= Solution
The <quantum state> is continuous through a sudden finite-force <quantum quench>, so it is still $|0\rangle$ at $t=0^+$. Define
$$
\lambda=\frac{F_0}{\sqrt{2M\hbar\omega^3}},\qquad b=a-\lambda,\qquad \theta=\omega t.
$$
The new <Hamiltonian operator> is $H=\hbar\omega(b^\dagger b+1/2)-\hbar\omega\lambda^2$. The old <Fock vacuum> obeys $b|0\rangle=-\lambda|0\rangle$, so the <coherent state after shifting an oscillator equilibrium> rotates with amplitude $-\lambda e^{-i\theta}$ relative to the new equilibrium. In the original oscillator basis, the full state is
$$
\boxed{|\psi(t)\rangle=e^{-i\theta/2+i\lambda^2(\theta-\sin\theta)}
\left|\lambda(1-e^{-i\theta})\right\rangle.}
$$
One way to obtain the phase is to write
$$
e^{-iHt/\hbar}|0\rangle
=e^{-i\theta/2+i\lambda^2\theta}D(\lambda)e^{-i\theta a^\dagger a}D(-\lambda)|0\rangle
$$
and apply the <displacement operator> composition rule. The <coherent-state oscillator expectation values> give
$$
\boxed{\langle x\rangle=\frac{F_0}{M\omega^2}(1-\cos\omega t),\qquad
\langle p\rangle=\frac{F_0}{\omega}\sin\omega t.}
$$
The <position uncertainty> stays $\sqrt{\hbar/(2M\omega)}$. These mean values obey the forced classical equation, while the nonzero <variance> retains the quantum width.
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