Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 61 / 2 / e / i / Solution

For real nonnegative envelopes, put $\Omega =\sqrt{{\Omega }_1^2+{\Omega }_2^2}$ and choose a continuous $\theta =atan2({\Omega }_1,{\Omega }_2)$. The state $|D⟩=\cos \theta |1⟩-\sin \theta |3⟩$ is dark because its two amplitudes feeding $|2⟩$ cancel: ${\Omega }_1\cos \theta -{\Omega }_2\sin \theta =0$. The orthogonal bright combination $|B⟩=\sin \theta |1⟩+\cos \theta |3⟩$ couples to $|2⟩$ with strength $\Omega$. Thus the other eigenstates are $(|B⟩\pm |2⟩)/\sqrt{2}$ with energies $\pm \Omega$.

Start with ${\Omega }_1=0$ and ${\Omega }_2\ne 0$, so $\theta =0$ and $|D⟩=|1⟩$. Slowly increase ${\Omega }_1/{\Omega }_2$ to infinity while maintaining a nonzero gap; end with ${\Omega }_2=0$ and ${\Omega }_1\ne 0$, so $|D⟩=-|3⟩$. This target-to-intermediate pulse first is the counterintuitive ordering of stimulated Raman adiabatic passage. Its quantitative slowness condition isIt follows from $|⟨\pm |\dot{D}⟩|=|\dot{\theta }|/\sqrt{2}$ 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 $⟨D|\dot{D}⟩=0$ as well. The ideal adiabatic quantum control therefore transfers $|1⟩$ to $-|3⟩$, physically the same target ray.