Past exam of the computer science course of the University of Oxford / 2022 / Quantum information (A16608W1)

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Solutions to all three questions.^[ref] System labels specify the order of each tensor product of quantum systems. In an operator product the rightmost operator acts first. In Question 3(d), the stated fidelity is the squared quantum fidelity of the full two-qubit output with the desired two-copy pure state.

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.

For normalized vector representatives in the same Hilbert space, two pure states are $\epsilon$-close in pure-state norm distance whenHere $\Vert \cdot {\Vert }_2$ is the Euclidean norm of the coefficient vector in any orthonormal basis. Equivalently, $2-2\mathrm{Re}⟨\psi |\varphi ⟩\le {\epsilon }^2$. If the convention identifies vector representatives differing by a global phase, first minimize over that phase, giving ${\min }_{\gamma }\Vert |\psi ⟩-e^{i\gamma }|\varphi ⟩{\Vert }_2=\sqrt{2-2|⟨\psi |\varphi ⟩|}$; the vector-norm convention above specifies the representatives.

Two unitary transformations $U,V$ on the same Hilbert space are $\epsilon$-close in operator-norm distance between unitary gates whenThus their outputs are $\epsilon$-close in pure-state norm distance for every normalized input. This is an operator norm condition, stronger than checking only one chosen input. If global phase is disregarded for the whole operation, use ${\min }_{\gamma }\Vert U-e^{i\gamma }V{\Vert }_{\mathrm{op}}$ instead, with one phase chosen for the operator rather than separately for each input.

An entangling gate is a unitary operator on two quantum systems for which some pure product input is mapped to an entangled state:It need not create quantum entanglement from every product state. For example, a controlled-NOT gate entangles $|+⟩|0⟩$ but leaves $|00⟩$ a product state.

The gate set is universal. Since the Hadamard gate satisfies $H^{†}=H$ and $H^2=I$, the supplied $U$ is a unitary operator, andTwo applications of $U$ therefore realize a Hadamard gate up to a global phase. Together with the supplied controlled-NOT gate and T gate, this simulates the given universal quantum gate set.

An exact matrix representative of $H$ can also be synthesized if desired. Put $A=U^2=-iH$ and $S=T^2$. Direct multiplication gives $(AS{)}^6=-iI$, henceThis finite product uses only $U$ and $T$, so the conclusion does not rely on discarding a global phase.

The gate set is not universal. The displayed two-qubit matrix is the SWAP gate:Both $V$ and the Hadamard gate act on one qubit. Thus every gate preserves a fully factored product state, and induction gives the same property for every quantum circuit built from them. This is the product-state obstruction to SWAP universality.

For a quantitative obstruction to even approximate universality, every two-qubit product state has squared overlap at most $1/2$ with a Bell state. Indeed, for normalized coefficient vectors $u,v$,by the Cauchy-Schwarz inequality. A controlled-NOT gate acting on $|+⟩|0⟩$ would instead produce that Bell state, so these circuits cannot approximate all unitary transformations. The fact that $V^2=e^{-i\pi /8}T$ supplies a T gate up to global phase does not supply an entangling gate.

Use the entanglement-assisted promised equality protocol. Alice applies the phase oracle $O_x|i⟩=(-1{)}^{x_i}|i⟩$ on $A$, and Bob applies $O_y|i⟩=(-1{)}^{y_i}|i⟩$ on $B$. These are unitary operators because their diagonal entries have modulus one. After Alice sends $A$ to Bob, the joint pure state isLet $|\Omega ⟩=n^{-1/2}{\sum }_i|ii⟩$ be the original maximally entangled state. If $d=d_H(x,y)$ is the Hamming distance, thenBob performs the projective measurement with projectors $P=|\Omega ⟩⟨\Omega |$ and $I-P$. Under the promise,He declares equal strings for the first outcome and distance $n/2$ for the other. This is exact because the promised outputs occupy orthogonal subspaces. A basic measurement also suffices: extend $|\Omega ⟩$ to an orthonormal basis and group all its other outcomes together. Alice transmits one $n$-dimensional system, realizable using $\lceil {\log }_{2}n\rceil$ qubits; $n$ need not be a power of two.

Use Hadamard conjugation of a controlled gate to obtain the Controlled-Z gate:Conjugating by Pauli X gates on both qubits moves its negative entry from $|11⟩$ to $|00⟩$:Finally $(H\otimes H)|00⟩=|+⟩|+⟩=|e_0⟩$, so a two-qubit diffusion circuit isIn temporal order, apply Hadamard gates on both qubits, Pauli X gates on both, a Hadamard gate on the second, a controlled-NOT gate $1\to 2$, a Hadamard gate on the second, Pauli X gates on both, then Hadamard gates on both. Every gate is permitted. The overall minus sign is the allowed global phase, so this implements the required inversion about the mean for Grover search algorithm.

The two conditions on a density matrix are positive semidefiniteness and unit trace:Here positive semidefinite includes being a Hermitian matrix, with $v^{†}\rho v\ge 0$ for every vector $v$. Equivalently, a density matrix is Hermitian with nonnegative eigenvalues summing to one.

The Pauli Z gate gives a rotation through $\pi$ about the $z$ axis of the Bloch sphere. In Bloch vector coordinates,For example, this follows from $ZXZ=-X$, $ZYZ=-Y$ and $ZZZ=Z$ in the Pauli matrix expansion $\rho =(I+r_{x}X+r_{y}Y+r_{z}Z)/2$.

Since $H=(X+Z)/\sqrt{2}$, the Hadamard gate is, up to a global phase, the rotation gateThus its Bloch sphere action is a rotation through $\pi$ about the axis $(1,0,1)/\sqrt{2}$, halfway between the positive $x$ and $z$ axes:The identities $HXH=Z$, $HYH=-Y$ and $HZH=X$ verify this Bloch vector transformation directly.

The state is separable. The identity matrix resolves into the four projectors of the computational basis, giving the explicit quantum state ensembleEach member is a pure state that is a product state, so this is a separable quantum state by its defining decomposition.

The state is separable. The off-diagonal terms of the two Bell state projectors have opposite signs and cancel:This is a quantum state ensemble of two pure states that are product states, with probability $1/2$ each, establishing a separable quantum state. Its correlations are classical; an ensemble decomposition into entangled states does not itself establish quantum entanglement of the density matrix.

For input $|0{⟩}_S$, the controlled-NOT gate fixes $|00{⟩}_{SC}$, which is the ideal two-copy pure state. Thus its global cloning fidelity is $F_0=1$. For input $|+{⟩}_S$, the actual output is the Bell state $|{\varphi }_{+}⟩$, whereas the desired output is $|++⟩$. Their overlap isWith the given equal input probabilities,These values use the specified squared quantum fidelity for the full output, rather than a one-copy marginal fidelity.

Assume perfect cloning of both specified pure states. Allowing irrelevant output global phases, this would requireA unitary operator preserves inner products, and therefore their magnitudes. But the input overlap and required output overlap have different magnitudes:This contradiction proves that no such two-qubit unitary exists, without assuming the no-cloning theorem.

Construct an economical two-state quantum cloner by moving both actual outputs slightly toward one another relative to the two desired copies. Set $|a⟩=|00⟩$ and $|b⟩=|++⟩$. Their overlap is $1/2$, so their sum and difference yield the orthonormal vectorsThe desired outputs have the symmetric expressionsChoose the actual outputs to beThey are normalized and satisfyConsequently the prescription $|00⟩\mapsto |c_0⟩$, $|+0⟩\mapsto |c_{+}⟩$ preserves the Gram matrix of the two inputs and defines a linear isometry on their span. The unitary extension of a finite-dimensional isometry gives a unitary operator on the same four-dimensional, two-qubit Hilbert space; no extra workspace is required.

For an explicit full-space construction, define $c=\cos (\pi /8)$, $s=\sin (\pi /8)$ and the orthonormal basisChoose two further orthonormal vectors perpendicular to $u,v$,Thenmaps one orthonormal basis to another and is therefore unitary. Since $|00⟩=c|p⟩+s|m⟩$ and $|+0⟩=c|p⟩-s|m⟩$, it has exactly the required two outputs.

Finally,Both inputs therefore have the same global cloning fidelity, and so does their average:This proves the requested existence with a two-qubit unitary gate.