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For the Schmidt-basis Pauli correlation tensor, choose Alice's axes ${\mathbf{e}}_z,{\mathbf{e}}_x$ and Bob's $({\mathbf{e}}_z\pm s{\mathbf{e}}_x)/\sqrt{1+s^2}$. The CHSH inequality expression is $2\sqrt{1+s^2}$. It exceeds two exactly when both Schmidt coefficients are nonzero, reaching $2\sqrt{2}$ for maximal quantum entanglement. A product state gives $s=0$ and saturates rather than violates the inequality.

For a normalized pure state of two qubits with coefficient matrix $C$, the concurrence is $2|\det C|$. It ranges from zero for a product state to one for a maximally entangled state.

Every rank-one projective measurement of one qubit of the three-qubit W state has at least one nonzero-probability outcome whose other two qubits remain entangled. To see this, contract the measured factor against $⟨u|$, where $|u⟩=a|0⟩+b|1⟩$. The unnormalized tensor-product coefficient matrix of the other pair isAt least one vector in a complete measurement basis has $a\ne 0$, giving nonzero determinant and an entangled conditional branch. Its probability $(2|a{|}^2+|b{|}^2)/3$ is positive. This property survives local unitary changes of basis. The GHZ state lacks it because computational-basis measurement of one qubit leaves a product state on each branch, so the property also separates the states under local-unitary equivalence.

A nonzero bipartite pure state is a product state if and only if all its partial inner products on the first subsystem lie in a common one-dimensional vector subspace of the second. For a product $|a⟩\otimes |b⟩$, every contraction equals $⟨u|a⟩|b⟩$. Conversely, contract against each vector of an orthonormal basis to write $|\Psi ⟩={\sum }_i|i⟩\otimes |{\chi }_i⟩$. If $|{\chi }_i⟩=c_i|b⟩$ for a common nonzero $|b⟩$, then $|\Psi ⟩=({\sum }_i{c}_i|i⟩)\otimes |b⟩$. Zero conditional vectors are allowed. A nontrivial linear dependence relation requires multipliers not both zero; allowing both multipliers to vanish would give no test at all.

The connected correlation of two observables is $⟨AB{⟩}_c=⟨AB⟩-⟨A⟩⟨B⟩$. It vanishes in a product state when $A$ and $B$ act on disjoint factors.

For a qubit control and finite-dimensional target with unitary operator V, the controlled unitary gate $C(V)=|0⟩⟨0|\otimes I+|1⟩⟨1|\otimes V$ is entangling if and only if V is not scalar. If $V=e^{i\theta }I$, the controlled operation is a phase gate on the control tensored with the target identity and preserves every product state.

Otherwise the spectral theorem supplies two orthonormal eigenvectors with distinct eigenvalues ${\lambda }_0,{\lambda }_1$. Set $|b⟩=(|v_0⟩+|v_1⟩)/\sqrt{2}$ and use product input $|+⟩|b⟩$. The two conditional target vectors $|b⟩$ and $V|b⟩$ are linearly independent because their eigenbasis coefficients would otherwise require ${\lambda }_0={\lambda }_1$. The conditional-vector product-state criterion shows that their coherent controlled superposition is entangled. Thus even a target gate arbitrarily close to scalar can be entangling, although the resulting entanglement need not be large.

A bipartite density operator is entangled when it is not a separable quantum state. For a pure state, this is equivalent to not being a product state, or to either reduced density matrix having rank greater than one.

A unitary gate on two quantum systems is entangling if it maps at least one pure state that is a product state to an entangled state. The controlled-NOT gate qualifies because it maps $|+⟩|0⟩$ to a Bell state. A gate that merely exchanges the two systems need not be a local unitary operation, but cannot create quantum entanglement from a product state.

When two masses occupy spatial superpositions, their branch-dependent Newtonian gravitational potential energy produces relative phases. If those phases cannot be separated into one phase for each mass, the interaction turns an initial product state into an entangled state.

To compute a single-qubit output probability of a Clifford circuit, propagate its measured Pauli observable backward through the circuit. Every conjugation remains a tensor-product Pauli, whose expectation on a product state factors into efficiently computable one-qubit expectations.

The Holevo capacity of a quantum channel is the maximum one-use output Holevo quantity,It is the unassisted classical communication limit for product state encodings with collective output measurements. Arbitrary entangled inputs require the regularized expression in the Holevo-Schumacher-Westmoreland theorem.

Two bipartite pure states on fixed subsystem spaces are local-unitary equivalent when related by the displayed product of unitary operators. Identity operators give reflexivity, inverse operators give symmetry, and products of local operators give transitivity. Thus this is an equivalence relation, whose classes are the orbits under local unitaries.

Every pair of normalized product states is local-unitary equivalent: extend each factor to an orthonormal basis and use a unitary operator carrying one basis to the other. Local operators preserve a product factorization, and their inverses do too, so quantum entanglement cannot be created or removed this way. More generally, the tensor-product coefficient matrix changes as $C\mapsto UC{V}^T$, which preserves its singular values and therefore its Schmidt coefficients. The transpose on the second factor is ordinary, not a conjugate transpose, because coefficients of a tensor product depend bilinearly on the two input vectors.

Write a source and target in Schmidt decomposition with squared coefficients $(1-s,s)$ and $(1-t,t)$, where $0<s,t\le 1/2$. If $s\ge t$, Nielsen's pure-state conversion theorem allows deterministic conversion. If $s<t$, the Schmidt-tail entanglement monotone gives $p\le s/t$. Alice attains this bound with a local two-outcome measurement in quantum mechanics whose success Kraus operator is $K_s=\mathrm{diag}(\sqrt{s(1-t)/(t(1-s))},1)$ and whose failure operator is $K_f=\mathrm{diag}(\sqrt{1-s(1-t)/(t(1-s))},0)$. They obey $K_s^{†}{K}_s+K_f^{†}{K}_f=I$. The successful unnormalized state is $\sqrt{s/t}$ times the target, and failure leaves a product state. A classical message tells Bob which branch occurred. Known local changes of Schmidt bases allow the same protocol for arbitrary two-qubit pure states with these coefficients.

Use $|m_1,m_2⟩$ for a product of the two individual spin-one spin eigenstates, with magnetic quantum numbers $m_i=-1,0,1$. Total $S_z=0$ requires $m_1+m_2=0$, leaving preciselyHere these are product labels, not labels of the total angular momentum. Write their coefficients in this order as $\alpha ,\beta ,\gamma$. With physical spin ladder operators $S_{\pm }=S_{1\pm }+S_{2\pm }$, each nonzero individual step has amplitude $\sqrt{2}\hslash$, soBoth must vanish for a singlet state. The independent product states therefore require $\alpha =\gamma =-\beta$. Wavefunction normalization gives the spin-one singlet stateup to an overall quantum phase. Conversely, $S_z|\psi ⟩=S_{+}|\psi ⟩=0$ implies $S^2|\psi ⟩=0$ through $S^2=S_{-}{S}_{+}+S_z^2+\hslash S_z$, so these conditions really select total spin zero. Distinguishability imposes no additional exchange restriction. The ladder hint uses dimensionless angular momentum units; restoring its common factor $\hslash$ does not change the coefficient ratios.

A bipartite unitary operator is an entangling gate if it maps at least one product state to an entangled state. It need not entangle every product input. For example, the controlled-NOT gate entangles $|+⟩|0⟩$ but leaves $|00⟩$ as a product. A gate that preserves every product input is not entangling, even if it exchanges the two subsystems.

A normalized bipartite pure state is entangled when it cannot be written as a product state $|a{⟩}_A\otimes |b{⟩}_B$. Equivalently, its Schmidt rank exceeds one. This definition concerns the specified partition into systems A and B; the mixed-state definition of separability requires a different convex-mixture condition.

Suppose $|W⟩=(U_A\otimes U_B\otimes U_C)|GHZ⟩$. Measure C in the orthonormal basis $\{U_C|0⟩,U_C|1⟩\}$. Its two AB conditional vectors would be proportional torespectively. Both are product states, because measuring the original GHZ state's C factor in the computational basis selects $|00⟩$ or $|11⟩$. This contradicts part (i), which applies to every basic measurement basis of the W state's C factor. Therefore no local unitary transformation maps GHZ to W.

An additional invariant makes the distinction visible without measurement records. Across the partition AB versus C, the GHZ reduced density matrix on C has eigenvalues $1/2,1/2$, whereas the W state's are $2/3,1/3$. By local-unitary invariance of reduced-state spectra, these spectra cannot be changed by local unitaries. This supplies the same impossibility and emphasizes that the proposed transformation is a product of unitaries, not arbitrary postselected local processing.

The controlled-NOT gate is its own inverse, so ${\mathrm{CNOT}}^2=I$. Every remaining nontrivial gate acts on one wire. By the tensor product rule, any circuit on a product input stays a product state; in particular, starting at $|00⟩$ it cannot produce a Bell state.

This also prevents arbitrary approximation, not merely exact preparation. For normalized single-qubit vectors $|u⟩,|v⟩$,by the Cauchy-Schwarz inequality. Hence the quantum fidelity with $|{\Phi }^{+}⟩$ is bounded away from one for every reachable state. This gate set is not universal. The obstruction is preservation of product states, a special case of the existing product-state obstruction to SWAP universality.

Write the product state as $|\Psi ⟩=|a⟩\otimes |b⟩$. The two partial inner products areBoth vectors therefore lie in the same one-dimensional vector subspace, possibly with one or both equal to zero. If $(c_1,c_2)\ne (0,0)$, choose ${\lambda }_1=c_2$ and ${\lambda }_2=c_1$; then ${\lambda }_1|{\psi }_1^{\prime }⟩={\lambda }_2|{\psi }_2^{\prime }⟩$, and the multipliers are not both zero. If both $c_j$ vanish, choose $({\lambda }_1,{\lambda }_2)=(1,0)$.

Thus the conditional vectors are linearly dependent, with a nontrivial pair of multipliers in every case. The printed existential condition does not explicitly exclude $(0,0)$, which would make it vacuous for arbitrary vectors; the argument establishes the stronger intended statement. These unnormalized vectors need not themselves be valid unit state vectors, and a zero vector corresponds to a measurement outcome of probability zero. This is the forward direction of the conditional-vector product-state criterion.

Factor the coefficients:Therefore the first state is a product state, not entangled. Both factors have unit norm, so this factorization also checks normalization.