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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.

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 $\mathrm{Tr}{\rho }^2$ and $\mathrm{Tr}{\rho }_d^2$ are constant. ThusCyclicity of the trace gives $\mathrm{Tr}([-i{H}_0,{\rho }_d]\rho )=-\mathrm{Tr}({\rho }_d[-i{H}_0,\rho ])$. These drift terms cancel in the derivative of the overlap, leavingHere $C$ 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 $f=\mathrm{Tr}({\rho }_d\rho )=1/2$, 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: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.