 Solution

The Von Neumann equation integrates to $\rho (t)=U(t)\rho (0)U(t{)}^{†}$, so the spectrum and purity of a density operator are invariant under any allowed open-loop control. These are the kinematic restrictions.

There is no additional orbit restriction here: the generators $i{\sigma }_z$ and $i{\sigma }_x$, together with their matrix commutator, span the special unitary Lie algebra:For freely chosen durations and piecewise continuous real control values, this gives density operator controllability. The connected reachable group is $SU(2)$; its action by conjugation is the same as that of $U(2)$, because a global phase cancels. Thus exactly the pairs ${\rho }_0\leftrightarrow {\rho }_2$ and ${\rho }_1\leftrightarrow {\rho }_3$ can be interconverted. No coherent control can turn either mixed state into either pure state, even though every orientation within each spectral orbit is reachable.