 Solution

For $|D⟩=(|1⟩-|3⟩)/\sqrt{2}$, direct multiplication gives $H_1|D⟩=0$ and $H_0|D⟩=-\omega |D⟩$. It is therefore a dark state of a driven Hamiltonian for every real control, andStarting from this state, the reachable physical ray is a singleton, so it cannot be transferred to $|2⟩$ or any distinct ray. More generally $P_D=|D⟩⟨D|$ commutes with both generators; hence $\mathrm{Tr}(P_D\rho )$ 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 $|B⟩=(|1⟩+|3⟩)/\sqrt{2}$ couples to $|2⟩$ with strength $\sqrt{2}df(t)$, while the dark line is isolated. The bright and dark state reduction of a star-coupled Hamiltonian makes that invariant decomposition explicit.