Optimality of entanglement witnesses constructed from arbitrary permutations

Abstract

The optimality of a class of entanglement witnesses constructed from any permutation $\pi$ for any $n\otimes n$ bipartite systems is investigated. A necessary and sufficient condition is presented for such entanglement witnesses to be optimal, and thus, a class of optimal entanglement witnesses constructed by permutations is obtained.

Appendix: A proof of main theorem

Here, we give a proof of the main theorem. Before doing it, let us recall some notions and give several lemmas needed.

Let $l$, $k\in \mathbb {N}$ (the set of all natural numbers), and let $A_{1},\ldots , A_{k}$, $C_{1},\ldots , C_{l}\in {\mathcal B}(H$, $K$). If, for each $|\psi \rangle \in H$, there exists an $l\times k$ complex matrix $(\alpha _{ij}(|\psi \rangle ))$ (depending on $|\psi \rangle$) such that $C_{i}|\psi \rangle =\sum _{j=1}^{k}\alpha _{ij}(|\psi \rangle )A_{j}|\psi \rangle$, $i=1,2,\ldots ,l,$ we say that $(C_{1},\ldots ,C_{l})$ is a locally linear combination of $(A_{1},\ldots ,A_{k})$ and $(\alpha _{ij}(|\psi \rangle ))$ is called a local coefficient matrix at $|\psi \rangle$. Furthermore, if for every $|\psi \rangle \in H$, the associated local coefficient matrix $(\alpha _{ij}(|\psi \rangle ))$ can be chosen so that its operator norm $\Vert (\alpha _{ij}(|\psi \rangle ))\Vert =\sup \{\Vert (\alpha _{ij}(|\psi \rangle )|x\rangle \Vert : |x\rangle \in {\mathbb C}^\mathrm{k}, \Vert |x\rangle \Vert \le 1\}\le 1$, we say that $(C_{1},\ldots ,C_{l})$ is a contractive locally linear combination of $(A_{1},\ldots ,A_{k})$; if there is a matrix $(\alpha _{ij})$ such that $C_{i}=\sum _{j=1}^{k}\alpha _{ij}A_{j}$ for all $i$, we say that $(C_{1},\ldots ,C_{l})$ is a linear combination of $(A_{1},\ldots ,A_{k})$ with coefficient matrix $(\alpha _{ij})$ (see [8, 9] for more details).

Lemma 1

([8, 9]) Let $H$ and $K$ be complex Hilbert spaces of any dimension and $\Phi : {\mathcal B}(H)\rightarrow {\mathcal B}(K)$ a linear map defined by $\Phi (X) =\sum _{i=1}^{k}C_{i}XC_{i}^{\dagger }-\sum _{j=1}^{l}D_{j}XD_{j}^{\dagger }$ for all $X\in {\mathcal B}(H)$, where $C_1,\ldots ,C_k; D_1,\ldots , D_l\in {\mathcal B}(H,K)$. Then, $\Phi$ is positive if and only if $(D_{1},\ldots ,D_{l})$ is a contractive locally linear combination of $(C_{1},\ldots ,C_{k})$. Furthermore, $\Phi$ is completely positive if and only if $(D_{1},\ldots ,D_{l})$ is a linear combination of $(C_{1},\ldots ,C_{k})$ with a contractive coefficient matrix and, in turn, if and only if there exist $E_1, E_2, \ldots , E_r$ in $\mathrm{span}\{C_{1},\ldots ,C_{k}\}$ such that $\Phi =\sum _{i=1}^r E_i(\cdot )E_i^\dagger .$

The following lemma comes from [23], which is useful to establish our key lemma, that is, Lemma 3 below, and is also useful in the proof of the main theorem.

Lemma 2

Let $s,M$ be positive numbers and $f(u_1,u_2,\ldots ,u_m)$ a function in $m$-variable defined by

$$\begin{aligned} f(u_1,u_2,\ldots ,u_m)=\frac{1}{s+u_1}+\frac{1}{s+u_2}+\cdots +\frac{1}{s+u_m} \end{aligned}$$

on the region $u_i>0$ with $u_1u_2\ldots u_m=M^m$, $i=1,2,\ldots ,m$. Then, $\sup f(u_1,u_2,\ldots ,u_m)=\max \{\frac{m-1}{s}, \frac{m}{s+M}\}.$ Moreover, we have

$$\begin{aligned} \max f(u_1,u_2,\ldots ,u_m)=f(M,M,\ldots ,M)=\frac{m}{s+M} \quad \text{ if }\ M\le \frac{s}{m-1} \end{aligned}$$

and

$$\begin{aligned} \sup f(u_1,u_2,\ldots ,u_m)= \frac{m-1}{s} \quad \text{ if }\ \ M>\frac{s}{m-1}. \end{aligned}$$

The next lemma is crucial for our purpose.

Lemma 3

Let $t$ be a fixed number with $0<t<1$ and $n$ a positive integer with $n\ge 2$. Let $x_1,x_2,\ldots ,x_n$ be any positive numbers with $\prod _{i=1}^nx_i=1$ and $(x_1,x_2,\ldots ,x_n)\not =(1,1,\ldots ,1)$. Then,

$$\begin{aligned} \frac{1-\sum _{i=1}^n\frac{1}{n-t+tx_i}}{\left( 1-\sum _{i=1}^n\frac{1}{n-t+tx_i}\right) \left( \frac{1}{n-t+tx_1}+\frac{1}{n-t+tx_2}\right) +\left( \frac{1}{n-t+tx_1}-\frac{1}{n-t+tx_2}||\right) ^2}\ge 1-t. \end{aligned}$$

Proof

Let $f$ be the function in $n$ variables defined by

$$\begin{aligned} \begin{array}{rl}f(x_1,x_2,\ldots ,x_n) =\frac{1-\sum _{i=1}^n\frac{1}{n-t+tx_i}}{\left( 1-\sum _{i=1}^n\frac{1}{n-t+tx_i}\right) \left( \frac{1}{n-t+tx_1}+\frac{1}{n-t+tx_2}\right) +\left( \frac{1}{n-t+tx_1}-\frac{1}{n-t+tx_2}\right) ^2}.\end{array} \end{aligned}$$

If $n=2$, then for any $x_1,x_2>0$ with $x_1x_2=1$ and $(x_1,x_2)\not =(1,1)$, we have

$$\begin{aligned} \begin{array}{rl}f(x_1,x_2) =&{}\frac{1-\sum _{i=1}^2\frac{1}{2-t+tx_i}}{\left( 1-\sum _{i=1}^2\frac{1}{2-t+tx_i}\right) \left( \frac{1}{2-t+tx_1}+\frac{1}{2-t+tx_2}\right) +\left( \frac{1}{2-t+tx_1}-\frac{1}{2-t+tx_2}\right) ^2}\\ =&{}\frac{1-\frac{1}{2-t+tx_1}-\frac{1}{2-t+tx_2}}{\frac{1}{2-t+tx_1}+\frac{1}{2-t+tx_2}-\frac{4}{(2-t+tx_1)(2-t+tx_2)}} =\frac{2(t-1)+(1-t)(x_1+x_2)}{x_1+x_2-2}=1-t.\end{array} \end{aligned}$$

Obviously, the lemma is true.

In the sequel, assume that $n\ge 3$. Since $(x_1,x_2,\ldots ,x_n)\not =(1,1,\ldots ,1)$, we have $1-\sum _{i=1}^n\frac{1}{n-t+tx_i}\not =0$, and

$$\begin{aligned} f(x_1,x_2,\ldots ,x_n)= \frac{1}{\frac{1}{n-t+tx_1}+\frac{1}{n-t+tx_2} +\frac{\left( \frac{1}{n-t+tx_1}-\frac{1}{n-t+tx_2}\right) ^2}{1-\sum _{i=1}^n\frac{1}{n-t+tx_i}}}, \end{aligned}$$

(3)

where $t$ is fixed with $0<t<1$, $x_1,x_2,\ldots ,x_n$ are any positive numbers with $\prod _{i=1}^nx_i=1$, and $(x_1,x_2,\ldots ,x_n)\not =(1,1,\ldots ,1)$.

Now fix $x_1$ and $x_2$; then $\prod _{i=3}^nx_i=\frac{1}{x_1x_2}$. Using Lemma 2 with $s=n-t$, $M=\frac{t}{\root n-2 \of {x_1x_2}}$ and $m=n-2$, we obtain

$$\begin{aligned} \max \sum _{i=3}^n\frac{1}{n-t+tx_i}= & {} \sum _{i=3}^n\frac{1}{n-t+\frac{t}{\root n-2 \of {x_1x_2}}}\nonumber \\= & {} \frac{n-2}{n-t+\frac{t}{\root n-2 \of {x_1x_2}}}\ \ \mathrm{for\ all}\ \ x_3,\ldots ,x_n>0 \end{aligned}$$

(4)

if $x_1x_2\ge (\frac{tn-3t}{n-t})^{n-2}$; and

$$\begin{aligned} \mathrm{sup}\sum _{i=3}^n\frac{1}{n-t+tx_i}=\frac{n-3}{n-t}\ \ \ \mathrm{for\ \ all}\ \ x_3,\ldots ,x_n>0 \end{aligned}$$

(5)

if $x_1x_2<(\frac{tn-3t}{n-t})^{n-2}$.

Assume that $0<x_1x_2<(\frac{tn-3t}{n-t})^{n-2}$; then Eq. (5) holds. Thus, for any positive numbers $x_1,x_2\ldots ,x_n$ satisfying $\prod _{i=1}^nx_i=1$ and $0<x_1x_2<(\frac{tn-3t}{n-t})^{n-2}$, Eqs. (5) and (3) yield

$$\begin{aligned} \begin{array}{rl}&{}f(x_1,x_2,\ldots ,x_n)\\ \ge &{}\frac{1}{\frac{1}{n-t+tx_1}+\frac{1}{n-t+tx_2} +\frac{\left( \frac{1}{n-t+tx_1}-\frac{1}{n-t+tx_2}\right) ^2}{1-\frac{1}{n-t+tx_1}-\frac{1}{n-t+tx_2}-\frac{n-3}{n-t}}}\\ &{}\quad =\frac{1-\frac{1}{n-t+tx_1}-\frac{1}{n-t+tx_2}-\frac{n-3}{n-t}}{\left( 1-\frac{1}{n-t+tx_1}-\frac{1}{n-t+tx_2}-\frac{n-3}{n-t}\right) \left( \frac{1}{n-t+tx_1}+\frac{1}{n-t+tx_2}\right) +\left( \frac{1}{n-t+tx_1}-\frac{1}{n-t+tx_2}\right) ^2}.\end{array} \end{aligned}$$

Define a function

$$\begin{aligned} g(x_1,x_2)=\frac{1-\frac{1}{n-t+tx_1}-\frac{1}{n-t+tx_2}-\frac{n-3}{n-t}}{\left( 1-\frac{1}{n-t+tx_1}-\frac{1}{n-t+tx_2}-\frac{n-3}{n-t}\right) \left( \frac{1}{n-t+tx_1}+\frac{1}{n-t+tx_2}\right) +\left( \frac{1}{n-t+tx_1}-\frac{1}{n-t+tx_2}\right) ^2}, \end{aligned}$$

where $x_1,x_2$ are any positive numbers with $x_1x_2<(\frac{tn-3t}{n-t})^{n-2}$. To prove that the lemma is true in this case, one only needs to check that $g(x_1,x_2)\ge 1-t$ for all $x_{i}>0$ $(i=1,2)$ with $x_1x_2<(\frac{tn-3t}{n-t})^{n-2}$.

In fact, since the denominator of $g(x_1,x_2)$ is not zero, a computation shows that

$$\begin{aligned} \begin{array}{rl}&{}g(x_1,x_2)\ge 1-t\\ \Leftrightarrow &{}1-\frac{1}{n-t+tx_1}-\frac{1}{n-t+tx_2}-\frac{n-3}{n-t}\ge \left[ \left( 1-\frac{1}{n-t+tx_1}-\frac{1}{n-t+tx_2}-\frac{n-3}{n-t}\right) \right. \\ &{}\left. \left( \frac{1}{n-t+tx_1}+\frac{1}{n-t+tx_2}\right) +\left( \frac{1}{n-t+tx_1}-\frac{1}{n-t+tx_2}\right) ^2\right] (1-t)\\ \Leftrightarrow &{}h(x_1,x_2)\ge 0,\end{array} \end{aligned}$$

where

$$\begin{aligned} \begin{array}{rl} h(x_1,x_2)=&(1-t)(n-t)(n-2+t)+t(2n-tn+2t-3)(x_1+x_2)+t^2(3-t)x_1x_2. \end{array} \end{aligned}$$

Note that $(1-t)(n-t)(n-2+t)$, $2nt-nt^2+2t^2-3t$ and $t^2(3-t)$ are all positive. So, $h(x_1,x_2)>0$ for all positive numbers $x_1$ and $x_2$. This implies that the lemma holds whenever $0<x_1x_2<(\frac{tn-3t}{n-t})^{n-2}$.

Now, consider the case $x_1x_2\ge (\frac{tn-3t}{n-t})^{n-2}$. In this case, Eq. (4) holds. Combining Eqs. (3–4), for any positive numbers $x_1,x_2\ldots ,x_n$ with $\prod _{i=1}^nx_i=1$ and $x_1x_2\ge (\frac{tn-3t}{n-t})^{n-2}$, we have

$$\begin{aligned} \begin{array}{rl}&{}f(x_1,x_2,\ldots ,x_n)\\ &{}\quad \ge \frac{1-\frac{1}{n-t+tx_1}-\frac{1}{n-t+tx_2}-\frac{n-2}{n-t+\frac{t}{\root n-2 \of {x_1x_2}}}}{\left( 1-\frac{1}{n-t+tx_1}-\frac{1}{n-t+tx_2} -\frac{n-2}{n-t+\frac{t}{\root n-2 \of {x_1x_2}}}\right) \left( \frac{1}{n-t+tx_1} +\frac{1}{n-t+tx_2}\right) +\left( \frac{1}{n-t+tx_1}-\frac{1}{n-t+tx_2}\right) ^2}.\end{array} \end{aligned}$$

Define a function

$$\begin{aligned} \begin{array}{rl}&{}g(x_1,x_2,x_3)\\ &{}\quad =\frac{1-\frac{1}{n-t+tx_1}-\frac{1}{n-t+tx_2}-\frac{n-2}{n-t+tx_3}}{\left( 1-\frac{1}{n-t+tx_1}-\frac{1}{n-t+tx_2}-\frac{n-2}{n-t+tx_3}\right) \left( \frac{1}{n-t+tx_1}+\frac{1}{n-t+tx_2}\right) +\left( \frac{1}{n-t+tx_1}-\frac{1}{n-t+tx_2}\right) ^2},\end{array} \end{aligned}$$

where $x_1,x_2,x_3$ are any positive numbers with $x_1x_2x_3^{n-2}=1$ and $x_1x_2\ge (\frac{tn-3t}{n-t})^{n-2}$. Still, we only need to check that the minimum of $g(x_1,x_2,x_3)$ is $1-t$ on the region $x_{i}>0$ with $x_{1}x_{2}x_3^{n-2}=1$ and $x_1x_2\ge (\frac{tn-3t}{n-t})^{n-2}$.

Also, note that the denominator of $g(x_1,x_2,x_3)$ is not zero whenever $(x_1,x_2,x_3)\not =(1,1,1)$. A computation shows that

$$\begin{aligned} g(x_1,x_2,x_3)\ge 1-t\Leftrightarrow h(x_1,x_2,x_3)\ge 0, \end{aligned}$$

where

$$\begin{aligned} \begin{array}{rl}h(x_1,x_2,x_3)=&{}(t^2-2t+2n-n^2)+(2t-2+n-nt)x_1+(2t-2+n-nt)x_2\\ &{}+\,(n^2-t^2-4n+4)x_3+(2t-t^2)x_1x_2\\ {} &{}+\,(n-2)tx_2x_3+(n-2)tx_1x_3+t^2x_1x_2x_3.\end{array} \end{aligned}$$

Let

$$\begin{aligned} L(x_1,x_2,x_3,\lambda )=h(x_1,x_2,x_3)+\lambda (x_{1}x_{2}x_3^{n-2}-1). \end{aligned}$$

By the method of Lagrange multipliers, we have the system

$$\begin{aligned} {\left\{ \begin{array}{ll}\mathrm{(i)}&{} L_{x_1}^\prime =(2t-2+n-nt)+(2t-t^2)x_2+(n-2)tx_3+t^2x_2x_3+\lambda x_2x_3^{n-2}=0,\\ \mathrm{(ii)} &{}L_{x_2}^\prime =(2t-2+n-nt)+(2t-t^2)x_1+(n-2)tx_3+t^2x_1x_3+\lambda x_1x_3^{n-2}=0,\\ \mathrm{(iii)} &{}L_{x_3}^\prime =(n^2-t^2-4n+4)+(n-2)tx_2+(n-2)tx_1+t^2x_1x_2+\lambda (n-2) x_1x_2x_3^{n-3}=0,\\ \mathrm{(iv)} &{}L_\lambda ^\prime =x_1x_2x_3^{n-2}-1=0.\end{array}\right. } \end{aligned}$$

By (i) and (ii), one obtains $(x_2-x_1)(2t-t^2+t^2x_3+\lambda x_3^{n-2})=0,$ which implies that

$$\begin{aligned} \mathrm{either}\ \ \ \ x_1=x_2\ \ \ \ \mathrm{or}\ \ \ \ 2t-t^2+t^2x_3+\lambda x_3^{n-2}=0. \end{aligned}$$

If $2t-t^2+t^2x_3+\lambda x_3^{n-2}=0$, by (ii), one gets $x_3=\frac{t-1}{t}<0$, a contradiction. Hence, we must have $x_1=x_2$. Thus, (ii) and (iii) become

$$\begin{aligned} (2t-2+n-nt)+(2t-t^2)x_1+(n-2)tx_3+t^2x_1x_3+\lambda x_1x_3^{n-2}=0 \end{aligned}$$

and

$$\begin{aligned} (n^2-t^2-4n+4)+2(n-2)tx_1+t^2x_1^2+\lambda (n-2)x_1^2x_3^{n-3}=0. \end{aligned}$$

Combining the above two equations and applying (iv), we get

$$\begin{aligned} \begin{array}{rl}&{}(t^2+4n-n^2-4)+(n^2t-6nt+8t)x_1+(nt^2-3t^2)x_1^2\\ &{}+\,(n^2-n^2t+4nt-4n-4t+4)\frac{x_1}{x_3}+(2nt-4t-nt^2+2t^2)\frac{x_1^2}{x_3}=0.\end{array} \end{aligned}$$

Letting $y>0$ and $y=\root n-2 \of {x_1}$, then $x_1=y^{n-2}$ and $x_3=y^{-2}$ as $x_1^2x_3^{n-2}=1$. Replacing $x_1$ and $x_3$ by $y$ in the above equation, one has

$$\begin{aligned} \begin{array}{rl}&{}(2nt-4t-nt^2+2t^2)y^{2n-2}+(nt^2-3t^2)y^{2n-4}\\ &{}+\,(n^2-n^2t+4nt-4n-4t+4)y^n+(n^2t-6nt+8t)y^{n-2}+(t^2+4n-n^2-4)=0.\end{array} \end{aligned}$$

Solving the equation, we achieve

$$\begin{aligned}&(y-1)[(2nt-4t-nt^2+2t^2)(y^{2n-3}+y^{2n-4})+(2nt-4t-t^2)\\&\quad (y^{2n-5}+y^{2n-6}+\cdots +y^n)\\&+\,(n^2-4n-t^2+4)(y^{n-1}+y^{n-2}+\cdots +y+1)]=0. \end{aligned}$$

Note that $2nt-4t-nt^2+2t^2$, $2nt-4t-t^2$ and $n^2-4n-t^2+4$ are all positive. So, the solution of the above equation is $y=1$, which implies $x_1=x_2=x_3=1$. It follows that the function $h(x_1,x_2,x_3)$ takes its extremum at the point $(1,1,1)$. Moreover, it is easy to check that $(1,1,1)$ is the minimal point of $h(x_1,x_2,x_3)$. Hence, $h(x_1,x_2,x_3)\ge h(1,1,1)=0$ for all $x_{i}>0$ satisfying $x_{1}x_{2}x_3^{n-2}=1$ and $x_1x_2\ge (\frac{tn-3t}{n-t})^{n-2}$, $i=1,2,3$. This implies that the lemma holds whenever $x_1x_2\ge (\frac{tn-3t}{n-t})^{n-2}$.

Therefore, the inequality in Lemma 3 holds for all $x_i>0$ ($i=1,2,\ldots ,n$) with $\prod _{i=1}^nx_i=1$ and $(x_1,x_2,\ldots ,x_n)\not =(1,1,\ldots ,1)$. The proof is finished. $\square$

Now, we are in a position to give our proof of the main theorem.

Proof of Main Theorem

For $\pi =(\pi _1)(\pi _2) \ldots (\pi _k)$, with no loss of generality, assume $l(\pi _j)=n_{j}-n_{j-1}$ and $\pi _j(n_{j-1}+i)=(n_{j-1}+i+1)\ \mathrm{mod}(n_{j}-n_{j-1})$, $i=1,2,\ldots ,n_{j}-n_{j-1}$, $j=1,\ldots ,k$. Here, set $n_0=0$ and $n_k=n$.

By a similar approach to that of [18], Theorem 3.2] using Proposition 1, one can prove that the “if” part holds.

In the rest part of this paper, we will show that, if one of the conditions (i–iii) in the theorem is not satisfied, then $W_{{n,t,\pi }}$ is not optimal, and therefore, the “only if” part holds. To do this, by Proposition 1, one has to find a matrix $C_0\in M_n({\mathbb C})$ such that the linear map $A\mapsto \Phi _{n,t,\pi }(A)-C_0AC_0^\dagger$ is positive.

For $\pi =(\pi _1)(\pi _2)\cdots (\pi _k)$, still assume that $l(\pi _j)=n_{j}-n_{j-1}$ and $\pi _j(n_{j-1}+i)=(n_{j-1}+i+1)\ \mathrm{mod}(n_{j}-n_{j-1})$, $i=1,2,\ldots ,n_{j}-n_{j-1}$ and $j=1,\ldots ,k$, with $n_0=0$ and $n_k=n$. Let $\Psi _{C_0}$ be the map defined by

$$\begin{aligned} \begin{array}{rl}\Psi _{C_0}(A)=&{}\Phi _{n,t,\pi }(A)-C_0AC_0^\dag \\ =&{}(n-t)\sum _{i=1}^nE_{ii}AE_{ii}^\dagger +t\sum _{j=1}^k\sum _{i=n_{j-1}+1}^{n_j}E_{i,\pi _j(i)}AE_{i,\pi _j(i)}^\dagger -A-C_0AC_0^\dag \end{array} \end{aligned}$$

for all $A\in M_n({\mathbb C})$. If $\Psi _{C_0}$ is positive, by Lemma 1, for any unit vector $|x\rangle \in {\mathbb C}^n$, there exist scalars $\{ \alpha _{i}^j\}$, $\{ \beta _{i}^j\}$, $\{\delta _i^j\}$, $\{\gamma _i^j\}$ depending on $|x\rangle$ such that

$$\begin{aligned} |x\rangle =I|x\rangle =\sum _{j=1}^k\sum _{i=n_{j-1}+1}^{n_j}\alpha _i^j(\sqrt{n-t}E_{ii})|x\rangle +\sum _{j=1}^k\sum _{i=n_{j-1}+1}^{n_j}\beta _i^j\sqrt{t}E_{i,\pi _j(i)}|x\rangle , \end{aligned}$$

(6)

$$\begin{aligned} C_0|x\rangle =\sum _{j=1}^k\sum _{i=n_{j-1}+1}^{n_j}\delta _i^j(\sqrt{n-t}E_{ii})|x\rangle +\sum _{j=1}^k\sum _{i=n_{j-1}+1}^{n_j}\gamma _i^j\sqrt{t}E_{i,\pi _j(i)}|x\rangle \end{aligned}$$

(7)

and the matrix

$$\begin{aligned} F_{x}=\left( \begin{array}{ccccccccccc} \alpha _1^1&{} \alpha _2^1 &{} \cdots &{} \alpha _{n_1}^1&{} \alpha _{n_1+1}^2&{}\cdots &{}\alpha _n^k &{}\beta _1^1 &{} \beta _2^1 &{} \cdots &{}\beta _n^k\\ \delta _1^1 &{} \delta _2^1 &{} \cdots &{}\delta _{n_1}^1&{}\delta _{n_1+1}^2&{}\cdots &{}\delta _n^k &{}\gamma _1^1 &{} \gamma _2^1 &{} \cdots &{}\gamma _n^k \end{array}\right) \end{aligned}$$

is contractive. Note that

$$\begin{aligned} F_xF_x^\dag =\left( \begin{array}{cc} \sum _{j=1}^k\sum _{i=n_{j-1}+1}^{n_j}(|\alpha _i^j|^2+|\beta _i^j|^2)&{}\sum _{j=1}^k\sum _{i=n_{j-1}+1}^{n_j}(\alpha _i^j\bar{\delta _i^j}+\beta _i^j\bar{\gamma _i^j})\\ \sum _{j=1}^k\sum _{i=n_{j-1}+1}^{n_j}(\bar{\alpha _i^j}\delta _i^j+\bar{\beta _i^j}\gamma _i^j)&{}\sum _{j=1}^k\sum _{i=n_{j-1}+1}^{n_j}(|\delta _i^j|^2+|\gamma _i^j|^2) \end{array}\right) ,\nonumber \\ \end{aligned}$$

(8)

and $\Vert F_x\Vert \le 1$ if and only if $\Vert F_xF_x^\dag \Vert \le 1$.

In the sequel, for any unit vector $|x\rangle \in {\mathbb C}^n$, write $|x\rangle =(|x_1|e^{i\theta _1},|x_2|e^{i\theta _2},\ldots , |x_n|e^{i\theta _n})^T$. Then, $\sum _{i=1}^n|x_i|^2=1$. We will complete the proof of the theorem by considering two cases.

Case 1 $\pi$ is cyclic.

In this case, $\pi =(\pi _1)$, $l(\pi )=l(\pi _1)=n$ and $\pi (i)=(i+1)$ mod $n$. We will show that, for any $0<t<1$ and any $C_0=\mathrm{diag}(c,-c,0,\cdots ,0)$ with $0<c\le \sqrt{1-t}$, $\Psi _{C_0}$ is positive, and so $W_{{n,t,\pi }}$ is not optimal.

In the present case, Eqs. (6–8) can be reduced to

$$\begin{aligned} |x\rangle= & {} \sum _{i=1}^n\alpha _i(\sqrt{n-t}E_{ii})|x\rangle +\sum _{i=1}^n\beta _i\sqrt{t}E_{i,i+1}|x\rangle , \qquad \qquad \qquad \qquad (4^\prime )\\ C_0|x\rangle= & {} \sum _{i=1}^n\delta _i(\sqrt{n-t}E_{ii})|x\rangle +\sum _{i=1}^n\gamma _i\sqrt{t}E_{i,i+1}|x\rangle \qquad \qquad \qquad \qquad (5^\prime )\\ \end{aligned}$$

and

$$\begin{aligned} F_xF_x^\dag =\left( \begin{array}{cc} \sum _{i=1}^n(|\alpha _i|^2+|\beta _i|^2)&{}\sum _{i=1}^n(\alpha _i\bar{\delta _i}+\beta _i\bar{\gamma _i})\\ \sum _{i=1}^n(\bar{\alpha _i}\delta _i+\bar{\beta _i}\gamma _i)&{}\sum _{i=1}^n(|\delta _i|^2+|\gamma _i|^2) \end{array}\right) ,\qquad \qquad \qquad \qquad (6^\prime ) \end{aligned}$$

where $\alpha _i=\alpha _i^1$, $\beta _i=\beta _i^1$,$\delta _i=\delta _i^1$ and $\gamma _i=\gamma _i^1$ for $i=1,2,\ldots ,n$.

Subcase 1.1 $|x_i|=\frac{1}{\sqrt{n}}$ for all $i=1,2,\ldots ,n$.

In Eqs. (4$^\prime$–5$^\prime$), by taking

$$\begin{aligned} (\alpha _1,\alpha _2,\ldots ,\alpha _n) =\left( \frac{\sqrt{n-t}}{n},\frac{\sqrt{n-t}}{n},\ldots , \frac{\sqrt{n-t}}{n}\right) \end{aligned}$$

and

$$\begin{aligned} (\delta _1,\delta _2,\ldots ,\delta _n)=\left( \frac{\sqrt{n-t}c}{n}, -\frac{\sqrt{n-t}c}{n},0,\cdots ,0\right) , \end{aligned}$$

we get

$$\begin{aligned}&(\beta _1,\beta _2, \ldots ,\beta _n;\gamma _1,\gamma _2,\ldots ,\gamma _n)=\left( \frac{ {\sqrt{t}}x_1}{nx_2}, \frac{{\sqrt{t}}x_2}{nx_3},\ldots ,\frac{{\sqrt{t}}x_n}{nx_1};\frac{ {\sqrt{t}c}x_1}{nx_2},\right. \nonumber \\&\qquad \left. -\frac{{\sqrt{t}}cx_2}{nx_3},0,\cdots , 0\right) . \end{aligned}$$

So, $\sum _{i=1}^n(|\alpha _i|^2+|\beta _i|^2)=1$, $\sum _{i=1}^n(|\delta _i|^2+|\gamma _i|^2)=\frac{2c^2}{n}$, and $\sum _{i=1}^n(\alpha _i\overline{\delta _i}+\beta _i\overline{\gamma _i}) =0$. It follows that

$$\begin{aligned} F_{x}F_{x}^\dag =\left( \begin{array}{cc} \sum _{i=1}^n(|\alpha _i|^2+|\beta _i|^2)&{}\sum _{i=1}^n(\alpha _i\bar{\delta _i}+\beta _i\bar{\gamma _i})\\ \sum _{i=1}^n(\bar{\alpha _i}\delta _i+\bar{\beta _i}\gamma _i)&{}\sum _{i=1}^n(|\delta _i|^2+|\gamma _i|^2) \end{array}\right) =\left( \begin{array}{cc} 1&{} 0\\ 0&{}\frac{2c^2}{n}\end{array}\right) , \end{aligned}$$

which implies that $\Vert F_xF_x^\dag \Vert \le 1\Leftrightarrow c^2\le \frac{n}{2}.$ Hence,

$$\begin{aligned} c^2\le 1-t\Rightarrow \Vert F_xF_x^\dag \Vert \le 1. \end{aligned}$$

Subcase 1.2 $x_i\not =0$ for all $i=1,2,\ldots ,n$ and $(|x_1|,|x_2|,\ldots ,|x_n|)\not =(\frac{1}{\sqrt{n}},\frac{1}{\sqrt{n}},\ldots ,\frac{1}{\sqrt{n}})$.

Let $r_i=|\frac{x_i}{x_{i+1}}|^2$ for $i=1,2,\ldots , n$. Then, $r_i>0$ with $\prod _{i=1}^nr_i=1$. Take

$$\begin{aligned} (\alpha _1,\alpha _2,\ldots ,\alpha _n)=\left( \frac{\sqrt{n-t}r_1}{t+(n-t)r_1}, \frac{\sqrt{n-t}r_2}{t+(n-t)r_2},\ldots , \frac{\sqrt{n-t}r_n}{t+(n-t)r_n}\right) \end{aligned}$$

and

$$\begin{aligned} (\delta _1,\delta _2,\ldots ,\delta _n) =\left( \frac{\sqrt{n-t}r_1c}{t+(n-t)r_1}, -\frac{\sqrt{n-t}r_2c}{t+(n-t)r_2},0,\cdots ,0\right) . \end{aligned}$$

By Eqs. (4$^\prime$–5$^\prime$), one gets

$$\begin{aligned} (\beta _1,\beta _2,\ldots , \beta _n) =\left( \frac{\sqrt{r_1}e^{i(\theta _1-\theta _2)}}{t+(n-t)r_1}, \frac{\sqrt{r_2}e^{i(\theta _2-\theta _3)}}{t+(n-t)r_2},\ldots , \frac{\sqrt{r_n}e^{i(\theta _n-\theta _1)}}{t+(n-t)r_n}\right) \end{aligned}$$

and

$$\begin{aligned} (\gamma _1,\gamma _2,\ldots ,\gamma _n) =\left( \frac{\sqrt{tr_1}ce^{i(\theta _1-\theta _2)}}{t+(n-t)r_1}, -\frac{\sqrt{tr_2}ce^{i(\theta _2-\theta _3)}}{t+(n-t)r_2},0,\cdots ,0\right) . \end{aligned}$$

So,

$$\begin{aligned} \sum _{i=1}^n(|\alpha _i|^2+|\beta _i|^2)= & {} \sum _{i=1}^n\frac{r_i}{t+(n-t)r_i},\ \ \sum _{i=1}^n(|\delta _i|^2+|\gamma _i|^2)\nonumber \\= & {} \frac{r_1c^2}{t+(n-t)r_1}+\frac{r_2c^2}{t+(n-t)r_2} \end{aligned}$$

and

$$\begin{aligned} \sum _{i=1}^n(\alpha _i\bar{\delta _i}+\beta _i\bar{\gamma _i})= \frac{r_1c}{t+(n-t)r_1}-\frac{r_2c}{t+(n-t)r_2}. \end{aligned}$$

It follows that

$$\begin{aligned} F_{x}F_{x}^\dag =\left( \begin{array}{cc} \sum _{i=1}^n\frac{r_i}{t+(n-t)r_i}&{} \frac{r_1c}{t+(n-t)r_1}-\frac{r_2c}{t+(n-t)r_2}\\ \frac{r_1c}{t+(n-t)r_1}-\frac{r_2c}{t+(n-t)r_2}&{}\frac{r_1c^2}{t+(n-t)r_1}+\frac{r_2c^2}{t+(n-t)r_2}\end{array}\right) . \end{aligned}$$

Note that $\Vert F_{x}F_{x}^\dag \Vert \le 1$ if and only if its maximal eigenvalue $\lambda _{\mathrm{max}}\le 1$. By a calculation, it is easily checked that $\lambda _{\mathrm{max}}\le 1$ holds if and only if

$$\begin{aligned} c^2\le \frac{1-\sum _{i=1}^n \frac{r_i}{t+(n-t)r_i}}{(1-\sum _{i=1}^n\frac{r_i}{t+(n-t)r_i}) \left( \frac{r_1}{t+(n-t)r_1}+ \frac{r_2}{t+(n-t)r_2}\right) +\left( \frac{r_1}{t+(n-t)r_1}- \frac{r_2}{t+(n-t)r_2}\right) ^2},\nonumber \\ \end{aligned}$$

(9)

where $r_1,r_2,\ldots ,r_n>0$ with $\prod _{i=1}^nr_i=1$ and $(r_1,r_2,\ldots ,r_n)\not =(1,1,\ldots ,1)$. Let

$$\begin{aligned} g(r_1,r_2,\ldots ,r_n)=\frac{1-\sum _{i=1}^n\frac{r_i}{t+(n-t)r_i}}{(1-\sum _{i=1}^n\frac{r_i}{t+(n-t)r_i}) \left( \frac{r_1}{t+(n-t)r_1}+\frac{r_2}{t+(n-t)r_2}\right) +\left( \frac{r_1}{t+(n-t)r_1}-\frac{r_2}{t+(n-t)r_2}\right) ^2}. \end{aligned}$$

Replacing $r_i$ by $\frac{1}{r_i}$ in $g(r_1,r_2,\ldots ,r_n)$ gives

$$\begin{aligned} g(r_1,r_2,\ldots ,r_n)=\frac{1-\sum _{i=1}^n\frac{1}{(n-t)+tr_i}}{(1-\sum _{i=1}^n\frac{1}{(n-t)+tr_i}) \left( \frac{1}{(n-t)+tr_1}+\frac{1}{(n-t)+tr_2}\right) +\left( \frac{1}{(n-t)+tr_1}-\frac{1}{(n-t)+tr_2}\right) ^2}. \end{aligned}$$

Now, applying Lemma 3 to the function $g$, we see that $g(r_1,r_2,\ldots ,r_n)\ge 1-t$ holds for all positive numbers $r_1,r_2,\ldots ,r_n$ with $\prod _{i=1}^nr_i=1$ and $(r_1,r_2,\ldots ,r_n)\not =(1,1,1\ldots ,1)$. This and Eq. (7) imply

$$\begin{aligned} c^2\le 1-t\Rightarrow \lambda _{\mathrm{max}}\le 1\Rightarrow \Vert F_{x}F_{x}^\dag \Vert \le 1. \end{aligned}$$

Combining Subcases 1.1–1.2 and by the proof in [8, 9] of Lemma 1, we have proved that, for any matrix $C_0=\mathrm{diag}(c,-c,0,\cdots ,0)$ with $0<c^2\le 1-t$, it holds that

$$\begin{aligned}&\Psi _{C_0}(|y\rangle \langle y|)\ge 0\ \mathrm{for \ all\ unit\ vectors}\ |y\rangle =(y_1,y_2,\ldots ,y_n)^T\ \mathrm{with} \ y_i\not =0\nonumber \\&\quad (i=1,\ldots ,n). \end{aligned}$$

(10)

Subcase 1.3 There exists at least one $x_{i_0}=0$ for some $i_0\in \{1,2,\ldots ,n\}$.

In this case, it is easily seen that there is a sequence of unit vectors

$$\begin{aligned} \{|y_m\rangle =(y_1^m,\ldots ,y_i^m,\ldots ,y_n^m)^T\}_{m=1}^\infty \ \ \mathrm{with}\ \ y_i^m\not =0\ \ \mathrm{for\ all}\ \ i,m \end{aligned}$$

such that $\lim _{m\rightarrow \infty }|y_m\rangle =|x\rangle$. So, $\lim _{m\rightarrow \infty }|y_m\rangle \langle y_m|=|x\rangle \langle x|$ under norm topology. Note that $\Psi _{C_0}(|y_m\rangle \langle y_m|)\ge 0$ for each $|y_m\rangle$ by Eq. (10) and $\Psi _{C_0}$ is continuous. It follows that

$$\begin{aligned} \lim _{m\rightarrow \infty }\Psi _{C_0}(|y_m\rangle \langle y_m|)=\Psi _{C_0}(|x\rangle \langle x|)\ge 0, \end{aligned}$$

that is, $\Psi$ is positive on any rank one projection $|x\rangle \langle x|$. Hence, $\Psi _{C_0}$ is positive. Then, by Proposition 1, $W_{{n,t,\pi }}$ is not optimal whenever $0<t<1$. The proof of Case 1 is finished.

Case 2 $\pi$ is not cyclic.

We will prove it by considering five subcases.

Subcase 2.1 $l=l(\pi _i)=1$ for all $i=1,2,\ldots ,k$.

In this case, $\pi =\mathrm{id}$, and so $\Phi _{n,t,\pi }$ is a completely positive linear map. It follows that $W_{n,t,\pi }$ is positive which is not an entanglement witness.

Subcase 2.2 $l(\pi )=2$.

In this case, $\pi ^2=\mathrm{id}$. By [20], $W_{\Phi _{n,t,\pi }}=Q_1+Q_2^T$ is decomposable with $Q_i\ge 0$ and $Q_i\not =0$, $i=1,2$. So, $W_{n,t,\pi }$ is not optimal because $Q_2^T$ is an entanglement witness [16].

Subcase 2.3 There exists some cycle $\pi _j$ such that $2\le l(\pi _j)<l$ with $l=l(\pi )\ge 3$.

For the simplicity, assume that $2\le l(\pi _1)<l$. We will prove that, for any $0<t\le \frac{n}{l}$, by taking $C_0=\mathrm{diag}(c,-c,0,\cdots ,0)$ with $0<c\le \sqrt{1-\frac{l(\pi _1)}{n}t}$, $\Psi _{C_0}$ is positive, and so $W_{{n,t,\pi }}$ is not optimal for any $0\le t\le \frac{n}{l}$. Indeed, similar to the argument of Subcase 1.3 in Case 1, one only needs to check whether there exists a coefficient matrix $F_x$ such that $\Vert F_x\Vert \le 1$ for those vectors $|x\rangle =(x_1,x_2,\ldots ,x_n)$ with $x_i\not =0$, $i=1,2,\ldots , n$.

Letting $r_{i}^{j}=|\frac{x_i}{x_{i+1}}|^2$ ($i=n_{j-1}+1,n_{j-1}+2,\ldots ,n_j$; $j=1,2,\ldots , k$), then $\prod _{i=n_{j-1}+1}^{n_j}r_{i}^{j}=1$. Take $\alpha _i^j=\frac{\sqrt{n-t}r_{i}^{j}}{t+(n-t)r_{i}^{j+1}}$ for all $i,j$, $\delta _1^1=\frac{c\sqrt{n-t}r_1^1}{t+(n-t)r_1^1}$, $\delta _2^1=\frac{-c\sqrt{n-t}r_2^1}{t+(n-t)r_2^1}$ and $\delta _i^j=0$ for other $i,j$. By Eqs. (6–7), one has $\beta _i^j=\frac{\sqrt{tr_{i}^{j}}e^{i(\theta _i-\theta _{i+1})}}{t+(n-t)r_{i}^{j}}$ for all $i,j$, $\gamma _1^1=\frac{c\sqrt{tr_1^1}e^{i(\theta _1-\theta _2)}}{t+(n-t)r_1^1}$, $\gamma _2^1=\frac{-c\sqrt{tr_2^1}e^{i(\theta _2-\theta _3)}}{t+(n-t)r_2^1}$ and $\gamma _i^j=0$ for other $i,j$, and so

$$\begin{aligned} \begin{array}{rl}F_{x}F_{x}^\dag =&{}\left( \begin{array}{cc} \sum _{j=1}^{k}\sum _{i=n_{j-1}+1}^{n_j}\frac{r_{i}^{j}}{t+(n-t)r_{i}^{j}}&{} \frac{r_1^{1}c}{t+(n-t)r_1^{1}}-\frac{r_2^{1}c}{t+(n-t)r_2^{1}}\\ \frac{r_1^{1}c}{t+(n-t)r_1^{1}}-\frac{r_2^{1}c}{t+(n-t)r_2^{1}}&{}\frac{r_1^{1}c^2}{t+(n-t)r_1^{1}}+\frac{r_2^{1}c^2}{t+(n-t)r_2^{1}}\end{array}\right) \\ =&{}F_1F_1^\dag +\left( \begin{array}{cc} \sum _{j=2}^{k}\sum _{i=n_{j-1}+1}^{n_j}\frac{r_{i}^{j}}{t+(n-t)r_{i}^{j}}&{} 0\\ 0&{}0\end{array}\right) ,\end{array} \end{aligned}$$

where

$$\begin{aligned} \begin{array}{rl}F_1F_1^\dag =&{}\left( \begin{array}{cc} \sum _{i=1}^{n_{1}}\frac{r_{i}^{1}}{t+(n-t)r_{i}^{1}}&{} \frac{r_1^{1}c}{t+(n-t)r_1^{1}}-\frac{r_2^{1}c}{t+(n-t)r_2^{1}}\\ \frac{r_1^{1}c}{t+(n-t)r_1^{1}}-\frac{r_2^{1}c}{t+(n-t)r_2^{1}}&{}\frac{r_1^{1}c^2}{t+(n-t)r_1^{1}}+\frac{r_2^{1}c^2}{t+(n-t)r_2^{1}}\end{array}\right) \\ =&{}\frac{l(\pi _1)}{n}\left( \begin{array}{cc} \sum _{i=1}^{n_{1}}\frac{r_i^{1}}{s+(l(\pi _1)-s)r_i^{1}}&{}\frac{r_1^{1}}{s+(l(\pi _1)-s)r_1^{1}}-\frac{r_2^{1}}{s+(l(\pi _1)-s)r_2^{1}}\\ \frac{r_1^{1}}{s+(l(\pi _1)-s)r_1^{1}}-\frac{r_2^{1}}{s+(l(\pi _1)-s)r_2^{1}}&{}\frac{r_1^{1}}{s+(l(\pi _1)-s)r_1^{1}} +\frac{r_2^{1}}{s+(l(\pi _1)-s)r_2^{1}}\end{array}\right) \end{array} \end{aligned}$$

and $s=\frac{l(\pi _1)}{n}t\le \frac{l(\pi _1)}{n}\frac{n}{l}=\frac{l(\pi _1)}{l}<1$. So, by taking $c$ with

$$\begin{aligned} c^2\le 1-s=1-\frac{l(\pi _1)}{n}t \end{aligned}$$

and using similar argument to Subcase 1.2, we obtain $\Vert F_1F_1^\dag \Vert \le \frac{l(\pi _1)}{n}$. Since $\prod _{i=n_{j-1}+1}^{n_j}r_{i}^{j}=1$, by Lemma 2, one gets

$$\begin{aligned} \sum _{i=n_{j-1}+1}^{n_j}\frac{r_{i}^{j}}{t+(n-t)r_{i}^{j}} =\sum _{i=n_{j-1}+1}^{n_j}\frac{1}{t\frac{1}{r_{i}^{j}}+(n-t)}\le \max \left\{ \frac{n_j-n_{j-1}-1}{n-t},\frac{n_j-n_{j-1}}{n}\right\} . \end{aligned}$$

Note that $n_j-n_{j-1}\le l(\pi )=l$ and $t\le \frac{n}{l}$, and we see that $\frac{n_{j}-n_{j-1}-1}{n-t}\le \frac{n_{j}-n_{j-1}}{n}$, and so

$$\begin{aligned} \sum _{i=n_{j-1}+1}^{n_j}\frac{r_{i}^{j}}{t+(n-t)r_{i}^{j}} \le \frac{n_{j}-n_{j-1}}{n}. \end{aligned}$$

(11)

This implies that

$$\begin{aligned} \Vert F_xF_x^\dag \Vert \le \Vert F_1F_1^\dag \Vert +\sum _{j=2}^{k}\sum _{i=n_{j-1}+1}^{n_j}\frac{r_{i}^{j}}{t+(n-t)r_{i}^{j}} \le \frac{n_1}{n}+\sum _{j=2}^{k}\frac{n_{j}-n_{j-1}}{n}= 1. \end{aligned}$$

Subcase 2.4 $t<\frac{n}{l(\pi )}$.

In this case, there exists some $\pi _i$ such that $l(\pi _i)=l(\pi )$, still, without loss of generality, assume $\pi _1$. By a similar argument to that of the above Subcase 2.3, one can show that $W_{n,t,\pi }$ is also not optimal.

Subcase 2.5 There exists at least two fixed points.

Without loss of generality, assume that $l(\pi _1)=l(\pi _2)=1$. Still, only consider the case that $x_i\not =0$ ($i=1,2,\ldots ,n$) for any $|x\rangle =(x_1,x_2,\ldots ,x_n)$. Let $C_0=\mathrm{diag}(c,-c,0,\cdots ,0)$. In Eqs. (6–7), by taking $\alpha _1^1=\alpha _2^2=\frac{\sqrt{n-t}}{n}$, $\delta _1^1=\frac{c\sqrt{n-t}}{n}$, $\delta _2^2=\frac{-c\sqrt{n-t}}{n}$ and $\alpha _i^j=\frac{\sqrt{n-t}r_{i}^{j}}{t+(n-t)r_{i}^{j}}$, $\delta _i^j=0$ for other $i,j$, one can obtain $\beta _1^1=\beta _2^2=\frac{\sqrt{t}}{n}$, $\gamma _1^1=\frac{c\sqrt{t}}{n}$, $\gamma _2^2=\frac{-c\sqrt{t}}{n}$ and $\beta _i^j=\frac{\sqrt{tr_{i}^{j}}e^{i(\theta _i-\theta _{i+1})}}{t+(n-t)r_{i}^{j}}$, $\gamma _i^j=0$ for other $i,j$. Thus,

$$\begin{aligned} \begin{array}{rl}F_{x}F_{x}^\dag =&{}\left( \begin{array}{cc} \frac{2}{n}+\sum _{j=3}^{k}\sum _{i=n_{j-1}+1}^{n_j}\frac{r_{i}^{j}}{t+(n-t)r_{i}^{j}}&{} 0\\ 0&{}\frac{2c^2}{n}\end{array}\right) .\end{array} \end{aligned}$$

By Eq. (9), we get

$$\begin{aligned} \frac{2}{n}+\sum _{j=3}^{k}\sum _{i=n_{j-1}+1}^{n_j}\frac{r_{i}^{j}}{t+(n-t)r_{i}^{j}}\le \frac{2}{n}+\sum _{j=3}^{k}\frac{n_{j}-n_{j-1}}{n}=1. \end{aligned}$$

This yields $0<c^2\le \frac{n}{2}\Rightarrow \Vert F_{x}F_{x}^\dag \Vert \le 1$.

Now, combining Subcases 2.1–2.5 and applying Lemma 1, $\Psi _{C_0}$ is positive, and so $W_{{n,t,\pi }}$ is not optimal. The proof of Case 2 is finished.

The proof of the main theorem is complete. $\square$