Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 324 / 1 / b

Put $P_1={\sigma }_1\otimes {\sigma }_2$ and $|\psi ⟩=|A{⟩}^{\otimes 2}$. After the first Hadamard gate and the two controlled Pauli gates, the joint state isThe final Hadamard gate changes this toConditioned on ancilla outcome $s\in \{0,1\}$, the normalized data state is therefore

Write the input in the two eigenspaces of $P_1$ aswhere the displayed eigenstates are normalized. A direct PBC measurement giveswith probabilities $|\alpha {|}^2$ and $|\beta {|}^2$. Hence the ancilla circuit and the Pauli measurement have identical outcome distributions and conditional data states after identifying the Pauli outcome with $(-1{)}^s$.

Use the standard ancilla-assisted Pauli measurement. To measure a Pauli $P$, reset the ancilla to $|0⟩$, apply $H$, apply the controlled version of every nonidentity factor of $P$ with the ancilla as control, apply $H$ again, and measure the ancilla in the computational basis. The preceding calculation shows that outcome $s$ projects the data with

First use this circuit with $P_1=Z\otimes I$, obtaining $s_1$. Since the measured ancilla is $|s_1⟩$, apply the classically controlled correction $X^{s_1}$ to reset it to $|0⟩$. Reuse it to measure $P_2=Z\otimes X$, obtaining $s_2$. All controlled Pauli gates and single-qubit corrections are Clifford gates. Since $P_1{P}_2=P_2{P}_1$, the final data state isThus one resettable ancilla implements both measurements of the PBC without disturbing the already measured Pauli eigenvalue.