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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 physics.SE

Cross-posted on quantumcomputing.SE

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  • $\begingroup$ Have you tried testing your conjecture on known families of entangled, PPT states? $\endgroup$ Commented Jul 12, 2020 at 14:58
  • $\begingroup$ @Omnomnomnom So the problem is that for a given entangled PPT state, I don't think there's an easy way to bring its witness in the canonical form. So the approach I've mentioned is a little hard to implement in practice. $\endgroup$
    – mathwizard
    Commented Jul 13, 2020 at 2:37
  • $\begingroup$ @Omnomnomnom However, if the state in question is already an edge state, we simply need to compute the kernel of the state itself and its partial transpose. I've tried this for a couple of states in dimension $d=d'=3$, but it doesn't work. But, given the witness, there's still hope for optimizing it (increasing the set of states that it detects) in such a way that it detects the partial transpose. $\endgroup$
    – mathwizard
    Commented Jul 13, 2020 at 2:52
  • $\begingroup$ By "testing your conjecture," I meant taking a non-entangled PPT state $\rho$ and checking whether $\rho + \rho^{\Gamma}$ is entangled, which if I recall correctly can be checked with semidefinite programming $\endgroup$ Commented Jul 13, 2020 at 7:32
  • $\begingroup$ @Omnomnomnom Well, if $\rho$ is non-entangled, i.e., separable, then $\rho^\Gamma$ is separable and so the sum $\rho + \rho^\Gamma$ is trivially separable (due to convexity). So there's nothing to test here. The non-trivial part is when $\rho$ is entangled because then, the sum can either be entangled or separable, depending on whether a common witness exists. $\endgroup$
    – mathwizard
    Commented Jul 13, 2020 at 8:33

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