How does the Berry curvature relate to the hopping strengths in the Haldane model?

Question

Take Haldane's Hamiltonian, as quoted from Fruchart et al.'s An Introduction to Topological Insulators:

3.5.3. Haldane's Hamiltonian

The first quantized Hamiltonian of Haldane's model can be written as: $$ \hat H = t \sum_{⟨i,j⟩} |i⟩⟨j| + t_2 \sum_{⟪i,j⟫} |i⟩⟨j| + M \left[ \sum_{i\in A} |i⟩⟨i| - \sum_{j\in B}|j⟩⟨j|\right] \tag{30} $$ where $|i⟩$ represents an electronic state localized at site $i$ (atomic orbital), $⟨i,j⟩$ represents nearest neighbors lattice sites $i$ and $j$, $⟪i,j⟫$ represents second nearest neighbors sites $i$ and $j$, $i\in A$ represents sites in the sublattice $A$ (resp. $i\in B$ in the sublattice $B$). This Hamiltonian is composed of a first nearest neighbors hopping term with a hopping amplitude $t$, a second neighbors hopping term with a hopping parameter $t_2$, and a last sublattice symmetry breaking term with on-site energies $+M$ for sites of sublattice $A$, and $-M$ for sublattice $B$, which thus breaks inversion symmetry. Moreover, the Aharonov-Bohm phases due to the time-reversal breaking local magnetic fluxes are taken into account through the Peierls substitution: $$ t_{ij} \to t_{ij} \exp\mathopen{}\left( -i \frac{e}{\hbar} \int_{\Gamma_{ij}} \vec A \cdot \mathrm d\vec \ell \right)\mathclose{} \tag{31} $$ where $t_{ij}$ is the hopping parameter between sites $i$ and $j$, and where $\Gamma_{ij}$ is the hop trajectory from site $i$ to site $j$ and $\vec A$ is a potential vector accounting for the presence of the magnetic flux.

How do the hopping parameters $t$ and $t_2$ of the model over the honeycomb lattice relate to the local Berry curvature of the unit cell, if at all? I do not think it is obvious, but the Berry connection and Hamiltonian depend on each other. This post discusses involving the Berry connection in the Hamiltonian.

I am more concerned about physical consequences of any potential relationship between the two. To take a wild guess as an example, does the local Berry curvature in k-space that corresponds to next-nearest-neighbor hopping paths in real space = 0? On a semi-related note, must the energy of the particle concerned be minimum on their hopping paths? Any references?

I would appreciate any advice or resources! Thanks!

Answer 1

this is something not that complicated. Assume you have an electron sitting in a lattice site, then it hops to a neighbour. while doing this electron feels electromagnetic potential obviouslay, so naturally when it reaches to neighbour site, it will gain that berry phase.

it is just a Ab effect to an electron while it is moving under electro magnetic field.

when electron hops, it means that it is moving under electromagnetic field, so naturally it must acquire AB phase. it is that simple.

let me give and example assume that we have a free electron system $$h=p^2/2m$$ now we turn on electro magnetic field the hamiltonian will be $$h=(p-A)^2/2m$$ so this is how you write an hamiltonian for a free elctron moving in a EM field.

what about lattice system in that case you can make actually the same thing. assume we have an hamiltonian in lattice w/o em field.

$$H=\sum_k c_{k,i}^\dagger h_{ij}(k)c_{k,j}$$

okay $k$ is momentum so we will do the same minimal coupling where i, j are some degree of freedom they are not lattice sites we are in momentum space.

$$H=\sum_k c_{k,i}^\dagger h_{ij}(k-A)c_{k,j}$$

so it is simple as that, if you perform a fourier transform and go to position space, you will exaclty get the terms in your question.

and qualitatively it means that, when you turn on $A$ electron acquire AB phase when they are hopping from one site to other.

edit: Let me give an example, asume we have an lattice hamiltonian given by; $$H=\sum_k c^\dagger_k\cos(k)\sigma_z c_k$$ let $k=k-A$ than we have $$H=\sum_k c^\dagger_k\cos(k-A)\sigma_z c_k$$

lets make a FT

$$c_k=\sum_j c_j \exp(ikj)$$

so by re writing $ck$ as above and substituting in hamiltonian we have

$$H=\frac{1}{2}\sum_{k,j,j'} c^\dagger_{j'}e^{(i(k-A)+i(j-j')k)}\sigma_z c_j+c^\dagger_{j'}e^{(-i(k-A)+i(j-j')k)}\sigma_z c_j$$ thus $$H=\frac{1}{2}\sum_{k,j,j'} c^\dagger_{j'}e^{-i(A)}e^{i(1+j-j')k)}\sigma_z c_j+c^\dagger_{j'}e^{i(A)}e^{i(j-j'-1)k)}\sigma_z c_j$$ by using $$\delta_{j,0}=\sum_{k}e^{ikj}$$ we have $$H=\frac{1}{2}\sum_{j,j'} c^\dagger_{j'}e^{-i(A)}\delta_{(1+j-j'),0}\sigma_z c_j+c^\dagger_{j'}e^{i(A)}\delta_{(j-j'-1),0}\sigma_z c_j$$ so $$H=\frac{1}{2}\sum_{j} c^\dagger_{j+1}e^{-i(A)}\sigma_z c_j+c^\dagger_{j-1}e^{i(A)}\sigma_z c_j$$