# 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?

• What do you mean by "differentiating eigenvectors are only differentiating the coefficients" ? What coefficients ? What is $\psi_{nk}$ in terms of $u_k$ ? In your opinion, what should you be differentiating ?
• If we are expressing $H(k)$ as a matrix, we are choosing a basis vector to form a representation. Eigenvectors of $H(k)$ matrix are only coefficients with respect to these basis vectors. Questions are edited to be more clear. In my opinion, both the coefficients and eigenvectors should be differentiated. Mar 6, 2016 at 11:16
Let me try to explain this using the Haldane model for Graphene. The tight binding Hamiltonian in Second Quantization looks like: $$H_{Total}=\underbrace{-t_{1}\sum_{r_i,\delta}(a_{r_i}^{\dagger}b_{r_{i}+\delta}+b_{r_{i}+\delta}^{\dagger}a_{r_i})}_\text{NN hopping}-\underbrace{t_{2}\sum_{i,j}(e^{i\phi}a_{i}^{\dagger}a_{j}+h.c)}_\text{A to A NNN hopping}-\underbrace{t_{2}\sum_{i,j}(e^{-i\phi}b_{i}^{\dagger}b_{j}+h.c)}_\text{B to B NNN hopping}.$$ Now we go to the fourier space representation of this Hamiltonian to get $\mathcal{H}(k)$. $$\bf h(k)=\sum_{i=1,2,3}d_{i}(k).\sigma_{i}$$ 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: $$F^{xy}_{n}(\bf k)=2i\sum_{m\neq n}Im\frac{\langle\ n|(\partial_{k_x}h(k))|m\rangle\langle\ m|(\partial_{k_y}h(k))|n\rangle}{(E_n-E_m)^2}$$ Now the eigenvectors of the Hamiltonian can be written as: $$|m\rangle=\frac{1}{\sqrt{2d(d+d_3)}}\left( \begin{array}{c} d_3+d \\ d_1-id_2 \end{array} \right)$$ $$|n\rangle=\frac{1}{\sqrt{2d(d-d_3)}}\left( \begin{array}{c} d_3-d \\ d_1-id_2 \end{array} \right)$$ 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.
• The $\mathbf{h(k)}$ matrix is not the full hamiltonian - it is called the kernel of the hamiltonian. The full hamiltonian is $\mathcal{H} = \mathbf{\Psi^\dagger h(k) \Psi}$. i think the OP is asking is why the derivatives of $\mathbf{\Psi}$ aren't considered when calculating the connection.