= Solution
{c}
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 $\phi=\delta T$. During the first pulse the <Bloch vector> is $(0,-\sin\theta,\cos\theta)$ for $0\leq\theta\leq\pi/2$. During free evolution,
$$
\mathbf r(t)=(-\sin\delta t,-\cos\delta t,0),\qquad 0\leq t\leq T,
$$
because the <Hamiltonian operator> generates $R_z(-\delta t)$. During the final pulse it is
$$
\mathbf r_{\rm last}(\theta)=(-\sin\phi,-\cos\phi\cos\theta,-\cos\phi\sin\theta),
\quad 0\leq\theta\leq\pi/2.
$$
These 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>:
$$
\begin{aligned}
U_R&=P e^{i\phi\sigma_z/2}P\\
&=-i\left[\cos(\phi/2)\sigma_x-\sin(\phi/2)\sigma_z\right].
\end{aligned}
$$
It is proportional to the <Hadamard gate> when $\cos(\phi/2)=-\sin(\phi/2)$, hence the shortest positive waiting time is
$$
\boxed{T_H=\frac{3\pi}{2\delta},\qquad U_R(T_H)=iH.}
$$
More generally $\delta T=-\pi/2\pmod{2\pi}$. With the alternative conventional ket definition $H_{01}$ specified in part (b)(i), the required matrix is $(\sigma_x-\sigma_z)/\sqrt2$, so $\boxed{T_{H_{01}}=\pi/(2\delta)}$ and $U_R=-iH_{01}$. 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 $r_z=-\cos\phi$, so the actual return population is
$$
\boxed{\rho_{11}(T)=\frac{1-\cos(\delta T)}2=\sin^2(\delta T/2).}
$$
\b[The printed plus sign is inconsistent with the two identical pulses.] At $T=0$, $U_R=P^2=-i\sigma_x$, so $|1\rangle$ becomes $|0\rangle$ 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.
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