Past exam of the computer science course of the University of Oxford / 2023 / Quantum information / 1 / b

Expand the density matrix of the Bell state and take the partial trace over Alice's qubit:The orthonormality of the computational basis removes the off-diagonal terms. Thus Bob's reduced density matrix is the maximally mixed state

Write the proposed quantum state ensemble as $\{(|u⟩,p),(|v⟩,1-p)\}$ with normalized pure states, so thatTaking the expectation values in $|u⟩$ and $|v⟩$, and setting $c=|⟨u|v⟩{|}^2$, givesConsequently $p\le 1/2$ and $p\ge 1/2$. Substitution then gives $c=0$. The two pure states are therefore orthogonal, with equal probabilities:Conversely, any orthonormal basis of a qubit has projectors summing to $I$, so such a quantum state ensemble does give $I/2$. This proves the characterization of two-state ensembles of a maximally mixed qubit.

Alice uses conjugate-basis steering of a Bell pair. Write the desired orthonormal basis for Bob as $|\psi ⟩=a|0⟩+b|1⟩$ and $|{\psi }^{\perp }⟩=c|0⟩+d|1⟩$. Entrywise complex conjugation in the computational basis gives Alice's orthonormal basisFor either $|u⟩$ in Bob's chosen orthonormal basis, the corresponding unnormalized conditional quantum state isIts squared norm is $1/2$. Thus Alice measures in the complex-conjugate basis, obtaining the required Bob pure states with probability $1/2$ each. Averaging over Alice's undisclosed outcome leaves Bob's density matrix equal to $I/2$, consistent with the no-communication theorem.