Show the Galilean covariance of Schrödinger equation 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?
 A: 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.
A: In his elementary book on Quantum Mechanis, "INTRODUCTION TO THE QUANTUM THEORY", David Park demonstrates the invariance of the Schroedinger equation under Galilean Transformations.  That material is presented in Appendix 3.
I suggest you give this a look.
M. L. Sloan
--- retired theoretical research physicist
