Consider a quantum particle of mass $$m$$ moving along a line, taken to be the $$x$$ axis, under the action of a spatially uniform time-dependent force $$F(t)$$. The time-evolution of the particle's wave function $$\Psi(x,t)$$ is governed by the Schrödinger equation, $i \hbar \frac{\partial \Psi(x,t)}{\partial t} = -\frac{\hbar^2}{2 m} \frac{\partial^2 \Psi(x,t)}{\partial x^2} - F(t) x \Psi(x,t) \,.$ The solution to this equation can be written as $\Psi(x,t) = \exp \left[ i \frac{p(t) x - s(t)}{\hbar} \right] \Psi_0 \big( x-q(t), t \big) \,,$ where $$\Psi_0$$ is the free-particle wave function that initially coincides with $$\Psi$$, $\Psi_0(x,0) = \Psi(x,0) \,,$ and the functions $$p(t)$$, $$q(t)$$, and $$s(t)$$ are defined as $p(t) = \int_0^t d\tau \, F(\tau) \,,$ $q(t) = \frac{1}{m} \int_0^{t} d\tau \, p(\tau) = \frac{1}{m} \int_0^{t} d\tau \int_0^{\tau} d\tau' \, F(\tau') \,,$ $s(t) = \frac{1}{2 m} \int_0^t d\tau \, p^2(\tau) = \frac{1}{2 m} \int_0^t d\tau \, \left[ \int_0^{\tau} d\tau' \, F(\tau') \right]^2 .$ Classically, $$p$$ and $$q$$ are the shifts in momentum and positions, respectively, caused by force $$F$$.
Proof: Let us verify that $$\Psi(x,t)$$ indeed satisfies the Schrödinger equation. We have $\frac{\partial \Psi}{\partial t} = i \frac{\dot{p} x - \dot{s}}{\hbar} \Psi +\left( - \dot{q} \frac{\partial \Psi_0}{\partial x} + \frac{\partial \Psi_0}{\partial t} \right) e^{i (p x - s) / \hbar} \,,$ or $\frac{\partial \Psi}{\partial t} = \frac{i}{\hbar} \left( F x - \frac{p^2}{2 m} \right) \Psi +\left( - \frac{p}{m} \frac{\partial \Psi_0}{\partial x} + \frac{\partial \Psi_0}{\partial t} \right) e^{i (p x - s) / \hbar} \,.$ The left-hand side of the Schrödinger equation reads $\text{LHS} = -F x \Psi + \frac{p^2}{2 m} \Psi - i \hbar \frac{p}{m} \frac{\partial \Psi_0}{\partial x} e^{i (p x - s) / \hbar} + i \hbar \frac{\partial \Psi_0}{\partial t} e^{i (p x - s) / \hbar} \,.$ Then, $\frac{\partial \Psi}{\partial x} = i \frac{p}{\hbar} \Psi + \frac{\partial \Psi_0}{\partial x} e^{i (p x - s) / \hbar}$ and $\frac{\partial^2 \Psi}{\partial x^2} = -\frac{p^2}{\hbar^2} \Psi + i \frac{2 p}{\hbar} \frac{\partial \Psi_0}{\partial x} e^{i (p x - s) / \hbar} + \frac{\partial^2 \Psi_0}{\partial x^2} e^{i (p x - s) / \hbar} \,.$ The right-hand side of the Schrödinger equation reads $\text{RHS} = -F x \Psi + \frac{p^2}{2 m} \Psi - i \hbar \frac{p}{m} \frac{\partial \Psi_0}{\partial x} e^{i (p x - s) / \hbar} - \frac{\hbar^2}{2 m} \frac{\partial^2 \Psi_0}{\partial x^2} e^{i (p x - s) / \hbar} \,.$ Hence, $\text{LHS} - \text{RHS} = \left( i \hbar \frac{\partial \Psi_0}{\partial t} + \frac{\hbar^2}{2 m} \frac{\partial^2 \Psi_0}{\partial x^2} \right) e^{i (p x - s) / \hbar} \,.$ Since $$\Psi_0$$ satisfies the free-particle Schrödinger equation, the expression inside the brackets vanishes.