aiqbench analysis
This paper establishes a rigorous separability rigidity theorem for finite-mode Gaussian states, demonstrating that dominance by a separable operator implies the state itself is separable across a fixed partition. These findings clarify the relationship between convex mixing of partitions and genuine multipartite entanglement within quantum information theory. The work advances fundamental understanding of quantum correlations without relying on classical machine learning techniques or AI applications.
Separability rigidity of Gaussian states under positive domination
arXiv:2609.36336v1 Announce Type: new Abstract: We prove a separability rigidity theorem for finite-mode Gaussian states. Let $\rho_G$ be Gaussian and let $\mathcal I$ be a fixed partition of the modes. If $\rho_G$ dominates a nonzero positive operator $T$ that is separable across $\mathcal I$, $0\neq T\preccurlyeq \rho_G$, then $\rho_G$ itself is separable across $\mathcal I$. The dominated operator is arbitrary and need not be Gaussian. As consequences, convex mixing among finitely many separability partitions creates no new Gaussian states: a Gaussian state belonging to such a convex class is already separable across one fixed partition. In particular, full inseparability and genuine multipartite entanglement coincide for finite-mode Gaussian states, and identical copies of a biseparable Gaussian state cannot activate genuine multipartite entanglement. We also show that the implication cannot be reversed by constructing a fully separable Gaussian state that dominates a positive multiple of a genuine multipartite entangled state.
Abstract:We prove a separability rigidity theorem for finite-mode Gaussian states. Let $\rho_G$ be Gaussian and let $\mathcal I$ be a fixed partition of the modes. If $\rho_G$ dominates a nonzero positive operator $T$ that is separable across $\mathcal I$, $0\neq T\preccurlyeq \rho_G$, then $\rho_G$ itself is separable across $\mathcal I$. The dominated operator is arbitrary and need not be Gaussian. As consequences, convex mixing among finitely many separability partitions creates no new Gaussian states: a Gaussian state belonging to such a convex class is already separable across one fixed partition. In particular, full inseparability and genuine multipartite entanglement coincide for finite-mode Gaussian states, and identical copies of a biseparable Gaussian state cannot activate genuine multipartite entanglement. We also show that the implication cannot be reversed by constructing a fully separable Gaussian state that dominates a positive multiple of a genuine multipartite entangled state.
arXiv:2609.36336 [quant-ph]
(or arXiv:2609.36336v1 [quant-ph] for this version)
https://doi.org/10.48550/arXiv.2609.36336 Focus to learn more arXiv-issued DOI via DataCite (pending registration)
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Full text under the publisher's open licence · 234 words
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