Kagome Lattice: Spin-orbit coupling Hamiltonian in tight-binding models

Question

Consider spin-orbit coupling (of strength $\lambda_1$) on lattice, with the below Hamiltonian

$$H = i \lambda_1 \sum_{<ij>} ~\frac{E_{ij} \times R_{ij}}{|E_{ij} \times R_{ij}|} \cdot \sigma ~c_i^\dagger c_j $$

with lattice sites $i, j$, nearest-neighbor connecting sites vector $R_{ij}$, E-field $E_{ij}$ and Paulis matrix $\sigma$.

Consider 2D plane, so $R_{ij} = (R_{ij}^x, R_{ij}^y, 0)$ and choose E-field $E_{ij} = (E_{ij}^x, E_{ij}^y, 0)$, with $E_{ij}^x, E_{ij}^y >0$. Factor in above Hamiltonian is

$$\frac{E_{ij} \times R_{ij}}{|E_{ij} \times R_{ij}|} \cdot \sigma = \sigma_z ~\text{sgn} (E_{ij}^x R_{ij}^y - E_{ij}^y R_{ij}^x)$$

Paper here considers 2d Kagome lattice, with Hamiltonian for spin orbit appearing in 1st line of Eq. (1). Going into k-space, the authors shown in Eq. (2) that spin-orbit Hamiltonian gives terms with cosines, like $\cos (k_x)$.

However, it looks to me like terms should be sines, like $\sin(k_x)$.

Consider the 2-d Kagome lattice shown in Fig 1 of the paper. There will be terms proportional to that below to make the horizontal part of the lattice along x direcion, where $R_{ij}^y = 0$:

$$\sum_n \text{sgn} (- E_{ij}^y R_{ij}^x) c_n^\dagger c_{n+1} \to \sum_k e^{-i k_x} c_k^\dagger c_k $$

and

$$ \sum_n \text{sgn} (- E_{ij}^y R_{ij}^x) c_n^\dagger c_{n-1} \to \sum_k - e^{+i k_x} c_k^\dagger c_k $$

But beacuse of opposite direction of $R_{ij}^x$ in top and bottom cases sgn function will be different, so that exponentials $\exp$ add to make a $\sin(k_x)$ and not a $\cos(k_x)$ as is put in second line of Eq. (2) of the paper.

Where is a gap in my understanding of spin orbit on 2-d lattice?

Answer 1

I think the authors got it right.

The subtlety lies in the definition of $\mathbf{E}_{ij}$ and $\mathbf{R}_{ij}$. The authors consider $\mathbf{E}_{ij}$ as the electric field felt by the electron during hopping from $j$ to $i$ (although the field is non-uniform, the direction of the field does not change throughout a bond). Thanks to Clara for pointing this out.

In the above, I have shown three unit cells along $x$ direction and enumerated them as '$-1$', '$0$', and '$+1$'. The charge centers are shown as red '+' symbol. The direction of the electric field (black arrows) at the center of each bond along the horizontal direction is shown (since the question concerns the hopping along $x$ direction, I left the other bonds to avoid clutter). Notice the staggering nature of the electric field along the $x$-direction.

1. If we focus on the hopping along the $x$ direction, and consider the term where an electron inside the unit cell '$0$' hops from site '2' to '1', then $\left(\mathbf{E}_{1,0;~2,0}\times\mathbf{R}_{1,0;~2,0}\right)$ points in the $+z$ direction and let the magnitude be $\alpha$. This hopping will contribute to the term $H_{12}$ of the Hamiltonian. Here I use a little different notation to distinguish the indices of the unit cells and the lattice sites. The term $\mathbf{R}_{a,b;~c,d}$ represents: vector that points to site $a$ of unit cell $b$, from site $c$ of unit cell $d$.
2. Site '$2$' of the unit cell '-1' also contributes to the wavefunction at site $1$ of the unit cell '$0$'. This hopping will contribute to the term $H_{12}$ of the Hamiltonian. Now, notice that the electric field at the bond between the two aforementioned sites is opposite to the previous case (where hopping happened entirely within the unit cell '0'). Moreover, the direction of hopping is also reversed. Therefore the cross-product $\left(\mathbf{E}_{1,0;~2,-1}\times\mathbf{R}_{1,0;~2,-1}\right)$ still points in the $+z$ direction with magnitude $\alpha$.

Since both the terms that contribute to site '$1$' have the same sign with different Bloch factors $\exp(ik_x)$ and $\exp(-ik_x)$, therefore the resulting term will be ~$\cos(k_x)$.

Answer 2

I think this would probably be more appropriate as a comment, but I want to include an image, so I am writing it as an answer. Sorry about that.

@Mehedi's explanation is appealing, but I don't think it corresponds to what the authors wrote. They did not say that $E_{ij}$ is the electric field felt by the electron at site $i$ due to an ionic core at site $j$, but that "$E_{ij}$ (is) the electric field from neighboring ions experienced along $R_{ij}$." Here is a figure from one of their earlier papers:

If we look at hopping processes between sites 1 and 2, for example, the electric field arises from the ion in the hexagon "beneath" this link, so the electric field points vertically upwards. I do not see how the 12 hopping sees a different direction of the field to the 21 hopping. Why would $E$ have a sign reversal?