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Nonlocal Boxes And All That Daniel Rohrlich Atom Chip Group, Ben Gurion University, Beersheba, Israel 21 January 2010 Axioms for special relativity • The laws of physics are the same in all inertial reference frames. • There is a maximum signaling speed, and it is the speed of light c. Axioms for quantum mechanics • Physical states are normalized vectors ψ(r), Ψ(r,t), ψ , (t) . • Measurable physical quantities – “observables” – correspond to Hermitian or (self-adjoint) operators on the state vectors. • If a system is an eigenstate a with eigenvalue a of an observable Â, then a measurement of  on a will yield a. • Conversely, if a measurement of  on any state yields a, the measurement leaves the system in an eigenstate a . • The probability that a system in a normalized state ψ can be 2 found in the state φ is φ ψ . Axioms for quantum mechanics • The time evolution of a quantum state (t ) is given by i (t ) Hˆ (t ) , t where Ĥ is the Hamiltonian (kinetic energy + potential energy) of the system in the state (t ) . • The wave function of identical fermions (spin ½, ¾,…) must ² be antisymmetric under exchange of any pair of them; the wave function of identical bosons (spin 0, 1,…) must be symmetric under exchange of any pair of them. “It’s like trying to derive special relativity from the wrong axioms.” – Yakir Aharonov • Fast objects contract in the direction of their motion. • Moving clocks slow down. • Twins cannot agree about who is the oldest. • The observer determines the results of measurements. •… Question: What, indeed, is so “special” about special relativity? Question: What, indeed, is so “special” about special relativity? Answer: The two axioms so nearly contradict each other that only a unique theory reconciles them. Question: What, indeed, is so “special” about special relativity? Answer: The two axioms so nearly contradict each other that only a unique theory reconciles them. Y. Aharonov and (independently) A. Shimony: Quantum mechanics, as well, reconciles two things that nearly contradict each other: Question: What, indeed, is so “special” about special relativity? Answer: The two axioms so nearly contradict each other that only a unique theory reconciles them. Y. Aharonov and (independently) A. Shimony: Quantum mechanics, as well, reconciles two things that nearly contradict each other: Question: What, indeed, is so “special” about special relativity? Answer: The two axioms so nearly contradict each other that only a unique theory reconciles them. Y. Aharonov and (independently) A. Shimony: Quantum mechanics, as well, reconciles two things that nearly contradict each other: Two axioms for quantum mechanics Axiom 1: There are nonlocal influences. (What are they?) Axiom 2: We cannot send superluminal signals. Can we derive quantum mechanics from these two axioms? 1. What is nonlocality? Can we derive quantum mechanics from these two axioms? 1. What is nonlocality? 2. What does “no signalling” mean? Can we derive quantum mechanics from these two axioms? 1. What is nonlocality? Nonlocal correlations? Aharonov-Bohm effect? “Modular” dynamical variables? 2. What does “no signalling” mean? Can we derive quantum mechanics from these two axioms? 1. What is nonlocality? Nonlocal correlations? Aharonov-Bohm effect? “Modular” dynamical variables? 2. What does “no signalling” mean? What is left of “no signalling” in the limit c → ∞ of nonrelativistic quantum mechanics? Can we derive quantum mechanics from these two axioms? 1. What is nonlocality? Nonlocal correlations 2. What does “no signalling” mean? What is left of “no signalling” in the limit c → ∞ of nonrelativistic quantum mechanics? Can we derive quantum mechanics from these two axioms? 1. What is nonlocality? Nonlocal correlations 2. What does “no signalling” mean? “No signalling” at any speed! Quantum mechanics does not allow “signalling” Introducing…Alice and Bob Quantum mechanics does not allow “signalling” Introducing…Alice and Bob Alice Bob drawings by Tom Oreb © Walt Disney Co. Quantum mechanics does not allow “signalling” Introducing…Alice and Bob, who share pairs in an entangled state and make spacelike separated measurements on members of each pair. Alice measures a fixed observable Aˆ ai ai ai on her member of each pair. Bob is free to choose any observable Bˆ bi bi bi to measure on his member of each pair. Alice Question: Does prob(ai) depend on what observableB̂ Bob chooses to measure? Bob drawings by Tom Oreb © Walt Disney Co. Quantum mechanics does not allow “signalling” Introducing…Alice and Bob, who share pairs in an entangled state and make spacelike separated measurements on members of each pair. Alice measures a fixed observable Aˆ ai ai ai on her member of each pair. Bob is free to choose any observable Bˆ bi bi bi to measure on his member of each pair. Alice Question: Does prob(ai) depend on what observableB̂ Bob chooses to measure? Answer: No, because prob(ai ) ai , b j 2 j Bob drawings by Tom Oreb © Walt Disney Co. Quantum mechanics does not allow “signalling” Introducing…Alice and Bob, who share pairs in an entangled state and make spacelike separated measurements on members of each pair. Alice measures a fixed observable Aˆ ai ai ai on her member of each pair. Bob is free to choose any observable Bˆ bi bi bi to measure on his member of each pair. Alice Question: Does prob(ai) depend on what observableB̂ Bob chooses to measure? Bob Answer: No, because prob(ai ) ai b j b j ai j ai , b j 2 j drawings by Tom Oreb © Walt Disney Co. Quantum mechanics does not allow “signalling” Introducing…Alice and Bob, who share pairs in an entangled state and make spacelike separated measurements on members of each pair. Alice measures a fixed observable Aˆ ai ai ai on her member of each pair. Bob is free to choose any observable Bˆ bi bi bi to measure on his member of each pair. Alice Question: Does prob(ai) depend on what observableB̂ Bob chooses to measure? Answer: No, because prob(ai ) Bob ai 2 ai , b j 2 j . drawings by Tom Oreb © Walt Disney Co. “Superquantum” correlations Back to the question: Can we derive quantum mechanics from the two axioms of nonlocal correlations and no “signalling” at any speed? This is a grandiose ambition. “Superquantum” correlations Back to the question: Can we derive quantum mechanics from the two axioms of nonlocal correlations and no “signalling” at any speed? This is a grandiose ambition. We could, however, entertain a more modest ambition: Could we derive a part of quantum mechanics from these two axioms? “Superquantum” correlations Back to the question: Can we derive quantum mechanics from the two axioms of nonlocal correlations and no “signalling” at any speed? This is a grandiose ambition. We could, however, entertain a more modest ambition: Namely, could we derive Tsirelson’s bound from these two axioms? Alice’s black box can nonlocally influence Bob’s. Alice Bob Alice’s black box can nonlocally influence Bob’s. A, A′, B, and B′ are physical quantities having values ±1. Alice measures A or A′ Bob measures B or B′ Alice’s black box can nonlocally influence Bob’s. A, A′, B, and B′ are physical quantities having values ±1. Alice measures A or A′ CHSH inequality |CL(A,B)+CL(A,B′)+CL(A′,B)−CL(A′,B′)| ≤ 2 Bob measures B or B′ Alice’s black box can nonlocally influence Bob’s. A, A′, B, and B′ are physical quantities having values ±1. Alice measures A or A′ CHSH inequality |CL(A,B)+CL(A,B′)+CL(A′,B)−CL(A′,B′)| ≤ 2 Bob measures B or B′ Tsirelson’s bound |CQ(A,B)+CQ(A,B′)+CQ(A′,B)−CQ(A′,B′)| ≤ 2√2 Alice’s black box can nonlocally influence Bob’s. A, A′, B, and B′ are physical quantities having values ±1. Alice measures A or A′ CHSH inequality (locality) |CL(A,B)+CL(A,B′)+CL(A′,B)−CL(A′,B′)| ≤ 2 Bob measures B or B′ Tsirelson's bound |CQ(A,B)+CQ(A,B′)+CQ(A′,B)−CQ(A′,B′)| ≤ 2√2 Alice’s black box can nonlocally influence Bob’s. A, A′, B, and B′ are physical quantities having values ±1. Alice measures A or A′ CHSH inequality (locality) |CL(A,B)+CL(A,B′)+CL(A′,B)−CL(A′,B′)| ≤ 2 Bob measures B or B′ Tsirelson's bound (“no signalling”?) |CQ(A,B)+CQ(A,B′)+CQ(A′,B)−CQ(A′,B′)| ≤ 2√2 How to beat Tsirelson’s bound in two easy steps: 1. For any measurement of A, A′, B, and B′, let the outcomes 1 and −1 be equally likely. 2. Let CSQ(A,B) = CSQ(A,B′) = CSQ(A′,B) = 1 = −CSQ(A′,B′). B A′ 45º A 45º 45º B′ Then CSQ(A,B)+CSQ(A,B′)+CSQ(A′,B)−CSQ(A′,B′) = 4. “Superquantum” correlations Example of CSQ(φ): CSQ(φ) 1 φ 0 45º 90º 135º 180º −1 0º S. Popescu and DR, Found. Phys. 24 (1994) 379 “Superquantum” correlations Q. Superquantum correlations do not imply quantum mechanics. Do they imply uncertainty? A. Yes! Let Bob measure σz(B). If σz(B) = 1, then σz(A) = 1. Then Δσx(A) = 1 = Δσy(A). This uncertainty is irreducible because, since CSQ(φ) violates the Tsirelson bound, it violates also the CHSH inequality. More recent results: • Superquantum correlations render all communication complexity problems trivial. W. van Dam, Thesis, U. Oxford (1999); W. van Dam, preprint quant-ph/0501159 (2005) • All correlations stronger than quantum correlations may render communication complexity problems trivial. G. Brassard et al., Phys. Rev. Lett. 96 (2006) 250401 From “superquantum correlations” to the “nonlocal box” “Mathematicians are like Frenchmen: whatever you say to them they translate into their own language and forthwith it is something entirely different.” ― Goethe From “superquantum correlations” to the “nonlocal box” { Nonlocal box: PNL(a,b|x,y) = Translation: ½ 0 a + b ↔ (a + b) mod 2 if a + b = xy otherwise From “superquantum correlations” to the “nonlocal box” { Nonlocal box: PNL(a,b|x,y) = Translation: ½ 0 if a + b = xy otherwise ↔ ↔ ↔ ↔ x=0 x=1 y=0 y=1 Alice’s result is 1 ↔ Alice’s result is –1 ↔ Bob’s result is 1 ↔ Bob’s result is –1 ↔ a=1 a=0 b=1 b=0 Alice measures A Alice measures A′ Bob measures B Bob measures B′ Axioms for quantum mechanics 1. What is nonlocality? We’ve tried nonlocal correlations. What about nonlocal equations of motion (nonlocal cause and effect)? 2. What does “no signalling” mean? “No signalling” at superluminal speeds Axioms for quantum mechanics Introducing…Jim “the Jammer” J. Grunhaus, S. Popescu and DR, Phys. Rev. A53 (1996) 3781 drawings by Tom Oreb © Walt Disney Co. Axioms for quantum mechanics Introducing…“jamming” Alice measures A or A′ Jim “the Jammer” Bob measures B or B′ drawings by Tom Oreb © Walt Disney Co. Q. Can “Jim the Jammer” jam the nonlocal correlations of Alice and Bob? A. Yes, if “jamming” satisfies two conditions: 1. The unary condition – For any measurement of A, A′, B, and B′, the outcomes 1 and −1 are equally likely, with or without jamming. 2. The binary condition – The intersection of the forward light cones of Alice’s and Bob’s measurements lies in the forward light cone of Jim’s action. A configuration that violates the binary condition: b a j A configuration that obeys the binary condition: b a j This configuration, too, obeys the binary condition! j b a Back to the “nonlocal box” { Nonlocal box: PNL(a,b|x,y) = Translation: ½ 0 if a + b = xy otherwise ↔ ↔ ↔ ↔ x=0 x=1 y=0 y=1 Alice’s result is 1 ↔ Alice’s result is –1 ↔ Bob’s result is 1 ↔ Bob’s result is –1 ↔ a=1 a=0 b=1 b=0 Alice measures A Alice measures A′ Bob measures B Bob measures B′ Back to the “nonlocal box” { Nonlocal box: PNL(a,b|x,y) = ½ 0 if a + b = xy otherwise Does this nonlocal box violate “information causality”? Information causality: If Alice sends Bob m (classical) bits, Bob has access to a range of at most m bits of Alice’s data. Case m = 0: “no signalling”. M. Pawłowski, T. Paterek, D. Kaszlikowski, V. Scarani, A. Winter and M. Żukowski, Nature 461 (2009) 1101 Back to the “nonlocal box” { Nonlocal box: PNL(a,b|x,y) = ½ 0 if a + b = xy otherwise Does this nonlocal box violate “information causality”? Information causality: If Alice sends Bob m (classical) bits, Bob has access to a range of at most m bits of Alice’s data. We will concentrate on the case m = 1. M. Pawłowski, T. Paterek, D. Kaszlikowski, V. Scarani, A. Winter and M. Żukowski, Nature 461 (2009) 1101 Back to the “nonlocal box” { Nonlocal box: PNL(a,b|x,y) = ½ 0 if a + b = xy otherwise Case m = 1: Alice has two fresh, independent bits a0 and a1. Bob receives a number, either 0 or 1, and his task is to obtain the corresponding bit from Alice, i.e. either a0 or a1. Alice sends him one classical bit. There is no other communication between them. Using classical physics, Bob cannot succeed. Back to the “nonlocal box” { Nonlocal box: PNL(a,b|x,y) = ½ 0 if a + b = xy otherwise Case m = 1: Alice has two fresh, independent bits a0 and a1. Bob receives a number, either 0 or 1, and his task is to obtain the corresponding bit from Alice, i.e. either a0 or a1. Alice sends him one classical bit. There is no other communication between them. Using classical physics, Bob cannot succeed. Using quantum correlations, Bob cannot succeed. Back to the “nonlocal box” { Nonlocal box: PNL(a,b|x,y) = ½ 0 if a + b = xy otherwise Case m = 1: Alice has two fresh, independent bits a0 and a1. Bob receives a number, either 0 or 1, and his task is to obtain the corresponding bit from Alice, i.e. either a0 or a1. Alice sends him one classical bit. There is no other communication between them. Using classical physics, Bob cannot succeed. Using quantum correlations, Bob cannot succeed. Using a nonlocal box, Bob can succeed. Back to the “nonlocal box” { Nonlocal box: PNL(a,b|x,y) = ½ 0 if a + b = xy otherwise Protocol for Alice and Bob: 1. Alice chooses, as input to the nonlocal box, x a0 a1. 2. Alice adds (mod 2) the output a from the nonlocal box to a0. She sends the result, i.e. a a0 , to Bob. 3. Bob, for his input to the nonlocal box, chooses y = 0 if he wants a0 and y = 1 if he wants a1. To the output b he adds the one bit from Alice, and this sum, b a a0, is his response. Back to the “nonlocal box” { Nonlocal box: PNL(a,b|x,y) = ½ 0 if a + b = xy otherwise Analysis: If Bob chooses y = 0, then a b xy 0. Thus Bob’s response reduces from b a a0 to a0 , as required. If Bob chooses y = 1, then a b xy x a0 a1. Thus b a a0 reduces to a0 a1 a0 a1, as required. Back to the “nonlocal box” { Nonlocal box: PNL(a,b|x,y) = ½ 0 if a + b = xy otherwise Analysis: If Bob chooses y = 0, then a b xy 0. Thus Bob’s response reduces from b a a0 to a0 , as required. If Bob chooses y = 1, then a b xy x a0 a1. Thus b a a0 reduces to a0 a1 a0 a1, as required. Back to the “nonlocal box” { Nonlocal box: PNL(a,b|x,y) = ½ 0 if a + b = xy otherwise Analysis: If Bob chooses y = 0, then a b xy 0. Thus Bob’s response reduces from b a a0 to a0 , as required. If Bob chooses y = 1, then a b xy x a0 a1. Thus b a a0 reduces to a0 a1 a0 a1, as required. Discussion • The (simplest) example involves the most extreme nonlocal box, the PR box. But any nonlocal box violating Tsirelson’s bound violates information causality. • So are we done? No, because the correlations of some nonlocal boxes (“noisy” nonlocal boxes) do not exceed Tsirelson’s bound yet are not quantum correlations. • In any case, “information causality” is a lovely name, but what does it mean? What is the connection with causality? Discussion To quote from Pawłowski et al.: “Note that the extreme violation of Information Causality observed here means that Bob can learn perfectly either bit, not that he can learn both Alice's bits simultaneously – the latter would imply signaling.” Is that so?