1. Nonlocal Boxes And All That Daniel Rohrlich Atom Chip Group, Ben Gurion University, Beersheba, Israel 21 January 2010

2. 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.

3. Axioms for quantum mechanics Physical states are normalized vectors ψ(r), Ψ(r,t),,. Measurable physical quantities – “observables” – correspond to Hermitian or (self-adjoint) operators on the state vectors. If a system is an eigenstate with eigenvalue a of an observable, then a measurement of on will yield a. Conversely, if a measurement of on any state yields a, the measurement leaves the system in an eigenstate. The probability that a system in a normalized state can be found in the state is.

4. Axioms for quantum mechanics The time evolution of a quantum state is given by where is the Hamiltonian (kinetic energy + potential energy) of the system in the state. The wave function of identical fermions (spin ½, ¾,…) must 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. ²

5. “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. …

6. Question: What, indeed, is so “special” about special relativity?

7. Answer: The two axioms so nearly contradict each other that only a unique theory reconciles them.

8. 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:

9. 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: Nonlocality

10. 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: No signalling Nonlocality

11. Two axioms for quantum mechanics Axiom 1: There are nonlocal influences. (What are they?) Axiom 2: We cannot send superluminal signals.

12. Can we derive quantum mechanics from these two axioms? 1. What is nonlocality?

13. Can we derive quantum mechanics from these two axioms? 1. What is nonlocality? 2. What does “no signalling” mean?

14. Can we derive quantum mechanics from these two axioms? 1. What is nonlocality? 2. What does “no signalling” mean? Nonlocal correlations? Aharonov-Bohm effect? “Modular” dynamical variables?

15. Can we derive quantum mechanics from these two axioms? 1. What is nonlocality? 2. What does “no signalling” mean? What is left of “no signalling” in the limit c → ∞ of nonrelativistic quantum mechanics? Nonlocal correlations? Aharonov-Bohm effect? “Modular” dynamical variables?

16. Can we derive quantum mechanics from these two axioms? 1. What is nonlocality? 2. What does “no signalling” mean? What is left of “no signalling” in the limit c → ∞ of nonrelativistic quantum mechanics? Nonlocal correlations

17. Can we derive quantum mechanics from these two axioms? 1. What is nonlocality? 2. What does “no signalling” mean? “No signalling” at any speed! Nonlocal correlations

18. Quantum mechanics does not allow “signalling” Introducing…Alice and Bob

19. Quantum mechanics does not allow “signalling” Introducing…Alice and Bob drawings by Tom Oreb © Walt Disney Co. Alice Bob

20. 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 on her member of each pair. Bob is free to choose any observable to measure on his member of each pair. drawings by Tom Oreb © Walt Disney Co. Question: Does prob(a i ) depend on what observable Bob chooses to measure? Alice Bob

21. 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 on her member of each pair. Bob is free to choose any observable to measure on his member of each pair. drawings by Tom Oreb © Walt Disney Co. Question: Does prob(a i ) depend on what observable Bob chooses to measure? Answer: No, because Alice Bob

22. 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 on her member of each pair. Bob is free to choose any observable to measure on his member of each pair. drawings by Tom Oreb © Walt Disney Co. Question: Does prob(a i ) depend on what observable Bob chooses to measure? Answer: No, because Alice Bob

23. 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 on her member of each pair. Bob is free to choose any observable to measure on his member of each pair. drawings by Tom Oreb © Walt Disney Co. Question: Does prob(a i ) depend on what observable Bob chooses to measure? Answer: No, because Alice Bob

24. “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.

25. “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?

26. “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?

27. Alice Bob Alice’s black box can nonlocally influence Bob’s.

28. 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.

29. 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. CHSH inequality |C L (A,B)+C L (A,B′)+C L (A′,B)−C L (A′,B′)| ≤ 2

30. 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. CHSH inequality |C L (A,B)+C L (A,B′)+C L (A′,B)−C L (A′,B′)| ≤ 2 Tsirelson’s bound |C Q (A,B)+C Q (A,B′)+C Q (A′,B)−C Q (A′,B′)| ≤ 2√2

31. 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. CHSH inequality (locality) |C L (A,B)+C L (A,B′)+C L (A′,B)−C L (A′,B′)| ≤ 2 Tsirelson's bound |C Q (A,B)+C Q (A,B′)+C Q (A′,B)−C Q (A′,B′)| ≤ 2√2

32. 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. CHSH inequality (locality) |C L (A,B)+C L (A,B′)+C L (A′,B)−C L (A′,B′)| ≤ 2 Tsirelson's bound (“no signalling”?) |C Q (A,B)+C Q (A,B′)+C Q (A′,B)−C Q (A′,B′)| ≤ 2√2

33. 45º 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 C SQ (A,B) = C SQ (A,B′) = C SQ (A′,B) = 1 = −C SQ (A′,B′). A A′ B B′ Then C SQ (A,B)+C SQ (A,B′)+C SQ (A′,B)−C SQ (A′,B′) = 4.

34. Example of C SQ (φ): φ C SQ (φ) 0 1 −1 180º90º 0º 45º135º S. Popescu and DR, Found. Phys. 24 (1994) 379 “Superquantum” correlations

35. Q. Superquantum correlations do not imply quantum mechanics. Do they imply uncertainty? A.Yes! Let Bob measure σ z (B). If σ z (B) = 1, then σ z (A) = 1. Then Δσ x (A) = 1 = Δσ y (A). This uncertainty is irreducible because, since C SQ (φ) violates the Tsirelson bound, it violates also the CHSH inequality. “Superquantum” correlations

36. 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

37. 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

38. From “superquantum correlations” to the “nonlocal box” Nonlocal box: P NL (a,b|x,y) = Translation: { ½ if a + b = xy 0 otherwise a + b ↔ (a + b) mod 2

39. From “superquantum correlations” to the “nonlocal box” Nonlocal box: P NL (a,b|x,y) = Translation: Alice measures A ↔ x = 0 Alice measures A′ ↔ x = 1 Bob measures B ↔ y = 0 Bob measures B′ ↔ y = 1 Alice’s result is 1 ↔ a = 1 Alice’s result is –1 ↔ a = 0 Bob’s result is 1 ↔ b = 1 Bob’s result is –1 ↔ b = 0 { ½ if a + b = xy 0 otherwise

40. Axioms for quantum mechanics 1. What is nonlocality? 2. What does “no signalling” mean? “No signalling” at superluminal speeds We’ve tried nonlocal correlations. What about nonlocal equations of motion (nonlocal cause and effect)?

41. 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.

42. Alice measures A or A′ Bob measures B or B′ Jim “the Jammer” drawings by Tom Oreb © Walt Disney Co. Axioms for quantum mechanics Introducing…“jamming”

43. 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.

44. a b j A configuration that violates the binary condition:

45. a b j A configuration that obeys the binary condition:

46. a b j This configuration, too, obeys the binary condition!

47. Back to the “nonlocal box” Nonlocal box: P NL (a,b|x,y) = Translation: Alice measures A ↔ x = 0 Alice measures A′ ↔ x = 1 Bob measures B ↔ y = 0 Bob measures B′ ↔ y = 1 Alice’s result is 1 ↔ a = 1 Alice’s result is –1 ↔ a = 0 Bob’s result is 1 ↔ b = 1 Bob’s result is –1 ↔ b = 0 { ½ if a + b = xy 0 otherwise

48. Back to the “nonlocal box” Nonlocal box: P NL (a,b|x,y) = 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 { ½ if a + b = xy 0 otherwise

49. Back to the “nonlocal box” Nonlocal box: P NL (a,b|x,y) = 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 { ½ if a + b = xy 0 otherwise

50. Back to the “nonlocal box” Nonlocal box: P NL (a,b|x,y) = Case m = 1: Alice has two fresh, independent bits a 0 and a 1. Bob receives a number, either 0 or 1, and his task is to obtain the corresponding bit from Alice, i.e. either a 0 or a 1. Alice sends him one classical bit. There is no other communication between them. Using classical physics, Bob cannot succeed. { ½ if a + b = xy 0 otherwise

51. Back to the “nonlocal box” Nonlocal box: P NL (a,b|x,y) = Case m = 1: Alice has two fresh, independent bits a 0 and a 1. Bob receives a number, either 0 or 1, and his task is to obtain the corresponding bit from Alice, i.e. either a 0 or a 1. 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. { ½ if a + b = xy 0 otherwise

52. Back to the “nonlocal box” Nonlocal box: P NL (a,b|x,y) = Case m = 1: Alice has two fresh, independent bits a 0 and a 1. Bob receives a number, either 0 or 1, and his task is to obtain the corresponding bit from Alice, i.e. either a 0 or a 1. 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. { ½ if a + b = xy 0 otherwise

53. Back to the “nonlocal box” Nonlocal box: P NL (a,b|x,y) = Protocol for Alice and Bob: 1. Alice chooses, as input to the nonlocal box, 2. Alice adds (mod 2) the output a from the nonlocal box to. She sends the result, i.e., to Bob. 3. Bob, for his input to the nonlocal box, chooses y = 0 if he wants and y = 1 if he wants a 1. To the output b he adds the one bit from Alice, and this sum,, is his response. { ½ if a + b = xy 0 otherwise

54. Back to the “nonlocal box” Nonlocal box: P NL (a,b|x,y) = Analysis: If Bob chooses y = 0, then Thus Bob’s response reduces from to as required. If Bob chooses y = 1, then Thus reduces to as required. { ½ if a + b = xy 0 otherwise

55. Back to the “nonlocal box” Nonlocal box: P NL (a,b|x,y) = Analysis: If Bob chooses y = 0, then Thus Bob’s response reduces from to as required. If Bob chooses y = 1, then Thus reduces to as required. { ½ if a + b = xy 0 otherwise

56. Back to the “nonlocal box” Nonlocal box: P NL (a,b|x,y) = Analysis: If Bob chooses y = 0, then Thus Bob’s response reduces from to as required. If Bob chooses y = 1, then Thus reduces to as required. { ½ if a + b = xy 0 otherwise

57. 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?

58. 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?