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Perfect Embezzlement of Entanglement

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 Added by Richard Cleve
 Publication date 2016
  fields Physics
and research's language is English




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Van Dam and Hayden introduced a concept commonly referred to as embezzlement, where, for any entangled quantum state $phi$, there is an entangled catalyst state $psi$, from which a high fidelity approximation of $phi otimes psi$ can be produced using only local operations. We investigate a version of this where the embezzlement is perfect (i.e., the fidelity is 1). We prove that perfect embezzlement is impossible in a tensor product framework, even with infinite-dimensional Hilbert spaces and infinite entanglement entropy. Then we prove that perfect embezzlement is possible in a commuting operator framework. We prove this using the theory of C*-algebras and we also provide an explicit construction. Next, we apply our results to analyze perfe



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We consider a bipartite transformation that we call emph{self-embezzlement} and use it to prove a constant gap between the capabilities of two models of quantum information: the conventional model, where bipartite systems are represented by tensor products of Hilbert spaces; and a natural model of quantum information processing for abstract states on C*-algebras, where joint systems are represented by tensor products of C*-algebras. We call this the C*-circuit model and show that it is a special case of the commuting-operator model (in that it can be translated into such a model). For the conventional model, we show that there exists a constant $epsilon_0 > 0$ such that self-embezzlement cannot be achieved with precision parameter less than $epsilon_0$ (i.e., the fidelity cannot be greater than $1 - epsilon_0$); whereas, in the C*-circuit model---as well as in a commuting-operator model---the precision can be $0$ (i.e., fidelity~$1$).
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This paper has been withdrawn by the authors, due a oversimplified decoherence model. It will be substituted by a new work.
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