Self-testing of binary observables based on commutation [ - PowerPoint PPT Presentation

Self-testing of binary observables based on commutation [ arXiv:1702.06845 , Phys. Rev. A 95 , 062323 (2017)] Jed Kaniewski QMATH, Department of Mathematical Sciences University of Copenhagen, Denmark Smolenice, Slovakia CEQIP ’17 31 May 2017

Outline What is nonlocality? What is self-testing? The CHSH inequality The biased CHSH inequality Multiple anticommuting observables Summary and open problems

Outline What is nonlocality? What is self-testing? The CHSH inequality The biased CHSH inequality Multiple anticommuting observables Summary and open problems

What is nonlocality? Bell scenario y x a b Pr[ a, b | x, y ]

What is nonlocality? Bell scenario y x a b Pr[ a, b | x, y ] Def.: Pr[ a, b | x, y ] is local if � Pr[ a, b | x, y ] = p ( λ ) p ( a | x, λ ) p ( b | y, λ ) . λ Otherwise = ⇒ nonlocal or it violates (some) Bell inequality

What is nonlocality? Assume quantum mechanics. . . what can I deduce about my system?

What is nonlocality? Assume quantum mechanics. . . what can I deduce about my system? Entanglement : separable states always produce local statistics � ρ AB = p λ σ λ ⊗ τ λ , λ � � � a ⊗ Q y · tr( Q y ( P x p λ · tr( P x Pr[ a, b | x, y ] = tr b ) ρ AB = a σ λ ) b τ λ ) � �� � � �� � λ p ( a | x,λ ) p ( b | y,λ )

What is nonlocality? Assume quantum mechanics. . . what can I deduce about my system? Entanglement : separable states always produce local statistics � ρ AB = p λ σ λ ⊗ τ λ , λ � � � a ⊗ Q y · tr( Q y ( P x p λ · tr( P x Pr[ a, b | x, y ] = tr b ) ρ AB = a σ λ ) b τ λ ) � �� � � �� � λ p ( a | x,λ ) p ( b | y,λ ) what else?

What is nonlocality? Assume quantum mechanics. . . what can I deduce about my system? Entanglement : separable states always produce local statistics � ρ AB = p λ σ λ ⊗ τ λ , λ � � � a ⊗ Q y · tr( Q y ( P x p λ · tr( P x Pr[ a, b | x, y ] = tr b ) ρ AB = a σ λ ) b τ λ ) � �� � � �� � λ p ( a | x,λ ) p ( b | y,λ ) what else? self-testing study you must

What is self-testing? � � a ⊗ Q y ( P x Given Pr[ a, b | x, y ] = tr b ) ρ AB a ) , ( Q y deduce properties of ρ AB , ( P x b )

What is self-testing? � � a ⊗ Q y ( P x Given Pr[ a, b | x, y ] = tr b ) ρ AB a ) , ( Q y deduce properties of ρ AB , ( P x b ) (don’t assume that ρ AB is pure or measurements are projective, deduce it instead!)

What is self-testing? � � a ⊗ Q y ( P x Given Pr[ a, b | x, y ] = tr b ) ρ AB a ) , ( Q y deduce properties of ρ AB , ( P x b ) (don’t assume that ρ AB is pure or measurements are projective, deduce it instead!) often only promised some Bell violation � c xy ab Pr[ a, b | x, y ] = β abxy

What is self-testing? � abxy c xy ab Pr[ a, b | x, y ] = β

What is self-testing? ? state ρ AB certification � abxy c xy ab Pr[ a, b | x, y ] = β

What is self-testing? ? state ρ AB certification � abxy c xy ab Pr[ a, b | x, y ] = β measurement a , Q y P x b certification ?

What is self-testing? ? state ρ AB certification � abxy c xy ab Pr[ a, b | x, y ] = β measurement a , Q y P x b certification ?

What is self-testing? ? state ρ AB certification � abxy c xy ab Pr[ a, b | x, y ] = β measurement a , Q y P x b certification Which measurements can be certified? ? ( P x a ) ?

What is self-testing? ? state ρ AB certification � abxy c xy ab Pr[ a, b | x, y ] = β measurement a , Q y P x b certification Which measurements can be certified? ? ( P x a ) ?

What is self-testing? Why care about self-testing of measurements? significantly less studied (particularly in the robust regime) relevant for (two-party) device-independent cryptography pinning down the optimal measurements immediately gives the optimal state

Outline What is nonlocality? What is self-testing? The CHSH inequality The biased CHSH inequality Multiple anticommuting observables Summary and open problems

The CHSH inequality Measurements with two outcomes F j = F † j , F j ≥ 0 , F 0 + F 1 = 1

The CHSH inequality Measurements with two outcomes F j = F † j , F j ≥ 0 , F 0 + F 1 = 1 Conveniently written as observables A = F 0 − F 1 One-to-one mapping, i.e. any A = A † and − 1 ≤ A ≤ 1 corresponds to a valid measurement [for projective measurements A 2 = 1 ]

The CHSH inequality The CHSH value � � β := tr for W := A 0 ⊗ ( B 0 + B 1 ) + A 1 ⊗ ( B 0 − B 1 ) Wρ AB √ Classically β ≤ 2 , but quantumly can reach up to 2 2

The CHSH inequality The CHSH value � � β := tr for W := A 0 ⊗ ( B 0 + B 1 ) + A 1 ⊗ ( B 0 − B 1 ) Wρ AB √ Classically β ≤ 2 , but quantumly can reach up to 2 2 What can we deduce from β > 2 ?

The CHSH inequality The CHSH value � � β := tr for W := A 0 ⊗ ( B 0 + B 1 ) + A 1 ⊗ ( B 0 − B 1 ) Wρ AB √ Classically β ≤ 2 , but quantumly can reach up to 2 2 What can we deduce from β > 2 ? square the Bell operator, fool!

The CHSH inequality If A 2 j = B 2 k = 1 , then W 2 = 4 · 1 ⊗ 1 − [ A 0 , A 1 ] ⊗ [ B 0 , B 1 ] .

The CHSH inequality If A 2 j = B 2 k = 1 , then W 2 = 4 · 1 ⊗ 1 − [ A 0 , A 1 ] ⊗ [ B 0 , B 1 ] . In general ( A 2 j , B 2 k ≤ 1 ) W 2 ≤ 4 · 1 ⊗ 1 − [ A 0 , A 1 ] ⊗ [ B 0 , B 1 ] . Simple upper bounds W 2 ≤ 4 · 1 ⊗ 1 + | [ A 0 , A 1 ] ⊗ [ B 0 , B 1 ] | = 4 · 1 ⊗ 1 + | [ A 0 , A 1 ] | ⊗ | [ B 0 , B 1 ] | ≤ 4 · 1 ⊗ 1 + 2 | [ A 0 , A 1 ] | ⊗ 1 .

The CHSH inequality W 2 ≤ 4 · 1 ⊗ 1 + 2 | [ A 0 , A 1 ] | ⊗ 1 .

The CHSH inequality W 2 ≤ 4 · 1 ⊗ 1 + 2 | [ A 0 , A 1 ] | ⊗ 1 . The Cauchy-Schwarz inequality � 2 ≤ tr( W 2 ρ AB ) · tr ρ AB = tr( W 2 ρ AB ) � tr( Wρ AB )

The CHSH inequality W 2 ≤ 4 · 1 ⊗ 1 + 2 | [ A 0 , A 1 ] | ⊗ 1 . The Cauchy-Schwarz inequality � 2 ≤ tr( W 2 ρ AB ) · tr ρ AB = tr( W 2 ρ AB ) � tr( Wρ AB ) leads to √ β ≤ 2 1 + t, � � where t := 1 2 tr | [ A 0 , A 1 ] | ρ A . Bell violation certifies incompatibility of observables!

The CHSH inequality The quantity t := 1 � � 2 tr | [ A 0 , A 1 ] | ρ A invariant under local unitaries and adding auxiliary systems easy to compute clear operational interpretation as “weighted average” t = 1 (max. value) implies UA 0 U † = σ x ⊗ 1 , UA 1 U † = σ y ⊗ 1 . [assuming ρ A is full-rank]

The CHSH inequality The quantity t := 1 � � 2 tr | [ A 0 , A 1 ] | ρ A invariant under local unitaries and adding auxiliary systems easy to compute clear operational interpretation as “weighted average” t = 1 (max. value) implies UA 0 U † = σ x ⊗ 1 , UA 1 U † = σ y ⊗ 1 . [assuming ρ A is full-rank] = ⇒ t = “ distance from the optimal arrangement ”

The CHSH inequality The relation √ β ≤ 2 1 + t, is non-trivial as soon as β > 2 is tight

The CHSH inequality The relation √ β ≤ 2 1 + t, is non-trivial as soon as β > 2 is tight CHSH violation certifies closeness to the optimal arrangement

The CHSH inequality The relation √ β ≤ 2 1 + t, is non-trivial as soon as β > 2 is tight CHSH violation certifies closeness to the optimal arrangement √ BONUS: β = 2 2 implies t = 1 and so UA 0 U † = σ x ⊗ 1 , UA 1 U † = σ y ⊗ 1 By symmetry the same applies to Bob, so W (up to local unitaries) is just a two-qubit operator tensored with identity = ⇒ finding the optimal state is easy

The CHSH inequality √ Complete rigidity statement: if β = 2 2 then there exists U = U A ⊗ U B and τ A ′ B ′ ρ AB = U (Φ AB ⊗ τ A ′ B ′ ) U † , where Φ AB = EPR pair and U A A 0 U † A = σ x ⊗ 1 , U A A 1 U † A = σ y ⊗ 1 , U B B 0 U † B = σ x ⊗ 1 , U B B 1 U † B = σ y ⊗ 1 .

The CHSH inequality √ Complete rigidity statement: if β = 2 2 then there exists U = U A ⊗ U B and τ A ′ B ′ ρ AB = U (Φ AB ⊗ τ A ′ B ′ ) U † , where Φ AB = EPR pair and U A A 0 U † A = σ x ⊗ 1 , U A A 1 U † A = σ y ⊗ 1 , U B B 0 U † B = σ x ⊗ 1 , U B B 1 U † B = σ y ⊗ 1 . very similar to the original proof by Popescu and Rohrlich

The CHSH inequality √ Complete rigidity statement: if β = 2 2 then there exists U = U A ⊗ U B and τ A ′ B ′ ρ AB = U (Φ AB ⊗ τ A ′ B ′ ) U † , where Φ AB = EPR pair and U A A 0 U † A = σ x ⊗ 1 , U A A 1 U † A = σ y ⊗ 1 , U B B 0 U † B = σ x ⊗ 1 , U B B 1 U † B = σ y ⊗ 1 . very similar to the original proof by Popescu and Rohrlich [generalises straightforwardly to multipartite inequalities: Mermin/MABK inequalities]

Outline What is nonlocality? What is self-testing? The CHSH inequality The biased CHSH inequality Multiple anticommuting observables Summary and open problems

The biased CHSH inequality For α ≥ 1 the biased CHSH value � � β := tr W α ρ AB for W α := α ( A 0 + A 1 ) ⊗ B 0 + ( A 0 − A 1 ) ⊗ B 1 . √ α 2 + 1 . Classically β ≤ 2 α , but quantumly we can reach up to 2 optimal state: maximally entangled of 2 qubits optimal observables of Bob: maximally incompatible optimal observables of Alice: non-maximally incompatible!