= Solution
The feedback law printed in the PDF is not sufficient for the asserted monotonicity. We first derive the exact identity, then give a counterexample and the corrected law. Both <density operators> evolve unitarily, so $\operatorname{Tr}\rho^2$ and $\operatorname{Tr}\rho_d^2$ are constant. Thus
$$
V=\|\rho-\rho_d\|_{\rm HS}^2=\operatorname{Tr}\rho^2+\operatorname{Tr}\rho_d^2-2\operatorname{Tr}(\rho_d\rho).
$$
Cyclicity of the <trace> gives $\operatorname{Tr}([-iH_0,\rho_d]\rho)=-\operatorname{Tr}(\rho_d[-iH_0,\rho])$. These drift terms cancel in the <derivative> of the overlap, leaving
$$
\frac d{dt}\operatorname{Tr}(\rho_d\rho)=fC,\qquad
C=\operatorname{Tr}(\rho_d[-iH_1,\rho]),\qquad
\boxed{\dot V=-2fC.}
$$
Here $C$ is real: $-i[H_1,\rho]$ is Hermitian, and the <trace> of the product of two <Hermitian matrices> is real. This is the <Hilbert-Schmidt Lyapunov feedback identity>.
For a direct counterexample to the source's overlap law, take $H_0=0$, $H_1=\sigma_y/2$, $\rho_d=(I+\sigma_z)/2$ and $\rho=(I+\sigma_x)/2$. All are admissible qubit operators and states. The printed overlap is $f=\operatorname{Tr}(\rho_d\rho)=1/2$, but $-i[H_1,\rho]=-\sigma_z/2$ gives $C=-1/2$. Hence $\dot V=1/2>0$: the distance initially increases. This proves that <feedback based on state overlap need not be stabilizing>.
The intended <Lyapunov quantum control> replaces the overlap by its control-direction <derivative>:
$$
\boxed{f=k\operatorname{Tr}(\rho_d[-iH_1,\rho]),\quad k>0,
\qquad\dot V=-2kC^2\leq0.}
$$
The conclusion is nonincrease, not strict decrease or automatic convergence. For example, opposite computational-basis <pure states>, diagonal $H_0$, and $H_1=\sigma_x$ give $C=0$ and zero control even though $V=2$. Target convergence requires further invariant-set and reachability conditions. A target with a different spectrum is also inaccessible under <Quantum Hamiltonian> evolution. The density-operator <derivatives> in the governing equations use the dots visible in the PDF, which the TeX loses.
Back to article page
