Past exam of the computer science course of the University of Oxford / 2022 / Quantum information / 2

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.