 Solution

The density operators are Hermitian, so their squared Hilbert-Schmidt distance is $2V=\mathrm{Tr}({\rho }^2)+\mathrm{Tr}({\rho }_d^2)-2\mathrm{Tr}(\rho {\rho }_d)$. Both purity of a density operator values are constant under their respective unitary time evolution: cyclicity gives $\mathrm{Tr}(\rho [H,\rho ])=0$ for every Hermitian $H$. ConsequentlyThe terms containing $H_0$ cancel because $\mathrm{Tr}([-i{H}_0,\rho ]{\rho }_d)=-\mathrm{Tr}(\rho [-i{H}_0,{\rho }_d])$. The quantity $F=\mathrm{Tr}({\rho }_d[-i{H}_1,\rho ])$ is real: $[-i{H}_1,\rho ]$ is Hermitian, and the trace of the product of two Hermitian matrices equals its conjugate. The Lyapunov quantum control choice $f=F$ therefore gives $\boxed{\dot{V}=-f^2\le 0}$. For $t_2\ge t_1$, integration yields $V(t_2)=V(t_1)-{\int }_{t_1}^{t_2}f(t{)}^{2}dt\le V(t_1)$. Since the distance is $\sqrt{2V}$, it is monotonically nonincreasing. The decrease need not be strict when $f=0$.