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?