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Assume piecewise-constant real controls with freely chosen duration, as in the usual finite-dimensional quantum controllability theorem. We can establish the strongest notion here: the Dynamical Lie algebra is $\mathfrak{u}(3)$, so the system has unitary operator controllability, and consequently both pure-state and fixed-spectrum density operator controllability.

To verify the hypotheses explicitly, use matrix units $E_{jk}=|j⟩⟨k|$ and put $K_0=-i{H}_0$, $X_{jk}=-i(E_{jk}+E_{kj})$, $Y_{jk}=E_{jk}-E_{kj}$. Then $K_1=-i{H}_1=d_1{X}_{12}+d_2{X}_{23}$ andSince $d_1{d}_2\ne 0$ and ${\omega }_1^2\ne {\omega }_2^2$, linear combinations of these two elements isolate $X_{12}$ and $X_{23}$. Their commutators with $K_0$ give ${\omega }_1{Y}_{12}$ and $-{\omega }_2{Y}_{23}$, respectively. Further commutators give two independent traceless diagonal generators, since $[X_{jk},Y_{jk}]=2i(E_{jj}-E_{kk})$, as well as the two $1$–$3$ generators. These eight generators span the special unitary Lie algebra $\mathfrak{su}(3)$.

Finally $\mathrm{Tr}{K}_0=i({\omega }_1+{\omega }_2)\ne 0$. Subtract its traceless component, already in the generated algebra, to obtain a nonzero multiple of $iI$. This proves the Lie algebra of the unitary group is all of $\mathfrak{u}(3)$, not merely $\mathfrak{su}(3)$. The compact-group controllability theorem then gives access to every unitary with a suitable control and some duration. This does not assert controllability at every predetermined short time or under additional bandwidth constraints. This proves that two adjacent transitions generate the qutrit unitary Lie algebra.