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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.

(22m2x2+f(t)x)Ψ(x,t)=iΨt

When I attempt to apply a boost in the form

Ψ(x,t)=eiF(t)x˜Ψ(x,t)

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

22m(2˜Ψx22iF(t)˜ΨxF(t)22˜Ψ2)=i˜Ψt

but I am unsure of how to proceed. If I attempt to find a separable solution in the form ˜Ψ(x,t)=X(x)Φ(t), I cannot resolve the cross term.

22m(¨XX2iF(t)˙XX)=i˙ΦΦF(t)22m

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+α(t), sucht that z=y and ddt=t+˙α(t)z by the chain rule. In consequence, your Schrödinger equation takes the following form : i˜Ψt=22m2˜Ψz2+i(F(t)m˙α(t))˜Ψz+F(t)22m˜Ψ Choosing α(t):=1mF(t)dt, we get
i˜Ψt=22m2˜Ψz2+F(t)22m˜Ψ, 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 ˜Ψ(k,t)=˜Ψ0(k)exp(i(2k22mt+t0F(t)22mdt)) where ˜Ψ0(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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  • 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?
    – Talmsmen
    Commented Jan 21, 2023 at 19:15
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    My attempt is available at this link. The final answers appear similar. physics.stackexchange.com/q/746519
    – Talmsmen
    Commented Jan 21, 2023 at 19:18
  • @Talmsmen Sure ;)
    – Abezhiko
    Commented Jan 21, 2023 at 19:26

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