Let's suppose that coordinates have to transform accoring to the Inhomogenous Galilean Group. Then
$$ x' = x + a + v(t+b) $$ $$ t' = t + b $$
Let's use a funtion $\psi(x,t)$ of $x$ and $t$ as the basis of a representation of the Galilean group.
$$G(a,b,v)\psi(x,t)=\psi(x',t')$$
where G(a,b,v) is a group element. Then, the generators of the group in this basis are given by
$$ X_{q} f(x,t) = -i\Bigg[\frac{\partial}{\partial q} G(a,b,v) \psi(x,t)\Bigg]_{a,b,v=0}$$
$$ = -i\Bigg[\frac{\partial}{\partial q} \psi(x',t')\Bigg]_{a,b,v=0}$$
$$ = -i\Bigg[(\frac{\partial x'}{\partial q}\frac{\partial}{\partial x'} + \frac{\partial t'}{\partial q}\frac{\partial}{\partial t'}) \psi(x',t')\Bigg]_{a,b,v=0}$$
where $q=a,b$ or $v$. This gives the generators
$$ X_a = -i\frac{\partial}{\partial x}, X_b = -i\frac{\partial}{\partial t}, X_v = -it\frac{\partial}{\partial x},$$
and the commutators
$$ [X_a,X_b]=0, [X_a,X_v]=0, [X_b,X_v]=iX_a$$
By Schur's Lemma, any operator that commutes with all of the generators is a constant. We can look at operators of the form
$$U_{l,m,n}(\alpha,\beta,\gamma)=\alpha X_a^l+\beta X_b^m+\gamma X_v^n$$
and find that they will be constant if $ \alpha = \beta = 0$ (you can also include cross terms). Do I understand correctly that then we would find all functions $\psi(x,t)$ that satisfy
$$ \sum_{l=1}^{\infty}\alpha_l\frac{\partial ^l}{\partial x^l}\psi(x,t)=U(\alpha_1, \alpha_2, ... )\psi(x,t)$$
have galilean covariance and thus can be used to represent physical observables? In particular, if the probability density was given by $\psi(x,t)$ rather than $|\psi(x,t)|^2$, a quantum state would have to satisfy the equation above. My understanding is if you assume a quantum state transforms like
$$ G(a,b,v)\psi(x,t)=e^{i\theta(a,b,v)} \psi(x',t')$$
instead, you would get the Schrödinger equation using the same approach (which is equivalent to saying that $|\psi(x,t)|^2$ is an observable). I am a bit sceptical about my result since there is no condition on the time dependence of $\psi(x,t)$. Is my logic correct or am I completely off?
