H({\bf R(t)})\ket{\psi(t)} = i\hbar \frac{d}{dt}\ket{\psi(t)}\end{aligned}$$ which translates into the differential equation, $$\begin{aligned}
E_n({\bf R}(t))\ket{n({\bf R}(t))} & = & \hbar \left( \frac{d}{dt} \theta(t) \right) \ket{n({\bf R}(t))} \nonumber \\ 
& & + i\hbar \frac{d}{dt} \ket{n({\bf R}(t))}\end{aligned}$$ Taking the scalar product with $\bra{n({\bf R}(t))}$ and assuming the state is normalised, we get, $$\begin{aligned}
E_n({\bf R}(t))- i\hbar \bra{n({\bf R}(t))} \frac{d}{dt} \ket{n({\bf R}(t))} = \hbar \left( \frac{d}{dt} \theta(t) \right) \nonumber\end{aligned}$$ $$\begin{aligned}
 \theta(t) &=& \frac{1}{\hbar} \int^t_0 E_n({\bf R}(t'))dt' \nonumber \\
&&- i \int^t_0  \bra{n({\bf R}(t'))} \frac{d}{dt'} \ket{n({\bf R}(t'))} dt'\end{aligned}$$ The first term of the phase is just the conventional dynamical phase that arises due to time evolution of the Hamiltonian. The negative of the second term is what is known as the Berry Phase $\gamma_n$, $$\begin{aligned}
\gamma_n = i \int^t_0  \bra{n({\bf R}(t'))} \frac{d}{dt'} \ket{n({\bf R}(t'))} dt'\end{aligned}$$ This term arises because the states at $t$ and $t+dt$ are not ‘identical’ and a phase is picked up that depends on the trajectory in the parameter space. From the previous expression, $$\begin{aligned}
\label{bphase}
\gamma_n &= & i \int^t_0  \bra{n({\bf R}(t'))} \nabla_{\bf R} \ket{n({\bf R}(t'))} \frac{d{\bf R}}{dt'} dt'\\
 &=& i \int^{{\bf R}_t}_{{\bf R}_0}  \bra{n({\bf R})} \nabla_{\bf R} \ket{n({\bf R})} d{\bf R}  \\
 &=&  \int^{{\bf R}_t}_{{\bf R}_0}  {\bf A}_n({\bf R}) d{\bf 