It is well known that the energy levels
$$
E_n = \hbar \omega\left(n+\frac{1}{2}\right)
$$
of the quantum harmonic oscillator verify the eigenvalue problem
$$
\hat{H}|\psi_n\rangle = E_n |\psi_n \rangle
$$
where
$$
\hat{H} = \frac{\hat{p}^2}{2m} +\frac{1}{2}m\omega^2 \hat{x}^2.
$$
Now let us assume that the harmonic potential is endowed with a uniform translation velocity, i.e.:
$$
\hat{H}_v = \frac{\hat{p}^2}{2m} +\frac{1}{2}m\omega^2 (\hat{x}-vt)^2.
$$
For an observer moving with velocity $v$, the harmonic potential will look at rest, and hence I expect that the energy levels $E_n$ should be the usual ones (see above). Conversely, for an observer in the laboratory frame, I expect that the energy levels will be different (in Classical Mechanics, it is well known that the kinetic energy is different in different inertial frames), and, in general, they depend on $v$. My question is therefore: which is the expression of the energy levels in the lab frame?
Best Answer
$\hat H_𝑣=\exp(−𝑖𝑣𝑡\hat p/ℏ)~\hat H\exp(𝑖𝑣𝑡\hat p/ℏ)$, so the two hamiltonians are canonically equivalent and have the same spectrum. You may also see this from $[\hat x -vt,\hat p]=i\hbar$, so $(\hat x,\hat p)\mapsto (\hat x -vt,\hat p)$ is a canonical transformation preserving the commutator, so defining $a$ and $a^\dagger$ with either pair will produce the same spectrum, $E_n$.
It is not clear to me what your classical limit argument is, since there is hardly an eigenvalue equation involved in your classical mechanics setup ("levels"?) expectations, which you may have to refine. In any case, classical physics is the same in all inertial frames.
Response to comments
Indeed, but you did not shift the momentum in concordance with the coordinate shift. What you wrote down in the OP is not the hamiltonian fo the TDSE of WP, $$ \breve H= 𝑈𝐻̂𝑈^\dagger + 𝑖ℏ\dot{U}𝑈^\dagger =\hat H_v + v\hat p= \frac{(\hat{p}+mv)^2}{2m}-\frac{mv^2}{2} +\frac{1}{2}m\omega^2 (\hat{x}-vt)^2. $$ The two hamiltonians have different spectra, differing by a constant shift term, amounting to a mere shift in the zero point energy, as you inferred. The level spacings are the same.