Past exam of the computer science course of the University of Oxford / 2021 / Quantum information / 2 / b

Put $D_{\theta }=\mathrm{diag}(1,e^{i\theta })$. Factoring the target operator giveswhereas $U^{\prime }=I\otimes V$. Their difference is a relative phase on the first qubit, not one overall global phase.

For an explicit witness, choose the product input $|+⟩\otimes |\chi ⟩$, with any normalized $|\chi ⟩$. The first-qubit outputs are respectively $(|0⟩+e^{i\theta }|1⟩)/\sqrt{2}$ and $|+⟩$. A Hadamard basis measurement returns $+$ with probabilitiesThese differ because $\theta$ is not a multiple of $2\pi$. The two unitary operators implement different quantum channels. The global phase of the conditional target operation has become a physically observable phase on its control.

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.