 Solution

True for the finite bound-level Morse model with nonzero adjacent control couplings. The Morse oscillator energies have the formOver the physical bound-level range these adjacent gaps are positive, and $B\ne 0$ makes them pairwise distinct. Label the retained levels consecutively. With $H_1={\sum }_j{d}_j(E_{j,j+1}+E_{j+1,j})$, all $d_j\ne 0$, the squared adjoint action of $i{H}_0$ has eigenvalue $-(E_{j+1}-E_j{)}^2$ on each adjacent skew-Hermitian coupling. A polynomial chosen by Lagrange interpolation can therefore isolate any one adjacent coupling from $i{H}_1$.

Commuting an isolated coupling once with the drift produces its independent antisymmetric quadrature. The bracket between the two quadratures yields a traceless diagonal generator; brackets along the connected chain produce every nonadjacent off-diagonal generator. Thus connected nondegenerate transition chain generates special unitary control proves $\mathfrak{su}(N)\subseteq \mathfrak{g}$, which gives density-state controllability. If the physical drift has nonzero trace, subtracting its traceless part also produces the identity direction, giving $\mathfrak{u}(N)$ and full operator controllability. Shifting the energy origin may remove that extra phase direction but does not change state controllability. The assertion concerns the finite-level model in the question, not the entire bound-plus-dissociation Hilbert space of the untruncated Morse potential.