times n})|\phi _{2n+m+2\lambda +1}\rangle$ and $(M_{k}\otimes I_{n\times n})|\phi _{2n+m+2\lambda +2}\rangle$ only on Alice’s side for $\sigma ,$ $\lambda$ = 0, 1, 2, $\cdots$, $d_{1}-3$ and $\sigma \ne \lambda$, i.e.,

$$\begin{aligned} \left\{ \begin{aligned} \langle \phi _{2n+m+2\sigma +1}|M_{k}^{\dag }M_{k}\otimes I_{n\times n}|\phi _{2n+m+2\lambda +1}\rangle =0\\ \langle \phi _{2n+m+2\sigma +1}|M_{k}^{\dag }M_{k}\otimes I_{n\times n}|\phi _{2n+m+2\lambda +2}\rangle =0\\ \langle \phi _{2n+m+2\sigma +2}|M_{k}^{\dag }M_{k}\otimes I_{n\times n}|\phi _{2n+m+2\lambda +1}\rangle =0\\ \langle \phi _{2n+m+2\sigma +2}|M_{k}^{\dag }M_{k}\otimes I_{n\times n}|\phi _{2n+m+2\lambda +2}\rangle =0\\ \end{aligned} \right. , \end{aligned}$$

so we have

$$\begin{aligned} \left\{ \begin{aligned} \sum _{i=2\lambda +4}^{2\lambda +5}a_{2\sigma +4,i}^{k}+ \sum _{i=2\lambda +4}^{2\lambda +5}a_{2\sigma +5,i}^{k}=0\qquad \qquad \quad \\ \sum _{i=2\lambda +4}^{2\lambda +5}(-1)^{i}a_{2\sigma +4,i}^{k}+ \sum _{i=2\lambda +4}^{2\lambda +5}(-1)^{i}a_{2\sigma +5,i}^{k} =0\quad \\ \sum _{i=2\lambda +4}^{2\lambda +5}a_{2\sigma +4,i}^{k}- \sum _{i=2\lambda +4}^{2\lambda +5}a_{2\sigma +5,i}^{k} =0\qquad \qquad \quad \\ \sum _{i=2\lambda +4}^{2\lambda +5}(-1)^{i}a_{2\sigma +4,i}^{k}+ \sum _{i=2\lambda +4}^{2\lambda +5}(-1)^{i+1}a_{2\sigma +5,i}^{k} =0\\ \end{aligned} \right. . \end{aligned}$$

Thus we get

$$\begin{aligned} \begin{aligned} a_{2\sigma +4,2\lambda +4}^{k}=a_{2\sigma +4,2\lambda +5}^{k}=0,\\ a_{2\sigma +5,2\lambda +4}^{k}=a_{2\sigma +5,2\lambda +5}^{k}=0, \end{aligned} \end{aligned}$$

(88)

where $\sigma ,$ $\lambda$ = 0, 1, 2, $\cdots$, $d_{1}-3$ and $\sigma \ne \lambda$. Similarly, because $(M_{k}\otimes I_{n\times n})|\phi _{n+2\sigma +3}\rangle$ and $(M_{k}\otimes I_{n\times n})|\phi _{n+2\sigma +4}\rangle$ are orthogonal to $(M_{k}\otimes I_{n\times n})|\phi _{n+2\lambda +3}\rangle$ and $(M_{k}\otimes I_{n\times n})|\phi _{n+2\lambda +4}\rangle$ for $\sigma ,$ $\lambda$ = 0, 1, 2, $\cdots$, $d_{1}-3$ and $\sigma \ne \lambda$, we have

$$\begin{aligned} \begin{aligned} a_{2\sigma +3,2\lambda +3}^{k}=a_{2\sigma +3,2\lambda +4}^{k}=0,\\ a_{2\sigma +4,2\lambda +3}^{k}=a_{2\sigma +4,2\lambda +4}^{k}=0, \end{aligned} \end{aligned}$$

(89)

where $\sigma ,$ $\lambda$ = 0, 1, 2, $\cdots$, $d_{1}-3$ and $\sigma \ne \lambda$.

By Eqs. (66), (69), (80), (82), (83), (86), (87), (88) and (89), we get

$$\begin{aligned} \begin{aligned} M_{k}^{\dag }M_{k}= \left[ \begin{array}{cccc} a_{00}^{k} &{}0 &{}\cdots &{}0\\ 0 &{}a_{00}^{k} &{}\cdots &{}0\\ \vdots &{}\vdots &{}\ddots &{}\vdots \\ 0 &{}0 &{}\cdots &{}a_{00}^{k}\\ \end{array} \right] _{m\times m.} \end{aligned} \end{aligned}$$

This means that Alice only can perform a trivial measurement to preserve the orthogonality of the post-measurement states. So does Bob because of the symmetry of the set of these states. Therefore, the set is nonlocal, i.e., these states cannot be perfectly discriminated by LOCC. This completes the proof. $\square$