# Galilean invariance of the free schroedinger equation [duplicate]

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?

## marked as duplicate by sammy gerbil, Qmechanic♦ quantum-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 12 '17 at 17:57

• 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