 Solution

During a short pulse, detuning is negligible relative to $\Omega$. Rotation through angle $\Omega \Delta t=\pi /2$ about $+x$ takes the initial north-pole Bloch vector toThe corresponding state is $(|1⟩-i|2⟩)/\sqrt{2}$, since $U_x=\exp (-i\pi {\sigma }_x/4)$. The phase of this equatorial superposition is defined relative to the oscillator that will supply the second pulse.