Significance of energy in a time dependent quantum box The Hamiltonian for a particle in a finite box is
$$H = \frac{p^2}{2m} + V(x)$$
which will give time evolution as 
$$ i\hbar d/dt|{\psi(t)}\rangle = H|{\psi(t)}\rangle \, .$$
However, if I do a Galilean transformation into a moving frame at some velocity $v$, then one expects a similar equation 
$$ i\hbar d/dt|{\psi''(t)}\rangle = H''|{\psi''(t)}\rangle $$
where $$ H'' = \frac{p^2}{2m} + V(x-vt) \, .$$
Indeed, this is true, because
$$ H'' = H' - i\hbar U \frac{dU^\dagger}{dt} = \frac{p^2}{2m} + V(x-vt) + \frac{1}{2}mv^2 $$
where $H'$ is the "transformed" operator given by $UHU^\dagger$ and 
$$ U= \exp\left(\frac{i}{\hbar}(pvt + xmv)\right) \, .$$
Therefore, the only difference between our expectations is a constant factor which contributes nothing to the eigenvectors except a shift of their respective eigenvalues.
Although the Hamiltonian operator for the stationary case is all well and good, and it clearly represents the observable of energy, what does $H''$ actually represent? It is time dependent so surely it can't represent the energy. Do we simply ignore any interpretation and say energy can only be defined well in the rest frame of the box where the Hamiltonian is $H$?
I've taken a year of undergraduate quantum, so that is about my level of expertise, although I have done quite a bit of independent learning beyond that.
It is clear that a $\frac{1}{2}mv^2$ makes no difference if the overall Hamiltonian is time independent, but I'm not sure if the two $H''$ I wrote out are indeed equivalent in the case of time dependence.
 A: Everything is the same in the new frame, inertial frames are equivalent for formulation of the equations. Hamiltonian is still sum of kinetic and potential energy. If in the original frame the potential energy is $V(x)$, and if the transformation to the new frame is
$$
x'' = x + vt
$$
then in the new frame the new Hamiltonian will depend on $x''$ and its conjugate $p''$:
$$
H'' = \frac{p''^2}{2m} + V(x'' - vt)
$$
which is explicitly time dependent.

It is time dependent so surely it can't represent the energy.

Where is this idea coming from? It is an operator, not energy value: it can depend on time and represent energy in the above sense, i.e. sum of kinetic and potential energy.
If you're worried that energy should be time-independent, then this is obeyed for expectation values of $H$ and $H''$:
$$
\frac{d}{dt}\langle H\rangle = 0.
$$
$$
\frac{d}{dt}\langle H''\rangle = 0.
$$
This is possible because besides $H''$, $\psi''$ depends on time in precisely such a way (it moves along with the potential energy function) that it cancels the dependence and the result is independent of time.
