I'm trying to show the Galilean covariance of the (time-dependent) Schrödinger equation by transforming as follows:

$$ \left\{\begin{eqnarray}\psi(\vec{r},t) &=& \psi(\vec{r}'-\vec{v}t,t),\\ \frac{\partial\psi(\vec{r},t)}{\partial t} & = & \frac{\partial\psi(\vec{r}'-\vec{v}t,t)}{\partial t}-\vec{v}\cdot\vec{\nabla}_{r'}\psi(\vec{r}'-\vec{v}t,t)\\ \vec{\nabla}_{r}\psi(\vec{r},t)&=&\nabla_{r'}\psi(\vec{r}'-\vec{v}t,t)\end{eqnarray}\right. $$ Which gives: $$ i\hbar\left[\frac{\partial}{\partial t}-\vec{v}\cdot\vec{\nabla}_{r'}\right]\psi(\vec{r}'-\vec{v}t,t)=\left[-\frac{\hbar^2}{2m}\nabla_{r'}^2+V(\vec{r}'-\vec{v}t,t)\right]\psi(\vec{r}'-\vec{v}t,t) $$

Now, supposedly, using the unitary transformation: $$ \psi(\vec{r}'-\vec{v}t,t)\to \psi(\vec{r}',t)\exp\left[\frac{i}{\hbar}m\vec{v}\cdot\vec{r}'+\frac{i}{\hbar}\frac{mv^2}{2}t\right], $$

it should be possible to recover the original form of the Schrödinger equation if there is no potential. When trying this, however, I can't seem to make it work. Is it just algebra or am I missing a crucial step here?

My lecturer also hinted that there was a physical reason that there may be no potential for Galilean invariance to hold. What is this reason and how can it intuitively be seen?

  • $\begingroup$ This approach is known as the Schrödinger group and is a low light of physics. The coordinate transformation depends on mass via a phase term. This overstretches the concept of a coordinate system. $\endgroup$
    – my2cts
    Aug 22, 2019 at 10:22
  • 1
    $\begingroup$ A comment related to the potential part. There is no potential invariant to a dynamical subgroup (i.e. containing Lorentzian or Galileian boosts) of the full dynamical group. That would make the potential velocity dependent, hence a dissipative system. Quantum theories are constructed around conservative systems. $\endgroup$
    – DanielC
    Aug 22, 2019 at 10:37
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/56024/2451 and links therein. $\endgroup$
    – Qmechanic
    Aug 22, 2019 at 10:39

1 Answer 1


I'm not convinced that your transformations are correct. This is one place where it is important, if $x=x'-vt$, to distinguish between $$ \left(\frac{\partial \psi}{\partial t}\right)_x $$ and $$ \left(\frac{\partial \psi}{\partial t}\right)_{x'}. $$ They are not the same thing. You notation is uclear as to what you are keeping fixed in your partials. I find it it easier to distinguish the derivatives by writing
$$ \xi=x-vt,\\ \tau=t. $$ Then the chain rule gives $$ \frac{\partial}{\partial t}= \frac{\partial}{\partial \tau}-v \frac{\partial}{\partial \xi}\\ \frac{\partial}{\partial x}= \frac{\partial}{\partial \xi} $$ where the derivative wrt $t$ keeps $x$ fixed, and the derivative wrt $\tau$ keeps $\xi$ fixed. Similarly the derivative wrt $x$ is at fixed $t$ and the derivative wrt $\xi$ is at fixed $\tau$. The equation in the frame in which the potential moves past you $$ i\hbar \frac{\partial \psi}{\partial t}=- \frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2}+V(x-vt)\psi $$ becomes, in the frame in which it is stationary $$ i\hbar\left( \frac{\partial }{\partial \tau}-v \frac{\partial }{\partial \xi}\right)\psi=- \frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial \xi^2}+V(\xi)\psi. $$ You can now absorb the extra $\xi$ derivative into the phase transformation that you mention to find a wavefunction $\tilde \psi(\xi,\tau)$ which obeys $$ i\hbar \frac{\partial\tilde \psi }{\partial \tau}\psi=- \frac{\hbar^2}{2m} \frac{\partial^2 \tilde \psi}{\partial \xi^2}+V(\xi)\tilde \psi, $$ in which the only change in the Schroedinger equation is in the time dependence of the potential.

I think I have a minus sign in the convective time derivative compared to your transformed equation.

This transformation is discussed in Baym's quantum mechanic textbook.


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