 Solution

Work with real positive pulse envelopes and define $\Omega =\sqrt{{\Omega }_{12}^2+{\Omega }_{23}^2}>0$. Then $\cos \theta ={\Omega }_{12}/\Omega$ and $\sin \theta ={\Omega }_{23}/\Omega$. In the ordered basis $|1⟩,|2⟩,|3⟩$, the candidate dark state of a driven Hamiltonian has coordinatesMultiplication by the given rotating-wave approximation matrix givesThusIts absence of an excited-state component is the destructive interference that makes it dark. The orthogonal bright ground-state combination is $|B⟩=\cos \theta |1⟩+\sin \theta |3⟩$. Since $H|B⟩=\Omega |2⟩$ and $H|2⟩=\Omega |B⟩$, the two remaining eigenstates are $(|B⟩\pm |2⟩)/\sqrt{2}$ with eigenvalues $\pm \Omega$.

For signed envelopes, use a continuous two-argument angle with $\cos \theta ={\Omega }_{12}/\Omega$ and $\sin \theta ={\Omega }_{23}/\Omega$, rather than an ambiguous arctangent branch. When both fields vanish, every state has zero eigenvalue and this formula does not select a unique dark vector.