Lyapunov quantum control ($\dot{V}=-f^2\le 0$)

For $\dot{\rho }=-i[H_0+f{H}_1,\rho ]$ and ${\dot{\rho }}_d=-i[H_0,{\rho }_d]$, the Hilbert-Schmidt distance gives $V=\Vert \rho -{\rho }_d{\Vert }^2/2$. The choice $f=\mathrm{Tr}({\rho }_d[-i{H}_1,\rho ])$ yields $\dot{V}=-f^2$. Decrease alone does not prove convergence: orthogonal pure eigenstates of a diagonal $H_0$ give zero input for a transverse Pauli control even though their distance remains nonzero. Such a law can be simulated to produce an open-loop control.

For $V=\Vert \rho -{\rho }_d{\Vert }_{\mathrm{HS}}^2$, with both states evolving under the same drift and only $\rho$ receiving $f{H}_1$, cyclicity of the trace cancels the drift terms and gives the displayed identity. The scalar $C$ is real because $-i[H_1,\rho ]$ is Hermitian. Choosing $f=kC$, $k>0$, makes $\dot{V}=-2k{C}^2\le 0$. This is nonincrease, not by itself convergence to the target: additional invariant-set and reachability conditions are needed.

Take $H_0=0$, $H_1={\sigma }_y/2$, ${\rho }_d=(I+{\sigma }_z)/2$ and $\rho =(I+{\sigma }_x)/2$. The overlap is $f=1/2$, but $C=\mathrm{Tr}({\rho }_d[-i{H}_1,\rho ])=-1/2$, so the Hilbert-Schmidt Lyapunov feedback identity gives $\dot{V}=1/2>0$. A target overlap and the derivative of that overlap along the control direction are different quantities.