 Solution

Keep the Pauli matrix and north-pole conventions of part (a), and neglect Rabi detuning during each short pulse. Write $P=R_x(\pi /2)$ and $\varphi =\delta T$. During the first pulse the Bloch vector is $(0,-\sin \theta ,\cos \theta )$ for $0\le \theta \le \pi /2$. During free evolution,because the Hamiltonian operator generates $R_z(-\delta t)$. During the final pulse it isThese formulas specify every arc in the two Bloch sphere sketches. For $T=\pi /(2\delta )$ the path is north $\to -y\to -x$, and the last pulse leaves $-x$ fixed. For $T=\pi /\delta$ the path is north $\to -y\to +y\to$ north.

To test the complete quantum gate, multiply its unitary operators:It is proportional to the Hadamard gate when $\cos (\varphi /2)=-\sin (\varphi /2)$, hence the shortest positive waiting time isMore generally $\delta T=-\pi /2(\mathrm{mod}2\pi )$. With the alternative conventional ket definition $H_{01}$ specified in part (b)(i), the required matrix is $({\sigma }_x-{\sigma }_z)/\sqrt{2}$, so $\boxed{T_{H_{01}}=\pi /(2\delta )}$ and $U_R=-i{H}_{01}$. The distinction concerns which basis order defines the named Hadamard gate; the evolution and the return population do not depend on that naming choice. Finally $r_z=-\cos \varphi$, so the actual return population isThe printed plus sign is inconsistent with the two identical pulses. At $T=0$, $U_R=P^2=-i{\sigma }_x$, so $|1⟩$ becomes $|0⟩$ and ${\rho }_{11}(0)=0$; the printed expression gives one. This counterexample is independent of Bloch sphere labeling. A plus sign would describe the complementary population ${\rho }_{00}$, or the return population with an opposite-sign final pulse. That changed sequence also changes the gate question and must not be silently substituted.