 Solution

Write the unitary conjugation as $\rho (t)=U(t)\rho (0)U(t{)}^{†}$. The identity component is unchanged, and the generalized Bloch vector transforms byThese coefficients are real, by the same trace argument as in part (a). Unitary conjugation preserves the Hilbert-Schmidt inner product, so the conjugated traceless generators form another orthonormal basis. It follows that $O(U{)}^{T}O(U)=I$. The unitary group is connected, $O(I)=I$, and the determinant of an orthogonal matrix is $\pm 1$; hence $\det O(U)=1$. Thus Hamiltonian evolution rotates the generalized Bloch vector by an element of $SO(N^2-1)$.

Equivalently, in units $\hslash =1$, the Von Neumann equation givesCyclicity of the trace shows $A_{k\ell }=-A_{\ell k}$, so $A$ is a real skew-symmetric matrix. This is the generator form of Hamiltonian rotations of Bloch vectors, valid also for a time-dependent Quantum Hamiltonian. For $N>2$, Hamiltonians generally realize only a proper subgroup of all rotations of this real space; they preserve the whole density-operator spectrum, not merely its purity of a density operator.