Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 324 / 4

The basic one-bit teleportation primitive uses the input as qubit 1 and a fresh $|+⟩$ quantum ancilla as qubit 2. Apply the Controlled-Z gate $E_{12}$ and measure qubit 1 in the equatorial qubit measurement basis with angle $\alpha$. If $s\in \{0,1\}$ is the result, the normalized post-measurement state on qubit 2 isBoth results have probability one half. For example, writing $|\psi ⟩=a|0⟩+b|1⟩$, the unnormalized branch is $(a|+⟩+(-1{)}^s{e}^{i\alpha }b|-⟩)/\sqrt{2}=X^{s}J(\alpha )|\psi ⟩/\sqrt{2}$. Thus $J(\alpha )$ is applied deterministically as a logical gate with known Pauli frame $X^s$; a permitted conditional Pauli X gate would remove this byproduct physically.

The literal resource restriction does not need an unstated direct Pauli X gate. A Pauli Z gate can be enacted by applying $E$ to the data and a fixed $|1⟩$ quantum ancilla. Two consecutive angle-zero one-bit teleportations, with results $p,q$, map an arbitrary current state $|\chi ⟩$ to $X^{q}H{X}^{p}H|\chi ⟩=X^q{Z}^p|\chi ⟩$. Apply the available $Z^p$ to remove the latter factor, up to an irrelevant global phase. This implements a heralded $X^q$: if $q=0$, the input is unchanged; if $q=1$, the required Pauli X gate has been applied.

If the original result is $s=1$ and a physically corrected output is required, repeat this heralded Pauli X correction using controlled-Z and measurements until $q=1$. Each attempt succeeds with probability one half independent of the input, so it terminates with probability one, using two attempts on average. Only $E$, fixed ancillary quantum states and single-qubit quantum measurements are used. This exact physical correction has no finite worst-case measurement bound; the standard finite deterministic realization is the logical Pauli frame version, which suffices for the later output-simulation parts.

Label the measured vertex of the four-cycle by $0$, its two neighbours by $1,3$, and the opposite vertex by $2$. The edges are $01,12,23,30$. By the definition of a graph state,All Controlled-Z gates commute. Conditioning on the computational basis value $r$ of vertex $0$, the factors on edges $01$ and $03$ act on their other endpoints as $Z_1^r$ and $Z_3^r$, since $E|r,a⟩=(-1{)}^{ra}|r,a⟩$. The remaining edges form the three-vertex path $1-2-3$:The Born rule gives probability one half for either result. Normalizing provesThis is computational-basis measurement of a graph-state vertex: remove the measured vertex and apply a known Pauli Z gate byproduct to each neighbour. The chosen labels specify precisely which factors occur; any vertex of the square can serve as vertex $0$ by relabelling.

The path graph state is $|{\psi }_3⟩=E_{23}{E}_{12}|+{⟩}_1|+{⟩}_2|+{⟩}_3$. The factor $Z_3^r$ commutes with both the Controlled-Z gates and the quantum measurement of vertex $1$. Treat $Z_1^r|+{⟩}_1$ as the input of one-bit teleportation. The equatorial qubit measurement result $s$ therefore gives the normalized stateHere the subscript on $Z_2^r$ expresses the teleported input factor; it is not an extra operation on the already measured vertex. Use $J(\alpha )Z^r=X^{r}J(\alpha )$ to obtainThe exclusive or exponent is equivalent to the printed sum $r+s$ because $X^2=I$. The conditional probability of $s$ is one half for either $r$, by the one-bit teleportation branch norm. This calculation concerns an equatorial measurement of a graph-state leaf; no third remaining qubit or extra $|-⟩$ factor is present.

Let the ideal two-qubit output before final quantum measurements beThe PDF places $J(\alpha )$ only on the upper wire, followed by the Controlled-Z gate and a quantum measurement in the computational basis on each output wire. Measure vertex $0$ of the square in the computational basis, obtaining $r$, then vertex $1$ of the surviving path in the fixed equatorial qubit measurement basis at angle $\alpha$, obtaining $s$. Put $t=r\oplus s$. The previous part gives $E_{23}{X}_2^{t}J(\alpha {)}_2{Z}_3^r|++⟩$.

Commute its Pauli frame through the Controlled-Z gate. Since $E_{23}{X}_2^t=X_2^t{Z}_3^t{E}_{23}$ and $E_{23}$ commutes with $Z_3^r$, the actual output isFinally measure vertices $2,3$ in the computational basis, with raw results $d_2,d_3$. A Pauli X gate flips a computational basis result, whereas a Pauli Z gate changes only its phase. Thus the purely classical correction isThis four-cycle graph-state simulation of an entangle-and-measure circuit reproduces the full joint output distribution, not just each marginal. All measurement bases are fixed beforehand, and no physical byproduct correction is necessary. As a check, the ideal Controlled-Z gate is diagonal in the output basis, so $P(k,l)=\frac{1}{2}{\cos }^2(\alpha /2)$ for $k=0$ and $\frac{1}{2}{\sin }^2(\alpha /2)$ for $k=1$, independently of $l$. The measurement procedure gives exactly these probabilities after its classical relabelling.

The logical depth of a measurement pattern counts sequential layers of quantum measurements forced by dependence of measurement bases on earlier outcomes. Depth one means every basis can be fixed before measurements start, allowing all quantum measurements on distinct qubits to be performed in parallel. This does not require depth-one resource-state preparation or constant-depth classical parity processing.

Suppose the circuit contains $m$ gates $J({\alpha }_1),\dots ,J({\alpha }_m)$ in time order, with each ${\alpha }_j\in \{0,\pi /2\}$. Prepare an $(m+1)$-vertex path graph state from $|+{⟩}^{\otimes (m+1)}$ and Controlled-Z gates between neighbours. Its first qubit supplies the specified input $|+⟩$. Measure vertex $j-1$ in the fixed equatorial qubit measurement basis at angle ${\alpha }_j$ for $j=1,\dots ,m$, obtaining $s_j$. Measure the final vertex in the computational basis, obtaining $d$.

Track the Pauli frame as $X^{a_j}{Z}^{b_j}$, starting with $a_0=b_0=0$. A hypothetical sequential reading of the same quantum measurements gives the step $X^{s_j}J({\alpha }_j)X^{a_{j-1}}{Z}^{b_{j-1}}$. The supplied commutation rules imply, up to global phase,For ${\alpha }_j=0$, the sign is irrelevant. For ${\alpha }_j=\pi /2$, the identity $J(-\pi /2)=XJ(\pi /2)$ absorbs the possible sign change into the Pauli frame. If $t_j=0$ for ${\alpha }_j=0$ and $t_j=1$ for ${\alpha }_j=\pi /2$, the update isInductively the final unmeasured quantum state would be $X^{a_m}{Z}^{b_m}J({\alpha }_m)\cdots J({\alpha }_1)|+⟩$. The final computational basis outcome is corrected by $a_m$ and is unaffected by $b_m$. The induction on a sequential interpretation proves the joint statistics; projectors on distinct vertices commute, so the identical fixed-basis pattern can actually be measured simultaneously, including its final vertex.

The first printed commutation formula has an incorrect exact scalar phase: with the printed matrix definition, $J(\alpha )X=e^{i\alpha }ZJ(-\alpha )$, rather than the negative-exponent prefactor. For example, at $\alpha =\pi /4$ the two versions differ by a factor $i$. They agree up to global phase, so this error does not affect the Pauli frame recurrence or any quantum measurement probability.

Consequently a fixed-basis path-state pattern of logical depth one simulates every such circuit. Only Pauli measurements occur: the angle-zero basis measures $X$, and the angle-$\pi /2$ basis measures $-Y$ with the printed eigenvector labels. Classical exclusive or processing suffices for all output corrections. These gates are Clifford gates, which explains why angle adaptation can be eliminated. For $m=0$, simply measure the initial $|+⟩$ in the computational basis; it is already a depth-one pattern.