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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.

By the spectral theorem, the nonscalar unitary V has orthonormal eigenvectors $|v_0⟩,|v_1⟩$ with distinct eigenvalues ${\lambda }_0\ne {\lambda }_1$. Choose $|b⟩=(|v_0⟩+|v_1⟩)/\sqrt{2}$. Then $V|b⟩=({\lambda }_0|v_0⟩+{\lambda }_1|v_1⟩)/\sqrt{2}$ is not proportional to $|b⟩$.

On the product input $|+⟩\otimes |b⟩$, the controlled unitary gate producesIts two conditional target vectors are linearly independent, so the conditional-vector product-state criterion proves that the output is entangled. Equivalently, in the control computational basis and target eigenbasis its coefficient matrix is $\frac{1}{2}\left(\begin{array}{cc}1& 1\\ {\lambda }_0& {\lambda }_1\end{array}\right)$, whose determinant is $({\lambda }_1-{\lambda }_0)/4\ne 0$. Thus U is an entangling gate. The excluded scalar case would only apply a phase gate to the control and could not entangle the factors; this is the controlled-unitary entangling criterion.

No: $U$ is not an entangling gate. Its factorization gives, for every product input,which remains a product state. More generally, a convex mixture of product density matrices remains separable under these local unitary operators. Thus being distinguishable from $I\otimes V$ does not require creating entanglement. The first qubit undergoes a genuine local phase gate, while the target undergoes the same $V$ in both branches.

An entangling gate is a unitary operator on two quantum systems for which some pure product input is mapped to an entangled state:It need not create quantum entanglement from every product state. For example, a controlled-NOT gate entangles $|+⟩|0⟩$ but leaves $|00⟩$ a product state.

The gate set is not universal. The displayed two-qubit matrix is the SWAP gate:Both $V$ and the Hadamard gate act on one qubit. Thus every gate preserves a fully factored product state, and induction gives the same property for every quantum circuit built from them. This is the product-state obstruction to SWAP universality.

For a quantitative obstruction to even approximate universality, every two-qubit product state has squared overlap at most $1/2$ with a Bell state. Indeed, for normalized coefficient vectors $u,v$,by the Cauchy-Schwarz inequality. A controlled-NOT gate acting on $|+⟩|0⟩$ would instead produce that Bell state, so these circuits cannot approximate all unitary transformations. The fact that $V^2=e^{-i\pi /8}T$ supplies a T gate up to global phase does not supply an entangling gate.

An entangling gate is a unitary transformation $U$ on two quantum systems for which at least one pure product input has an entangled output:It need not create quantum entanglement on every product state. For example, the controlled-NOT gate leaves $|00⟩$ a product state but maps $|+⟩|0⟩$ to a Bell state, so it is an entangling gate.