Yesterday, I asked a question about solving the 1D Schrodinger equation in a time varying potential $f(t)x$ using a method solely in configuration space. Although this approach does not directly answer my original question — hence my asking it separately, I would be greatly appreciative if someone could verify my solution.

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

Then, I used the cosmetic exchanges $x\mapsto i\hbar \frac{\partial}{\partial p}$ and $-i\hbar \frac{\partial}{\partial x} \mapsto p$ to produce the equation

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

Then, changing coordinates corresponds to a time-dependent boost in momentum space where $F(t)$ is antiderivative of $f(t)$

$$ \begin{align} p' &= p + F(t) \\ t' &= t \end{align} $$

producing the relationships

$$ \begin{align} \frac{\partial}{\partial p} &= \frac{\partial}{\partial p'}\\ \frac{\partial}{\partial t} &= \frac{\partial}{\partial t'} + f(t) \frac{\partial}{\partial p} \end{align} $$

and yielding the final equation

$$ -\frac{(p' - F(t))^2}{2m} \hat\Psi(p', t') = i\hbar \frac{\partial}{\partial t'} \hat\Psi(p', t') $$

The solution then becomes

$$ \hat \Psi(p', t') = C \exp\left(-\frac{i}{\hbar} \int_0^{t'} \frac{(p' - F(\tau))^2}{2m} d\tau\right) $$

or removing the primes by relabeling

$$\hat \Psi(p, t) = C \exp\left(-\frac{i}{\hbar} \int_0^{t} \frac{(p - F(\tau))^2}{2m} d\tau\right)$$

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    $\begingroup$ Hi again @Talmsmen ! Seems correct to me, apart from the typo in $p'$, which should be $p' = p + F(t)$, hence $\partial_t = \partial_{t'} + f(t)\partial_p$, I guess. But the final answers are not impacted. $\endgroup$
    – Abezhiko
    Jan 21, 2023 at 19:29
  • $\begingroup$ Thanks so much! I'll get rid of that extra $x$. $\endgroup$
    – Talmsmen
    Jan 21, 2023 at 19:40