1
$\begingroup$

I have a Bloch Hamiltonian H(k) which describe a triangular lattice with six atoms inside one unite cell. This lattice is nothing but group six atoms instead of two in a honeycomb lattice. So as you can see in the band structure, the Dirac cones at K and K' points are folded back to the center of first Brillouin zone.

$$H(k)= \begin{bmatrix}0 & e^{-i\left(-\frac12 k_x +\frac{\sqrt3}{2}k_y\right)} & 0 &e^{-k_x} &0 & e^{i\left(\frac12 k_x +\frac{\sqrt3}{2}k_y\right)}\\ e^{i\left(-\frac12 k_x +\frac{\sqrt 3}2k_y\right)} & 0 & e^{ik_x} & 0 & e^{-i\left(\frac12 k_x + \frac{\sqrt 3}2k_y\right)} & 0\\ 0 & e^{-k_x} & 0 & e^{-i\left(-\frac12 k_x -\frac{\sqrt 3}2k_y\right)} & 0 & e^{-i\left(-\frac12 k_x +\frac{\sqrt 3}2k_y\right)}\\ e^{ik_x} & 0& e^{i\left(-\frac12 k_x -\frac{\sqrt 3}2k_y\right)} & 0 & e^{-i\left(\frac12 k_x -\frac{\sqrt 3}2k_y\right)} & 0 \\ 0 & e^{i\left(\frac12 k_x+ \frac{\sqrt 3}2k_y\right)} & 0 & e^{i\left(\frac12 k_x- \frac{\sqrt 3}2k_y\right)} & 0 & e^{ik_x}\\ e^{-i\left(\frac12 k_x \frac{\sqrt 3}2k_y\right)} & 0 & e^{i\left(-\frac12 k_x \frac{\sqrt 3}2k_y\right)} & 0 & e^{ik_x} & 0\end{bmatrix}$$

Here $k_x$ and $k_y$ are the crystal momentum. The band structure is as following: enter image description here

What I try to do is to expand the full 6 by 6 Hamiltonian around \Gamma point to get a 4 by 4 effective Hamiltonian.

I tried to find the four degenerate states at $\Gamma$ point, $\psi_{1,2,3,4}$ and get a 4 by 4 effective matrix in this way: $\langle\psi_m|H(k)|\psi_n\rangle, m,n=1,2,3,4$, however, the linear dispersion relationship lost in the 4 by 4 effective Hamiltonian.

I think this may due to $\psi_{1,2,3,4}$ are not eigenstates of H(k), so the approximation $\langle\psi_m|H(k)|\psi_n\rangle, m,n=1,2,3,4$ fails.

I am wondering if there are some ways to do this job?

Thanks.

$\endgroup$
  • $\begingroup$ The 4 states with zero energy (at the $\Gamma$ point) are eigenstate of $H$. Why do you say they are not ? They may not form an orthonormal basis of that subspace (depending on your diagonalization algorithm) but they are at least linearly independent. $\endgroup$ – PinkFloyd Oct 6 '16 at 18:10
  • $\begingroup$ @PinkFloyd The 4 states are the eigenstates when k = 0 for the H, away from k = 0, they are not the eigenstates for H(k). $\endgroup$ – Aaron Oct 7 '16 at 2:49
  • $\begingroup$ @Aaron All the states in the band structure are by definition Eigenstates of the system. Why don't you just diagonalize the Hamiltonian and throw away the first and last row and column i.e. take the subspace of the 4 medium energy eigenstates? $\endgroup$ – Jannick Oct 10 '16 at 13:17
  • $\begingroup$ @Jannick. I want to write down a 4 by 4 effective Hamiltonian with two 2 by 2 block described by Dirac equation, one for K valley and one for K' valley. If I diagonalize it and take the subspace of 4 medium energy states, I will not get this information. $\endgroup$ – Aaron Oct 11 '16 at 1:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.