CHSH rigidity Proof and POVMs

CHSH rigidity Proof and POVMs Ask Question Asked today Modified today Viewed 4 times 0 I have a small question about the usual proof for CHSH rigidity, as detailed in Cleve Lecture Notes. The CHSH rigidity assumes an optimal strategy of Alice and Bob, this strategy consists in observables A 0 𝐴 0 and A 1 𝐴 1 (two outcomes) for Alice and B 0 𝐵 0 and B 1 𝐵 1 for Bob, and a shared entangled state | ψ AB ⟩ | 𝜓 𝐴 𝐵 ⟩ . The proof usually begins by prooving that a two outcomes POVM { E 0 , E 1 } { 𝐸 0 , 𝐸 1 } is equivalent to an ensemble of binary observables { A x } x { 𝐴 𝑥 } 𝑥 in which x 𝑥 is a sample of Bernouilli laws whose expectations are the eigenvalues of E 0 𝐸 0 . The CHSH rigidity proof continues then by focusing on each binary observable A x 𝐴 𝑥 . I understand that POVM are more general than PVM, e.g. state discrimination does not lead to PVMs, but I just wonder why the CHSH rigidity proof cares about POVM ? The initial assumption is just that Alice and Bob have observables A 0 𝐴 0 , A 1 𝐴 1 , B 0 𝐵 0 and B 1 𝐵 1 which check optimality : 1 4 ⟨ ψ AB | A 0 ⊗ B 0 + A 0 ⊗ B 1 + A 1 ⊗ B 0 − A 1 ⊗ B 1 | ψ AB ⟩= 1 2 – √ 1 4 ⟨ 𝜓 𝐴 𝐵 | 𝐴 0 ⊗ 𝐵 0 + 𝐴 0 ⊗ 𝐵 1 + 𝐴 1 ⊗ 𝐵 0 − 𝐴 1 ⊗ 𝐵 1 | 𝜓 𝐴 𝐵 ⟩ = 1 2 Those observables are then replaced in the proof by an ensemble of binary observables { A x 0,1 } x { 𝐴 0 , 1 𝑥 } 𝑥 and { B x 0,1 } x { 𝐵 0 , 1 𝑥 } 𝑥 for Alice and Bob. Does it mean that we have following equality ? ∑ x p(x)⟨ ψ AB | A x 0 ⊗ B x 0 + A x 0 ⊗ B x 1 + A x 1 ⊗ B x 0 − A x 1 ⊗ B x 1 | ψ AB ⟩ =⟨ ψ AB | A 0 ⊗ B 0 + A 0 ⊗ B 1 + A 1 ⊗ B 0 − A 1 ⊗ B 1 | ψ AB ⟩ ∑ 𝑥 𝑝 ( 𝑥 ) ⟨ 𝜓 𝐴 𝐵 | 𝐴 0 𝑥 ⊗ 𝐵 0 𝑥 + 𝐴 0 𝑥 ⊗ 𝐵 1 𝑥 + 𝐴 1 𝑥 ⊗ 𝐵 0 𝑥 − 𝐴 1 𝑥 ⊗ 𝐵 1 𝑥 | 𝜓 𝐴 𝐵 ⟩ = ⟨ 𝜓 𝐴 𝐵 | 𝐴 0 ⊗ 𝐵 0 + 𝐴 0 ⊗ 𝐵 1 + 𝐴 1 ⊗ 𝐵 0 − 𝐴 1 ⊗ 𝐵 1 | 𝜓 𝐴 𝐵 ⟩ entanglementbell-experimentchsh Share Improve this question Follow asked 1 hour ago deb2014 6233 3 silver badges 8 8 bronze badges Add a comment Know someone who can answer? 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