Past exam of the computer science course of the University of Oxford / 2022 / Quantum information / 3

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.