Monogamy of entanglement

'''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 measuresE, as follows:

'''For any tripartite state of systems  A,  B1,  B2 we have

E(AB1) + E(AB2) ≤ E(AB1B2).''' 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(AB1) + E(AB2) + … + E(ABN) ≤ E(AB1B2…BN).

Notice that the entanglement measuresEC 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.


Last modified: 
Monday, October 26, 2015 - 17:56