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

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!

• Also, I would appreciate any insight on appropriate values for the hopping parameters. In all references I've seen, $t_2$ is some small fraction of $t$. Jan 19, 2019 at 13:30
• For future reference: please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines; for formulae, use MathJax instead. I've transcribed your text in this question but you should do it yourself in future posts. Jan 20, 2019 at 18:06
• As for your comment: there is a strict limit of $t_2 \leq t/3$, which is explained in the original Haldane paper. If this is violated then the bands will overlap in the energy domain (if I remember correctly). Jan 20, 2019 at 18:07
• @EmilioPisanty berry curvature is defined on some parameter space it is not necesarrily a K space. Jan 22, 2019 at 11:24
• Are you talking about Peierls substitution? If so, you can have a look at the Wikipedia page of "Peierls substitution", there are some arguments about how the tight binding parameters changes when there is a magnetic field. While the "rigorous derivation" in that page is wrong. The correct calculation can be found in an old article written by Luttinger in 1951. See PhysRev.84.814 Jan 25, 2019 at 2:00

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 $$H=\frac{1}{2}\sum_{j} c^\dagger_{j+1}e^{-i(A)}\sigma_z c_j+c^\dagger_{j-1}e^{i(A)}\sigma_z c_j$$

• Thank you for your answer. However, I am having trouble digesting the relationship between Berry curvature (or phase, from your answer), and the hopping terms $t$ and $t_2$ from the Haldane model. Would you mind clarifying this? From your answer, I got that the Hamiltonian allows for comparing the hopping strengths with Berry phase via Fourier transforms. However, I had known this from the SE post I linked to in the question - and I was hoping for a more specific answer. Please let me know if I’m missing something! Thanks! Jan 22, 2019 at 12:38
• okay, make a foruier transform to you haldane model without magnetic field. then you will have momentum psace hamiltonian than change k to k-A in that hamiltonian and again go back to position space with fourier transform than you will understand, everything Jan 22, 2019 at 13:57
• Thank you for the suggestion. The Fruchart et al. reference provides a sublattice basis of the Hamiltonian post-Fourier Transform. However, all terms involve a trigonometric function involving the product of $k$ with a Pauli matrix. For example: $h_x (k) = t (1 + cos(k.b1) + cos(k.b2))$. It is not clear to me how I could take the inverse Fourier transform with the $b_i$ Pauli matrices in the way. When I used Mathematica with $k \rightarrow k-A$ and all $b_i = 1$, however, I got something promising. Is setting $b_i = 1$ valid? How can I use this new info to understand Berry curvature? Jan 25, 2019 at 10:39
• $c_k=\sum_j c_j \exp(ikj)$ and $\delta_{j,0}=\sum_k \exp(ikj)$ use these identities also wrtie the cos sin terms in terms of exponantieals. Jan 25, 2019 at 12:42
• @TribalChief see the edit Jan 25, 2019 at 14:42