Multipartite Systems

A multipartite system is a composite system with more than two named subsystems. The Hilbert space has the form

$$\mathcal H_{1\cdots n} = \mathcal H_1\otimes\mathcal H_2 \otimes\cdots\otimes \mathcal H_n.$$

This looks like a straightforward extension of the bipartite tensor product, but the conceptual structure changes sharply. There is no longer only one split $A\vert B$. There are many ways to group subsystems into blocks, many reduced density operators, many possible notions of locality, and many inequivalent patterns of entanglement.

Helpful background. Tensor products fixes the subsystem spaces; partial trace computes a state after some factors are discarded. We use finite-dimensional named subsystems below.

Tensor Products of Many Hilbert Spaces

For $n$ distinguishable subsystems, the composite Hilbert space is the tensor product of the subsystem Hilbert spaces:

$$\mathcal H = \bigotimes_{j=1}^{n}\mathcal H_j.$$

The order of factors is part of the convention. If

$$\lvert\psi_j\rangle\in\mathcal H_j,$$

then a fully product vector is

$$\lvert\psi_1\rangle\otimes \lvert\psi_2\rangle\otimes\cdots\otimes \lvert\psi_n\rangle,$$

often abbreviated as

$$\lvert\psi_1\psi_2\cdots\psi_n\rangle$$

when the subsystem ordering is clear.

For $n$ qubits, a standard product basis is

$$\lvert x_1x_2\cdots x_n\rangle, \qquad x_j\in\{0,1\}.$$

For example,

$$\lvert0101\rangle = \lvert0\rangle_1\otimes \lvert1\rangle_2\otimes \lvert0\rangle_3\otimes \lvert1\rangle_4.$$

A general pure state is a linear combination

$$\lvert\Psi\rangle = \sum_{x_1,\ldots,x_n} c_{x_1\cdots x_n} \lvert x_1\cdots x_n\rangle$$

for qubits, or the analogous expansion in product bases for higher-dimensional subsystems. The coefficients form a multi-index tensor rather than a matrix. This is the first reason multipartite entanglement is more complex than bipartite entanglement: there is no single matrix singular-value decomposition that simultaneously solves all subsystem splits.

Subsystem Labels and Supports

Let

$$N=\{1,2,\ldots,n\}$$

be the set of subsystem labels. For a subset $S\subseteq N$, define

$$\mathcal H_S = \bigotimes_{j\in S}\mathcal H_j, \qquad \bar S=N\setminus S.$$

With a fixed ordering convention, the full Hilbert space can be regarded as

$$\mathcal H_N \cong \mathcal H_S\otimes\mathcal H_{\bar S}.$$

An operator supported on $S$ has the form

$$O_S\otimes I_{\bar S},$$

where $O_S$ acts on $\mathcal H_S$. The identity on all other factors is often left implicit after the support has been stated.

This support language is essential in many-body physics and quantum information. A one-site observable, a two-spin correlator operator, and a stabilizer generator have specified supports. A block entropy is instead a nonlinear function of the state reduced to that block, not an observable with an operator support.

Partitions

A partition of the subsystem labels is a collection of nonempty disjoint blocks

$$\Pi = \{S_1,S_2,\ldots,S_m\}$$

whose union is all of $N$:

$$S_a\cap S_b=\varnothing \quad(a\ne b), \qquad \bigcup_{\alpha=1}^{m}S_\alpha=N.$$

The partition says which groups of subsystems are being treated as the parties. For example, for three subsystems $A,B,C$, the partition

$$\{A,B,C\}$$

treats all three as separate parties, while

$$\{A,BC\}$$

treats $B$ and $C$ as one composite block.

A pure state is product with respect to the partition $\Pi$ if it can be written as

$$\lvert\Psi\rangle = \bigotimes_{\alpha=1}^{m} \lvert\psi_\alpha\rangle_{S_\alpha}.$$

The strongest product condition is full productness:

$$\lvert\Psi\rangle = \lvert\psi_1\rangle\otimes \lvert\psi_2\rangle\otimes\cdots\otimes \lvert\psi_n\rangle,$$

which is product with respect to the singleton partition

$$\{\{1\},\{2\},\ldots,\{n\}\}.$$

Productness with respect to a coarse partition is weaker. The state

$$\lvert0\rangle_A\otimes \frac{1}{\sqrt2} \bigl( \lvert00\rangle_{BC} + \lvert11\rangle_{BC} \bigr)$$

is product across $A\vert BC$, but it is not fully product because $B$ and $C$ are entangled with each other.

Bipartitions

A bipartition is a partition into two blocks:

$$S\vert\bar S.$$

Every nonempty proper subset $S\subset N$ determines a bipartition, but $S\vert\bar S$ and $\bar S\vert S$ describe the same split. Therefore the number of nontrivial bipartitions of $n$ labeled subsystems is

$$2^{n-1}-1.$$

For $n=3$, the three nontrivial bipartitions are

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

For a pure state, each bipartition reduces to an ordinary bipartite problem. One may write a Schmidt decomposition across the grouped split:

$$\lvert\Psi\rangle = \sum_r s_r^{(S)} \lvert r\rangle_S \otimes \lvert r\rangle_{\bar S}, \qquad s_r^{(S)}\ge0.$$

Thus every bipartition has its own Schmidt rank, Schmidt coefficients, and entanglement entropy:

$$S_S(\Psi) = - \operatorname{Tr}(\rho_S\log_2\rho_S), \qquad \rho_S=\operatorname{Tr}_{\bar S} \lvert\Psi\rangle\langle\Psi\rvert.$$

Entropy is in bits here. These bipartite cuts are useful, but they do not fully classify multipartite entanglement. A list of one-vs-rest entropies can miss distinctions between different kinds of tripartite and many-party correlations.

Four qubits: every single-qubit reduction can mislead

Consider two independent Bell pairs,

$$|\psi\rangle=|\Phi^+\rangle_{12}\otimes|\Phi^+\rangle_{34} =\tfrac12(|0000\rangle+|0011\rangle+|1100\rangle+|1111\rangle).$$

Every single-qubit reduction is $I/2$, so every one-versus-three cut is entangled. Yet the state is a product across $12:34$, and thus is not genuinely four-party entangled. Across $13:24$ it has four equal Schmidt coefficients $1/2$, because both Bell pairs cross the cut.

Cut	Schmidt rank	Entropy in bits
Any one qubit versus the other three	$2$	$1$
$12:34$	$1$	$0$
$13:24$ or $14:23$	$4$	$2$

These are all seven nontrivial bipartitions. Testing only one-versus-rest cuts would miss the product split. For a pure state, genuine multipartite entanglement requires entanglement across every nontrivial bipartition. For mixed states, even that collection of tests is insufficient in general; multipartite separability explains mixtures of different splits.

Reduced States for Many Subsystems

For a density operator $\rho_N$ on the full system, the reduced state on a subset $S$ is

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

It gives all expectation values of observables supported on $S$:

$$\langle O_S\rangle_\rho = \operatorname{Tr}_S(\rho_S O_S).$$

For three systems,

$$\rho_{AB} = \operatorname{Tr}_C\rho_{ABC}, \qquad \rho_A = \operatorname{Tr}_{BC}\rho_{ABC}.$$

The reductions are consistent under further tracing:

$$\rho_A = \operatorname{Tr}_B\rho_{AB} = \operatorname{Tr}_{BC}\rho_{ABC}.$$

This consistency is useful but not complete information. A collection of one-body reduced states usually does not determine the global state. Even collections of low-body marginals can miss global phase coherence or topological constraints.

The GHZ example is the standard warning. For

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

tracing out one qubit gives

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

independent of the phase $\phi$. The two-qubit marginal remembers perfect $Z$-basis correlation but forgets the three-body coherence.

Multipartite Entanglement Complexity

For two subsystems, pure-state entanglement is controlled by the Schmidt coefficients. For three or more subsystems, no single list of coefficients plays the same universal role.

Several distinctions become necessary:

• A state may be entangled across one bipartition but product across another.
• A state may have pairwise entanglement without genuine all-party entanglement.
• A state may have genuine multipartite entanglement even though all two-party marginals are separable.
• Different multipartite entangled states may be inequivalent under local operations even when their bipartite Schmidt ranks look similar.
• Mixed-state multipartite separability has a hierarchy rather than one yes-or-no test.

For example, the three-qubit state

$$\lvert0\rangle_A\otimes \lvert\Phi^+\rangle_{BC}$$

is entangled across $B\vert AC$ and $C\vert AB$, but product across $A\vert BC$. It is not genuinely tripartite entangled.

By contrast, GHZ States have global coherence among all parties, while W States distribute a single excitation coherently over many parties. These are not merely two examples of the same bipartite phenomenon; they represent different multipartite patterns.

References

• W. Dür, G. Vidal, and J. I. Cirac, “Three qubits can be entangled in two inequivalent ways”, Physical Review A 62, 062314 (2000).
• 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).
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press (2010).

Exercises

1. Count the nontrivial bipartitions of $n$ labeled subsystems.

Solution

Every nonempty proper subset $S\subset N$ determines a split $S\vert\bar S$. There are $2^n-2$ such subsets. Since $S\vert\bar S$ and $\bar S\vert S$ are the same bipartition, divide by two:

$$\frac{2^n-2}{2} = 2^{n-1}-1.$$

2. Let

$$\rho_N = \rho_1\otimes\rho_2\otimes\cdots\otimes\rho_n.$$

Show that the reduced state on $S\subseteq N$ is

$$\rho_S = \bigotimes_{j\in S}\rho_j.$$

Solution

Trace over each factor in $\bar S$. Since every density operator has trace one,

$$\operatorname{Tr}_{\bar S} \left( \bigotimes_{j=1}^{n}\rho_j \right) = \left( \bigotimes_{j\in S}\rho_j \right) \prod_{k\in\bar S}\operatorname{Tr}\rho_k = \bigotimes_{j\in S}\rho_j.$$

3. For two Bell pairs $|\Phi^+\rangle_{12}|\Phi^+\rangle_{34}$, write a Schmidt decomposition across $13:24$.

Solution

Reorder each basis ket explicitly:

$$|\psi\rangle=\frac12\sum_{a,b=0}^1 |ab\rangle_{13}|ab\rangle_{24}.$$

Both local families are orthonormal. There are four Schmidt coefficients $1/2$, so the rank is four and entropy is two bits. Regrouping the tensor factors reveals existing correlations; it does not physically apply a SWAP gate.

4. Trace out qubit $C$ from

$$\lvert\mathrm{GHZ}_3(\phi)\rangle = \frac{1}{\sqrt2} \bigl( \lvert000\rangle + e^{i\phi}\lvert111\rangle \bigr).$$

Why does the answer not depend on $\phi$?

Solution

The density operator contains diagonal terms and cross terms:

$$\frac12 \lvert000\rangle\langle000\rvert + \frac12 \lvert111\rangle\langle111\rvert + \frac{e^{-i\phi}}{2} \lvert000\rangle\langle111\rvert + \frac{e^{i\phi}}{2} \lvert111\rangle\langle000\rvert.$$

When tracing over $C$, the cross terms contain $\langle1\vert0\rangle_C$ or $\langle0\vert1\rangle_C$, so they vanish. The result is

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

The phase $\phi$ is stored in three-body coherence, not in this two-body marginal.

5. Why is “all two-party marginals are separable” not enough to conclude that a pure three-party state has no entanglement?

Solution

The three-qubit GHZ state is the standard counterexample. Its two-party reduced states are

$$\frac12 \lvert00\rangle\langle00\rvert + \frac12 \lvert11\rangle\langle11\rvert,$$

which are separable. Nevertheless the full pure state

$$\frac{1}{\sqrt2} \bigl( \lvert000\rangle+\lvert111\rangle \bigr)$$

is entangled across every one-vs-two bipartition. Entanglement can live in genuinely multipartite coherence that disappears from pairwise marginals.