Monogamy of Entanglement

Entanglement is monogamous when strong entanglement between two subsystems limits how much entanglement either subsystem can share with a third. The slogan is useful, but the precise statement is always quantitative: it depends on the chosen tensor-product structure, the entanglement measure, the subsystem dimensions, and the class of states under discussion.

The simplest intuition is exact. If $A$ and $B$ form a pure Bell pair, then no third system $C$ can be entangled or even correlated with $AB$. The more general three-qubit statement is the Coffman–Kundu–Wootters inequality: squared concurrence between $A$ and the rest bounds the sum of squared pairwise concurrences involving $A$.

Monogamy does not mean multipartite entanglement is impossible. It means that entanglement cannot usually be decomposed into independently shareable pairwise links. GHZ states and W states show two sharply different ways this can happen.

Helpful background. Two-qubit concurrence defines the pairwise quantity; multipartite partitions specifies which subsystem is held fixed.

A pure pair decouples from every extension

Classical correlations can be shared broadly. Three parties can receive the same random bit $R$, giving the fully separable state

$$\rho_{ABC} = \frac12 \lvert000\rangle\langle000\rvert + \frac12 \lvert111\rangle\langle111\rvert.$$

Every pair is perfectly correlated in the computational basis, yet the state is a mixture of product states. The correlation is copied classical information.

A Bell pair behaves differently. If

$$\rho_{AB} = \lvert\Phi^+\rangle\langle\Phi^+\rvert, \qquad \lvert\Phi^+\rangle = \frac{1}{\sqrt2} \bigl( \lvert00\rangle+\lvert11\rangle \bigr),$$

then any extension to a larger system must factorize:

$$\rho_{ABC} = \lvert\Phi^+\rangle\langle\Phi^+\rvert_{AB} \otimes \rho_C.$$

The reason is that $\rho_{AB}$ has rank one. A positive operator $\rho_{ABC}$ whose $AB$ marginal has support only on the line spanned by $\lvert\Phi^+\rangle$ must have support only on

$$\operatorname{span}\{\lvert\Phi^+\rangle_{AB}\} \otimes \mathcal H_C.$$

So $C$ can be in any state, but it cannot be correlated with the Bell pair.

The factorization follows from purity of the pair, not from maximal entanglement: it also holds if $AB$ is any pure product or partially entangled state. Quantitative monogamy inequalities constrain more general cases where the pair is mixed.

A nearly pure pair approximately decouples

There is also a quantitative version that does not depend on a particular entanglement measure. Work in finite dimensions and suppose the pair has large overlap with a normalized pure target $|\phi\rangle_{AB}$:

$$F=\langle\phi|\rho_{AB}|\phi\rangle\ge1-\epsilon>0.$$

Define the normalized conditional state

$$\sigma_C=\frac{\langle\phi|\rho_{ABC}|\phi\rangle}{F}.$$

The matrix element here acts only on $AB$, leaving an operator on $C$. With trace distance defined by $D(\rho,\omega)=\tfrac12\|\rho-\omega\|_1$, where $\|M\|_1=\operatorname{Tr}\sqrt{M^\dagger M}$, one has

$$D\!\left(\rho_{ABC},|\phi\rangle\langle\phi|_{AB}\otimes\sigma_C\right) \le\sqrt{1-F}\le\sqrt{\epsilon}.$$

To see this, purify $\rho_{ABC}$ to $|\Psi\rangle_{ABCE}$ and project onto $|\phi\rangle$ on $AB$. The normalized projected vector is

$$|\Omega\rangle= \frac{(|\phi\rangle\langle\phi|_{AB}\otimes I_{CE})|\Psi\rangle}{\sqrt F} =|\phi\rangle_{AB}\otimes|\chi\rangle_{CE}, \qquad |\langle\Psi|\Omega\rangle|^2=F.$$

Two pure states with this overlap have trace distance $\sqrt{1-F}$. Taking the partial trace over $E$ cannot increase that distance and gives the claimed product comparison state. These trace-norm identities are proved in Watrous, section 1.1.3, equations (1.183) and (1.186).

No projection need actually be performed: it constructs a product state close to the original state. The comparison uses the conditional $\sigma_C$, which need not equal the original marginal $\rho_C$. Any measurement event has probabilities differing by at most $\sqrt\epsilon$ between these two joint states. For a Bell target this quantifies the restriction on outside correlations; deriving a cryptographic security guarantee still requires the protocol’s additional assumptions.

Three-Qubit Concurrence Monogamy

For a two-qubit density operator $\rho_{AB}$, let $C_{AB}$ denote the concurrence of $\rho_{AB}$. For a pure three-qubit state $\lvert\psi\rangle_{ABC}$, also define the concurrence across the bipartition $A\vert BC$ by treating $A$ as one qubit and $BC$ as a four-dimensional system:

$$C_{A\vert BC} = 2\sqrt{\det\rho_A} = \sqrt{ 2\bigl(1-\operatorname{Tr}\rho_A^2\bigr) }, \qquad \rho_A = \operatorname{Tr}_{BC} \lvert\psi\rangle\langle\psi\rvert.$$

The Coffman–Kundu–Wootters monogamy inequality says

$$C_{A\vert BC}^2 \ge C_{AB}^2 + C_{AC}^2.$$

The residual quantity

$$\tau_A = C_{A\vert BC}^2 - C_{AB}^2 - C_{AC}^2 \ge 0$$

is the residual tangle. For pure three-qubit states it is the three-tangle: although the formula singles out $A$, the resulting residual multipartite contribution is symmetric under permutations of the three qubits.

The inequality compares the entanglement across $A:BC$ with two reduced-pair quantities. The squares are essential: it is not an additive decomposition of the state into independent entangled pairs. In particular, the three-tangle does not detect every genuinely tripartite entangled state.

Bell Pair with a Spectator

Consider

$$\lvert\Psi\rangle_{ABC} = \lvert\Phi^+\rangle_{AB} \otimes \lvert0\rangle_C.$$

The reduced state $\rho_{AB}$ is a Bell state, so

$$C_{AB}=1.$$

The reduced state $\rho_{AC}$ is product:

$$\rho_{AC} = \frac{I_A}{2} \otimes \lvert0\rangle\langle0\rvert_C,$$

so

$$C_{AC}=0.$$

Since $\rho_A=I_A/2$,

$$C_{A\vert BC} = 2\sqrt{\det(I_A/2)} = 1.$$

Thus the inequality is saturated:

$$1 = 1 + 0.$$

This state is not genuinely tripartite entangled. It is separable across the cut $AB\vert C$ and illustrates exact monogamy of a maximally entangled pair.

GHZ State: Global Entanglement Without Pairwise Concurrence

For

$$\lvert\mathrm{GHZ}_3\rangle = \frac{1}{\sqrt2} \bigl( \lvert000\rangle+\lvert111\rangle \bigr),$$

the one-qubit reduced state is maximally mixed:

$$\rho_A = \frac{I_A}{2}, \qquad C_{A\vert BC}=1.$$

However, the two-qubit marginal is

$$\rho_{AB} = \frac12 \lvert00\rangle\langle00\rvert + \frac12 \lvert11\rangle\langle11\rvert.$$

This is separable, so $C_{AB}=0$. By symmetry, $C_{AC}=0$. Therefore

$$\tau_A = 1 - 0 - 0 = 1.$$

The GHZ state is genuinely tripartite entangled, but that entanglement is not visible as pairwise concurrence. It lives in a global coherence between the two branches $\lvert000\rangle$ and $\lvert111\rangle$.

This is a common point of confusion: zero pairwise concurrence does not mean no multipartite entanglement.

W State: Pairwise Entanglement With Zero Three-Tangle

For

$$\lvert W_3\rangle = \frac{1}{\sqrt3} \bigl( \lvert100\rangle + \lvert010\rangle + \lvert001\rangle \bigr),$$

the one-qubit reduced state of $A$ is

$$\rho_A = \frac23 \lvert0\rangle\langle0\rvert + \frac13 \lvert1\rangle\langle1\rvert.$$

Thus

$$C_{A\vert BC} = 2 \sqrt{ \frac23\frac13 } = \frac{2\sqrt2}{3}, \qquad C_{A\vert BC}^2 = \frac89.$$

Tracing out $C$ gives

$$\rho_{AB} = \frac13 \lvert00\rangle\langle00\rvert + \frac23 \lvert\Psi^+\rangle\langle\Psi^+\rvert, \qquad \lvert\Psi^+\rangle = \frac{1}{\sqrt2} \bigl( \lvert01\rangle+\lvert10\rangle \bigr).$$

Using the two-qubit concurrence formula gives

$$C_{AB} = \frac23.$$

By symmetry, $C_{AC}=2/3$. Hence

$$C_{AB}^2+C_{AC}^2 = \frac49+\frac49 = \frac89 = C_{A\vert BC}^2,$$

and

$$\tau_A=0.$$

The W state is still genuinely tripartite entangled. Its squared concurrences saturate this particular inequality, while the state remains inseparable across every cut. Numerical saturation does not express the state as independent pair resources.

Mixed-state scope of the inequality

For a mixed state on a qubit $A$ and the remaining systems, $4\det\rho_A$ is not an entanglement measure: a fully separable state can have $\rho_A=I/2$. The relevant mixed-state extension is the convex-roof tangle,

$$\tau_{A:BC}(\rho)=\min_{\rho=\sum_k p_k|\psi_k\rangle\langle\psi_k|} \sum_k p_k\,4\det\rho_A^{(k)}.$$

The mixed three-qubit inequality is $\tau_{A:BC}(\rho)\ge C(\rho_{AB})^2+C(\rho_{AC})^2$. Do not replace its left side by $4\det\rho_A$ and interpret the result as total entanglement. The corresponding squared-concurrence/tangle inequality extends to arbitrary numbers of qubits (Osborne and Verstraete), with a sum over all pairs involving the chosen qubit. Larger local dimensions require separate theorems.

Relation to Security and Communication

Monogamy is one reason entanglement is useful for cryptography. The near-pure-pair bound makes precise one step: large Bell overlap implies proximity to a state with no outside correlations. Entanglement-based security proofs combine such bounds with entropy inequalities, uncertainty relations, or error-correction arguments.

The same idea also appears in quantum communication networks. A single qubit cannot form independent Bell pairs with several neighbors at once. A laboratory holding several qubits can: each local qubit can be paired with a different neighbor. The local dimension and subdivision of the node are therefore essential. Protocols such as entanglement swapping, repeater chains, and network routing manage this scarcity by moving and converting entanglement rather than treating it as a freely broadcast resource.

This page gives only the structural preview. Full cryptographic security and network-capacity statements require additional operational definitions.

Many-Body Implications Preview

In many-body systems, monogamy helps explain why entanglement patterns are constrained by geometry and locality. If a spin is strongly entangled with one neighbor, there is less room for it to be independently entangled with many others. This intuition appears in spin chains, frustration of local singlet formation, tensor-network states, and area-law discussions.

The warning is equally important: monogamy alone does not prove an area law, determine a phase of matter, or classify many-body entanglement. Entropy inequalities, locality, spectral gaps, symmetries, and dimensionality all matter. Monogamy is one organizing principle among several.

Why the measure and its power matter

For $W_3$, the unsquared expression fails:

$$C_{A:BC}=\frac{2\sqrt2}{3}<\frac43=C_{AB}+C_{AC}.$$

Squaring each concurrence gives the valid equality $8/9=4/9+4/9$. One cannot replace concurrence squared with a preferred entropy, negativity, or mutual information and assume the same theorem. Mutual information includes freely shareable classical correlations and obeys different entropy constraints.

References

• V. Coffman, J. Kundu, and W. K. Wootters, “Distributed entanglement”, Physical Review A 61, 052306 (2000).
• W. Dür, G. Vidal, and J. I. Cirac, “Three qubits can be entangled in two inequivalent ways”, Physical Review A 62, 062314 (2000).
• M. Koashi and A. Winter, “Monogamy of quantum entanglement and other correlations”, Physical Review A 69, 022309 (2004).
• T. J. Osborne and F. Verstraete, “General monogamy inequality for bipartite qubit entanglement”, Physical Review Letters 96, 220503 (2006).
• John Watrous, The Theory of Quantum Information, Cambridge University Press, 2018, section 1.1.3, equations (1.183) and (1.186). Author’s text.

Exercises

1. Pure marginal decoupling. Suppose a tripartite state $\rho_{ABC}$ has $\rho_{AB}=\lvert\phi\rangle\langle\phi\rvert_{AB}$. Show that $\rho_{ABC}=\lvert\phi\rangle\langle\phi\rvert_{AB}\otimes\rho_C$.

Solution

Let $P=\lvert\phi\rangle\langle\phi\rvert_{AB}$ and $Q=I_{AB}-P$. Since

$$\operatorname{Tr}_{ABC} \bigl[ (Q\otimes I_C)\rho_{ABC} \bigr] = \operatorname{Tr}_{AB}(Q\rho_{AB}) = 0,$$

and $(Q\otimes I_C)\rho_{ABC}(Q\otimes I_C)$ is positive, the state has no support outside $P\otimes I_C$. Therefore

$$\rho_{ABC} = (P\otimes I_C)\rho_{ABC}(P\otimes I_C) = P\otimes\rho_C.$$

2. Coherent leakage saturates the distance bound. For $0\le\epsilon<1$, let

$$|\Psi_\epsilon\rangle =\sqrt{1-\epsilon}\,|\Phi^+\rangle_{AB}|0\rangle_C +\sqrt\epsilon\,|\Phi^-\rangle_{AB}|1\rangle_C, \qquad |\Phi^-\rangle=\frac{|00\rangle-|11\rangle}{\sqrt2}.$$

Using the target $|\Phi^+\rangle$, find $F$, $\sigma_C$, and the distance to the product comparison state. Can the bound $\sqrt\epsilon$ be replaced by $\epsilon$ for every extension?

Solution

The $AB$ marginal is $(1-\epsilon)|\Phi^+\rangle\langle\Phi^+|+\epsilon|\Phi^-\rangle\langle\Phi^-|$, so $F=1-\epsilon$. Projection onto the target gives $\sigma_C=|0\rangle\langle0|$. Both the original joint state and its product comparison are pure, with overlap squared $1-\epsilon$. Thus

$$D\!\left(|\Psi_\epsilon\rangle\langle\Psi_\epsilon|, |\Phi^+\rangle\langle\Phi^+|\otimes|0\rangle\langle0|\right) =\sqrt\epsilon.$$

The bound is saturated: coherent error amplitudes can produce distance of order $\sqrt\epsilon$, even though the target infidelity is $\epsilon$. For example, $\epsilon=0.01$ gives distance $0.1$. At $\epsilon=0$ the state returns to an exactly decoupled Bell pair.

3. Unequal GHZ amplitudes. For $a|000\rangle+b|111\rangle$, with $|a|^2+|b|^2=1$, find the three-tangle.

Solution

Every pair marginal is the separable mixture $|a|^2|00\rangle\langle00|+|b|^2|11\rangle\langle11|$, so pair concurrences vanish. The one-qubit marginal is $\operatorname{diag}(|a|^2,|b|^2)$, giving $\tau=4|ab|^2$. It is one for equal magnitudes and zero at the product endpoints.

4. Asymmetric single-excitation state. For $a|100\rangle+b|010\rangle+c|001\rangle$ with $|a|^2+|b|^2+|c|^2=1$, verify CKW saturation using $C_{AB}=2|ab|$ and $C_{AC}=2|ac|$.

Solution

$\rho_A=\operatorname{diag}(|b|^2+|c|^2,|a|^2)$, so $C_{A:BC}^2=4|a|^2(|b|^2+|c|^2)$. The sum of pairwise squares is $4|ab|^2+4|ac|^2$, exactly the same. The three-tangle is zero. If all three amplitudes are nonzero, the pure state is nevertheless entangled across all three cuts and genuinely tripartite entangled.

5. Classical sharing versus entanglement sharing. Explain why

$$\rho_{ABC} = \frac12 \lvert000\rangle\langle000\rvert + \frac12 \lvert111\rangle\langle111\rvert$$

does not contradict monogamy, even though every pair is perfectly correlated in the computational basis.

Solution

The state is fully separable:

$$\rho_{ABC} = \sum_{r=0}^{1} \frac12 \lvert r\rangle\langle r\rvert_A \otimes \lvert r\rangle\langle r\rvert_B \otimes \lvert r\rangle\langle r\rvert_C.$$

The correlations come from a shared classical label $r$. Each two-qubit marginal has nonzero classical mutual information but zero concurrence. Monogamy constrains quantum entanglement, not all classical correlation.

6. Measure dependence. Does the CKW inequality prove that every entanglement measure is monogamous in every finite-dimensional quantum system?

Solution

No. CKW is a theorem about squared concurrence for qubits, with extensions to multiqubit systems. Other dimensions and other measures require separate statements. Some measures obey monogamy only after taking suitable powers, some satisfy different inequalities, and some are not monogamous in the CKW sense.