Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 51 / 3 / c / Solution

Label the chain cluster state from top to bottom by $1,2,3,4$, and let $k,l,m$ be the phase-measurement outcomes on the first three vertices, with raw computational basis outcome $s$ on vertex four. Repeated one-bit teleportation leaves the last qubit, up to global phase, inThis sequential description applies even though the whole graph state is prepared first: a later Controlled-Z gate acts only on unmeasured qubits and commutes with earlier single-qubit measurement operators. It can therefore be moved past those measurements for the purpose of calculating each conditional branch.

Take $\alpha =\theta$ and $\beta =0$, so $U(0)=H$. After the second measurement the logical state isFor binary $p,q$, the identities proved above imply the Pauli-frame propagation along a measurement wire rulewhere $\simeq$ means equality up to global phase. Choose the third angle adaptively as $\gamma =(-1{)}^{l\oplus 1}\varphi$. Then $(-1{)}^l\gamma =-\varphi$, andReading the target circuit in temporal order gives its premeasurement stateConsequentlyThe Pauli Z gate changes only phases in the computational basis, while the Pauli X gate exchanges its two measurement outcomes. Thus the explicit choices areIndeed, for every earlier outcome branch,Hence averaging over the earlier random outcomes leaves exactly the target probability distribution. This implements the desired measurement-based quantum computation without physically removing the final Pauli frame.