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My question follows this question: Naive interpretation of Galilean invariance of the TDSE

Essentially, I'm not sure how to proceed mathematically.

We have the transformations:


Then, the TDSE for a free particle is

$$i\hbar\frac{\partial\psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}\tag{2}$$

which after transforming the coordinates becomes

$$i\hbar\left(\frac{\partial\psi}{\partial t'}-v\frac{\partial\psi}{\partial x'}\right)=-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x'^2}\tag{3}$$

For simplification, we look at the plane wave




we get

$$p'=p-mv$$ so that $$p'x'-E't' = p'x' - \frac{p'^2}{2m}t' =(p-mv)(x-vt) - \frac{(p-mv)^2}{2m}t=px-Et-mvx+\frac{mv^2}{2}t$$

then the wavefunction in the new frame is


My question is how do I show that the new wavefunction satisfies schroedinger's equation. I tried plugging it into both of (2) or (3), it doesn't satisfy either, so what am I doing wrong?


marked as duplicate by sammy gerbil, Qmechanic quantum-mechanics Sep 12 '17 at 17:57

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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/56024/2451 and links therein. $\endgroup$ – Qmechanic Sep 12 '17 at 14:31
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    $\begingroup$ You are starting from the wrong hypothesis that a wavefunction is a scalar under Galilean transformation. It is not. There is a further phase depending on $t$ and $x$, $\psi'(t',x') = e^{i m\gamma(t,x)} \psi(t,x)$. You should be able to find it just requiring that the new wavefunction satisfies the S. equation with respect to the new Hamiltonian...$m$ is the mass of the particle and its presence gives rise to a famous superselection rule due to Bargmann... $\endgroup$ – Valter Moretti Sep 12 '17 at 15:09
  • $\begingroup$ @valter if you look towards the end of my question you will see that I did transform the wave function, I just don't know where to plug it in, the original SE or the "new" one. And for some reason neither works $\endgroup$ – user35687 Sep 12 '17 at 16:35
  • $\begingroup$ Pass to momentum representation... $\endgroup$ – Valter Moretti Sep 12 '17 at 16:41
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    $\begingroup$ Possible duplicate of Galilean invariance of the Schrodinger equation $\endgroup$ – sammy gerbil Sep 12 '17 at 17:23