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

Factor the sum before making an entanglement claim:The state is not entangled: it is a product state. A sum written with several tensor-product terms need not be entangled; those terms may combine into a single tensor product.

The displayed computational basis expansion is already a Schmidt decomposition, with coefficients $\sqrt{1-p}$ and $\sqrt{p}$. For $0<p<1$ both are nonzero, so the Schmidt rank is two. A pure product state has Schmidt rank one. The state is therefore entangled throughout the specified open interval.

Equivalently, the reduced density matrix isThe reduced state is a mixed state, which is impossible for one factor of a pure product state. At the excluded endpoints the Schmidt rank drops to one, giving $|00⟩$ or $|11⟩$.

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.

Either experiment is possible. In the sequential experiment choose $U=X$. The first computational basis measurement on $I/2$ returns a uniform bit $j$ and prepares $|j⟩$. The Pauli X gate changes it to $|1-j⟩$, so the second result is always opposite. The two possible records have probabilities $\mathrm{Pr}(0,1)=\mathrm{Pr}(1,0)=1/2$.

In the two-qubit experiment choose the fixed pure stateThe Born rule gives the same two possible records, again with probability $1/2$ each. The observed computational basis anticorrelation alone does not identify whether the correlation was generated sequentially or shared between two qubits.

Either experiment is again possible. The sequential experiment with $U=I$ repeats its first measurement outcome in every basis: the first projective measurement has prepared one of that basis's eigenstates, and the identity leaves it unchanged.

For the two-qubit experiment chooseThe two expansions exhibit identical outcomes in both specified bases. Each allowed equal-label record has probability $1/2$. These are fixed choices of $U$ and $|\psi ⟩$, respectively; neither depends on which of the two bases Alice selects.

Only the two-qubit experiment is possible. Choose the spin singlet stateIt has eigenvalue $-1$ under each of $X\otimes X$, $Y\otimes Y$ and $Z\otimes Z$. Thus measuring either qubit in the same Pauli basis as the other always gives opposite eigenvalue labels. Each opposite-label pair occurs with probability $1/2$. More generally, collective-unitary covariance of the two-qubit singlet ensures opposite outcomes in any common orthonormal qubit basis.

For the sequential experiment, the first outcome is uniformly distributed and the Lüders rule prepares the corresponding eigenstate. Certain opposite outcomes in the computational basis force the fixed unitary matrix to have the formCertain opposite Hadamard basis outcomes require $⟨+|U|+⟩=(a+b)/2=0$, hence $b=-a$. Therefore $U=iaY$. But this operator preserves the two Pauli Y eigenstate rays: it gives the same Y-basis label on the second measurement, rather than the opposite one. This contradicts the third set of observations.

Equivalently, flipping all three bases would require $U^{†}XU=-X$, $U^{†}YU=-Y$, and $U^{†}ZU=-Z$. Conjugating $XY=iZ$ in these two ways gives both $iZ$ and $-iZ$, an impossibility. This three-axis obstruction to unitary qubit orthogonalization is a finite test of the obstruction behind the universal NOT gate.