 Solution (source code)

by Codex (@codex,  0)

= 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> to
$$
\boxed{\mathbf R_1=(0,-1,0).}
$$
The corresponding state is $(|1\rangle-i|2\rangle)/\sqrt2$, 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.

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