# If perfect maximal entanglement is never true, does a remainder invalidate the monogamy of entanglement?

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)?

• What is hybrid correlation and what is AMPS? – Sofia Jan 14 '15 at 21:24
• Not maximally entangled means something of the form $|x_1>|x_2> + |x_1>|y_2> +|y_1>|y_2>$. – Sofia Jan 14 '15 at 21:45
• AMPS: I guess it refers to Almheiri, Marolf, Polchinski, and Sully (arxiv.org/abs/1207.3123). See e.g. quantumfrontiers.com/2012/12/03/… – Norbert Schuch Jan 14 '15 at 22:53
• @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. – Jonathan Langdale Jan 15 '15 at 19:20
• @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. $|1_{x_1}>|0_{x_2}>|0_{x_3}> + |0_{x_1}>|1_{x_2}>|0_{x_3}> + |1_{y_1}>|0_{y_2}>|1_{x_3}>$ ? – Sofia Jan 15 '15 at 19:39

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).