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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:

$$\begin{cases}x'=x-vt\\t'=t\end{cases}.\tag{1}$$

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

$$\psi\left(x,t\right)=e^{i(px-Et)/\hbar}$$

using

$$x'=x-vt$$

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

$$\psi'(x',t')=e^{i(p'x'-E't')/\hbar}=e^{i(px-Et)/\hbar}e^{-i(mvx-\frac{mv^2}{2}t)/\hbar}$$

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?

Qmechanic
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user35687
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    Possible duplicates: http://physics.stackexchange.com/q/56024/2451 and links therein. – Qmechanic Sep 12 '17 at 14:31
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    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... – Valter Moretti Sep 12 '17 at 15:09
  • @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 – user35687 Sep 12 '17 at 16:35
  • Pass to momentum representation... – Valter Moretti Sep 12 '17 at 16:41
  • I mean, you have to remove the last phase you wrote before inserting the wavefunction into the new Schroedinger equation. If you deal with in momentum representation and intepret actively the transformation, the found non-scalar transformation rule appears much more natural... – Valter Moretti Sep 12 '17 at 16:46
  • Another approach is assuming Galilean invariance of the two involved reference frame. So you know a priori that the wavevefunction with fixed momentum must be of the standard form also in the new reference frame with the transformed value of the momentum obviously. This forces the transformation law to be that i pointed out: transform as a scalar and then multiply with a certain phase depending on the mass to recover the standard form. – Valter Moretti Sep 12 '17 at 16:54
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    Possible duplicate of Galilean invariance of the Schrodinger equation – sammy gerbil Sep 12 '17 at 17:23
  • @valter thank you, but sorry I don't get it – user35687 Sep 12 '17 at 23:32

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