'''Monogamy ''' is one of the most fundamental properties of entanglement and can, in its extremal form, be expressed as follows: If two qubits A and B are maximally quantumly correlated they cannot be correlated at all with a third qubit C. In general, there is a trade-off between the amount of entanglement between qubits A and B and the same qubit A and qubit C. This is mathematically expressed by the Coffman-Kundu-Wootters (CKW) monogamy inequality:
CAB2 + CAC2 ≤ CA(BC)2,
where CAB, CAC are the concurrences between A and B respectively between A and C, while CA(BC) is the concurrence between subsystems A and BC.
It was proved that the above inequality can be extended to the case of n qubits.
More generally, the monogamy inequality can be expressed in terms of entanglement measures E, as follows:
'''For any tripartite state of systems A, B1, B2 we have
E(A∣B1) + E(A∣B2) ≤ E(A∣B1B2).
''' If the above inequality holds in general, i.e. not only for qubits, then it can be immediately generalized by induction to the multipartite case:
E(A∣B1) + E(A∣B2) + … + E(A∣BN) ≤ E(A∣B1B2…BN).
Notice that the entanglement measures EC and EF do not satisfy the monogamy inequality, whereas squashed-entanglement does.
Moreover, is was proved that the Bell-CHSH inequality is monogamous: if three parties A, B and C share a quantum state $\; \varrho$ and each chooses to measure one of two observables, then the trade-off between AB’s and AC’s violation of the CHSH inequality is given by
$$\; |Tr(\mathcal{B}_{CHSH}^{AB}\varrho)| + |Tr(\mathcal{B}_{CHSH}^{AC}\varrho)| \leq 4 .$$
This means that if AB violate the CHSH inequality then AC cannot.
Related papers
- V. Coffman et al., Phys. Rev. A 61, 052306 (2000)
- B. M. Terhal, Linear Algebra Appl. 323, 61 (2001)
- B. M. Terhal, IBM J. Res. Dev. '48, 71 (2004). Available at arXiv:quant-ph/0307120
- M. Koashi, A. Winter, Phys. Rev. A 69, 022309 (2004)
- T. J. Osborne, F. Verstraete, Phys. Rev. Lett. 96, 220503 (2006)
- B. Toner, Proc. R. Soc. A 465, 59-69 (2009)