Past exam of the computer science course of the University of Oxford / 2022 / Quantum information / 1

A basic measurement of a finite-dimensional quantum system is specified by an orthonormal basis $\{|b_j⟩{\}}_j$ of its Hilbert space. For a normalized input pure state $|\chi ⟩$, the Born rule givesThe post-measurement state is therefore $|b_j⟩$ up to a global phase. An outcome of zero probability has no conditional post-measurement state. For a general input density matrix $\rho$, the same basic measurement has $p_j=⟨b_j|\rho |b_j⟩$ and conditional density matrix $|b_j⟩⟨b_j|$ whenever $p_j>0$.

Use the Pauli X eigenstates $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$ and $\cos \theta +i\sin \theta =e^{i\theta }$. Their probability amplitudes areThe Born rule givesThe two probabilities sum to one. Their dependence on the relative quantum phase is the quantum interference between the two components in the computational basis.

If Alice's basic measurement returns a normalized vector $|u{⟩}_A$, form Bob's unnormalized conditional vectorThe Born rule and normalization giveThe joint post-measurement state is $|u{⟩}_A\otimes |{\chi }_u{⟩}_B$. For $|u{⟩}_A=\alpha |0{⟩}_A+\beta |1{⟩}_A$ with real coefficients and ${\alpha }^2+{\beta }^2=1$, contraction of the Bell state givesThus Bob's normalized conditional state is $\alpha |0{⟩}_B+\beta |1{⟩}_B$. For complex coefficients the contraction instead contains their complex conjugates; this is the same conjugation underlying conjugate-basis steering of a Bell pair.

Let $p_{ab}$ denote the conditional winning probability in the CHSH game. With both settings zero, computational-basis measurements of the Bell state always agree, so $p_{00}=1$. When just one setting is one, Bob's conditional pure state from part (i), or the corresponding calculation with Alice and Bob exchanged, gives agreement probability ${\cos }^2\theta$. Hence $p_{01}=p_{10}={\cos }^2\theta$.

For $a=b=1$, winning requires different output bits. Alice obtains each of her two outcomes with probability $1/2$. Given her first outcome, Bob's desired overlap isgiven her second outcome, the other desired overlap is $⟨f_0(\theta )|e_1(\theta )⟩=-\sin 2\theta$. Thus $p_{11}={\sin }^2(2\theta )$. Averaging over the four equally likely inputs givesThe maximum winning probability for noncommunicating classical players is $3/4$, the game form of the CHSH inequality. Setting $t=\cos 2\theta$, strict improvement requires $t-t^2>0$, or $0<t<1$. On the stipulated interval this givesThe endpoints give equality with the classical bound. Within this reflected real-basis CHSH strategy, the maximum is $13/16$ at $t=1/2$, namely $\theta =\pi /6$; this follows by completing the square in $t$.

Use entanglement-assisted discrimination of a Pauli basis. Define the known single-qubit unitary operator $V$ by $V|0⟩=|\psi ⟩$ and $V|1⟩=|{\psi }^{\perp }⟩$. Prepare the device input $A$ and a reference ancilla qubit $R$ inApply the unknown device once to $A$, then apply the known inverse $V^{†}$ to $A$. Sincethe possible two-qubit outputs areThese are the four mutually orthogonal Bell states. A Bell-basis measurement therefore identifies the device with certainty. Explicitly, apply a controlled-NOT gate $A\to R$, then a Hadamard gate on $A$, followed by computational-basis measurements. The readouts $00,10,01,11$ identify $U_0,U_1,U_2,U_3$, respectively. Thus one use of the device and one reference qubit suffice.