Past exam of the laser science and quantum information processing course of the University of Oxford / 2018 / 4 / b / Solution

Since ${\sigma }_{\varphi }^2=I$, the rotation gate isFor two chronological $\pi$ pulses, later operations multiply on the left:Thus equatorial pulses generate arbitrary rotations about the $z$-axis. The minus sign is an overall global phase.

For the other chronological sequence, $T^4=Z$, andThis equality has no additional overall phase under the standard definitions of $H$ and $T$. Rotations about two nonparallel axes, such as $x$ and $y$, generate all single-qubit operations by an Euler angles decomposition. Equivalently, the available equatorial rotations include $R_y(\pi /2)$ and arbitrary $R_z$ rotations constructed above; for example $H=R_y(\pi /2)Z$ and $T=e^{i\pi /8}{R}_z(\pi /4)$. Hence these rotations provide a universal single-qubit set. Overall phases do not affect any probability or isolated-register observable; a phase controlled by an additional qubit would instead be relative and could not be discarded without accounting for it.

The minimum number of distinct fixed rotation gates for approximate universality is two. One fixed gate and its powers generate a commuting subgroup and cannot approximate all noncommuting single-qubit operations. Two suitable equatorial gates suffice: for example $R_x(\pi /\sqrt{2})$ and $R_y(\pi /\sqrt{3})$. Each has an angle irrational relative to $2\pi$, so its powers densely approximate rotations about its axis; combining the two supplies the Euler decomposition. Noncommutation alone is insufficient: two Clifford quarter-turns can generate only a finite subgroup. If continuous angles are regarded as available controls, two nonparallel rotation axes likewise suffice, now for exact Euler synthesis.