Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 49 / 2 / a / Solution

The Pauli Z gate exchanges the two Hadamard basis states: $Z|+⟩=|-⟩$ and $Z|-⟩=|+⟩$. Multiplying the displayed gate by the Pauli X gate givesOn the other hand,The two operators agree, provingAlso $U(\theta )=H\mathrm{diag}(1,e^{-i\theta })=J(-\theta )$ in the J gate convention. Keeping this angle sign explicit avoids confusing the two conventions for equatorial qubit measurements.