Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 49 / 2

The Pauli Z gate exchanges the two Hadamard basis states: $Z|+⟩=|-⟩$ and $Z|-⟩=|+⟩$. Multiplying the displayed gate by the Pauli X gate givesOn the other hand,The two operators agree, provingAlso $U(\theta )=H\mathrm{diag}(1,e^{-i\theta })=J(-\theta )$ in the J gate convention. Keeping this angle sign explicit avoids confusing the two conventions for equatorial qubit measurements.

Prepare the three-vertex graph-state wire: start all three qubits in $|+⟩$ and apply Controlled-Z gates on edges $1-2$ and $2-3$. In this paper's equatorial qubit measurement convention the basis vectors can be chosen asThe given identity means that measuring an input qubit at angle $\theta$, with outcome $r$, teleports its state to the next qubit as $X^{r}U(\theta )|\psi ⟩$. Each outcome has probability $1/2$, since $X^{r}U(\theta )$ is unitary. This is one-bit teleportation with a known Pauli frame.

Measure vertex 1 at angle $\alpha$ and record $r$. Then measure vertex 2 at angle $\gamma =(-1{)}^r\beta$, taken modulo $2\pi$, and record $s$. The controlled-Z on edge $2-3$ commutes with the first measurement, so the two teleportation steps can be applied successively. The remaining normalized state isFor $r=0$ this is $X^{s}U(\beta )U(\alpha )|+⟩$. For $r=1$, the relation proved above gives $U(-\beta )X=e^{i\beta }ZU(\beta )$. Thus in every branch the remaining state isThe scalar is a global phase and has no effect on probabilities.

Finally measure vertex 3 in the computational basis, obtaining $z$, and reportThe $Z^r$ byproduct changes only computational-basis phases; $X^s$ changes the output label by $s$. Hence the corrected bit has exactly the ideal circuit's probability distribution in every branch, not merely on average. The procedure uses only single-qubit measurements after preparing the graph state, adaptive classical angle selection, and deterministic classical output processing.

First expand the result of the first gate in the computational basis:Using $1+e^{-i\alpha }=2e^{-i\alpha /2}\cos (\alpha /2)$ and $1-e^{-i\alpha }=2i{e}^{-i\alpha /2}\sin (\alpha /2)$, this becomesApply $U(\beta )|0⟩=|+⟩$ and $U(\beta )|1⟩=e^{-i\beta }|-⟩$ to obtainFor example, its quantum measurement in the computational basis has probabilities $\mathbb{P}(k=0)=(1+\sin \alpha \sin \beta )/2$ and $\mathbb{P}(k=1)=(1-\sin \alpha \sin \beta )/2$, obtained by squaring the two basis amplitudes.

Use two-bit remote state preparation on a graph-state path. Alice performs the first two measurements from the preceding measurement-based quantum computation: angle $\alpha$ on her first qubit, giving $r$, then angle $(-1{)}^r\beta$ on her second, giving $s$. She sends the pair $(r,s)$ to Bob, using exactly two classical communication bits.

The branch calculation already gives Bob's state as $e^{ir\beta }{X}^s{Z}^r|\chi ⟩$, where $|\chi ⟩=U(\beta )U(\alpha )|+⟩$. Bob applies $X^s$ first and $Z^r$ second, so his correction operator is $Z^r{X}^s$. Since $X^2=Z^2=I$,ThusThe final state is exactly the desired state up to the irrelevant global phase. No postselection is used: all four measurement branches have probability $1/4$ and are corrected. Bob does not need the angles themselves; Alice uses them in her measurement choices, while the two bits specify his Pauli correction.