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If something is only very nearly (and/or observed to be) maximally entangled, does that remainder allow for a menage trois of hybrid correlation (as it relates to AMPS)?

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  • $\begingroup$ What is hybrid correlation and what is AMPS? $\endgroup$ – Sofia Jan 14 '15 at 21:24
  • $\begingroup$ Not maximally entangled means something of the form $|x_1>|x_2> + |x_1>|y_2> +|y_1>|y_2>$. $\endgroup$ – Sofia Jan 14 '15 at 21:45
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    $\begingroup$ AMPS: I guess it refers to Almheiri, Marolf, Polchinski, and Sully (arxiv.org/abs/1207.3123). See e.g. quantumfrontiers.com/2012/12/03/… $\endgroup$ – Norbert Schuch Jan 14 '15 at 22:53
  • $\begingroup$ @Sofia hybrid correlation as in breaking the monogamy of entanglement, an aggregate being partially entangled to 2 or more separate sets of particles. This distinguishes a single fundamental particle entanglement from an aggregate. $\endgroup$ – Jonathan Langdale Jan 15 '15 at 19:20
  • $\begingroup$ @JonathanLangdale There is a theorem that says that a given particle cannot be engaged in more than one entanglement. Now, the word aggregate doesn't mean much to me. Can you tell me how that aggregate looks like? Could is be that the number of particles in it is not constant, e.g. $|1_{x_1}>|0_{x_2}>|0_{x_3}> + |0_{x_1}>|1_{x_2}>|0_{x_3}> + |1_{y_1}>|0_{y_2}>|1_{x_3}>$ ? $\endgroup$ – Sofia Jan 15 '15 at 19:39
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Simple answer: Yes it does.

The monogamy of entanglement is often stated as "when two particles are maximally entangled, they cannot be entangled by a third particle", while really it means that the more two particles are entangled, the less they can be entangled with a third particle. However, since "more" and "less" are ill-defined, phrasing it in the first way makes it a correct statement, which show you the essence of what's going on.

Now, making the statement precise requires measures of entanglement. Some such measures are various entropies - not all of which satisfy monogamy in a strict sense. The concurrence (for qubits) does, so does the squashed entanglement. See here for one in the long list of papers related to the problem: http://arxiv.org/abs/quant-ph/0502176

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Maximal Entanglement theoretically cannot be achieved on particular $n$-qu$d$it systems for a specific $(n,d)$ pair. For example, the paper demonstrates this taking the case of a four-qubit system and extends Concurrence to pure states of arbitrary dimensions (aka multi-qudit pure states) in a very intuitive way by looking at the geometry of the tensor products with increasing dimensionality. A pedagogic explanation to reach the generalization from the basic notion of separability may be found in our recent article here: arXiv:1607.00164 (2016).

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  • $\begingroup$ What would prevent two maximally entangled virtual micro black hole from forming? $\endgroup$ – Jonathan Langdale Oct 17 '16 at 4:51
  • $\begingroup$ Nothing. I meant to say that for particular n-qudit systems, if you work out a simple math, you'll find that the entanglement measure like concurrence is upper open-bounded, which means you can get arbitrarily close to it but not perfect maximal entanglement (e.g. four qubit systems). Whereas, there are also other systems (n,d) where the bound is a closed one, thereby possible to attain "perfect" maximal entanglement (eg. two qubit systems, for instance). Please have a look at this paper: link $\endgroup$ – Vineeth Bhaskara Oct 17 '16 at 6:07

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