# 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?

• nice question - wonder if the time-dependent isn't invariant, but the time-independent SE is. – tom Dec 3 '14 at 22:00
• Possible duplicates: physics.stackexchange.com/q/56024/2451 and links therein. – Qmechanic Dec 3 '14 at 22:07
• @tom The time-independent certainly is invariant using this analysis since the asymmetry comes from $\frac{\partial}{\partial t}$ and, of course, the free TISE depends only on $\frac{\partial}{\partial x}$. – theage Dec 3 '14 at 22:15
• You need to transform $\psi$ as well to get Schroedinger's equation from the point of view of the primed frame. – Ján Lalinský Dec 3 '14 at 22:21

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.)
• Yes you can derive this more generally. No operator-algebraic considerations needed. Let $\psi$ transform with an arbitrary phase $\psi'(x',t') = e^{i\alpha(x,t)}\psi(x+ut,t)$. Then isolate $\alpha(x,t)$ by insisting that the schrödinger equation is solved in the new frame as well. This should cancel the extra term in OP's post. – doublefelix Dec 19 '19 at 14:56