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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. Does this imply that the sum $\rho + \rho^\Gamma$ (ignoring trace normalization) is also entangled? If not, can we impose restrictions on $\rho$ which guarantee that the above proposition holds?

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. But is this always true? Can we use optimization of entanglement witness to ensure this condition?


Cross posted on math.SE

Cross posted on quantumcomputing.SE

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  • $\begingroup$ Do you have an example for a state where this does hold? $\endgroup$ Commented Jul 23, 2020 at 15:40
  • $\begingroup$ @NorbertSchuch Well, PPT entangled states which are invariant under partial transposition, i.e., $\rho = \rho^\Gamma$ trivially satisfy the stated proposition. Unfortunately, I don't have any other examples at the moment. $\endgroup$
    – mathwizard
    Commented Jul 23, 2020 at 19:26
  • $\begingroup$ Are there examples of PPT states which are invariant under partial transpose? $\endgroup$ Commented Jul 23, 2020 at 20:44
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    $\begingroup$ @NorbertSchuch Yes. Take a look at the class of local diagonal orthogonal invariant bipartite matrices from a recent paper arxiv.org/pdf/2007.11219.pdf $\endgroup$
    – mathwizard
    Commented Jul 24, 2020 at 4:19

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I believe this is not true, based on https://arxiv.org/abs/quant-ph/9903012, Eq. (8), where a given $\rho\in\mathcal{M}_2(\mathbb{C})\otimes\mathcal{M}_N(\mathbb{C})$ can be always written as $$\rho=\frac{\rho+\rho^{T_A}}{2}+\frac{\rho-\rho^{T_A}}{2}=\rho_s+\sigma_y^A\otimes B,$$ i.e., a separable part $2\rho_s=\rho+\rho^{T_A}$ and a part that might be inseparable. Thus, if I am not mistaken, even if $\rho$ is PPT entangled, $\frac{\rho+\rho^{T_A}}{2}$ is separable.

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  • $\begingroup$ So in $\mathcal{M}_2(\mathbb{C}) \otimes \mathcal{M}_d(\mathbb{C})$, since invariance under partial transposition guarantees separability, $\rho + \rho^\Gamma$ is separable for all positive semi-definite $\rho \in \mathcal{M}_2(\mathbb{C}) \otimes \mathcal{M}_d(\mathbb C)$. Very nice! I'll modify my question based on your finding. Thank you very much! $\endgroup$
    – mathwizard
    Commented Jul 20, 2020 at 4:58
  • $\begingroup$ Btw, I was also able to find counterexamples in $\mathcal{M}_3(\mathbb{C}) \otimes \mathcal{M}_3(\mathbb{C})$, namely in the class of PPT entangled states from arxiv.org/pdf/quant-ph/9703004.pdf. $\endgroup$
    – mathwizard
    Commented Jul 20, 2020 at 5:07
  • $\begingroup$ Nice! Maybe you find this interesting: arxiv.org/abs/1009.4385 . The author has some nice works on PPT entanglement. $\endgroup$ Commented Jul 20, 2020 at 18:08
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    $\begingroup$ @SatvikSingh If you found an answer to your own question, please post it as such, for the benefit of future visitors! $\endgroup$ Commented Jul 21, 2020 at 10:45

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