 Solution

The second pulse has the same phase and area as the first, so another $+x$ rotation by $\pi /2$ givesAt $T=2\pi /\delta$, the final vector is the south pole and all population is in $|2⟩$. 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 $|1⟩$. Finite-pulse detuning corrections are small in $\delta /\Omega$; the result is the stated short-pulse Ramsey limit.