= Solution
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: \b[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> $E_{jk}=|j\rangle\langle k|$ and put $K_0=-iH_0$, $X_{jk}=-i(E_{jk}+E_{kj})$, $Y_{jk}=E_{jk}-E_{kj}$. Then $K_1=-iH_1=d_1X_{12}+d_2X_{23}$ and
$$
\operatorname{ad}_{K_0}^2K_1=-\omega_1^2d_1X_{12}-\omega_2^2d_2X_{23}.
$$
Since $d_1d_2\ne0$ 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 $\omega_1Y_{12}$ and $-\omega_2Y_{23}$, 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 $\operatorname{Tr}K_0=i(\omega_1+\omega_2)\ne0$. 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>.
Back to article page
