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

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.