Bloch's theorem and Bloch's state The question is not so much about the theorem, but more about what it means in this context: see this link.
So yes, because of Bloch's theorem the Hamiltonian eigenstates in a crystalline system can be written as
\begin{align}
\psi_{n,\vec{k}}(\vec{r})=e^{i\vec{k}\cdot\vec{r}}u_{n,\vec{k}}(\vec{r}),
\end{align}
and so the Berry connection can be defined:
\begin{align}
A_{n}(\vec{k})=i\langle n(\vec{k})|\nabla_{\vec{k}}|n(\vec{k})\rangle,
\end{align}
but what in the world is $|n(\vec{k})\rangle$?
I've read a few articles on topological insulators and they always seem to start off with the Bloch wavefunction $e^{i\vec{k}\cdot\vec{r}} u_k(\vec{r})$, and then somehow they magically get the ket $|u(\vec{k})\rangle$ from which the Berry connection is defined... is $|u(\vec{k})\rangle$ the column vector comprised of the Fourier coefficients of $u_\vec{k}(\vec{r})$ w.r.t. $e^{i\vec{G}\cdot\vec{r}}$ or what?
 A: I have discussed this question with my friend, and he gives a satisfying answer:
the Bloch state is $|\psi_k\rangle$, then we define a new state $|u_k\rangle$ by :
$|u_k\rangle=e^{-ik\cdot\hat{r}}|\psi_k\rangle$. Notice that $\hat{r}$ in phase factor is the position operator, not a parameter in real space. You can use this definition and the Bloch theorem: $T(R)|\psi_k\rangle=e^{ik\cdot R}|\psi_k\rangle$ (where $T(R)=e^{i\frac{\hat{p}\cdot R}{\hbar}}$, $p$ is the momentum operator) to check the wavefunction of $|u_k\rangle$ is periodic.
The way to understand the definition is that: $e^{-ik\cdot\hat{r}}$ is a translation operator in momentum space, so if we use the wavefunction in momentum space, we will get :$u_k(k')=\psi_k(k'+k)$.
But one important thing to remember: $|u_k\rangle$ is not the eigenstate of the origional Hamiltonian $H$, but it's the eigenstate of the modified Hamiltonian: $H(k)=e^{-ik\hat\cdot{r}}He^{ik\cdot\hat{r}}$. When we talk about the band theory, we usually talk about Hamiltonian $H(k)$ and states $|u_k\rangle$.
A: First recall that when talking about Berry phases we are working in the adiabatic approximation where the Hamiltonian is allowed to be controlled by some slow varying variables, which in this case are $\vec{k}(t)$. In this case the underlying assumption is that all of the time dependence is encapsulated in these variables $H(\vec{r}, \vec{p}, \vec{k}(t))$.
$|n(\vec{k})\rangle$ are the eigenstates of the Hamiltonian which are allowed to depend on $\vec{k}$ since we are working in the adiabatic approximation. Recall now that in the Berry phases we look at accumulated phase from a cyclic process $\gamma(0) = \gamma(T)$, and assume w.l.o.g that we start with the eigenstate $n(\vec{k})$ which will not change under the adiabatic approximation. In this case the connection is given by the infinitesmal change of the state due to changes in $k$
\begin{equation}
i A_n(\vec{k}) = \langle n(\vec{k}) | \nabla_k | n(\vec{k})\rangle
\end{equation}
