2012
DOI: 10.1103/physrevlett.109.180401
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Device-Independent Bounds for Hardy’s Experiment

Abstract: In this Letter we compute an analogue of Tsirelson's bound for Hardy's test of nonlocality, that is, the maximum violation of locality constraints allowed by the quantum formalism, irrespective of the dimension of the system. The value is found to be the same as the one achievable already with two-qubit systems, and we show that only a very specific class of states can lead to such maximal value, thus highlighting Hardy's test as a device-independent self-test protocol for such states. By considering realistic… Show more

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Cited by 70 publications
(99 citation statements)
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“…We now need an operator X B such that it is hermitian, unitary and behaves almost similar to X (21) and condition (16) we have that…”
Section: B Proofmentioning
confidence: 99%
See 1 more Smart Citation
“…We now need an operator X B such that it is hermitian, unitary and behaves almost similar to X (21) and condition (16) we have that…”
Section: B Proofmentioning
confidence: 99%
“…al. in [16] have illustrated a method to self test Hardy inequality, which involves partially entangled qubit states. Their method however, are not known to be robust and only limited to Hardy's inequality.…”
Section: Introductionmentioning
confidence: 99%
“…Reference [11] proves that, for 2-qutrit systems, maximum achievable value of Hardy's success probability is same as that of 2-qubit system and conjectures that it will remain same for arbitrary dimension n. Recently, reference [20] provides a proof of this conjecture. This result tells that for showing the contradiction of quantum mechanics with the local realism higher dimensional systems give no advantage in experimental implementation of such a test.…”
Section: Non-signaling Conditionsmentioning
confidence: 99%
“…This in effect can provide us with an upper bound in the paradoxical probability of this argument that is allowed by relativistic causality alone, in absence of any further constraints. This study is also important because the optimal success probability of Hardy's argument is deeply connected with the security proof of several information processing tasks [20][21][22]. Here we deal with these questions.…”
Section: Introductionmentioning
confidence: 99%
“…case, we simply have B 1 = 0 for any h ∈ [1/4, 1/3) [11]. In that case, the inequality can be violated by any quantum probability distribution that demonstrates a Hardy-type paradox [28][29][30], where P (01|01) = P (10|10) = P (00|11) = 0 but P (00|00) > 0. This gives M (l, h) = lP (00|00)P XY (00), thereby violating the inequality for any l > 0.…”
Section: The Iid MDL Inequalitymentioning
confidence: 99%