Berry phase disappearing in 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: $$ |\psi(t)\rangle =e^{-\frac{i}{\hbar}\int_0^t \varepsilon_n(t^\prime)dt^\prime} (|n(t) \rangle+i\hbar \sum_{m\neq n} |m(t)\rangle \frac{\langle m(t) |\frac{d}{dt}|n(t)\rangle}{E_m - E_n}). $$ My question is, why does the Berry phase term disappear? The derivation in Shen's book is confusing, can anyone give a clear derivation?

Answer 1

Here is a possible explanation for what you have observed. In your first equation the Berry phase is, as you say,

$$ \gamma_n(t) = i\int_0^t dt' \, \langle n(t') |\dot{n}(t')\rangle. $$

However the term $e^{i\gamma_n}$ (as well as the other, dynamical phase) is simply a phase and so it is un-important in quantum mechanics and can safely be discarded.

It can also be discarded in the following sense. The eigenvector $|n(t)\rangle$ is only defined up to a phase. I could redefine $|n(t)\rangle$ for each $t$ in such a way as to absorb the Berry phase (and the dynamical phase too).

However now imagine the following situation. Your system (or if you want your Hamiltonian) depends on some number $n$ of real parameters $X_1,\ldots, X_n$. You can imagine to change these parameters very slowly, i.e. adiabatically, along a certain path $\mathbf{r}(t)=(X_1(t),\ldots, X_n(t)$ with $t\in[0,\tau]$. Now imagine that the path is closed, that is $\mathbf{r}(0)=\mathbf{r}(\tau)$. In other words you traveled somewhere (very slowly) in parameter space and came back to the initial point. In this case you can quickly see that the Berry phase becomes a geometrical object:

\begin{align} \gamma_n(t) &= i\int_0^\tau dt' \, \langle n(t') |\dot{n}(t')\rangle \\ &= i\oint dX\, \langle n(r) |\nabla_X {n}(r)\rangle \end{align}

It also turns out that this term can no longer be reabsorbed by a continuous reparametrization of the eigenstate $|n(t)\rangle$. In some situations it is even possible to observe this term.

Sometimes it is this specific term that is called Berry phase.

In short:

The Berry phase does not disappear when including the first order corrections. Simply for an open adiabatic path the Berry phase can be discarded and this is probably what was intended in the paper you linked. For a cyclic adiabatic  transformation the Berry phase becomes a geometrical object and can even be observed.

Answer 2

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.