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Does local realism imply separability?

  1. Aug 27, 2015 #1
    Hi.

    Bell's formulation of local realism is $$P(a,b)=\int\ d\lambda\cdot\rho(\lambda)p_A(a,\lambda)p_B(b,\lambda)\enspace.$$
    Let's for simplicity assume there's only a finite number of states, so this becomes $$P(a,b)=\sum_{i} p_i\cdot\ p_A(a,i)p_B(b,i)\enspace.$$
    I'm trying to translate this into density operator notation and then show that it implies that the state needs to be separable. So my ansatz is
    $$P(a,b)=tr(\hat{\rho}\hat{A}(a)\otimes\hat{B}(b))\enspace,$$
    where ##\hat{A}(a)## and ##\hat{B}(b)## are observables with spectrum ##\{1,0\}## (detecting or not detecting a photon). I'm trying to show by comparing the last two equations that ##\hat{\rho}## must have the form
    $$\hat{\rho}=p_i\cdot\hat{\rho}_A ^i\otimes\hat{\rho}_B ^i$$
    where ##\hat{\rho}_A ^i## and ##\hat{\rho}_B ^i## are density operators on their respective subsystems. However I can't see how to do this.

    Showing that separable states satisfiy local realism is trivial, is the converse even true in general? If yes, how do you do this? Or is my ansatz nonsense? I'm unsure because I had to pick observables with eigenvalues and if ##\{1,0\}## was the right choice.
     
    Last edited: Aug 27, 2015
  2. jcsd
  3. Aug 27, 2015 #2

    atyy

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    Yes, assuming "local realism" one can derive separability. There are a number of different definitions of "local realism", so there are different derivations:

    http://arxiv.org/abs/1503.06413

    I like the derivation given by Wood and Spekkens (Fig.19):

    http://arxiv.org/abs/1208.4119

    Edit: This post does not use the definition of separability used in the OP. I misread. See jfizzix's answer below.
     
    Last edited: Aug 28, 2015
  4. Aug 27, 2015 #3

    jfizzix

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    No. There are non-separable states that satisfy local realism. See, for example:
    http://journals.aps.org/pra/abstract/10.1103/PhysRevA.40.4277

    However, if along with local realism, you consider the additional assumption that all measurement probabilities are completely described by, and reducible to quantum measurements, then we can say the following:
    If one's measurement statistics obey local realism, and one's measurement probabilities are reducible to quantum measurements, then the joint quantum state must be separable.

    Indeed, it is for this reason that violating a Bell inequality proves the state is entangled; separable states are a strict subset of states admitting a local hidden variable model.
     
  5. Aug 28, 2015 #4

    atyy

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    Although my answer differs from jfizzix's, I don't disagree with him. I gave the opposite answer because I misread the definition of separability in the OP. The answer in post #2 is correct for a different definition of separability than used in the OP. jfizzix's answer is correct for the definition of separability in the OP.
     
  6. Sep 20, 2015 #5
    Hi,

    The paper mentions that every pure state that satisfies local realism is separable (it actually says that every pure state admitting a hidden-variable model is classically correlated, but as far as I can see the definitions are the same).

    I have no access to the referenced paper, but is there an easy way to show this?
     
  7. Sep 20, 2015 #6

    jfizzix

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    I don't know of an easy way to show it, but if you want a source to look up, it's Gisin's theorem that says every entangled pure state must violate some sort of bell inequality. So, if a pure state does satisfy local realism, it must not be entangled.
     
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