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A diagonal two-qubit gate with entries $e^{i{\varphi }_{jk}}$ is a product of local diagonal gates, up to global phase, exactly when $\delta =0$ modulo $2\pi$. This follows by factoring the two-by-two array of phases; its determinant vanishes precisely in that case. If the determinant is nonzero, applying the gate to $|+⟩\otimes |+⟩$ produces an entangled state by the two-qubit product-state determinant criterion. In particular $\mathrm{diag}(-1,1,1,-1)=-Z\otimes Z$ is local, whereas $\mathrm{diag}(1,1,1,-1)$ is the entangling Controlled-Z gate.

Consider one normalized measurement vector $|u⟩=a|0⟩+b|1⟩$ on C. The unnormalized partial inner product on AB isIts tensor-product coefficient matrix and determinant areThe two-qubit product-state determinant criterion therefore says that this branch is entangled whenever $a\ne 0$. Its probability is $(|b{|}^2+2|a{|}^2)/3>0$, so such a branch is physically possible and can be normalized without changing whether its determinant vanishes.

For an orthonormal basis $\{|u_0⟩,|u_1⟩\}$, completeness gives $|⟨0|u_0⟩{|}^2+|⟨0|u_1⟩{|}^2=1$. At least one vector has $a\ne 0$. Hence at least one nonzero-probability outcome leaves AB entangled, proving conditional entanglement of the three-qubit W state. Some bases give two entangled branches; the computational basis gives one entangled and one product branch. The claim is not that every outcome must entangle AB.

Taking partial inner products with $⟨0|$ and $⟨1|$ on the first wire gives $(|0⟩+|1⟩)/2$ and $(-|0⟩+|1⟩)/2$. They are nonzero and orthogonal, hence linearly independent. This contradicts the necessary product-state property proved in part (a), so the second state is entangled.

Equivalently, its tensor-product coefficient matrix is $C=\frac{1}{2}\left(\begin{array}{cc}1& 1\\ -1& 1\end{array}\right)$, with $\det C=1/2\ne 0$. The two-qubit product-state determinant criterion gives the same conclusion. In fact $C{C}^{†}=I/2$, so its Schmidt coefficients are both $1/\sqrt{2}$: it is maximally entangled.

Grouping by the first qubit givesThe vectors on both sides form orthonormal bases, so this is a Schmidt decomposition with two nonzero Schmidt coefficients $1/\sqrt{2}$. Hence the state is maximally entangled. Independently, its coefficient matrix has determinantwhich excludes a product state by the two-qubit product-state determinant criterion. Each reduced density matrix is $I/2$.

A pure bipartite state is entangled when it is not a product state $|u{⟩}_A\otimes |v{⟩}_B$.

Sincewe haveIts coefficient matrix is proportional towhose determinant is $-4\ne 0$. By the two-qubit product-state determinant criterion, a product state would have determinant zero. Therefore $|\alpha ⟩$ is entangled.

A quantum state on the tensor product of two two-dimensional complex state spaces. A normalized pure state has squared coefficient moduli summing to one. It is a product state exactly when $\alpha \delta =\beta \gamma$, by the two-qubit product-state determinant criterion. The weaker equality of moduli only characterizes independence of the computational-basis outcomes; phases can preserve quantum entanglement despite those independent statistics.