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Following the sketch given in this answer, I hoped to solve the 1+1 dimensional Schrodinger equation under a potential $f(t)x$ by using a time dependent boost.

$$\left(\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + f(t)x \right) \Psi(x,t) = i\hbar \frac{\partial \Psi}{\partial t}$$

When I attempt to apply a boost in the form

$$ \Psi(x,t) = e^{-i\frac{F(t)}{\hbar} x} \tilde{\Psi}(x,t) $$

where $F(t)$ is the antiderivative of $f(t)$, I get

$$ -\frac{\hbar^2}{2m}\left(\frac{\partial^2 \tilde{\Psi}}{\partial x^2} - 2i \frac{F(t)}{\hbar}\frac{\partial \tilde{\Psi}}{\partial x} -\frac{F(t)^2}{\hbar^2} \tilde \Psi^2\right) = i\hbar \frac{\partial \tilde \Psi}{\partial t} $$

but I am unsure of how to proceed. If I attempt to find a separable solution in the form $\tilde \Psi(x,t) = X(x) \Phi (t)$, I cannot resolve the cross term.

$$ -\frac{\hbar^2}{2m}\left(\frac{\ddot X}{X} - 2i \frac{F(t)}{\hbar}\frac{\dot X}{X} \right) = i\hbar \frac{\dot \Phi}{\Phi} - \frac{F(t)^2}{2m} $$

Did I choose the wrong ansatz for a boost? Or, is my derivation mistaken?

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You took the right boost. In order to simplify the resulting equation further, we need a change of variable, namely $z := x + \alpha(t)$, sucht that $\frac{\partial}{\partial z} = \frac{\partial}{\partial y}$ and $\frac{\mathrm{d}}{\mathrm{d}t} = \frac{\partial}{\partial t} + \dot{\alpha}(t)\frac{\partial}{\partial z}$ by the chain rule. In consequence, your Schrödinger equation takes the following form : $$ i\hbar\frac{\partial\tilde{\Psi}}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2\tilde{\Psi}}{\partial z^2} + i\hbar\left(\frac{F(t)}{m}- \dot{\alpha}(t)\right)\frac{\partial\tilde{\Psi}}{\partial z} + \frac{F(t)^2}{2m}\tilde{\Psi} $$ Choosing $\alpha(t) := \frac{1}{m}\int F(t)\mathrm{d}t$, we get
$$ i\hbar\frac{\partial\tilde{\Psi}}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2\tilde{\Psi}}{\partial z^2} + \frac{F(t)^2}{2m}\tilde{\Psi}, $$ which can now be solved in several ways (see e.g. here). For example, taking the Fourier transform with respect to the variable $z$, we end with $$ \tilde{\Psi}(k,t) = \tilde{\Psi}_0(k)\exp\left(-\frac{i}{\hbar}\left(\frac{\hbar^2k^2}{2m}t+\int_0^t\frac{F(t)^2}{2m}\mathrm{d}t\right)\right) $$ where $\tilde{\Psi}_0(k)$ is an initial condition. Finally, it is to be noticed that, even if the final result will obviously depend on the initial condition, linear potential tend to produce solutions involving Airy functions.

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  • $\begingroup$ Thank you very much. I am just posting another question where I am asking for solution verification of my work in momentum space. When I finish it in around 5 minutes would it be too much for me to ask for your reviewal? $\endgroup$
    – Talmsmen
    Jan 21, 2023 at 19:15
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    $\begingroup$ My attempt is available at this link. The final answers appear similar. physics.stackexchange.com/q/746519 $\endgroup$
    – Talmsmen
    Jan 21, 2023 at 19:18
  • $\begingroup$ @Talmsmen Sure ;) $\endgroup$
    – Abezhiko
    Jan 21, 2023 at 19:26

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