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.

Helpful background. Multipartite partitions identifies the parties and cuts; separable mixtures explains convex decomposition. We assume finitely many finite-dimensional subsystems, so the displayed finite convex decompositions suffice.

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

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

Products such as $\rho_B\otimes\rho_{AC}$ are embedded in the fixed $A,B,C$ order by the corresponding factor permutation. 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.

A biseparable state entangled across every fixed cut

Define three normalized vectors in the fixed $A,B,C$ order,

$$|\psi_A\rangle=\frac{|000\rangle+|011\rangle}{\sqrt2},\quad |\psi_B\rangle=\frac{|000\rangle+|101\rangle}{\sqrt2},\quad |\psi_C\rangle=\frac{|000\rangle+|110\rangle}{\sqrt2},$$

and let $\rho=\tfrac13\sum_{j=A,B,C}|\psi_j\rangle\langle\psi_j|$. Each term has one separated party and a Bell pair on the other two, so the mixture is biseparable by construction.

Nevertheless, its partial transpose on $A$ has the following invariant block in the basis $|001\rangle,|010\rangle,|100\rangle$:

$$\frac16\begin{pmatrix}0&0&1\\0&0&1\\1&1&0\end{pmatrix}.$$

This block has eigenvalues $0,\sqrt2/6,-\sqrt2/6$. Therefore $\rho$ is entangled across $A:BC$. Permutation symmetry gives the same conclusion for the other two cuts. This concrete example shows why certifying entanglement across every fixed cut does not certify genuine multipartite entanglement for a mixed state.

Witnessing genuine multipartite entanglement

A genuine multipartite entanglement witness must have nonnegative expectation on every biseparable state, including mixtures over different cuts. For a pure target $|\psi\rangle$, let $\alpha$ be the largest squared Schmidt coefficient, maximized over all nontrivial bipartitions. Then

$$W=\alpha I-|\psi\rangle\langle\psi|$$

is nonnegative on each cut-product vector, hence on all their mixtures. For GHZ, $\alpha=1/2$; for $W_n$, $\alpha=(n-1)/n$. Thus target overlaps above $1/2$ and $(n-1)/n$, respectively, prove genuine $n$-party entanglement. Failing either threshold is inconclusive. Entanglement witnesses explains the overlap bound and measurement assumptions.

Number of groups versus size of entangled groups

Two related conventions answer different questions. A pure state is $k$-separable if it factors into at least $k$ groups. It is $k$-producible if it factors into groups containing at most $k$ parties. For mixed states, take convex mixtures with these properties, allowing the grouping to vary between terms.

For example, four qubits in two independent Bell pairs are two-separable and two-producible, but not fully separable. A three-qubit GHZ state with one product spectator is also two-separable, but needs a three-party entangled group. Its entanglement depth is three: the smallest $k$ for which the state is $k$-producible. These definitions distinguish the number of independent groups from the number of parties entangled together.

References

• A. Acín, D. Bruß, M. Lewenstein, and A. Sanpera, “Classification of mixed three-qubit states”, Physical Review Letters 87, 040401 (2001).
• O. Gühne and G. Tóth, “Entanglement detection”, Physics Reports 474, 1–75 (2009).
• M. Seevinck and J. Uffink, “Partial separability and entanglement criteria for multiqubit quantum states”, Physical Review A 78, 032101 (2008).

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. A three-qubit state has GHZ overlap $F=0.62$. What follows from the witness $W=I/2-|\mathrm{GHZ}_3\rangle\langle\mathrm{GHZ}_3|$? Compare a $W_3$ overlap of $0.62$ with its corresponding target witness.

Solution

For GHZ, $\langle W\rangle=0.5-0.62=-0.12$, which proves genuine tripartite entanglement if the overlap is established with sufficient precision. For the $W_3$ target, the biseparable bound is $2/3$; $0.62$ does not exceed it, so that witness is inconclusive. A target overlap is meaningful only relative to the appropriate bound.

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.