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.

Informal Idea

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.

This exact decoupling is the cleanest form of monogamy. Quantitative monogamy inequalities measure what remains true when $A$ and $B$ are not exactly maximally entangled.

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 has a simple reading. The total entanglement of $A$ with everything else sets a budget. Pairwise entanglement between $A$ and $B$ consumes part of that budget; pairwise entanglement between $A$ and $C$ consumes another part; the remainder, when present, is not stored in either pair alone.

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 entanglement pattern is different from GHZ entanglement: the entanglement visible from $A$ is exhausted by pairwise concurrences with $B$ and $C$.

Relation to Security and Communication

Monogamy is one reason entanglement is useful for cryptography. If two honest parties share a state close to a pure Bell pair, then any outside system is close to decoupled from that pair. Entanglement-based security proofs turn this intuition into quantitative statements using trace distance, entropy inequalities, uncertainty relations, or error-correction arguments.

The same idea also appears in quantum communication networks. A node cannot generally be maximally entangled with many independent neighbors at once. 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.

What Monogamy Does Not Say

• It does not say classical correlations are monogamous in the same way.
• It does not say multipartite entanglement is absent.
• It does not say every entanglement measure obeys the same inequality.
• It does not say every dimension behaves like three qubits.
• It does not say pairwise entanglement is impossible in a multipartite state.
• It does not identify the best operational resource for a protocol by itself.

For example, mutual information counts total correlation, including classical correlation. It is constrained by entropy inequalities, but it is not the same object as concurrence or distillable entanglement.

Common Mistakes

• Treating monogamy as the statement “if $A$ is entangled with $B$, then $A$ cannot be entangled with anything else.”
• Assuming GHZ states have pairwise Bell entanglement because they are strongly multipartite entangled.
• Assuming W states have no genuine tripartite entanglement because their three-tangle is zero.
• Confusing concurrence monogamy with a universal theorem for all entanglement measures.
• Forgetting that monogamy is defined relative to a chosen subsystem decomposition.
• Using pairwise marginals alone to decide genuine multipartite entanglement.
• Turning cryptographic intuition into a security proof without quantitative distance or entropy bounds.

Cross-Links

• Multipartite Systems
• Multipartite Separability
• GHZ States
• W States
• Graph States
• Entanglement Sharing
• Bell States
• Classical Correlation versus Entanglement
• Concurrence for Two Qubits
• Entanglement Entropy
• Mutual Information
• LOCC Preview
• Marginals and Correlations

References

• V. Coffman, J. Kundu, and W. K. Wootters, “Distributed Entanglement,” Physical Review A 61, 052306, 2000.
• W. K. Wootters, “Entanglement of Formation of an Arbitrary State of Two Qubits,” Physical Review Letters 80, 2245-2248, 1998.
• T. J. Osborne and F. Verstraete, “General Monogamy Inequality for Bipartite Qubit Entanglement,” Physical Review Letters 96, 220503, 2006.
• M. Koashi and A. Winter, “Monogamy of Quantum Entanglement and Other Correlations,” Physical Review A 69, 022309, 2004.
• W. Dur, G. Vidal, and J. I. Cirac, “Three Qubits Can Be Entangled in Two Inequivalent Ways,” Physical Review A 62, 062314, 2000.
• R. Horodecki, P. Horodecki, M. Horodecki, and K. Horodecki, “Quantum Entanglement,” Reviews of Modern Physics 81, 865-942, 2009.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.

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. Bell pair with spectator. For $\lvert\Phi^+\rangle_{AB}\otimes\lvert0\rangle_C$, compute $C_{AB}$, $C_{AC}$, and $C_{A\vert BC}$.

Solution

The $AB$ marginal is a Bell state, so $C_{AB}=1$. The $AC$ marginal is

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

which is separable, so $C_{AC}=0$. Finally, $\rho_A=I_A/2$, hence

$$C_{A\vert BC} = 2\sqrt{\det\rho_A} = 1.$$

The CKW inequality is saturated.

3. GHZ residual tangle. Show that $\lvert\mathrm{GHZ}_3\rangle$ has $C_{AB}=C_{AC}=0$ and $\tau_A=1$.

Solution

Tracing out $C$ gives

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

which is separable, so $C_{AB}=0$. By symmetry, $C_{AC}=0$. Since $\rho_A=I_A/2$,

$$C_{A\vert BC}=1.$$

Thus

$$\tau_A = C_{A\vert BC}^2-C_{AB}^2-C_{AC}^2 = 1.$$

4. W-state saturation. Verify that the three-qubit W state saturates the CKW inequality with zero residual tangle.

Solution

For $\lvert W_3\rangle$,

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

so

$$C_{A\vert BC}^2 = 4\det\rho_A = \frac89.$$

The two-qubit reductions have concurrence

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

Therefore

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

and $\tau_A=0$.

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.