2015
DOI: 10.1088/1367-2630/18/1/013007
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Can different quantum state vectors correspond to the same physical state? An experimental test

Abstract: A century after the development of quantum theory, the interpretation of a quantum state is still discussed. If a physicist claims to have produced a system with a particular quantum state vector, does this represent directly a physical property of the system, or is the state vector merely a summary of the physicist's information about the system? Assume that a state vector corresponds to a probability distribution over possible values of an unknown physical or 'ontic' state. Then, a recent no-go theorem shows… Show more

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Cited by 71 publications
(46 citation statements)
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References 34 publications
(88 reference statements)
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“…Notably, Pusey, Barrett, and Rudolph showed that no ψ-epistemic models satisfying some natural independence condition for composite systems can reproduce quantum predictions [3]. A number of other no-go theorems have since been proven, under various assumptions [4][5][6][7][8][9][10][11][12][13][14][15][16][17][18]. Explicit ψ-epistemic models have nevertheless been constructed for quantum measurement statistics in Hilbert spaces of any dimension [14,19], which get around the assumptions of these no-go theorems.…”
mentioning
confidence: 99%
“…Notably, Pusey, Barrett, and Rudolph showed that no ψ-epistemic models satisfying some natural independence condition for composite systems can reproduce quantum predictions [3]. A number of other no-go theorems have since been proven, under various assumptions [4][5][6][7][8][9][10][11][12][13][14][15][16][17][18]. Explicit ψ-epistemic models have nevertheless been constructed for quantum measurement statistics in Hilbert spaces of any dimension [14,19], which get around the assumptions of these no-go theorems.…”
mentioning
confidence: 99%
“…Indeed, the best such a Bell-local model can do is to satisfy three of the four. An easy way to see the contradiction is by multiplying all four equations, such that the left-hand side is a product of squares, which must be +1, while the right-hand side equals −1 [81]. Using instead a shared a GHZ-state 1 2 ( 000⟩ + 111⟩), to generate the outputs as the results of measurements of σ x (for input 0) and σ y (for input 1), all four requirements ideally satisfied exactly.…”
Section: Ghz Paradox and Multipartite Entanglementmentioning
confidence: 99%
“…A violation could instead be obtained from measurements on a statistical mixture of bipartite entangled states with a classically correlated third particle [83]. To demonstrate genuine tripartite Bell-nonlocality, Svetlichny derived an inequality that cannot be violated by any states that are only bipartite entangled [81,83,84]…”
Section: Ghz Paradox and Multipartite Entanglementmentioning
confidence: 99%
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