= Solution
For real nonnegative envelopes, put $\Omega=\sqrt{\Omega_1^2+\Omega_2^2}$ and choose a continuous $\theta=\operatorname{atan2}(\Omega_1,\Omega_2)$. The state $|D\rangle=\cos\theta|1\rangle-\sin\theta|3\rangle$ is dark because its two amplitudes feeding $|2\rangle$ cancel: $\Omega_1\cos\theta-\Omega_2\sin\theta=0$. The orthogonal bright combination $|B\rangle=\sin\theta|1\rangle+\cos\theta|3\rangle$ couples to $|2\rangle$ with strength $\Omega$. Thus the other <eigenstates> are $(|B\rangle\pm|2\rangle)/\sqrt2$ with energies $\pm\Omega$.
Start with $\Omega_1=0$ and $\Omega_2\ne0$, so $\theta=0$ and $|D\rangle=|1\rangle$. Slowly increase $\Omega_1/\Omega_2$ to infinity while maintaining a nonzero gap; end with $\Omega_2=0$ and $\Omega_1\ne0$, so $|D\rangle=-|3\rangle$. This target-to-intermediate pulse first is the counterintuitive ordering of <stimulated Raman adiabatic passage>. Its quantitative slowness condition is
$$
\boxed{\hbar|\dot\theta|\ll\Omega,\qquad
\dot\theta=\frac{\Omega_2\dot\Omega_1-\Omega_1\dot\Omega_2}{\Omega_1^2+\Omega_2^2}.}
$$
It follows from $|\langle\pm|\dot D\rangle|=|\dot\theta|/\sqrt2$ and the gap $\Omega$. Avoid switching both envelopes off while $\theta$ is changing; switch off together only after the transfer angle has become stationary. The dark <eigenvalue> is zero, so there is no dynamical phase; with this real <eigenvector> $\langle D|\dot D\rangle=0$ as well. The ideal <adiabatic quantum control> therefore transfers $|1\rangle$ to $-|3\rangle$, physically the same target ray.
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