Multipartite Entanglement

Three or more subsystems can share entanglement in several distinct ways. A Bell pair accompanied by an uncorrelated third qubit is entangled, but its entanglement does not involve all three parties. Conversely, a GHZ state involves all three parties even though every two-qubit reduced state is separable. This chapter explains how to distinguish these possibilities and what happens when a party is measured or lost.

Helpful background. Use bipartite entanglement to test a specified cut and partial traces to describe the parties that remain.

Enter this chapter

Question	Start here
Which groups of subsystems are being compared?	Multipartite systems
What does entanglement involving every party mean?	Multipartite separability
How does losing a qubit change correlations?	GHZ states and W states
How can a large state have a compact description?	Graph states and stabilizer states
How much entanglement can different pairs share?	Monogamy and entanglement sharing

Separability depends on the partition

Consider three finite-dimensional systems with ordered Hilbert space

$$\mathcal H_{ABC}=\mathcal H_A\otimes\mathcal H_B\otimes\mathcal H_C.$$

A pure state is fully product when it is $|a\rangle_A|b\rangle_B|c\rangle_C$. It is biseparable when it factorizes across at least one of the three cuts $A|BC$, $B|AC$, or $C|AB$. For example, $|\Phi^+\rangle_{AB}|0\rangle_C$ is biseparable: $A$ and $B$ share a Bell pair, while $C$ is independent of them.

A pure state is genuinely multipartite entangled when it is entangled across every nontrivial bipartition. For three parties, this is equivalent to all three single-party reduced states being mixed. For four parties, checking only single-party reductions is insufficient: two independent Bell pairs have mixed single-qubit reductions but factorize across the cut separating the pairs.

Mixed-state separability concerns possible preparations. A fully separable state admits

$$\rho_{ABC}=\sum_k p_k\, \rho_A^{(k)}\otimes\rho_B^{(k)}\otimes\rho_C^{(k)}, \qquad p_k\geq0,\quad\sum_kp_k=1.$$

A biseparable mixed state is a mixture of states separable across bipartitions, and different terms may use different cuts. Thus a preparation may randomly choose to distribute an entangled pair to $AB$, $AC$, or $BC$, leaving the third party separate each time. Such a mixture is biseparable even if no single fixed cut describes all its terms. Genuine multipartite entanglement rules out every such mixture; entanglement across each fixed cut alone is not a sufficient mixed-state test. See multipartite separability for the convex-set definitions and examples.

GHZ and W states behave differently under loss

The three-qubit states

$$\begin{aligned} |\mathrm{GHZ}_3\rangle&=\frac{|000\rangle+|111\rangle}{\sqrt2},\\ |W_3\rangle&=\frac{|100\rangle+|010\rangle+|001\rangle}{\sqrt3} \end{aligned}$$

are both genuinely tripartite entangled. Discarding $C$ nevertheless gives very different states:

$$\begin{aligned} \rho_{AB}^{\mathrm{GHZ}} &=\tfrac12|00\rangle\langle00|+\tfrac12|11\rangle\langle11|,\\ \rho_{AB}^{W} &=\tfrac23|\Psi^+\rangle\langle\Psi^+|+\tfrac13|00\rangle\langle00|, \end{aligned}$$

where $|\Psi^+\rangle=(|01\rangle+|10\rangle)/\sqrt2$. The first is explicitly a mixture of product states. The second remains entangled, as its partial transpose has a negative eigenvalue (Exercise 2). The loss of a party therefore removes the accessible entanglement in the GHZ example, whereas pair entanglement survives in the W example. It does not make either state universally more useful: the answer depends on the intended task. Dür, Vidal, and Cirac describe their distinct three-qubit entanglement classes.

For $n\geq3$, the equal superposition of $|0\rangle^{\otimes n}$ and $|1\rangle^{\otimes n}$, and the equal superposition of the $n$ single-excitation basis states, generalize GHZ and W respectively. Both are genuinely multipartite entangled, but their reduced states still carry different correlations.

Measuring a party differs from discarding it

Rewrite the GHZ state using $|\pm\rangle_C=(|0\rangle_C\pm|1\rangle_C)/\sqrt2$:

$$|\mathrm{GHZ}_3\rangle =\frac{|\Phi^+\rangle_{AB}|+\rangle_C +|\Phi^-\rangle_{AB}|-\rangle_C}{\sqrt2}.$$

If $C$ measures in this basis and communicates the outcome, $A$ and $B$ know which Bell state they share. A local phase correction converts $|\Phi^-\rangle$ to $|\Phi^+\rangle$. If the outcome is ignored, they still describe their pair by the separable marginal above. This distinction motivates localizable entanglement: measurements by other parties can make entanglement available to a selected pair when their outcomes are used.

Graph states and limits on sharing

A graph state gives a compact construction for many qubits. For a finite simple graph, prepare $|+\rangle$ at each vertex and apply a controlled-$Z$ gate on each edge. Equivalently, the state is the unique simultaneous $+1$ eigenvector of the independent commuting operators

$$K_v=X_v\prod_{u\in N(v)}Z_u,$$

where $N(v)$ denotes the neighbors of vertex $v$. A graph edge specifies a gate in this construction; it is not a claim that the corresponding two-qubit marginal is entangled. The graph-state and stabilizer pages explain the relation.

For a pure three-qubit state, the Coffman–Kundu–Wootters inequality is

$$C_{AB}^2+C_{AC}^2\leq4\det\rho_A.$$

Here $C_{AB}$ and $C_{AC}$ are the two-qubit concurrences of the reduced pairs, while $2\sqrt{\det\rho_A}$ measures pure-state entanglement across $A|BC$. It expresses a limit on how these particular pair entanglements can be shared. Other dimensions, mixed global states, and other measures require their own statements; “monogamy” is not a universal inequality for every entanglement measure. Continue to monogamy of entanglement for the definitions and extensions.

Exercises

Mixed marginals without genuine four-party entanglement

Compute the single-qubit marginals of $|\Phi^+\rangle_{AB}|\Phi^+\rangle_{CD}$. Why do they not establish genuine four-party entanglement?

Solution

Each marginal is $I_2/2$, since each qubit belongs to a Bell pair. Nevertheless the given state is explicitly product across $AB|CD$. For a pure four-party state, the two-versus-two cuts must also be tested.

Entanglement remaining after loss

In basis $|00\rangle,|01\rangle,|10\rangle,|11\rangle$, partially transpose $\rho_{AB}^{W}$ on $B$ and find a negative eigenvalue.

Solution

The partial transpose is

$$(\rho_{AB}^{W})^{T_B} =\frac13\begin{pmatrix}1&0&0&1\\0&1&0&0\\0&0&1&0\\1&0&0&0\end{pmatrix}.$$

Two eigenvalues are $1/3$; the remaining block has eigenvalues $(1\pm\sqrt5)/6$. The negative value $(1-\sqrt5)/6$ proves that the reduced pair is entangled.

The missing measurement record

After $C$ measures the GHZ state in the $x$ basis, suppose $A$ and $B$ receive no outcome record. Average their two conditional states and explain why this procedure cannot send a message to them instantaneously.

Solution

The mixture $\tfrac12|\Phi^+\rangle\langle\Phi^+|+\tfrac12|\Phi^-\rangle\langle\Phi^-|$ cancels the off-diagonal $|00\rangle\langle11|$ terms. It equals the original $\rho_{AB}^{\mathrm{GHZ}}$. Their joint statistics are unchanged unless they condition on the communicated record.

References

• V. Coffman, J. Kundu, and W. K. Wootters, “Distributed entanglement,” Physical Review A 61, 052306 (2000). Article and preprint.
• W. Dür, G. Vidal, and J. I. Cirac, “Three qubits can be entangled in two inequivalent ways,” Physical Review A 62, 062314 (2000). Article and preprint.
• O. Gühne and G. Tóth, “Entanglement detection,” Physics Reports 474, 1–75 (2009). DOI.
• M. Hein, W. Dür, J. Eisert, R. Raussendorf, M. Van den Nest, and H.-J. Briegel, “Entanglement in graph states and its applications,” in Quantum Computers, Algorithms and Chaos, edited by G. Casati, D. L. Shepelyansky, P. Zoller, and G. Benenti, IOS Press, 2006, pp. 115–218. Preprint.