Question
When going from the Hamiltonian of an inertial frame $H$ to that of a non-inertial frame $H'$ they are related by the canonical transformation:
$$ H = H' + \frac{\partial F}{\partial t}$$
I suspect in quantum mechanics the phase factor that one picks up when doing this transformation is nothing but the generating function.
$$ \psi_{i}(r,t) = \exp \big(- \frac{i}{\hbar} F \big){U'}_{ni}^\dagger(t) \psi'_{ni}(r',t)$$
where $\psi$ is the wave function in different frames, the subscripts $i$ and $ni$ are inertial and non-inertial and $r$,$t$, $r'$ and $t'$ are the non-inertial coordinates. How does one prove this? or is this some strange coincidence? Also I haven't seen this in the literature? (Which makes me think my suspicions are misplaced)
Background and Example
In a system free of potential we have the Hamiltonian $H$:
$$ H = \frac{p^2}{2m}$$
with momentum $p$ and mass $m$. But let's go into a non-inertial linearly accelerated from with:
$$ x' = x +\frac{1}{2} gt^2 $$ $$ p' = p + mgt $$
where $x'$ is the new position coordinate, $p'$ is the new momentum, $g$ is our linear acceleration. By squaring the above above equation and dividing by $2m$:
$$ \frac{p^2}{2m} = \frac{p'^2}{2m} + \frac{mg^2 t^2}{2} + p'gt$$
This is conveniently in the form of generating functions for Hamiltonian mechanics:
$$ H = H' +\frac{\partial F}{\partial t}$$
Let us explicitly write $F$:
$$ F = \frac{p'g t^2}{2} + \frac{mg^2t^3}{3!}$$
But wait! I've seen this $F$ somewhere? Here:
$$\Psi'(r,t)=\Psi\left(r-\frac{At^2}2,t\right)\exp\left[\frac {im}\hbar\left(Atr-\frac {A^2t^3}6\right)\right].$$
It seems to me under a change of coordinates the phase factor becomes:
$$ \exp(\frac {im}\hbar (Atr-\frac {A^2t^3}6) = \exp (- \frac{i}{\hbar}F) \hat U'^\dagger(t) $$
where $\hat U'$ is the unitary operator of $H'$. Let us explicitly check by comparing acceleration and setting $A = g$. The second term $\frac{mg^2t^3}{3!}$ is obviously a match. What about the first?
Well,
$$ At = gt = v'$$
But I also know:
$$ p' = m v'$$
Thus using the conservation of energy in classical mechanics for the first term:
$$ \frac{1}{2}m v'^2 t = (E' - mgx')t$$
We can throw out the constant $E'$ since it corresponds to the unitary operator.
