= Solution
For $|D\rangle=(|1\rangle-|3\rangle)/\sqrt2$, direct multiplication gives $H_1|D\rangle=0$ and $H_0|D\rangle=-\omega|D\rangle$. It is therefore a <dark state of a driven Hamiltonian> for every real control, and
$$
\boxed{U(t)|D\rangle=e^{i\omega t/\hbar}|D\rangle.}
$$
Starting from this state, the reachable physical ray is a singleton, so it cannot be transferred to $|2\rangle$ or any distinct ray. More generally $P_D=|D\rangle\langle D|$ commutes with both generators; hence $\operatorname{Tr}(P_D\rho)$ is conserved. The system thus fails full-qutrit <pure-state controllability>, <density operator controllability>, and <unitary operator controllability>.
This is the appropriate full-space meaning of the PDF's phrase about all senses of <controllability>. It does not preclude control within a restricted invariant subspace: the bright state $|B\rangle=(|1\rangle+|3\rangle)/\sqrt2$ couples to $|2\rangle$ with strength $\sqrt2df(t)$, while the dark line is isolated. The <bright and dark state reduction of a star-coupled Hamiltonian> makes that invariant decomposition explicit.
Back to article page
