Naive interpretation of Galilean invariance of the TDSE I was told today by someone smarter than myself that the time-dependent Schroedinger equation in one dimension was invariant under a Galilean transformation of $(x,t)$, namely under
$$\begin{cases}x'=x+ut\\t'=t\end{cases}.\tag{1}$$
Going to check this, I looked at the time dependent Schroedinger equation of a free particle.
$$i\hbar\frac{\partial\psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}\tag{2}$$
Computing the transformation of the differential operators via the chain rule:
$$\begin{cases}
\frac{\partial}{\partial x}=\frac{\partial t'}{\partial x}\frac{\partial}{\partial t'}+\frac{\partial x'}{\partial x}\frac{\partial}{\partial x'} = \frac{\partial}{\partial x'}
\\
\frac{\partial}{\partial t}=\frac{\partial t'}{\partial t}\frac{\partial}{\partial t'}+\frac{\partial x'}{\partial t}\frac{\partial}{\partial x'} = \frac{\partial}{\partial t'}+u\frac{\partial}{\partial x'}
\end{cases}$$
and plugging all of this back into $(2)$ gives the TDSE in the relatively inertial frame $(x',t')$.
$$i\hbar\left(\frac{\partial\psi}{\partial t'}+u\frac{\partial\psi}{\partial x'}\right)=-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x'^2}\tag{3}$$
This would imply that there's some additional term like $i\hbar u\frac{\partial\psi}{\partial x'}$ in the equation that represents an asymmetry under $(1)$. We have that said term is not zero (for that would imply that the wavefunction is space-independent in the relative frame, which is clearly not the case). Clearly I've misunderstood something here - is $(2)$ not Galilean invariant after all?
 A: In your derivation you've implicitly assumed that the wavefunction does not change its values when you go to the Galilean boosted frame.  In other words, you've assumed $\psi'(x', t') = \psi \bigl(x(x',t'), t(x',t') \bigr)$.  However, this isn't right.
The wavefunction encodes information about a particle's momentum, so when you go to a different frame the wavefunction must change to represent the momentum the particle has in the new frame.  For example, in the case of a plane wave, $\psi(x, t) = e^{i(kx - \omega t)}$. When you boost by velocity $u$, the wave's $k, \omega$ must change to match the new momentum and energy, like $k' = k + mu/\hbar$ and $\omega' = \hbar k'^2 / 2m$.  So it's not the same function anymore; it has different wavelength and frequency, above and beyond the simple change of coordinates.
In other words, the Schrödinger equation is Galilean-invariant not in the sense that the same solution works after a boost, but in the sense that there exist solutions representing waves traveling at all different speeds.  A boost maps a solution with $k, \omega$ to another solution with $k', \omega'$.  (And for solutions that are not plane waves, we can use the Fourier transform to decompose them into plane waves, boost each one, and recompose them.)
This argument might be a bit unsatisfying since it relies on physical intuition about the meaning of wavelength and frequency, rather than being a purely mathematical derivation.  I wonder if you might be able to derive it more rigorously from some operator-algebraic considerations, or some such.
