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Consider an entangled bipartite quantum state $\rho \in \mathcal{M}_d(\mathbb{C}) \otimes \mathcal{M}_{d'}(\mathbb{C})$ which is positive under partial transposition, i.e., $\rho^\Gamma \geq 0$. As separability of $\rho$ is equivalent to separability of its partial transpose $\rho^\Gamma$, we know that $\rho^\Gamma$ is entangled. What are the conditions on $\rho$ which will guarantee that the sum $\rho + \rho^\Gamma$ (ignoring trace normalization) is also entangled?

It turns out that the above proposition does not hold for arbitrary PPT entangled states. Easiest counterexamples can be found in $\mathcal{M}_2(\mathbb{C}) \otimes \mathcal{M}_{d}(\mathbb{C})$, where $\rho + \rho^\Gamma$ is separable for all quantum states $\rho \in \mathcal{M}_2(\mathbb{C}) \otimes \mathcal{M}_d(\mathbb{C})$ (see separability in 2xN systems).

In the language of entanglement witnesses, the problem reduces to finding a common witness that detects both $\rho$ and $\rho^\Gamma$. Let $W$ be the entanglement witness detecting $\rho$, i.e., $\text{Tr} (W\rho) < 0$. Then $W$ is non-decomposable (as $\rho$ is PPT) and is of the canonical form $P+Q^\Gamma - \epsilon \mathbb{I}$, where $P, Q \geq 0$ are such that $\text{range}(P) \subseteq\text{ker}(\delta)$ and $\text{range}(Q) \subseteq \text{ker}(\delta^\Gamma)$ for some bipartite edge state $\delta$ (these are special states that violate the range criterion for separability in an extreme manner, see edge states) and $0 < \epsilon \leq \text{inf}_{|e,f\rangle} \langle e,f | P+Q^\Gamma | e,f \rangle$. If $\delta$ is such that $\text{ker}(\delta) \cap \text{ker}(\delta^\Gamma)$ is not empty, then we can choose $P=Q$ to be the orthogonal projector on $\text{ker}(\delta) \cap \text{ker}(\delta^\Gamma)$, in which case $W=W^\Gamma$ is the common witness. Can we find a class of PPT entangled states for which the previous statement holds? Can optimization of entanglement witnesses be somehow used to ensure this condition?


Cross-posted on math.SE

Cross-posted on physics.SE

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  • $\begingroup$ Maybe I've forgotten the details but I thought entanglement witnesses can always be written as $I\otimes \Lambda$ for a positive map $\Lambda$. If that's true, then an entanglement witness for $\rho+\rho^\Gamma$ would be $(I\otimes\Lambda)(\rho+\rho^\Gamma)<0$ - some rearranging shows that this is equivalent to $I\otimes (\Lambda+\Lambda^\gamma)\rho < 0$. So the condition is equivalent to: $\rho$ has an entanglement witness $\Lambda$ such that $\Lambda + \Lambda^\Gamma$ is also an entanglement witness. $\endgroup$
    – Sam Jaques
    Jul 21 '20 at 14:18
  • $\begingroup$ In the language of positive maps, if $\Lambda$ is an entanglement witness for $\rho$, then we know that $\Lambda^\Gamma = \Lambda \circ \text{transp}$ is an entanglement witness for $\rho^\Gamma$, since $(I \otimes \Lambda)(\rho) < 0$ implies $(I \otimes \Lambda^\Gamma)(\rho^\Gamma) < 0$. The problem is that $\Lambda + \Lambda^\Gamma$ is not necessarily an entanglement witness for $\rho + \rho^\Gamma$, because we have no handle on how the cross terms $(I \otimes \Lambda)(\rho^\Gamma)=(I \otimes \Lambda^\Gamma)(\rho)$ look like. $\endgroup$
    – mathwizard
    Jul 22 '20 at 11:46
  • $\begingroup$ Right, I should have said "$\Lambda + \Lambda^\Gamma$ is also an entanglement witness for $\rho$". $\endgroup$
    – Sam Jaques
    Jul 23 '20 at 13:03

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