Multipartite Entanglement

Multipartite entanglement begins when a system has three or more named subsystems. The main conceptual change is not simply a larger tensor product. It is the proliferation of ways to group those subsystems into parties, cuts, and partitions.

For $n$ subsystems,

$$\mathcal H = \bigotimes_{i=1}^{n}\mathcal H_i.$$

A state can be product across one cut, entangled across another, fully separable, biseparable, or genuinely multipartite entangled. Pairwise entanglement may coexist with global entanglement, vanish despite global entanglement, or be constrained by monogamy relations.

The reliable order of analysis is

$$\begin{gathered} \text{label all subsystems} \\ \downarrow \\ \text{specify the relevant partitions} \\ \downarrow \\ \text{classify separability} \\ \downarrow \\ \text{study correlations and sharing}. \end{gathered}$$

Chapter Map

Task	Canonical page	Main output
define many-party Hilbert spaces and partitions	Multipartite Systems	subsystem labels, bipartitions, reductions, and partition bookkeeping
study all-or-nothing coherence	GHZ States	global coherence, separable proper-subset reductions, and loss fragility
study robust single-excitation entanglement	W States	entangled reductions and contrast with GHZ structure
construct states from a graph	Graph States	controlled-Z preparation, graph connectivity, and cluster-state preview
use commuting Pauli constraints	Stabilizer States Preview	compact state descriptions and a bridge to error correction
distinguish full, bi-, and genuine separability	Multipartite Separability	a precise hierarchy for pure and mixed states
understand restrictions on pairwise sharing	Monogamy of Entanglement	measure- and dimension-dependent monogamy inequalities
connect global structure to networks	Entanglement Sharing	localizable entanglement, swapping, cuts, and distribution

Partitions Are Part of the Question

For three subsystems $A$, $B$, and $C$,

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

There are three nontrivial bipartitions:

$$A|BC, \qquad B|AC, \qquad C|AB.$$

There is also the finest partition $A|B|C$. These partitions answer different questions.

Partition question	Example conclusion
Is $A$ entangled with the joint system $BC$?	analyze the cut separating $A$ from $BC$
Are all three parties independently prepared?	test full product or full separability
Can the state be built by mixing states separable across some bipartition?	test biseparability
Does every preparation require genuinely three-party entanglement?	test genuine multipartite entanglement

For more parties, partitions are set partitions of the subsystem labels. The number of possibilities grows rapidly, so every claim should name the relevant grouping.

Pure-State Separability

A three-party pure state is fully product if

$$\lvert\psi\rangle_{ABC} = \lvert a\rangle_A \otimes \lvert b\rangle_B \otimes \lvert c\rangle_C.$$

It is biseparable if it factorizes across at least one bipartition, for example

$$\lvert\psi\rangle_{ABC} = \lvert a\rangle_A \otimes \lvert\phi\rangle_{BC}.$$

The state $\lvert\phi\rangle_{BC}$ may itself be entangled. Thus biseparable does not mean unentangled; it means that at least one party is separable from the others for that pure state.

A pure state is genuinely multipartite entangled if it is entangled across every bipartition. For three parties, Schmidt analysis can be applied separately to the cuts $A|BC$, $B|AC$, and $C|AB$.

For a pure state,

$$\begin{gathered} \text{genuine three-party entanglement} \\ \iff \\ \operatorname{rank}\rho_A>1, \quad \operatorname{rank}\rho_B>1, \\ \text{and}\quad \operatorname{rank}\rho_C>1. \end{gathered}$$

This reduced-rank criterion is exact because the global state is pure.

Mixed-State Separability

A three-party mixed state is fully separable if it admits a decomposition

$$\rho_{ABC} = \sum_k p_k\, \rho_A^{(k)}\otimes \rho_B^{(k)}\otimes \rho_C^{(k)}.$$

A mixed state is biseparable if it is a convex mixture of states separable across bipartitions. Different terms may use different cuts:

$$\begin{aligned} \rho_{\mathrm{bisep}} =& \sum_k p_k^{A|BC} \rho_A^{(k)}\otimes\rho_{BC}^{(k)} \\ &+ \sum_\ell p_\ell^{B|AC} \rho_B^{(\ell)}\otimes\rho_{AC}^{(\ell)} \\ &+ \sum_m p_m^{C|AB} \rho_C^{(m)}\otimes\rho_{AB}^{(m)}, \end{aligned}$$

with nonnegative weights whose total is one.

A mixed state is genuinely multipartite entangled if it is not biseparable.

This definition contains an important subtlety: a biseparable mixed state need not be separable across one fixed bipartition. Its decomposition may mix different cuts. Consequently, testing each fixed cut independently does not automatically solve the mixed-state genuine-entanglement problem.

The canonical definitions and examples belong to Multipartite Separability.

The GHZ State

The $n$-qubit Greenberger–Horne–Zeilinger state is

$$\lvert\mathrm{GHZ}_n\rangle = \frac{ \lvert0\rangle^{\otimes n} + \lvert1\rangle^{\otimes n} }{\sqrt2}.$$

For $n\geq3$, it is genuinely multipartite entangled. Across every nontrivial bipartition it has Schmidt rank two with equal Schmidt coefficients.

For three qubits, tracing out any one qubit gives

$$\rho_{AB}^{\mathrm{GHZ}} = \frac12 \left( \lvert00\rangle\langle00\rvert + \lvert11\rangle\langle11\rvert \right).$$

The two-qubit reduction is separable. The global phase coherence between $\lvert000\rangle$ and $\lvert111\rangle$ is lost when one qubit is discarded.

This does not mean the original state lacked entanglement. It means GHZ entanglement is stored globally rather than as surviving pairwise entanglement in the two-qubit reductions.

The W State

The $n$-qubit W state is the equal superposition of all computational-basis states with one excitation:

$$\lvert W_n\rangle = \frac{1}{\sqrt n} \sum_{r=1}^{n} \lvert0\cdots010\cdots0\rangle,$$

where the $1$ appears at position $r$. It is also genuinely multipartite entangled.

For three qubits,

$$\lvert W_3\rangle = \frac{ \lvert100\rangle + \lvert010\rangle + \lvert001\rangle }{\sqrt3}.$$

Tracing out $C$ gives

$$\rho_{AB}^{W} = \frac23 \lvert\Psi^+\rangle\langle\Psi^+\rvert + \frac13 \lvert00\rangle\langle00\rvert,$$

where

$$\lvert\Psi^+\rangle = \frac{\lvert01\rangle+\lvert10\rangle}{\sqrt2}.$$

Unlike the GHZ reduction, this two-qubit state remains entangled. W-type entanglement therefore has a different loss and sharing structure.

Feature	Three-qubit GHZ state	Three-qubit W state
genuine multipartite entanglement	yes	yes
one-qubit marginals	maximally mixed	spectrum two-thirds and one-third
two-qubit reductions	separable but correlated	entangled
loss of one qubit	destroys remaining pair entanglement	leaves pair entanglement
pairwise concurrence	zero	nonzero

The comparison does not make one state universally “more entangled” than the other. They occupy different multipartite structures, and different operational tasks favor different features.

Graph and Stabilizer States

A graph state associates one qubit with each vertex of a simple graph. Starting from $\lvert+\rangle^{\otimes n}$, apply a controlled-Z gate along every edge:

$$\lvert G\rangle = \prod_{(u,v)\in E} \operatorname{CZ}_{uv} \lvert+\rangle^{\otimes n}.$$

The same state is characterized as the simultaneous $+1$ eigenstate of commuting stabilizer generators

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

where $N(v)$ is the neighborhood of vertex $v$.

Graph connectivity provides a compact guide to the state’s entanglement structure, but graph edges are not ordinary pairwise-entanglement links. Local-unitary transformations and local Clifford operations can change the graph representation while preserving important state properties.

Use Graph States for the controlled-Z construction and Stabilizer States Preview for the commuting-operator language.

Monogamy Is Measure Dependent

Entanglement cannot always be shared with arbitrarily many partners in the way classical correlation can. A standard three-qubit result is the Coffman–Kundu–Wootters inequality

$$C_{A|BC}^{2} \geq C_{AB}^{2} + C_{AC}^{2},$$

where the concurrences are defined in the appropriate two-qubit and qubit-versus-pair settings.

For the GHZ state, the pairwise concurrences vanish while $A$ is entangled with $BC$. For the W state, pairwise concurrences are nonzero and the sharing pattern differs.

Monogamy is not a universal slogan that every entanglement measure obeys in every dimension. The inequality, exponent, local dimensions, and state class matter. Monogamy of Entanglement states the three-qubit result and its limits carefully.

Pairwise, Global, and Localizable Entanglement

Pairwise reduced-state entanglement does not fully characterize a multipartite state. The GHZ state is the simplest counterexample: every two-qubit reduction is separable, yet the pure three-qubit state is genuinely multipartite entangled.

Measurements on other parties can also change what is available to a chosen pair. If $C$ measures a GHZ state in the $x$ basis and reports the outcome, $A$ and $B$ are projected onto one of two Bell states up to a known local phase correction. The average state without the record remains the separable two-qubit marginal.

This motivates localizable entanglement: the entanglement that can be concentrated between selected parties through local measurements on the others and classical communication.

Network analysis adds another layer. Entanglement may be distributed across edges, swapped through intermediate nodes, or bounded across a cut separating groups of nodes. The Entanglement Sharing page connects these ideas without moving the full protocol theory out of Quantum Information.

Reduced States Across Many Parties

For a subset $S$ of parties with complement $\bar S$, define

$$\rho_S = \operatorname{Tr}_{\bar S}\rho.$$

If the global state is pure, the nonzero spectra of $\rho_S$ and $\rho_{\bar S}$ agree. Schmidt decomposition applies to the bipartition $S|\bar S$, even when each side contains many elementary subsystems.

This cut-by-cut use of bipartite tools is indispensable but incomplete. It can diagnose pure-state genuine multipartite entanglement by checking all bipartitions. It does not by itself provide a unique scalar measure of multipartite entanglement or solve the mixed-state convex problem.

An Analysis Protocol

1. Name every elementary subsystem

Write the ordered tensor product and explain what each factor represents. Do not infer physical parties from matrix dimensions alone.

2. State the partition being tested

Distinguish an elementary-party partition from a coarse-grained bipartition such as $AB|CD$. If several cuts matter, list them.

3. Decide whether the state is pure or mixed

For pure states, Schmidt ranks across bipartitions give exact factorization tests. For mixed states, use explicit decompositions, witnesses, semidefinite criteria, or other tools with their scope stated.

4. Separate pairwise from global questions

Compute reduced pair states only when pairwise entanglement is the target. A separable pair reduction does not rule out genuine multipartite entanglement.

5. Specify the resource task

State whether the goal is loss tolerance, state conversion, measurement-assisted localization, network distribution, metrology, error correction, or foundational correlation. Different structures answer different tasks.

Common Mistakes

• Calling every nonproduct three-party state genuinely multipartite entangled. A Bell pair tensored with a spectator is biseparable.
• Using one bipartition to settle every partition. Different cuts can have different Schmidt ranks and operational meanings.
• Treating pure- and mixed-state biseparability identically. Mixed biseparable decompositions may combine different cuts.
• Inferring no global entanglement from separable pair reductions. GHZ states refute this inference.
• Reading graph edges as pairwise entanglement amounts. Graph edges define controlled-Z preparation and stabilizers, not an edge-by-edge entanglement measure.
• Calling GHZ universally more or less entangled than W. Their order depends on the task and measure.
• Treating monogamy as dimension independent. The measure, exponent, and local dimensions matter.
• Ignoring outcome records in localizable entanglement. Conditional Bell states and the unconditioned marginal are different descriptions.

Reading Paths

Foundations and classification: Multipartite Systems → Multipartite Separability → Schmidt Decomposition.

Canonical state comparison: GHZ States → W States → Monogamy of Entanglement.

Structured many-qubit states: Graph States → Stabilizer States Preview → Entanglement Witnesses.

Networks and sharing: Entanglement Sharing → LOCC Preview → Entanglement and Quantum Information.

References

• D. M. Greenberger, M. A. Horne, and A. Zeilinger, “Going Beyond Bell’s Theorem,” in Bell’s Theorem, Quantum Theory and Conceptions of the Universe, Kluwer, 1989.
• W. Dür, G. Vidal, and J. I. Cirac, “Three Qubits Can Be Entangled in Two Inequivalent Ways,” Physical Review A 62, 062314, 2000.
• V. Coffman, J. Kundu, and W. K. Wootters, “Distributed Entanglement,” Physical Review A 61, 052306, 2000.
• H. J. Briegel and R. Raussendorf, “Persistent Entanglement in Arrays of Interacting Particles,” Physical Review Letters 86, 910–913, 2001.
• 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, IOS Press, 2006.
• O. Gühne and G. Tóth, “Entanglement Detection,” Physics Reports 474, 1–75, 2009.
• R. Horodecki, P. Horodecki, M. Horodecki, and K. Horodecki, “Quantum Entanglement,” Reviews of Modern Physics 81, 865–942, 2009.
• N. Brunner, D. Cavalcanti, S. Pironio, V. Scarani, and S. Wehner, “Bell Nonlocality,” Reviews of Modern Physics 86, 419–478, 2014.

Exercises

Exercise 1: Classify four pure states

Classify each three-qubit pure state as fully product, biseparable but not fully product, or genuinely multipartite entangled:

$$\begin{aligned} \lvert\psi_1\rangle&=\lvert000\rangle, \\ \lvert\psi_2\rangle&=\lvert\Phi^+\rangle_{AB}\lvert0\rangle_C, \\ \lvert\psi_3\rangle&=\lvert\mathrm{GHZ}_3\rangle, \\ \lvert\psi_4\rangle&=\lvert W_3\rangle. \end{aligned}$$

Solution

$\lvert\psi_1\rangle$ is fully product. $\lvert\psi_2\rangle$ is product across $AB|C$ but entangled inside $AB$, so it is biseparable and not fully product.

Both $\lvert\mathrm{GHZ}_3\rangle$ and $\lvert W_3\rangle$ are entangled across every bipartition and are therefore genuinely multipartite entangled.

Exercise 2: Reduced state of GHZ

Trace qubit $C$ out of $\lvert\mathrm{GHZ}_3\rangle$. Is the resulting state on $AB$ entangled?

Solution

Expanding the projector gives diagonal terms and two cross terms. The cross terms contain $\lvert0\rangle\langle1\rvert_C$ or its adjoint, whose trace is zero. Therefore

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

This is an explicit convex mixture of product states, so it is separable. The original three-party pure state is nevertheless genuinely multipartite entangled.

Exercise 3: Reduced state of W

Trace qubit $C$ out of $\lvert W_3\rangle$ and show that the result has the form

$$\rho_{AB} = \frac23\lvert\Psi^+\rangle\langle\Psi^+\rvert + \frac13\lvert00\rangle\langle00\rvert.$$

Solution

Group the state by the value of qubit $C$:

$$\begin{aligned} \lvert W_3\rangle &= \sqrt{\frac23}\, \lvert\Psi^+\rangle_{AB}\lvert0\rangle_C \\ &\quad+ \sqrt{\frac13}\, \lvert00\rangle_{AB}\lvert1\rangle_C. \end{aligned}$$

The two $C$ states are orthogonal, so the cross terms vanish under the partial trace. The displayed mixture follows. Its partial transpose has a negative eigenvalue, so the reduction is entangled.

Exercise 4: Why mixed biseparability is subtler

Consider

$$\begin{aligned} \rho = \frac12 \left( \lvert0\rangle\langle0\rvert_A \otimes \lvert\Phi^+\rangle\langle\Phi^+\rvert_{BC} \right) \\ + \frac12 \left( \lvert0\rangle\langle0\rvert_B \otimes \lvert\Phi^+\rangle\langle\Phi^+\rvert_{AC} \right). \end{aligned}$$

Why is $\rho$ biseparable even though its two terms use different bipartitions?

Solution

The first term is product across $A|BC$, and the second is product across $B|AC$. A mixed state is biseparable when it is a convex mixture of states separable across bipartitions; the cut may vary from one term to another. The displayed expression is already such a decomposition, so $\rho$ is not genuinely multipartite entangled.

This example also shows why searching for one fixed cut that separates the entire mixed state is too restrictive a definition of biseparability.

Exercise 5: GHZ localization by measurement

Rewrite $\lvert\mathrm{GHZ}_3\rangle$ in the $x$ basis of qubit $C$ and identify the conditional states of $AB$ after measuring $C$.

Solution

Using

$$\lvert0\rangle = \frac{\lvert+\rangle+\lvert-\rangle}{\sqrt2}, \qquad \lvert1\rangle = \frac{\lvert+\rangle-\lvert-\rangle}{\sqrt2},$$

one obtains

$$\begin{aligned} \lvert\mathrm{GHZ}_3\rangle =\frac{1}{\sqrt2}\bigl(& \lvert\Phi^+\rangle_{AB}\lvert+\rangle_C \\ &+ \lvert\Phi^-\rangle_{AB}\lvert-\rangle_C \bigr). \end{aligned}$$

Each outcome occurs with probability $1/2$. Conditioned on the reported outcome, $AB$ is in $\lvert\Phi^+\rangle$ or $\lvert\Phi^-\rangle$. If the outcome is ignored, their equal mixture is the separable computational-basis correlation obtained by tracing out $C$.