Foundations of quantum mechanics · Theory and numerical analysis
Network nonlocality with tripartite entangled sources
arXiv:2610.35838v1 ↗PDF ↗Submitted 26 October 2026
- Question
- Network nonlocality has mostly been studied with sources that emit pairs of particles. A three-qubit source has two distinct notions of nonlocality: the Mermin inequality excludes fully local models, and the Svetlichny inequality certifies genuine tripartite nonlocality. When independent sources emit three-qubit states and central nodes measure without any input, how is a network violation related to these two single-source tests?
- Result
- It depends on the wiring. In the closed network, where two input-free central nodes each hold one particle from every source, Bell-state measurements give a two-copy value at least the normalized Svetlichny value, \(\mathcal N_{\rm c}\ge\mathcal S/4\); every entangled generalized GHZ state, and even some biseparable states, violate the network inequality. In the open network, with one central node, the value is at most the normalized Mermin value, \(\mathcal N_{\rm o}\le\mathcal M/2\). With \(n\) sources, the largest violation of these two-correlator inequalities is \(\sqrt2\) (closed) or \(2\) (open) for even \(n\), but only \(2^{1/2-1/(2n)}\) or \(2^{1-1/(2n)}\) for odd \(n\).
The paper in detail
For one three-qubit source with two settings per party, the Mermin operator is \(M=A_0B_0C_1+A_0B_1C_0+A_1B_0C_0-A_1B_1C_1\) and \(M^\prime\) swaps the settings. Local models obey \(|\langle M\rangle|\le2\); models in which two parties share arbitrary nonlocal resources still obey \(|\langle M+M^\prime\rangle|\le4\), so violating the second certifies genuine tripartite nonlocality. The GHZ state reaches \(4\) and \(4\sqrt2\). All results are written in terms of the correlation tensor \(T_{\mu k\nu}=\operatorname{Tr}[\rho\,\sigma_\mu\otimes\sigma_k\otimes\sigma_\nu]\), whose middle index belongs to the particle sent to an end node.
In the closed network, Alice and Charlie each receive one particle from every source and always perform the same measurement; each end node receives the third particle and chooses between two settings. Classically, two central nodes that see the same hidden variables are no stronger than one merged node, because any randomness the merged node uses can travel with one of the sources (Lemma 1). The closed network therefore inherits the star-network bound \[|I|^{1/n}+|J|^{1/n}\le1,\] where \(I\) and \(J\) correlate fixed sign functions of the central outcomes with the sum and the alternating sum of the end-node settings.
With Bell-state measurements at both central nodes, the two-copy value \(\mathcal N_{\rm c}\) has a closed form, and Theorem 2 gives \(\mathcal N_{\rm c}\ge\mathcal S/4\). Noisy GHZ sources saturate it, \(\mathcal N_{\rm c}=\sqrt2\,v\). Generalized GHZ states \(\cos\alpha\ket{000}+\sin\alpha\ket{111}\) give \(\mathcal N_{\rm c}\ge\sqrt{1+\sin^22\alpha}>1\) for every \(0<\alpha<\pi/2\), although they violate the Svetlichny inequality only for \(\sin2\alpha>1/\sqrt2\) and the Mermin inequality only for \(\sin2\alpha>1/2\). The W state reaches \(\sqrt{13}/3\approx1.202\), with critical visibility 0.832 against 0.919 for Svetlichny. With GHZ sources the two end nodes are left in a Bell state whose relative sign is the product of one sign from each central node, so either outcome alone leaves them separable. With noise, the herald \(c_z=a_z\) occurs with probability \((1+v^2)/2\) and leaves a Werner state of visibility \(2v^2/(1+v^2)\); for \(0.707<v\le0.740\) the network inequality is violated while the swapped state violates no CHSH inequality.
In the open network, a single central node Charlie holds one particle from each source and the other two particles of every source go to two end nodes. Using the halves \(U_0=\tfrac12(A_0B_0-A_1B_1)\) and \(U_1=\tfrac12(A_0B_1+A_1B_0)\) of the Mermin operator, two GHZ sources reach the algebraic maximum \(2\), a network form of the GHZ paradox. Theorem 3 gives \[\mathcal N_{\rm o}^2=\max\bigl[|\mathbf u_0|^2+|\mathbf u_1|^2+2|\mathbf u_0\times\mathbf u_1|\bigr]\le\bigl(\mathcal M/2\bigr)^2,\] so a violation requires a Mermin violation of one source. With more sources, Theorem 4 shows that an input-free central node cannot herald two anticommuting products at once, which lowers the maximal violation for odd \(n\): for three sources the closed value is \(2^{1/3}\approx1.26\), below the \(\sqrt2\) of two sources. Whether network tests with tripartite sources can certify genuine multipartite entanglement remains open.
Transferable methods
Two perpendicular directions from one end node
An end node with two dichotomic qubit settings can be rewritten as measuring along two perpendicular directions. Maximizing a two-correlator expression then reduces to how much of two Bloch-like vectors those directions can capture, which gives closed forms and the comparison with the Svetlichny test.
With \(\mathbf b_0\pm\mathbf b_1\propto\mathbf n,\mathbf n^\prime\) and \(\mathbf n\perp\mathbf n^\prime\), the two-source closed-network value is governed by \[\Lambda(\mathbf\tau,\mathbf\tau^\prime)=\max_{\mathbf n\perp\mathbf n^\prime}\bigl[(\mathbf\tau\cdot\mathbf n)^2+(\mathbf\tau^\prime\cdot\mathbf n^\prime)^2\bigr]=\tfrac12\Bigl[s+\sqrt{s^2-4(\mathbf\tau\cdot\mathbf\tau^\prime)^2}\Bigr],\] with \(s=|\mathbf\tau|^2+|\mathbf\tau^\prime|^2\) and \(\mathbf\tau_{\mu\nu}=(\operatorname{Tr}[\rho\,A_\mu\otimes\sigma_k\otimes C_\nu])_{k}\). Orthogonal vectors give the sum of their squared lengths, parallel ones only the larger. Sorted by the end node’s two perpendicular directions, the Svetlichny expression contains four correlators entering with weights of total size at most one, so \((\mathcal S/4)^2\le\max(X^2,Y^2)+\max(Z^2,W^2)\); a Bell-state measurement reads all parities in every run and can take the larger correlator in each bracket.
Editorial check: for \(\alpha=0.3\), direct computation gives \(\mathbf\tau_{30}=\hat{\mathbf z}\) and \(\mathbf\tau_{11}=\sin2\alpha\,\hat{\mathbf x}=0.565\,\hat{\mathbf x}\), orthogonal as required for \(\sqrt{1+\sin^22\alpha}\). For the W state, \(\mathbf\tau_{33}=-\hat{\mathbf z}\) and \(\mathbf\tau_{13}=\tfrac23\hat{\mathbf x}\), giving \(\sqrt{13}/3\). Simulating two noisy GHZ sources with standard-frame Bell measurements at \(v=0.8\), the herald \(c_z=a_z\) occurs with probability 0.82 \(=(1+v^2)/2\) and leaves a Werner state of visibility 0.7805 \(=2v^2/(1+v^2)\), as stated.
How to reuse it
For any Bell-type expression in which one party enters through the sum and difference of two qubit observables, replace its settings by a perpendicular pair and optimize the quadratic form \(\Lambda\). This yields closed-form quantum values and shows immediately whether the competing correlators can be read at once or must be traded off.
Anticommutation bound for input-free central nodes
When a node without an input must herald two correlators from one fixed measurement, the algebra of the end-node operators being heralded, not the sources, limits the violation. Counting anticommuting factors gives tight bounds that depend only on the parity of the number of sources.
The sum and difference of an end node’s two qubit observables are proportional to Pauli operators along perpendicular directions, which anticommute. With \(P=\prod_i\mathbf n_i\cdot\mathbf\sigma\) and \(Q=\prod_i\mathbf n_i^\prime\cdot\mathbf\sigma\), moving \(Q\) through \(P\) produces one sign per factor, \(PQ=(-1)^nQP\). For even \(n\) the two products commute and can both be heralded with definite values. For odd \(n\) they anticommute, so \(\langle P\rangle^2+\langle Q\rangle^2\le1\) in every state, and the maximum of \(|I|^{1/n}+|J|^{1/n}\) drops from \(\sqrt2\) to \(2^{1/2-1/(2n)}\) in the closed network and from \(2\) to \(2^{1-1/(2n)}\) in the open one.
The bound uses neither the source states nor their independence, so it applies to any star-like inequality with only two correlators and central nodes without input. Tavakoli and coauthors’ star inequality with \(2^{n-1}\) correlators, one per GHZ stabilizer with an even number of \(\sigma_2\) factors, avoids it and reaches \(\sqrt2\) for every \(n\).
How to reuse it
Before designing a network test with fixed central measurements, check whether the observables each correlator heralds commute. If they anticommute, no choice of state helps, and the test needs more correlators or an input at the central node. The same count bounds any protocol that must certify two complementary properties from a single fixed measurement.
Context & limitations
Why it matters. The same source looks more or less nonlocal depending only on how its particles are routed, and the paper identifies why: an input-free node reads all parities of a Bell-state measurement in every run, which beats the setting-switching of a Svetlichny test but loses to the free choice of directions in a Mermin test. The parity theorem also caps existing trilocal inequalities for three tripartite sources at \(2^{1/3}\).
Limitations. The closed and open values are computed for Bell-state measurements at the central nodes. For W sources, product measurements do better numerically (1.316 against 1.202), and the optimal general strategy is unknown. A violation certifies neither genuine tripartite entanglement nor genuine tripartite nonlocality of the sources, and the closed-network parity bound assumes qubit end nodes with projective measurements.