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


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?

  • $\begingroup$ nice question - wonder if the time-dependent isn't invariant, but the time-independent SE is. $\endgroup$
    – tom
    Commented Dec 3, 2014 at 22:00
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/56024/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Dec 3, 2014 at 22:07
  • $\begingroup$ @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 $\frac{\partial}{\partial x}$. $\endgroup$
    – theage
    Commented Dec 3, 2014 at 22:15
  • 2
    $\begingroup$ You need to transform $\psi$ as well to get Schroedinger's equation from the point of view of the primed frame. $\endgroup$ Commented Dec 3, 2014 at 22:21

1 Answer 1


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.

  • $\begingroup$ Algebraic considerations exist leading to that result but are quite technical: the structure of the second cohomology group of the Lie algebra of the Galilean group. A celebrated paper by Bargmann of 1954 (I think) clarified everithing at mathematical level :) $\endgroup$ Commented Sep 12, 2017 at 17:09
  • $\begingroup$ 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. $\endgroup$ Commented Dec 19, 2019 at 14:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.