 Solution

Let $U(t,t_0)$ be the unitary Schrödinger propagator. The Heisenberg state is fixed at its initial value, while $A_H=U^{†}{A}_{S}U$. Since ${\psi }_S(t)=U{\psi }_H$, matrix elements and probabilities agree exactly in the two pictures. Differentiating, for an operator with no explicit Schrödinger-time dependence, givesFor constant Hamiltonian operator, $U=e^{-iH(t-t_0)/\hslash }$ and $H_H=H$.

A spatial rotation is represented by $U_R(\theta )=\exp [-i\theta n\cdot J/\hslash ]$, where $J$ is the total angular momentum operator. For a particle $J=L+S$. The commutators $[J_i,x_j]=i\hslash {ϵ}_{ijk}{x}_k$ and $[J_i,S_j]=i\hslash {ϵ}_{ijk}{S}_k$ giveThese are the correct infinitesimal vector rotations, verifying the generator and sign.

For the specified Hamiltonian operator, $p^2$ and $x^2$ commute with every $J_i$. Moreoverby antisymmetry. Thus $[J_i,H]=0$ and every total-angular-momentum component is constant in the Heisenberg picture. The separate torques areThey cancel but need not vanish individually, so orbital and spin angular momentum are generally not separately conserved.