 Solution

Read the source network as a Hadamard test: a Hadamard gate on the first qubit, a controlled unitary gate with the first qubit as control, and another Hadamard gate on that control. Starting from $|0⟩|\varphi ⟩$, the successive quantum states areApply b to the last line. Since $U^{†}U=I$ by unitarity,For the Pauli Z gate observable on the first qubit, the two eigenvalues are $+1$ and $-1$. ThusAs a check, $U=I$ returns outcome zero certainly, while $U=-I$ returns one certainly. The controlled unitary gate makes this relative phase measurable even though multiplying an isolated system's quantum state by $-1$ is only a global phase.