 Solution

Let $U(t)=e^{-iHt/\hslash }$ and write ${\psi }_S(t)=U(t){\psi }_0$ in the Schrödinger picture, with observables $A_S$ constant unless they have an explicit time dependence. In the Heisenberg picture the state is ${\psi }_H={\psi }_0$ andDifferentiate the two unitary factors to obtain the equation. Unitarity gives $⟨{\psi }_S,A_S{\psi }_S⟩=⟨{\psi }_0,A_H{\psi }_0⟩$.

For the interaction picture choose a reference split $H=H_0+V$, put $U_0=e^{-i{H}_{0}t/\hslash }$, and define ${\psi }_I=U_0^{\ast }{\psi }_S$, $A_I=U_0^{\ast }{A}_S{U}_0$. Then $i\hslash {\dot{\psi }}_I=V_I{\psi }_I$, $V_I=U_0^{\ast }V{U}_0$, while $A_I$ evolves with $H_0$ (and its explicit time dependence). AgainThese are different unitary allocations of the same dynamics between states and observables, so measured expectations agree.

The canonical commutation relation now givesMultiplication of the last equation by $e^{i\omega t}$ and integration givesSince $A_t^{\ast }-A_t=2im\omega X_t$ and $A_0{\psi }_0=0$, both initial expectations of $A_0,A_0^{\ast }$ vanish. ThereforeThe printed positive displacement endpoint has the wrong sign for the Hamiltonian's positive $\mathcal{E}X$ term. The two turning values are $0$ and $-2\mathcal{E}/(m{\omega }^2)$; their ordering depends on the sign of $\mathcal{E}$. The result oscillates about the displaced equilibrium $-\mathcal{E}/(m{\omega }^2)$, exactly as the force $-\mathcal{E}$ predicts.