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Quantumness of Product States

  1. Apr 10, 2012 #1

    DrChinese

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    I saw this today, thought some of you might find it interesting (as I did).

    Product states do not violate Bell inequalities. Entangled states do, and this draws a clear line between the quantum world and the classical world.

    So imagine my surprise with this: a paper that shows that even Product States can cross the line between the quantum world and the classical world.

    Quantumness of Product States
    Jing-Ling Chen, Hong-Yi Su, Chunfeng Wu, C. H. Oh
    http://arxiv.org/abs/1204.1798

    From the paper:

    Abstract: Product states do not violate Bell inequalities. In this work, we investigate the quantumness of product states by violating a certain classical algebraic models. Thus even for product states, statistical predictions of quantum mechanics and classical theories do not agree. An experiment protocol is proposed to reveal the quantumness...

    ...a classical model must satisfy the AR [Alicki-Van Ryn] inequality


    [itex](1) <A>\: ≥\: 0,[/itex]
    [itex](2) <B>\: ≥\: 0,[/itex]
    [itex](3) <B − A>\: ≥\: 0, [/itex]
    [itex](4) <B^2 − A^2>\: ≥\: 0[/itex]

    However, in quantum mechanics there exist noncommutative observables that violate the fourth constraint, namely, one can find positive-definite observables A and B satisfying [the first three but not the last]. This violation is called quantumness. Experimental tests have been performed for the case of one qubit...
    [itex]\:[/itex]
     
    Last edited: Apr 10, 2012
  2. jcsd
  3. Apr 10, 2012 #2
    browsed through the 2 page paper

    it would help, some of us, if you can give the layman definition of "product state" and compare/contrast from quantum entangled state (or any other relevant state), thanks
     
  4. Apr 10, 2012 #3

    zonde

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    Well, I do not understand QM formalisms so well to say it's crap but it certainly looks like that.
    For one thing [itex]A\otimes B[/itex] is not [itex]A-B[/itex] and [itex]A^2[/itex] (whatever that means) is not [itex]A\otimes A[/itex].
    If A and B are two-dimensional matrices then [itex]A-B[/itex] is two-dimensional matrice as well but [itex]A\otimes B[/itex] is four-dimensional matrice (combination of everything with everything).
     
  5. Apr 11, 2012 #4
    ...is described in: "R. Alicki and N. Van Ryn, J. Phys. A: Math. Theor. 41, 062001 (2008).", which does not seem to be available online. Google does not help. Any ideas?

    Also: http://www.nist.gov/manuscript-publication-search.cfm?pub_id=900940
     
  6. Apr 11, 2012 #5

    DrChinese

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  7. Apr 11, 2012 #6

    martinbn

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    It seems trivial, or am I missing something? Of course non-commutativity is not classical.
     
  8. Apr 11, 2012 #7

    DrChinese

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    When I try to put my socks on after I put on my shoes, I am not so sure.
     
  9. Apr 11, 2012 #8

    martinbn

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    Non-commutativity of observables, that was clear, no?
     
  10. Apr 11, 2012 #9

    DrChinese

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    I don't believe there is a proof that the existence of non-commuting observables violates classical realism.
     
  11. Apr 12, 2012 #10

    martinbn

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    I don't know what classical realism is, and whether it has to be violated in order for me to say that something is not classical. For example the uncertainty relations for momentum and coordinates is not classical, and that is a clear statement no matter what kind of realism is violated.
     
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