eigenvectors of tight binding Hamiltonian I am trying to calculate berry connection using tight binding method. The most important part is to calculate $\partial_k u_k(x)$, where $u_k(x)$ is the periodic part of bloch waves, i.e. $\psi_{nk}(x) = e^{ikx}u_k(x)$.
Suppose tight binding hamiltonian is $H(k)$, in my opinion, $\partial_k u_k(x)$ cannot be obtained by differentiating $H(k)$'s eigenvector. This is because to express Hamiltonian $\hat{H}$ as a matrix form, we use $k$ dependent basis vectors, i.e. $\psi_{nk} = \sum_R e^{ikR}|Rn\rangle$, where $|Rn\rangle$ is atom orbitals. Eigenvectors of $H(k)$ is just coefficients of these basis vectors. Thus, differentiating eigenvectors are only differentiating the coefficients and if we need $\partial_k u_k(x)$, we need to also differentiate basis vectors.
However, it seems many literature are only differentiating eigenvectors to get $\partial_k u_k(x)$. For example, pythtb package is doing this (http://www.physics.rutgers.edu/pythtb/examples.html). Also, a lecture note on the web claims eigenvectors of $H(k)$ is $u_k$, (http://www-personal.umich.edu/~sunkai/teaching/Fall_2013/chapter5.pdf) but I cannot understand its arguments.
Can anyone help me solve such a problem?
 A: Let me try to explain this using the Haldane model for Graphene. The tight binding Hamiltonian in Second Quantization looks like:
\begin{equation}
H_{Total}=\underbrace{-t_{1}\sum_{r_i,\delta}(a_{r_i}^{\dagger}b_{r_{i}+\delta}+b_{r_{i}+\delta}^{\dagger}a_{r_i})}_\text{NN hopping}-\underbrace{t_{2}\sum_{i,j}(e^{i\phi}a_{i}^{\dagger}a_{j}+h.c)}_\text{A to A NNN hopping}-\underbrace{t_{2}\sum_{i,j}(e^{-i\phi}b_{i}^{\dagger}b_{j}+h.c)}_\text{B to B NNN hopping}.
\end{equation}
Now we go to the fourier space representation of this Hamiltonian to get $\mathcal{H}(k)$.
\begin{equation}
\bf h(k)=\sum_{i=1,2,3}d_{i}(k).\sigma_{i}
\end{equation}
Where the constant term is not mentioned as it just shifts the energy level and no interesting topology arises from it. The berry curvature is given by the following formula:
\begin{equation}
F^{xy}_{n}(\bf k)=2i\sum_{m\neq n}Im\frac{\langle\ n|(\partial_{k_x}h(k))|m\rangle\langle\ m|(\partial_{k_y}h(k))|n\rangle}{(E_n-E_m)^2}
\end{equation}
Now the eigenvectors of the Hamiltonian can be written as:
\begin{equation}
|m\rangle=\frac{1}{\sqrt{2d(d+d_3)}}\left( \begin{array}{c}
d_3+d \\
d_1-id_2
\end{array} \right) 
\end{equation}
\begin{equation}
|n\rangle=\frac{1}{\sqrt{2d(d-d_3)}}\left( \begin{array}{c}
d_3-d \\
d_1-id_2
\end{array} \right) 
\end{equation}
So, it's just a matter of finding the coefficients of the pauli matrices and then taking their derivatives. One nice way is to actually write the hamiltonian in terms of pauli matrices.
