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?

  • $\begingroup$ 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 ? $\endgroup$
    – Adam
    Mar 6, 2016 at 9:56
  • $\begingroup$ 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. $\endgroup$
    – atbug
    Mar 6, 2016 at 11:16

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – Sidd
    Mar 24, 2020 at 3:09

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