Where does the lower limit of the integral for the dynamic phase factor come from? I'm working on a problem right now where we have to figure out the transition probability between arbitrary excited states of the harmonic oscillator under a small time-dependent perturbation. Its time dependence is as $e^{-t^2}$, and it's applied at initial time $t_i = -\infty$. (It's not explicitly called small and gradually-applied, per se, but I think that's a reasonable interpretation of the conditions given in the problem.) The system is more or less a variation on the driven harmonic oscillator.
After much internet-scouring, I've determined that the best way to approach this is to use the near-adiabatic approximation, which involves the dynamic phase factor:
$$
\theta_n(t) = -\int^t_0\omega_n(t')dt' = -\frac1\hbar\int^t_0E_n(t')dt'
$$
Every definition of this equation I can find online sets its lower limit at $0$. Problem is, that's almost certainly under the implicit assumption that $t_i = 0$, which it isn't for me. What makes things particularly challenging is the fact that my expression for $E_n(t)$ contains a constant term, so if I set the lower limit to $-\infty$ the integral just evaluates as that and breaks everything.
Is that $0$ not actually related to the initial time, and rather just... part of the expression? And if I'm right, i.e. if the lower limit should indeed be $-\infty$, how in the world could I get around that?
(Disclaimer: It is distinctly possible that I might be totally off the mark and this isn't even relevant to the problem, but I don't... think that's the case. At any rate, I haven't yet found any better options for solving it. I can provide more of the details if necessary, but I'm kinda new here and don't 100% know where 'asking for help' ends and 'asking for answers' begins - so I'm erring on the side of caution for now)
 A: Caveat, the adiabatic approximation works only in the limit of a slow perturbation. In your case, when you write the time dependence of the perturbation generally in the form:
$$
V(t) = e^{-(t/\tau)^2}V_0
$$
in a harmonic oscillator of frequency $\omega$, the treatment is valid in the limit:
$$
\tau\omega\gg 1
$$
Also, you forgot the geometric phase:
$$
\gamma_n = \int \langle n(s)|i\frac{d}{ds}|n(s)\rangle ds
$$
This problem is not due to the adiabatic theorem. Without the perturbation, $E_n$ is rigorously constant and you get oscillating phase at $t\to\pm\infty$. This is not a problem since you are interested in transition probabilities, so you are only interested in the modulus square.
Now, add the perturbation that vanishes at the beginning and the end. The adiabatic theorem tells you that the final transition probabilities between energy eigenstates is the same as in the unperturbed case, namely an incoming energy eigenstate exits as the same energy eigenstate (up to a phase). This not surprising as it is built into the adiabatic theorem. Differences appear when you look at observables that are not conserved.
In order to make the difference more apparent, it is best to "factor out" the unperturbed contribution. Formally, this amounts to looking at the interaction picture, and in practice this means that you subtract the constant unperturbed part of energy eigenvalue in $E_n$ so that you may have a finite contribution to the phase.
I think that in your case, you typically interested in something like the the forced harmonic oscillator. The unperturbed Hamiltonian is (setting $\omega=1$):
$$
H_0 = a^\dagger a
$$
with $a$ the destruction operator satisfying $[a,a^\dagger]=1$. The perturbation is:
$$
V = -f(t)^*a-f(t)a^\dagger
$$
where you can take $f$ of the form $f(t) = f_0e^{-(t/\tau)^2}$. For the application of the adiabatic theorem, you'll need:
$$
E_n = n-|f|^2
$$
and the fact that the instantaneous eigenstates are obtained by the displacement operator $D(\alpha) = e^{\alpha a^\dagger-\alpha^*a}$:
$$
|n(t)\rangle = D(f(t))|n\rangle
$$
so using:
$$
\frac{d}{dt}D(f) = D(f)\left(\frac{\dot f f^*-\dot f^*f}{2}+\dot f a^\dagger-\dot f^* a\right)
$$
you get:
$$
\langle n(t)|i\frac{d}{dt}|n(t)\rangle = i\frac{\dot f f^*-\dot f^*f}{2}
$$
Note that in either case, the perturbation gives a constant additional term (independent of $n$). The adiabatic theorem therefore says that the final evolution is trivial, i.e. is merely the multiplication by a global phase.
This is consistent with the exact result as the total Hamiltonian induces a displacement in $a$:
$$
\alpha = \int_{-\infty}^{+\infty}ds e^{i(s-t_f)}f(s)
$$
In the adiabatic limit $\tau\to\infty$, you indeed get $\alpha\to 0$  thanks to the Rieman-Lebesgue theorem.
Out of curiosity, why don't you try by first order perturbation theory? It is more common starting point.
Hope this helps.
