# Finding a class $C$ of bipartite PPT states such that entanglement of $\rho \in C$ implies entanglement of $\rho + \rho^{\Gamma}$

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

• 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. Jul 21 '20 at 14:18
• 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. Jul 22 '20 at 11:46
• Right, I should have said "$\Lambda + \Lambda^\Gamma$ is also an entanglement witness for $\rho$". Jul 23 '20 at 13:03