= Solution
The second pulse has the same <phase> and area as the first, so another $+x$ rotation by $\pi/2$ gives
$$
\mathbf R_3(T)=(\sin\delta T,0,-\cos\delta T),\qquad
\boxed{p_2(T)=\frac{1-R_{3z}}2=\frac{1+\cos\delta T}{2}.}
$$
At $T=2\pi/\delta$, the final vector is the south pole and all population is in $|2\rangle$. The complete path consists of a north-to-equator quarter turn, a full equatorial loop, and an equator-to-south quarter turn. At $\delta T=\pi$ the pulses instead return the population to $|1\rangle$. Finite-pulse detuning corrections are small in $\delta/\Omega$; the result is the stated short-pulse Ramsey limit.
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