= Solution
Let $U(t)=e^{-iHt/\hbar}$ 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 $\psi_H=\psi_0$ and
$$
A_H(t)=U(t)^*A_SU(t),\qquad\boxed{\dot A_H=\frac{i}{\hbar}[H,A_H]+U^*(\partial_tA_S)U.}
$$
Differentiate the two <unitary> factors to obtain the equation. <Unitarity> gives $\langle\psi_S,A_S\psi_S\rangle=\langle\psi_0,A_H\psi_0\rangle$.
For the <interaction picture> choose a reference split $H=H_0+V$, put $U_0=e^{-iH_0t/\hbar}$, and define $\psi_I=U_0^*\psi_S$, $A_I=U_0^*A_SU_0$. Then $i\hbar\dot\psi_I=V_I\psi_I$, $V_I=U_0^*VU_0$, while $A_I$ evolves with $H_0$ (and its explicit time dependence). Again
$$
\boxed{\langle\psi_I,A_I\psi_I\rangle=\langle\psi_S,A_S\psi_S\rangle=\langle\psi_0,A_H\psi_0\rangle.}
$$
These are different <unitary> allocations of the same dynamics between states and <observables>, so measured <expectations> agree.
The <canonical commutation relation> now gives
$$
\boxed{\dot X_t=P_t/m,\quad\dot P_t=-m\omega^2X_t-\mathcal E,\quad\dot A_t=-i\omega A_t-\mathcal E.}
$$
Multiplication of the last equation by $e^{i\omega t}$ and integration gives
$$
\boxed{A_t=e^{-i\omega t}A_0-\frac{i\mathcal E}{\omega}(e^{-i\omega t}-1).}
$$
Since $A_t^*-A_t=2im\omega X_t$ and $A_0\psi_0=0$, both initial <expectations> of $A_0,A_0^*$ vanish. Therefore
$$
\boxed{\langle X_t\rangle=-\frac{\mathcal E}{m\omega^2}(1-\cos\omega t).}
$$
\b[The printed positive displacement endpoint has the wrong sign] for the <Hamiltonian>'s positive $\mathcal EX$ 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.
Back to article page
