Self-testing of qutrit systems Jdrzej Kaniewski QMATH, Department - PowerPoint PPT Presentation

Self-testing of qutrit systems Jędrzej Kaniewski QMATH, Department of Mathematical Sciences University of Copenhagen, Denmark joint work with Antonio Acín, Remigiusz Augusiak, Flavio Baccari, Alexia Salavrakos, Ivan Šupić, Jordi Tura CEQIP ’18 15 June 2018

Outline Bell nonlocality Self-testing Sum-of-squares decomposition Example: CHSH inequality Result 1: SATWAP inequality Result 2: generalised CHSH inequality

Outline Bell nonlocality Self-testing Sum-of-squares decomposition Example: CHSH inequality Result 1: SATWAP inequality Result 2: generalised CHSH inequality

Bell nonlocality Bell scenario j k P ( a, b | j, k ) a b

Bell nonlocality Bell scenario j k P ( a, b | j, k ) a b Assume that P ∈ Q is quantum � � ( F j a ⊗ G k P ( a, b | j, k ) = tr b ) ρ AB .

Bell nonlocality Bell scenario j k P ( a, b | j, k ) a b Assume that P ∈ Q is quantum � � ( F j a ⊗ G k P ( a, b | j, k ) = tr b ) ρ AB . Def.: P ∈ L is local if � P ( a, b | j, k ) = p ( λ ) p A ( a | j, λ ) p B ( b | k, λ ) . λ Bell: L � Q ⇐ ⇒ “ quantum mechanics is (Bell) nonlocal ”

Bell nonlocality Given some P ∈ Q , how to show that P �∈ L ?

Bell nonlocality Given some P ∈ Q , how to show that P �∈ L ? Real vector C = ( c abjk ) define � � C, P � := c abjk P ( a, b | j, k ) abjk and β L := max P ∈L � C, P � (local value) β Q := max P ∈Q � C, P � (quantum value) (suppose β L < β Q )

Bell nonlocality Given some P ∈ Q , how to show that P �∈ L ? Real vector C = ( c abjk ) define � � C, P � := c abjk P ( a, b | j, k ) abjk and β L := max P ∈L � C, P � (local value) β Q := max P ∈Q � C, P � (quantum value) (suppose β L < β Q ) Bell violation: � C, P � > β L = ⇒ P �∈ L

Bell nonlocality Obs.: Separable states give local statistics (for all measurements) � ρ AB = p λ σ λ ⊗ τ λ , λ

Bell nonlocality Obs.: Separable states give local statistics (for all measurements) � ρ AB = p λ σ λ ⊗ τ λ , λ � � � � � � � ( F j a ⊗ G k F j G k P ( a, b | j, k ) = tr b ) ρ AB = p λ · tr a σ λ · tr b τ λ . � �� � � �� � λ p A ( a | j,λ ) p B ( b | k,λ ) Nonlocality = ⇒ entanglement

Bell nonlocality Obs.: Separable states give local statistics (for all measurements) � ρ AB = p λ σ λ ⊗ τ λ , λ � � � � � � � ( F j a ⊗ G k F j G k P ( a, b | j, k ) = tr b ) ρ AB = p λ · tr a σ λ · tr b τ λ . � �� � � �� � λ p A ( a | j,λ ) p B ( b | k,λ ) Nonlocality = ⇒ entanglement can we make this connection more explicit/rigorous?

Outline Bell nonlocality Self-testing Sum-of-squares decomposition Example: CHSH inequality Result 1: SATWAP inequality Result 2: generalised CHSH inequality

Self-testing � � ( F j a ⊗ G k Given P ( a, b | j, k ) = tr b ) ρ AB deduce properties of ρ AB , { F j a } , { G k b }

Self-testing � � ( F j a ⊗ G k Given P ( a, b | j, k ) = tr b ) ρ AB deduce properties of ρ AB , { F j a } , { G k b } (i) we do not assume that ρ AB is pure or that the measurements are projective (we want to rigorously deduce it!)

Self-testing � � ( F j a ⊗ G k Given P ( a, b | j, k ) = tr b ) ρ AB deduce properties of ρ AB , { F j a } , { G k b } (i) we do not assume that ρ AB is pure or that the measurements are projective (we want to rigorously deduce it!) (ii) often only promised some Bell violation � C, P � = β

Self-testing � � ( F j a ⊗ G k Given P ( a, b | j, k ) = tr b ) ρ AB deduce properties of ρ AB , { F j a } , { G k b } (i) we do not assume that ρ AB is pure or that the measurements are projective (we want to rigorously deduce it!) (ii) often only promised some Bell violation � C, P � = β might seem like a hopeless task... ...but often can deduce essentially everything !

Outline Bell nonlocality Self-testing Sum-of-squares decomposition Example: CHSH inequality Result 1: SATWAP inequality Result 2: generalised CHSH inequality

Sum-of-squares decomposition Given a Bell functional C , how to compute β Q = max P ∈Q � C, P � ? Easy to provide lower bounds, what about upper bounds?

Sum-of-squares decomposition Given a Bell functional C , how to compute β Q = max P ∈Q � C, P � ? Easy to provide lower bounds, what about upper bounds? 1 Construct Bell operator � c abjk F j a ⊗ G k W := b abjk

Sum-of-squares decomposition Given a Bell functional C , how to compute β Q = max P ∈Q � C, P � ? Easy to provide lower bounds, what about upper bounds? 1 Construct Bell operator � c abjk F j a ⊗ G k W := b abjk 2 Prove that for all measurements W ≤ c 1 for c ∈ R

Sum-of-squares decomposition Given a Bell functional C , how to compute β Q = max P ∈Q � C, P � ? Easy to provide lower bounds, what about upper bounds? 1 Construct Bell operator � c abjk F j a ⊗ G k W := b abjk 2 Prove that for all measurements W ≤ c 1 for c ∈ R 3 Then β Q ≤ c because for all quantum realisations � C, P � = tr( Wρ AB ) ≤ c tr( ρ AB ) = c

Sum-of-squares decomposition Q: How to show that W ≤ c 1 for all measurements?

Sum-of-squares decomposition Q: How to show that W ≤ c 1 for all measurements? A: Write difference as sum of squares � L † c 1 − W ≥ j L j . j (operators L j depend on measurement operators) if β Q = c = ⇒ sum-of-squares (SOS) decomposition is tight

Outline Bell nonlocality Self-testing Sum-of-squares decomposition Example: CHSH inequality Result 1: SATWAP inequality Result 2: generalised CHSH inequality

Example: the CHSH inequality the CHSH operator W := A 0 ⊗ ( B 0 + B 1 ) + A 1 ⊗ ( B 0 − B 1 ) where − 1 ≤ A j ≤ 1 and − 1 ≤ B k ≤ 1

Example: the CHSH inequality the CHSH operator W := A 0 ⊗ ( B 0 + B 1 ) + A 1 ⊗ ( B 0 − B 1 ) where − 1 ≤ A j ≤ 1 and − 1 ≤ B k ≤ 1 define L 0 = A 0 ⊗ 1 − 1 ⊗ B 0 + B 1 √ 2 L 1 = A 1 ⊗ 1 − 1 ⊗ B 0 − B 1 √ 2

Example: the CHSH inequality the CHSH operator W := A 0 ⊗ ( B 0 + B 1 ) + A 1 ⊗ ( B 0 − B 1 ) where − 1 ≤ A j ≤ 1 and − 1 ≤ B k ≤ 1 define L 0 = A 0 ⊗ 1 − 1 ⊗ B 0 + B 1 √ 2 L 1 = A 1 ⊗ 1 − 1 ⊗ B 0 − B 1 √ 2 check 1 � � 1 ) − ( L † 0 L 0 + L † ( A 2 0 + A 2 1 ) ⊗ 1 + 1 ⊗ ( B 2 0 + B 2 W = √ 1 L 1 ) 2

Example: the CHSH inequality the CHSH operator W := A 0 ⊗ ( B 0 + B 1 ) + A 1 ⊗ ( B 0 − B 1 ) where − 1 ≤ A j ≤ 1 and − 1 ≤ B k ≤ 1 define L 0 = A 0 ⊗ 1 − 1 ⊗ B 0 + B 1 √ 2 L 1 = A 1 ⊗ 1 − 1 ⊗ B 0 − B 1 √ 2 check 1 � � 1 ) − ( L † 0 L 0 + L † ( A 2 0 + A 2 1 ) ⊗ 1 + 1 ⊗ ( B 2 0 + B 2 W = √ 1 L 1 ) 2 √ √ W ≤ 2 2 1 and β Q = 2 2 , so the SOS decomposition is tight

Example: the CHSH inequality 1 � � 1 ) − ( L † 0 L 0 + L † ( A 2 0 + A 2 1 ) ⊗ 1 + 1 ⊗ ( B 2 0 + B 2 W = √ 1 L 1 ) 2 √ observing tr( Wρ AB ) = 2 2 implies:

Example: the CHSH inequality 1 � � 1 ) − ( L † 0 L 0 + L † ( A 2 0 + A 2 1 ) ⊗ 1 + 1 ⊗ ( B 2 0 + B 2 W = √ 1 L 1 ) 2 √ observing tr( Wρ AB ) = 2 2 implies: 1 all measurements are projective on the local supports: tr( A 2 j ρ A ) = tr( B 2 k ρ B ) = 1

Example: the CHSH inequality 1 � � 1 ) − ( L † 0 L 0 + L † ( A 2 0 + A 2 1 ) ⊗ 1 + 1 ⊗ ( B 2 0 + B 2 W = √ 1 L 1 ) 2 √ observing tr( Wρ AB ) = 2 2 implies: 1 all measurements are projective on the local supports: tr( A 2 j ρ A ) = tr( B 2 k ρ B ) = 1 2 observables of Alice and Bob satisfy L j ρ AB = 0 � � 1 ⊗ B 0 + B 1 ( A 0 ⊗ 1 ) ρ AB = √ ρ AB 2

Example: the CHSH inequality 1 � � 1 ) − ( L † 0 L 0 + L † ( A 2 0 + A 2 1 ) ⊗ 1 + 1 ⊗ ( B 2 0 + B 2 W = √ 1 L 1 ) 2 √ observing tr( Wρ AB ) = 2 2 implies: 1 all measurements are projective on the local supports: tr( A 2 j ρ A ) = tr( B 2 k ρ B ) = 1 2 observables of Alice and Bob satisfy L j ρ AB = 0 � � 1 ⊗ B 0 + B 1 ( A 0 ⊗ 1 ) ρ AB = √ ρ AB 2 if ρ A and ρ B are full-rank, then � B 0 + B 1 � 2 A 2 0 = 1 = ⇒ √ = 1 = ⇒ { B 0 , B 1 } = 0 2

Example: the CHSH inequality the relation determines form of observables B 0 = U B ( σ x ⊗ 1 ) U † B B 2 0 = B 2 1 = 1 and { B 0 , B 1 } = 0 = ⇒ B 1 = U B ( σ z ⊗ 1 ) U † B

Example: the CHSH inequality the relation determines form of observables B 0 = U B ( σ x ⊗ 1 ) U † B B 2 0 = B 2 1 = 1 and { B 0 , B 1 } = 0 = ⇒ B 1 = U B ( σ z ⊗ 1 ) U † B the inequality is symmetric, so A 0 and A 1 have the same form √ construct W and determine the eigenspace corresponding to 2 2

Example: the CHSH inequality the relation determines form of observables B 0 = U B ( σ x ⊗ 1 ) U † B B 2 0 = B 2 1 = 1 and { B 0 , B 1 } = 0 = ⇒ B 1 = U B ( σ z ⊗ 1 ) U † B the inequality is symmetric, so A 0 and A 1 have the same form √ construct W and determine the eigenspace corresponding to 2 2 √ Self-testing (rigidity) statement for CHSH: if β = 2 2 then A 0 = U A ( σ x ⊗ 1 ) U † B 0 = U B ( σ x ⊗ 1 ) U † A B A 1 = U A ( σ z ⊗ 1 ) U † B 1 = U B ( σ z ⊗ 1 ) U † A B and ρ AB = U (Φ A ′ B ′ ⊗ τ A ′′ B ′′ ) U † for U := U A ⊗ U B