Remark: The preparation Hamiltonian is not K θ(t)

Let

Then $G$ is Hermitian and $e^{-i\theta G}|00⟩=|{\chi }_{\theta }⟩$. A differentiable path is prepared by

In contrast, $K_{\theta (t)}{\Omega }_{\theta (t)}=0$. Unless the vector is constant, it does not solve the Schrödinger equation generated by that instantaneous parent Hamiltonian. The ground-state family and its preparation dynamics are different data.