# Momentum Space Solutions of the 1D Schrodinger equation in the Potential $f(t)x$ [closed]

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

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

• 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. Commented Jan 21, 2023 at 19:29
• Thanks so much! I'll get rid of that extra $x$. Commented Jan 21, 2023 at 19:40