We show that there is some entangled states that are far from any separable and maximally entangled states. As a consequence, there exists entanglement witness that cannot witness any maximally entangled states. At the first sight it seems to be the opposite: the maximally entangled states contains the ‘most’ entanglement in some sense, so at least one of them should be detected by a fixed entanglement witness. Unfortunately it does not hold.

Let $\operatorname{MAXE}$ be the set of all maximally entangled state, $\operatorname{SEP}$ be the set of separable states. We first show that the convex hull of $\operatorname{MAXE}\cup\operatorname{SEP}$ is closed.

Lemma The convex hull \(\operatorname{cov}\{\operatorname{MAXE}\cup\operatorname{SEP}\}:=\{\rho|\rho=\sum_jp_j\rho_j,\rho_j\in\operatorname{MAXE}\cup\operatorname{SEP},p_j\ge0,\sum_jp_j=1\}\) is closed.

Proof Note that $\operatorname{MAXE}$ is an orbit of the group action the unitary group $U(d)$ which is compact, so $\operatorname{MAXE}$ is compact. It is well-known that $\operatorname{SEP}$ is compact. Therefore $\operatorname{MAXE}\cup\operatorname{SEP}$ is compact. Since $\operatorname{MAXE}\cup\operatorname{SEP}\subseteq\mathcal{B}(\mathbb{C}^d\otimes\mathbb{C}^d)$ is finite-dimensional, its convex hull is also compact, thus closed. $\blacksquare$

We can also show that any partially entangled pure state $\sigma=| \psi\rangle\langle \psi |, |\psi\rangle\notin\operatorname{MAXE}$ is not in the convex hull: note that $\sigma$ is an extreme point in the set of all density matrices. Suppose for contradiction that $\sigma\in\operatorname{cov}{\operatorname{MAXE}\cup\operatorname{SEP}}$, then it must be also extreme in $\operatorname{cov}{\operatorname{MAXE}\cup\operatorname{SEP}}$. So it must be either maximally entangled $|\psi\rangle\in\operatorname{MAXE}$ or separable $|\psi\rangle=|\psi\rangle_A\otimes|\psi\rangle_B$, a contradiction.

So we have a closed convex set $\operatorname{cov}{\operatorname{MAXE}\cup\operatorname{SEP}}$, and a point $\sigma$ which is not in the set. By Hahn–Banach theorem, there is a linear functional denoted by hermitian operator $H$ such that $\operatorname{Tr}[H\sigma]<0$ and $\operatorname{Tr}[H\rho]\ge0$ for all $\rho\in\operatorname{cov}{\operatorname{MAXE}\cup\operatorname{SEP}}$. Obviously $\operatorname{SEP}\subseteq\operatorname{cov}{\operatorname{MAXE}\cup\operatorname{SEP}}$, so $H$ separates $\operatorname{SEP}$ and $\sigma$, therefore is an entanglement witness. Also, $\operatorname{MAXE}\subseteq\operatorname{cov}{\operatorname{MAXE}\cup\operatorname{SEP}}$, so $H$ cannot witness any maximally entangled states.

Acknowledgement {#acknowledgement .unnumbered}

We thank Marek Miller for valuable discussion.