{"schema":"pubphys.bundle/1","record":{"author":{"account_ref":null,"orcid":null},"builds_on":[],"content_schema":"pubphys.content.revision/1","content_sha256":"a9b8f73655013738edab0d6d8c7f7083a8707aa7fea586d66d1b87e88146cac5","created":"2026-10-03T07:18:09Z","files":[],"origin":{"assisted_by":[],"kind":"seed"},"parents":["5475cf5dc5b1a72283d26dc95dc6e00d7ea4e00096a7c6492e409a1816eeb011"],"salt":"3e011b1bf1818109d4d504d0d1b8c0636122e537bd5f6354570b1f51100fc2de","schema":"pubphys.record/2","site":"pubphys.com","target":null,"type":"revision"},"content":{"answer_type":"yes-no","assisted_by":[],"external_id":"qi.contextuality-nonlocality.many-copy-nonlocality","kind":"well-posed","literature_status":"open","n":"1","parents":[],"plain":"Some entangled states never violate a Bell inequality when measured one pair at a time. The question is whether measuring many copies together, with local filtering, always reveals nonlocality for every entangled state.","posed_since":"","precise":"For an entangled bipartite state $\\rho$, determine whether some $k$ and local measurements on $k$ copies of $\\rho$ (possibly after local filtering, with no auxiliary entangled states and no extra shared entanglement) violate a Bell inequality. Superactivation for some states was shown by Palazuelos (2012) and Bell violation by bound-entangled states by Vertesi and Brunner (2014). An answer is a proof for all entangled states or an entangled state with a local model for every $k$.","problem_ref":null,"references":"","settled_by":"A general activation proof, or a family of local models covering all numbers of copies.","status_note":"With auxiliary entangled states in a network, every entangled state can be certified device-independently (Bowles, Supic, Cavalcanti, Acin, PRL 2018, doi:10.1103/PhysRevLett.121.180503); the many-copy question without such help is open.","title":"Is every entangled state nonlocal given many 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