Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 15D (Quantum Information and Computation)

Apply the gates from left to right. The first Hadamard gate and the first controlled-NOT gate produce the Bell stateThe parallel $Z\otimes H$ gates then produceThe final controlled-NOT has the lower qubit as control; it interchanges $|01⟩$ and $|11⟩$, whose amplitudes here are equal. Thus the output isBy quantum measurement in the computational basis, every basis outcome has probability equal to the squared modulus of its amplitude:

Use the inverse quantum circuit rule. Every gate in $C$ is self-inverse, so reverse their order. From left to right, the circuit for $V^{-1}$ is:

1. a controlled-NOT with the lower qubit controlling the upper;
2. $Z$ on the upper qubit and $H$ on the lower qubit;
3. a controlled-NOT with the upper qubit controlling the lower;
4. $H$ on the upper qubit.

This is the requested circuit description, with the two wires and control directions unchanged under each individual gate's adjoint.

Use three controlled-NOT gates with control-target directionsOn computational-basis bits they act asThey therefore implement the Three-CNOT decomposition of the SWAP gate, and linearity proves the identity on every two-qubit state.

After the first Hadamard gate and the controlled unitary gate, the state isThe final Hadamard gate givesHence the Hadamard test returns one with probabilityIf $U|\psi ⟩=|\psi ⟩$, then $\theta =0$ and the outcome is zero with certainty. If $U|\psi ⟩=-|\psi ⟩$, then $\theta =\pi$ modulo $2\pi$ and the outcome is one with certainty. The single measurement therefore distinguishes the two promised cases exactly.