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

Using the printed measurement-vector convention, the unitary operator isThus it is a phase gate followed by a Hadamard gate, with angle sign opposite to the parameter in the catalogued J gate. Its rows are the conjugate transposes of the orthonormal measurement vectors, so it is unitary. Right multiplication by the Pauli Z gate changes the sign of the second column, which is exactly the effect of left multiplication by the Pauli X gate, exchanging the two rows. Explicitly,Similarly, right multiplication by the Pauli X gate exchanges columns, givingTherefore the exact identities, retaining the global phase in the second one, areThese supply Pauli-frame propagation along a measurement wire.

Write the normalized input quantum state as $|\psi ⟩=a|0⟩+b|1⟩$. The fresh quantum ancilla is $|+⟩$, so the Controlled-Z gate createsApplying the measurement bra $⟨v_k(\theta )|$ to the first qubit leaves the unnormalized second-qubit vectorSince $|+⟩,|-⟩$ form an orthonormal basis,Also $U(\theta )|\psi ⟩=a|+⟩+e^{-i\theta }b|-⟩$, while the Pauli X gate fixes $|+⟩$ and changes the sign of $|-⟩$. Dividing by the branch norm therefore givesThis is one-bit teleportation with the present positive-exponent measurement vectors; the transferred logical gate is $U(\theta )=J(-\theta )$.

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.

Let $\varphi =j\pi /2$ for an integer $j$, and put $q=j\mathrm{mod}2$. Use the fixed phase-measurement angles $(\theta ,0,-\varphi )$ and a fixed final computational basis measurement. Denote their outcomes by $k,l,m,s$. The adaptive construction would instead use $-\varphi$ when $l=0$ and $+\varphi$ when $l=1$.

For the printed measurement vectors,Therefore $+\varphi$ and $-\varphi$ define the same unordered pair of rank-one linear projections; the labels are exchanged if and only if $q=1$. In the branch $l=1$, the outcome that would be called $m_{\mathrm{ad}}$ in the adaptive experiment is thus $m\oplus q$. In both branches,The output correction from the preceding part becomesIf $j$ is even this reduces to $s\oplus k\oplus m$; if $j$ is odd an additional $l$ is included. This equatorial measurement relabeling at Clifford angles changes only the classical interpretation, not any physical measurement basis.

All four fixed single-qubit projective measurements act on different vertices, so their operators commute. They can be performed simultaneously, with the displayed classical correction applied afterward. The first angle $\theta$ may be arbitrary: it was never adaptive. Only the sign choice for the third angle needed to be replaced. Thus the simulation is nonadaptive for every $\varphi \in (\pi /2)\mathbb{Z}$.