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Does entanglement of a bipartite PPT state $rho$ imply entanglement of $rho + rho^{Gamma}$?

Physics Asked by Satvik Singh on August 2, 2020

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., $text{Tr} (Wrho) < 0$. 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 $text{range}(P) subseteqtext{ker}(delta)$ 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 $0 < epsilon leq text{inf}_{|e,frangle} langle e,f | P+Q^Gamma | e,f rangle$. 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

One Answer

I believe this is not true, based on https://arxiv.org/abs/quant-ph/9903012, Eq. (8), where a given $rhoinmathcal{M}_2(mathbb{C})otimesmathcal{M}_N(mathbb{C})$ can be always written as $$rho=frac{rho+rho^{T_A}}{2}+frac{rho-rho^{T_A}}{2}=rho_s+sigma_y^Aotimes B,$$ i.e., a separable part $2rho_s=rho+rho^{T_A}$ 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.

Answered by Karl Pilkington on August 2, 2020

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