Adiabatic approximation

Question

The celebrated adiabatic theorem states that for a system initially in the eigenstate $|psi(0)rangle = |n(0)rangle$ for $t=0$, it will stay in that state afterward under adiabatic evolution:
$$
|psi(t)rangle = e^{i gamma_n(t)}e^{-frac{i}{hbar}int_0^t varepsilon_n(t^prime)dt^prime}|n(t)rangle.
$$
where $|n(t)rangle $is the instantaneous eigenstate of $H(t)$ and $gamma_n$ is the Berry phase.
However, in this paper
it states that up to first order, the adiabatic evolution of the state is:
$$
|psi(t)rangle =e^{-frac{i}{hbar}int_0^t varepsilon_n(t^prime)dt^prime} 
(|n(t) rangle+ihbar sum_{mneq n} |m(t)rangle frac{langle m(t) |frac{d}{dt}|n(t)rangle}{E_m - E_n}).
$$
My question is, why does the Berry phase term disappear? The derivation in Shen's book is confusing, can anyone give a clear derivation?

T. ssP · Answer

I think that if you express the wavefunction as an infinite series of some coefficients $c_n(t)$, your question is easier to answer. I made some notes on the adiabatic theorem that I will attach here for you to check if it answers your doubt (see below).
Equation (10) is equivalent to the eq. of the paper that you are referring to. Here you can see that apart from the term of the temporal variation of the Hamiltonian $dot{H}$, which in the adiabatic approximation is assumed to be very small compared with the other terms, the term that defines the Berry Phase is still present: $c_m(t)<psi_m|dot{psi}_m>$.

At the end of the day what really matters to have a non-zero Berry phase is not the velocity the system have in a process along the evolution path. The Berry Phase only depends on the path taken, for this reason is usually denoted as a topological phase and is also present in classical mechanics.
More information in: https://en.wikipedia.org/wiki/Berry_connection_and_curvature
I hope this to be useful to you.
T.

Adiabatic approximation

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