I'm doing numerical simulations. I have the Haldane model in a honeycomb lattice where

$$ H = \sum \limits_{<ij>}a^\dagger_i b_j + h.c $$

Where $i$ belongs to sublattice $A$, and $j$ to sublattice $B$, which comes from the fact that there are two atoms per unit cell. $a$ is the annihilator on sublattice $A$ and $b$ on sublattice $B$.

This operator is Hermitean. I can use its Fourier representation

$$ H \sim \sum \limits_k H(k) e^{ikr} $$

Where $r$ denotes lattice vectors and $k$ the corresponding lattice momentum. The amplitudes $H(k)$ are not Hermitean but they satisfy $H(k)^\dagger = H(-k)$ obviously, so the above sum remains Hermitean if I take out pairs of $(k,-k)$. I cannot figure if by discretizing the elementary cell in $k$-space, thus effectively taking out some of the $k$'s in the above sum, the Hermiticity of this truncated Hamiltonian would get lost.

  • $\begingroup$ Doesn't the discretized sum have both $k$ and $-k$? $\endgroup$ – leongz May 27 '13 at 8:16
  • 1
    $\begingroup$ I have a 2D lattice. So I have two reciprocal vectors $\vec{G}_1$ and $\vec{G}_2$. I construct an $a\times b$ discretized lattice where $\{(k_i,k_j)\} = i/a \cdot \vec{G}_1 + j/b \cdot \vec{G}_2$ with $i,j \in [0,a-1]\times[0,b-1]$ If I take any discretized $k$-point, send it to $-k$, i want to be sure that by there exists $c,d \in \mathbb{Z}$ such that $-\vec{k} + c\vec{G}_1 + d\vec{G}_2 = k$, which ammounts to saying that the -k is mapped correctly back into the brillouin zone onto k, by symmetry. After reading what I just wrote, it seems obvious this is the case, so hermiticity is ok $\endgroup$ – Mathusalem May 27 '13 at 8:35

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