Linear monogamy of entanglement in three-qubit systems.
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For any three-qubit quantum systems ABC, Oliveira et al. numerically found that both the concurrence and the entanglement of formation (EoF) obey the linear monogamy relations in pure states.More specifically, when the amount of entanglement between A and B equals to that of A and C, we show that the sum of EoF itself saturates the linear monogamy relation, while the sum of the squared EoF is minimum.Different from EoF, the concurrence and the squared concurrence both saturate the linear monogamy relations when the entanglement between A and B equals to that of A and C.
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Affiliation: State Key Laboratory of Networking and Switching Technology, Beijing University of Posts and Telecommunications, Beijing, 100876, China.
ABSTRACT
For any three-qubit quantum systems ABC, Oliveira et al. numerically found that both the concurrence and the entanglement of formation (EoF) obey the linear monogamy relations in pure states. They also conjectured that the linear monogamy relations can be saturated when the focus qubit A is maximally entangled with the joint qubits BC. In this work, we prove analytically that both the concurrence and EoF obey linear monogamy relations in an arbitrary three-qubit state. Furthermore, we verify that all three-qubit pure states are maximally entangled in the bipartition A/BC when they saturate the linear monogamy relations. We also study the distribution of the concurrence and EoF. More specifically, when the amount of entanglement between A and B equals to that of A and C, we show that the sum of EoF itself saturates the linear monogamy relation, while the sum of the squared EoF is minimum. Different from EoF, the concurrence and the squared concurrence both saturate the linear monogamy relations when the entanglement between A and B equals to that of A and C. No MeSH data available. |
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Mentions: Furthermore, it is not hard to determine that the implicit function p(x) is a concave function of x if the second-order derivative d2p(x)/dx2 ≤ 0. The details for proving the above results are all shown in Methods. Then we have max p(x) = p(0.7071) ≈ 1.4142, and derive the monogamy inequality of Eq. (13), such that we have completed the whole proof showing that the concurrence is linearly monogamous in three-qubit mixed states. Here, x = C(ρAB) ≈ 0.7071 comes from and Eq. (2). These results can be easily verified by a Mathematica program for the binary function p, and they can also be intuitively observed from Fig. 2(a). Thus we obtain the conclusion that the saturation of the upper bound 1.4142 also comes from both entangled pairs AB and AC with equal intensity, i.e., C(ρAB) = C(ρAC). |
View Article: PubMed Central - PubMed
Affiliation: State Key Laboratory of Networking and Switching Technology, Beijing University of Posts and Telecommunications, Beijing, 100876, China.
No MeSH data available.