Monogamy of Entanglement

Entanglement is monogamous: if Alice is maximally entangled with Bob, she has nothing left to share with Charlie. Unlike classical correlation, which can be freely broadcast, entanglement obeys a strict trade-off that constrains how it can be distributed among many parties. This is one of the deepest structural features distinguishing quantum from classical correlations.

The CKW inequality

Coffman, Kundu, and Wootters proved the first quantitative monogamy relation, for three qubits. Using the squared concurrence — the tangle $\tau_{XY} = C_{XY}^2$ — the inequality reads

C_{AB}^2 + C_{AC}^2 \ \leq\ C_{A(BC)}^2,

where:

$C_{AB}$ is the concurrence of the reduced two-qubit state $\rho_{AB}$ (A with B alone),
$C_{AC}$ likewise for $\rho_{AC}$ ,
$C_{A(BC)}$ is the concurrence across the bipartition $A$ vs. the pair $BC$ .

For a pure tripartite state $C_{A(BC)}^2 = 4\det\rho_A = 2(1 - \operatorname{Tr}\rho_A^2)$ measures the entanglement of $A$ with everything else. The inequality says: the entanglement $A$ shares pairwise with $B$ and with $C$ separately can never exceed the total entanglement $A$ has with the rest. A finite resource cannot be handed out twice.

The residual is the three-tangle

The slack in the CKW inequality is exactly the three-tangle met in the previous lesson:

\tau_{ABC} = C_{A(BC)}^2 - C_{AB}^2 - C_{AC}^2 \ \geq 0.

A foundational result is that $\tau_{ABC}$ is invariant under permutations of the three qubits, despite the asymmetric-looking definition — it is a genuine property of the triple, not of the chosen focal party $A$ . It quantifies the irreducibly tripartite entanglement:

$|\mathrm{GHZ}\rangle$ : $C_{AB} = C_{AC} = 0$ but $C_{A(BC)}^2 = 1$ , so $\tau = 1$ — all the entanglement is tripartite, none pairwise.
$|\mathrm{W}\rangle$ : $C_{AB}^2 = C_{AC}^2 = \tfrac49$ and $C_{A(BC)}^2 = \tfrac89$ , so $\tau = \tfrac89 - \tfrac49 - \tfrac49 = 0$ — the inequality is saturated, all entanglement is pairwise.

W maximizes pairwise sharing precisely by driving the tripartite residual to zero; GHZ does the opposite.

Generalizations

The three-qubit result is the tip of a large structure:

Osborne–Verstraete extended the tangle inequality to $N$ qubits: $\sum_{j\neq 1} C_{1j}^2 \leq C_{1(2\cdots N)}^2$ — one party's pairwise entanglements with all others are bounded by its entanglement with the rest.
Monogamy holds for some measures and fails for others: squared concurrence (tangle) and squared negativity are monogamous, while concurrence and entanglement of formation themselves are not monogamous in general. Choosing a squared measure is what makes the additive trade-off work.
Monogamy underlies the security of quantum key distribution: if Alice and Bob are strongly entangled, an eavesdropper Eve is necessarily weakly correlated with them — she cannot share in their entanglement, which bounds her information.