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A Bell's theorem+Holographic entanglement entropy

  1. Jan 18, 2017 #1
    Bell's theorem debunks theories concerning local hidden variables.
    Many people interpret that as the complete absence of local hidden variables.
    Hidden variable theories were espoused by some physicists who argued that the state of a physical system, as formulated by quantum mechanics, does not give a complete description for the system.

    Entanglement entropy proves the presence of hidden variables in the 'correlation function'.
    Entanglement entropy is a measurement of quantum states (2 dimensional info) in entanglement.


    Doesn't this prove that Bell's theorem debunks only theories concerning local hidden variables?
    Entanglement entropy shows there are local hidden variables.
     
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  3. Jan 18, 2017 #2

    Simon Phoenix

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    Bell's theorem says that local hidden variable theories cannot reproduce all of the predictions of quantum mechanics - so if that's what you mean by 'debunked', then I guess so :smile:

    I have absolutely no idea what you mean by this. Could you explain why you think entanglement entropy proves the presence of hidden variables? Entanglement entropy is just an information theoretic, and basis-independent, measure of the correlation between 2 quantum systems. What feature of 'entanglement entropy' shows that there are hidden variables? Furthermore, what do you mean by 'correlation function'? How does a correlation function have hidden variables?

    If you could explain things here a bit more precisely then I might be able to understand what it is you are asking
     
  4. Jan 18, 2017 #3
    Entanglement entropy measures quantum states/2 dimensional info. Aren't quantum states (hidden) variables?
    With 'correlation function' I mean a function of 2 entangled particles...a single wave.
     
  5. Jan 18, 2017 #4

    DrChinese

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    If you look closely, you will see that you have just assumed that which you seek to prove.

    So no, quantum states are not hidden variables (unless they turn out to be such, based on something we learn in the future). :smile:
     
  6. Jan 18, 2017 #5

    Strilanc

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    I'm going to disagree with DrChinese here. Yes, the quantum state vector is a big 'ol bunch of hidden variables. And those variables determine the experimental probabilities that we measure.

    But those hidden variables aren't local. The state vector isn't associated with single parts of the system. It specifies weights for configurations of the entire system. You have an amplitude for "all of the qubits are OFF", but you don't have an amplitude for "qubit 1 is OFF" (except in simple separable cases).

    Since the quantum state vector doesn't rely on local variables, it's not ruled out by Bell's theorem.
     
  7. Jan 18, 2017 #6
    Aren't they local because we know they are not local or because our science can't make valid assumptions yet?
     
  8. Jan 18, 2017 #7

    Strilanc

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    I'm not sure what you mean. Amplitudes aren't local for the same reason that probabilities aren't: they talk about multiple parts of the system. I have no idea why you'd dive into esoteric philosophy of science issues about assumptions being valid over something that simple.

    For example, consider the probability distribution "Alice and Bob each have a coin. There's a 50% chance both coins are tails, otherwise both coins are heads.". Notice that it's impossible to factor this distribution into two independent probability distributions, one that talks about just Alice and another that talks about just Bob. The fact that you can't split the situation into an Alice piece and a Bob piece, without losing either the 50% detail or the agreement detail, is why it's not a local description.

    (Note that even though a probability distribution is not necessarily local, they can't violate Bell inequalities. Being non-local (or not counterfactually-definite) may be necessary for violating the inequalities, but it's not sufficient.)
     
  9. Jan 18, 2017 #8

    DrChinese

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    I understood itoero to say the the quantum state determines the specific outcome, not the probabilities of an outcome. And I also understood itoero for them to be local as well. I don't think either of those would be requirements/descriptions as we understand the quantum state. Do you see differently?
     
  10. Jan 18, 2017 #9

    Strilanc

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    I think we're in agreement that the quantum state is not a local hidden variable. But I do consider it to be a non-local hidden variable... unless by "hidden" we literally mean "can't get any information about the variable". Partially-hidden would be more appropriate, I suppose.
     
    Last edited: Jan 18, 2017
  11. Jan 19, 2017 #10
    If you measure one entangled particle, then the other will also acquire a definite state...that's a causal relationship.
    The quantum states/2-dimensional info build space time.(there is a paper from Ooguri Hiroshi https://arxiv.org/abs/1412.1879)
    Isn't it logic that the same quantum states cause the relationship between entangled particles?

    English is not my native language, I hope I'm a bit clear. :)
     
  12. Jan 19, 2017 #11

    DrChinese

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    Your conclusion is strictly by assumption. There is no evidence whatsoever that there is a causal relationship. Other than saying the earlier measurement causes the second result. (Which of course is the definition of causation, therefore why your argument is circular.)

    QM, on the other hand, does not call for a causal relationship here. We don't know what is going on past what QM says.
     
  13. Jan 19, 2017 #12
    I think there is something fishy going on.
    Does non local causality exist?
     
  14. Jan 19, 2017 #13

    DrChinese

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    No one knows. But there are theories that are non-local causal. Bohmian Mechanics is the best known example.
     
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