Atom Chip Group, Ben Gurion University, Beersheba, Israel Stronger-than-quantum correlations violate relativistic causality in the classical limit Daniel Rohrlich Atom Chip Group, Ben Gurion University, Beersheba, Israel 26 February 2012

Outline What is a PR-box Why I hate the PR-box The PR-box is rotten Killing off the PR-box Killing off all stronger-than-quantum boxes A paradox A miraculous derivation of Tsirelson’s bound From quantum correlations to Hilbert space

Outline What is a PR-box Why I hate the PR-box The PR-box is rotten Killing off the PR-box Killing off all stronger-than-quantum boxes A paradox A miraculous derivation of Tsirelson’s bound From quantum correlations to Hilbert space

Axioms for special relativity The laws of physics are the same in all inertial reference frames. There is a maximum signalling speed, and it is the speed of light c.

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 .

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

“It’s like trying to derive special relativity from the wrong axioms “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. …

H. Poincaré (1909) “La mécanique nouvelle” [“The new mechanics”] “One needs to make still a third hypothesis, much more surprising, much more difficult to accept, one which is of much hindrance to what we are currently used to. A body in translational motion suffers a deformation in the direction in which it is displaced.”

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: Nonlocality

No signalling Nonlocality 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 No signalling

Can we derive a part quantum mechanics from these two axioms? 1. Nonlocal correlations 2. Relativistic causality

Alice and Bob share entangled pairs. They measure a, a′, b, and b′ – 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 Tsirelson’s bound |CQ(a,b)+CQ(a,b′)+CQ(a′,b)−CQ(a′,b′)| ≤ 2 Bob measures b or b′ drawings by Tom Oreb © Walt Disney Co.

Alice and Bob share entangled pairs. They measure a, a′, b, and b′ – 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 Tsirelson’s bound |CQ(a,b)+CQ(a,b′)+CQ(a′,b)−CQ(a′,b′)| ≤ 2 Bob measures b or b′ drawings by Tom Oreb © Walt Disney Co.

Alice and Bob share entangled pairs. They measure a, a′, b, and b′ – 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 Tsirelson’s bound (relativistic causality?) |CQ(a,b)+CQ(a,b′)+CQ(a′,b)−CQ(a′,b′)| ≤ 2 Bob measures b or b′ drawings by Tom Oreb © Walt Disney Co.

For any measurement of a, a′, b, and b′, let the outcomes 1 and −1 be equally likely. Let CPR(a,b) = CPR(a,b′) = CPR(a′,b) = 1 = −CPR(a′,b′). Superquantum/maximally nonlocal box/PR-box correlations A A′ B B′ 45º 45º 45º Then CPR(a,b)+CPR(a,b′)+CPR(a′,b)−CPR(a′,b′) = 4. S. Popescu and DR, Found. Phys. 24 (1994) 379

Outline What is a PR-box Why I hate the PR-box The PR-box is rotten Killing off the PR-box Killing off all stronger-than-quantum boxes A paradox A miraculous derivation of Tsirelson’s bound From quantum correlations to Hilbert space

It should not be so easy to disprove such a lovely conjecture!

More recent results: PR-box renders all communication complexity problems trivial. W. van Dam, Thesis, U. Oxford (1999); W. van Dam, preprint quant-ph/0501159 (2005) Stronger-than-quantum correlations may render communication complexity problems trivial. G. Brassard et al., Phys. Rev. Lett. 96 (2006) 250401 Most stronger-than-quantum correlations violate “information causality”. M. Pawłowski et al., Nature 461 (2009) 1101

Stronger-than-quantum correlations do not survive in the macroscopic (classical) limit. M. Navascués and H. Wunderlich, Proc. R. Soc. A 466 (2010) 881

Outline What is a PR-box Why I hate the PR-box The PR-box is rotten Killing off the PR-box Killing off all stronger-than-quantum boxes A paradox A miraculous derivation of Tsirelson’s bound From quantum correlations to Hilbert space

PR-box correlations Suppose Alice measures a. She knows that if Bob measures b, he obtains b = a. If he measures b′, he obtain b′ = a.

PR-box correlations Suppose Alice measures a′. She knows that if Bob measures b, he obtains b = a′. If he measures b′, he obtain b′ = –a′.

PR-box correlations Suppose Alice measures a. She knows that b = b′. Suppose Alice measures a′. She knows that b = –b′.

PR-box correlations Suppose Alice measures a. She knows that b = b′. Suppose Alice measures a′. She knows that b = –b′. We might call a, a′, b and b′ (maximally) nonlocal hidden variables.

PR-box correlations All that stops Alice from signalling to Bob is complementarity between Bob’s measuring b and his measuring b′, even though (from Alice’s point of view) no uncertainty principle governs b and b′. Alice can even prepare ensembles (e.g. by measuring a and postselecting a = 1) in which Δb = 0 = Δb′. Complementarity is a

PR-box correlations All that stops Alice from signalling to Bob is complementarity between Bob’s measuring b and his measuring b′, even though (from Alice’s point of view) no uncertainty principle governs b and b′. Alice can even prepare ensembles (e.g. by measuring a and postselecting a = 1) in which Δb = 0 = Δb′. Complementarity is a for the PR-box.

Outline What is a PR-box Why I hate the PR-box The PR-box is rotten Killing off the PR-box Killing off all stronger-than-quantum boxes A paradox A miraculous derivation of Tsirelson’s bound From quantum correlations to Hilbert space

Minimal, physical additional axiom: a classical limit Quantum mechanics has a classical limit. In this limit there are no noncommuting quantum observables; there are only jointly measurable macroscopic observables. This classical limit – our direct experience – is an inherent constraint, a kind of boundary condition, on quantum mechanics and on any generalization of quantum mechanics. Thus stronger-than-quantum correlations, too, must have a classical limit.

PR-box correlations Now suppose Alice measures just a or just a′ on N pairs. Define macroscopic observables B and B′: Alice already knows the values of B and B′, and there must be “weak” measurements that Bob can make to obtain partial information about both B and B′: there is no complementarity in the classical limit! On average both B and B′ vanish, but if Alice measures a, B and B′ will of order 1/√N and correlated; if she measures a′, B and B′ will of order 1/√N and anti-correlated.

Alice and Bob can measure exponentially many pairs (in groups of N) Alice and Bob can measure exponentially many pairs (in groups of N). Their expenses and exertions don’t concern us. Ultimately, Alice will be able to signal to Bob by consistently measuring a or a′. What matters is only that when Bob detects a correlation, it is more likely that Alice measured a than when he detects an anti-correlation. If not, Bob's measurements yield zero information about B or about B′, contradicting the axiom of a classical limit in which B and B′ are jointly measurable. drawings by Tom Oreb © Walt Disney Co.

Outline What is a PR-box Why I hate the PR-box The PR-box is rotten Killing off the PR-box Killing off all stronger-than-quantum boxes A paradox A miraculous derivation of Tsirelson’s bound From quantum correlations to Hilbert space

What about weaker-than-PR-box correlations? In the classical limit, Bob has some information about both B and B′, hence some information about B ± B′. However, B + B′ = 0 if Alice measures only a′, while B + B′ fluctuates binomially if Alice measures only a. Either way , but the standard deviation Δa(B + B′) or Δa′(B + B′) reveals to him Alice’s message. What about weaker-than-PR-box correlations? What about a correlation C(a,b) such that C(a,b) = C(a,b′) = C(a′,b) = C = −C(a′,b′) for some constant –1 < C < 1? drawings by Tom Oreb © Walt Disney Co.

If C is close enough to 1, it will still be possible for Alice to signal to Bob. But for some critical C, it will not be possible. Is the critical C equal to CQ = /2? What about weaker-than-PR-box correlations? What about a correlation C(a,b) such that C(a,b) = C(a,b′) = C(a′,b) = C = −C(a′,b′) for some constant –1 < C < 1? drawings by Tom Oreb © Walt Disney Co.

If C is close enough to 1, it will still be possible for Alice to signal to Bob. But for some critical C, it will not be possible. Is the critical C equal to CQ = /2? Note that C does not determine B + B′ . It only determines a range for B + B′ . drawings by Tom Oreb © Walt Disney Co.

Question: Alice measures a. If Bob measures b, what are the probabilities p± that he obtains b = ±a? Answer: p+ + p– = 1 and p+ – p– = C, hence p± = (1 ± C)/2. If Bob measures b′, the probabilities that he obtains b′ = ±a are the same. Likewise if Alice measures a′, the probabilities that Bob obtains b = ±a′ are the same. But the probabilities that Bob obtains b′ = ±a′ are (1 C)/2.

p+ p− maximize Δa(B + B′) minimize Δa(B + B′) Δa(B + B′ ) ≥ 2 C

p+ p− maximize Δa(B + B′) minimize Δa(B + B′) Δa(B + B′ ) ≥ 2 C minimize Δa′(B + B′) maximize Δa′(B + B′) Δa′(B + B′ ) ≤ 2 C

If Δa′(B + B′ ) ≠ Δa′(B + B′ ), then Alice can signal to Bob.

If Δa′(B + B′ ) ≠ Δa′(B + B′ ), then Alice can signal to Bob. If C > 1/2, then Δa′(B + B′ ) ≤ 2 ≠ 2 ≤ Δa(B + B′ ).

Outline What is a PR-box Why I hate the PR-box The PR-box is rotten Killing off the PR-box Killing off all stronger-than-quantum boxes A paradox A miraculous derivation of Tsirelson’s bound From quantum correlations to Hilbert space

Bell’s theorem Hidden variables Locality C ≤ 1/2

Relativistic causality Bell’s theorem? Relativistic causality Classical limit Hidden variables C ≤ 1/2

Relativistic causality Bell’s theorem? Relativistic causality Classical limit ? Hidden variables Locality C ≤ 1/2

Outline What is a PR-box Why I hate the PR-box The PR-box is rotten Killing off the PR-box Killing off all stronger-than-quantum boxes A paradox A miraculous derivation of Tsirelson’s bound From quantum correlations to Hilbert space

The PR-box tells us nothing about Δa(B + B′ ) and Δa′(B + B′ ). But the classical limit tells us to think “outside the PR-box”.

The PR-box tells us nothing about Δa(B + B′ ) and Δa′(B + B′ ). But the classical limit tells us to think “outside the PR-box”.

Lemma Therefore and hence relativistic causality requires

Back to the correlations C(a,b), C(a,b′), C(a′,b) and C(a′,b′) . Suppose Alice measures a on sets of N pairs. If we define then A takes values –1, –1+2/N, –1+4/N,…,1–2/N, 1 in a perfect binomial distribution. If A = 1 – 2n/N, then B + B′ ≈ (1 – 2n/N)[C(a,b)+C(a,b′)]

Back to the correlations C(a,b), C(a,b′), C(a′,b) and C(a′,b′) . Suppose Alice measures a on sets of N pairs. If we define then A takes values –1, –1+2/N, –1+4/N,…,1–2/N, 1 in a perfect binomial distribution. If A = 1 – 2n/N, then B + B′ ≈ (1 – 2n/N)[C(a,b)+C(a,b′)].

[C(a,b)+C(a,b′)]/ ≤ Δa(B + B′ ) Back to the correlations C(a,b), C(a,b′), C(a′,b) and C(a′,b′) . Suppose Alice measures a on sets of N pairs. If we define then A takes values –1, –1+2/N, –1+4/N,…,1–2/N, 1 in a perfect binomial distribution. If A = 1 – 2n/N, then B + B′ ≈ (1 – 2n/N)[C(a,b)+C(a,b′)]. The minimum Δa(B + B′) (a convolution of ΔaA and “noise”) is for B + B′ = (1 – 2n/N)[C(a,b)+C(a,b′)], i.e. [C(a,b)+C(a,b′)]/ ≤ Δa(B + B′ ) [C(a′,b)–C(a′,b′)]/ ≤ Δa′(B – B′ ) Similarly

[C(a,b)+C(a,b′)]/ ≤ Δa(B + B′ ) Now [C(a,b)+C(a,b′)]/ ≤ Δa(B + B′ ) [C(a′,b)–C(a′,b′)]/ ≤ Δa′(B – B′ )

Now

|C(a′,b)–C(a′,b′)|2 = 4 – |C(a,b)+C(a,b′)|2 i.e. To maximize |C(a,b) + C(a,b′) + C(a′,b) – C(a′,b′)| we should saturate this bound, so that |C(a′,b)–C(a′,b′)|2 = 4 – |C(a,b)+C(a,b′)|2 and so we maximize with z = |C(a,b) + C(a,b′)|,

to derive Tsirelson’s bound, from the axioms of nonlocality, relativistic causality and the existence of a classical limit!

to derive Tsirelson’s bound, from the axioms of nonlocality, relativistic causality and the existence of a classical limit!

Outline What is a PR-box Why I hate the PR-box The PR-box is rotten Killing off the PR-box Killing off all stronger-than-quantum boxes A paradox A miraculous derivation of Tsirelson’s bound From quantum correlations to Hilbert space

By applying the constraint Δa(B + B′ ) = Δa′(B + B′ ) so that there will be no signalling, we have found that the maximal correlation is the quantum correlation, CQ= /2. The maximal correlation implies that when Alice measures a and obtains ±1, Bob measures b+b′ (not the same as measuring b or b′) and obtains ±2CQ.

By applying the constraint Δa(B + B′ ) = Δa′(B + B′ ) so that there will be no signalling, we have found that the maximal correlation is the quantum correlation, CQ= /2. The maximal correlation implies that when Alice measures a and obtains ±1, Bob measures b+b′ (not the same as measuring b or b′) and obtains ±2CQ. The fact that the linear combination (b + b′)/2CQ = (b + b′)/ of variables taking values ±1 is another variable taking values ±1 shows that b and b′ add as vectors, and not as scalars.