Multipartite Separability

Multipartite separability is not a single yes-or-no property. With three or more subsystems, one must specify whether the question concerns full product structure, separability across a particular partition, separability across some bipartition, or genuine entanglement shared by all parties.

For bipartite systems, the central contrast is separable versus entangled. For multipartite systems, the hierarchy is richer:

$$\text{fully separable} \subset \text{biseparable} \subset \text{all states}.$$

States outside the biseparable set are genuinely multipartite entangled. They cannot be explained as mixtures of states that are separable across some bipartition.

Fully Product and Fully Separable States

For $n$ named subsystems, a pure state is fully product if

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

This is product with respect to the singleton partition

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

A mixed state is fully separable if it can be written as a classical mixture of fully product states, or more generally product density operators:

$$\rho_{1\cdots n} = \sum_r p_r\, \rho_1^{(r)} \otimes \rho_2^{(r)} \otimes\cdots\otimes \rho_n^{(r)}, \qquad p_r\ge0, \qquad \sum_r p_r=1.$$

Fully separable states may have classical correlations. The shared classical label $r$ can correlate the local preparations, just as in the bipartite case. What is absent is entanglement across any grouping of the named parties.

Separability with Respect to a Partition

Let

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

be a partition of the subsystem labels. A pure state is product with respect to $\Pi$ if

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

A mixed state is separable with respect to $\Pi$ if

$$\rho = \sum_r p_r \bigotimes_{\alpha=1}^{m} \rho_{S_\alpha}^{(r)}.$$

This permits entanglement inside a block. For three parties, the state

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

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

Thus “separable” is incomplete unless the partition is named. The phrase “separable across $A\vert BC$” is precise; the phrase “separable” alone is often ambiguous in multipartite settings.

Biseparable States

A pure state is biseparable if it is product across at least one nontrivial bipartition:

$$\lvert\Psi\rangle = \lvert\psi\rangle_S \otimes \lvert\phi\rangle_{\bar S}$$

for some nonempty proper subset $S$.

For mixed states, biseparability allows mixtures over different bipartitions. For three parties,

$$\rho_{ABC} = \sum_r p_r\, \rho_A^{(r)}\otimes\rho_{BC}^{(r)} + \sum_s q_s\, \rho_B^{(s)}\otimes\rho_{AC}^{(s)} + \sum_t r_t\, \rho_C^{(t)}\otimes\rho_{AB}^{(t)},$$

with all coefficients nonnegative and with total weight

$$\sum_r p_r+\sum_s q_s+\sum_t r_t=1.$$

The states $\rho_{BC}^{(r)}$, $\rho_{AC}^{(s)}$, and $\rho_{AB}^{(t)}$ may themselves be entangled inside their two-party blocks. What matters is that each term is separable across some one-vs-two split.

This mixed-state definition is more subtle than checking one fixed cut. A mixed state can be a convex mixture of terms separable across different cuts. Therefore “not separable across any one displayed cut” is not automatically the same as “genuinely multipartite entangled.”

Genuine Multipartite Entanglement

A pure state is genuinely multipartite entangled if it is not product across any nontrivial bipartition. Equivalently, for every bipartition $S\vert\bar S$, the Schmidt rank satisfies

$$\operatorname{SR}_{S\vert\bar S}(\Psi)>1.$$

For mixed states, a state is genuinely multipartite entangled if it is not biseparable. That is, it cannot be written as a mixture of states that are separable across some bipartition.

The three-qubit GHZ and W states are standard pure-state examples:

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

and

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

Both are entangled across every one-vs-two cut, but they organize their entanglement differently. The separability hierarchy says they are both genuinely tripartite entangled; it does not say they are equivalent resources.

Bipartitions as Diagnostics

For pure multipartite states, bipartitions are powerful. To test whether a pure state is genuinely multipartite entangled, check every nontrivial bipartition:

$$S\vert\bar S.$$

If the state has Schmidt rank one across some cut, it is biseparable. If every cut has Schmidt rank greater than one, it is genuinely multipartite entangled.

For mixed states, bipartition tests are necessary but not always sufficient. A state separable across one cut is certainly biseparable. But a state that fails a separability test across every fixed cut may still be a mixture of states separable across different cuts. Proving genuine multipartite entanglement usually requires ruling out the whole biseparable convex set.

This is why multipartite entanglement witnesses are useful. A genuine multipartite entanglement witness is an observable $W$ such that

$$\operatorname{Tr}(W\sigma)\ge0 \quad \text{for every biseparable state }\sigma,$$

but

$$\operatorname{Tr}(W\rho)<0$$

for at least one genuinely multipartite entangled state $\rho$.

The idea is the same convex-separation logic used for bipartite Entanglement Witnesses, but the separable set is replaced by the biseparable set.

Examples

The state

$$\lvert000\rangle$$

is fully product and therefore fully separable.

The state

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

is biseparable. It is separable across $A\vert BC$, but not fully separable.

The mixed state

$$\rho = \frac13 \lvert0\rangle\langle0\rvert_A \otimes \lvert\Phi^+\rangle\langle\Phi^+\rvert_{BC} + \frac13 \lvert0\rangle\langle0\rvert_B \otimes \lvert\Phi^+\rangle\langle\Phi^+\rvert_{AC} + \frac13 \lvert0\rangle\langle0\rvert_C \otimes \lvert\Phi^+\rangle\langle\Phi^+\rvert_{AB}$$

is biseparable by construction: each term is separable across one one-vs-two cut, although the cut changes from term to term.

The state $\lvert\mathrm{GHZ}_3\rangle$ is genuinely tripartite entangled. Its two-party marginals are separable, so pairwise marginals alone do not decide genuine multipartite entanglement.

Common Mistakes

• Saying “separable” in a multipartite problem without naming the partition or hierarchy level.
• Confusing fully separable with biseparable.
• Assuming a state with one entangled pair is genuinely tripartite entangled.
• Assuming separable two-party marginals imply absence of genuine multipartite entanglement.
• Checking only one bipartition and drawing a conclusion about all-party entanglement.
• Forgetting that mixed biseparable states may mix different bipartitions.
• Treating GHZ and W states as equivalent merely because both are genuinely multipartite entangled.

Cross-Links

• Multipartite Systems
• Separable Mixed States
• Entangled States
• GHZ States
• W States
• Graph States
• Monogamy of Entanglement
• Entanglement Sharing
• Schmidt Rank
• Entanglement Entropy
• Negativity and PPT Criterion
• Entanglement Witnesses
• Reduced Density Operators
• Marginals and Correlations
• Certification of Entanglement turns biseparable and $k$-producible null sets into experimental witness, depth, dimensionality, and reporting claims.

References

• W. Dur, G. Vidal, and J. I. Cirac, “Three Qubits Can Be Entangled in Two Inequivalent Ways,” Physical Review A 62, 062314, 2000.
• A. Acin, D. Bruss, M. Lewenstein, and A. Sanpera, “Classification of Mixed Three-Qubit States,” Physical Review Letters 87, 040401, 2001.
• M. Seevinck and J. Uffink, “Partial Separability and Entanglement Criteria for Multiqubit Quantum States,” Physical Review A 78, 032101, 2008.
• O. Guhne and G. Toth, “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, Cambridge University Press, 2010.

Exercises

1. Show that every fully separable three-party mixed state is biseparable.

Solution

A fully separable state has the form

$$\rho_{ABC} = \sum_r p_r\, \rho_A^{(r)}\otimes \rho_B^{(r)}\otimes \rho_C^{(r)}.$$

Regroup each product as

$$\rho_A^{(r)} \otimes \bigl( \rho_B^{(r)}\otimes\rho_C^{(r)} \bigr).$$

This is separable across $A\vert BC$ term by term, so the state is biseparable. The same argument works for any chosen one-vs-rest cut.

2. Classify

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

as fully product, biseparable, or genuinely tripartite entangled.

Solution

It is product across $A\vert BC$, so it is biseparable. It is not fully product because $\lvert\Phi^+\rangle_{BC}$ is entangled across $B\vert C$. It is not genuinely tripartite entangled because it is product across one bipartition.

3. Show that $\lvert\mathrm{GHZ}_3\rangle$ is entangled across $A\vert BC$.

Solution

Across $A\vert BC$,

$$\lvert\mathrm{GHZ}_3\rangle = \frac{1}{\sqrt2} \bigl( \lvert0\rangle_A\lvert00\rangle_{BC} + \lvert1\rangle_A\lvert11\rangle_{BC} \bigr).$$

This is already a Schmidt decomposition with two nonzero coefficients. Therefore the Schmidt rank across $A\vert BC$ is $2$, so the state is entangled across that cut. By symmetry the same holds for the other one-vs-two cuts.

4. Explain why a mixture of states separable across different cuts can still be biseparable.

Solution

Biseparability for mixed states is defined by a convex mixture over all bipartitions. Each term only has to be separable across some cut; the cut may depend on the term. Therefore a state such as

$$\sum_\lambda p_\lambda\, \rho_{S_\lambda}^{(\lambda)} \otimes \rho_{\bar S_\lambda}^{(\lambda)}$$

is biseparable even when the subsets $S_\lambda$ are not all the same.

5. Why do separable two-party marginals not rule out genuine tripartite entanglement?

Solution

The GHZ state is the counterexample. Its two-party marginals are mixtures such as

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

which are separable. Nevertheless the full pure state is entangled across every one-vs-two bipartition. The entanglement is stored in global coherence that is lost when any one qubit is traced out.