 Solution

For a Hermitian Quantum Hamiltonian, $[{\sigma }_m,{\sigma }_n{]}^{†}=-[{\sigma }_m,{\sigma }_n]$. ConsequentlySwapping $m,n$ changes the sign of the commutator, so $L_{nm}=-L_{mn}$. The generator is therefore real and antisymmetric. Its last column vanishes because the identity commutes with $H$, so $c=0$ and the traceless block satisfies $A^T=-A$.

For constant $H$, the Bloch vector evolves by $s(t)=R(t)s(0)$ with $R(t)=e^{At}$. Direct differentiation givesHence $R^{T}R=I$. Its determinant is continuously connected to one and cannot change between the two possible values $\pm 1$, so $\det R=1$. This proves that the evolution is a rotation in the special orthogonal group. A time-dependent Hermitian Quantum Hamiltonian has the same conclusion from $\dot{R}=A(t)R$.

These Hamiltonian rotations of Bloch vectors preserve the Euclidean norm and purity of a density operator. For $N>2$, not every rotation of the ambient sphere is a physical unitary conjugation: the higher-dimensional positivity constraints and full spectrum of the density operator impose further restrictions.