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 $R_{ij}^y = 0$:

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


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

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?

  • 1
    $\begingroup$ Hi, we've noticed that you have made a large number of minor edits to this post. Please be mindful that every edit bumps the post in the "active" tab of the site and try to make your edits substantial. If you foresee improving this post repeatedly, maybe collect several edits and make them in one go instead of submitting them individually. $\endgroup$ – Chris Jan 10 at 19:41
  • 1
    $\begingroup$ From what I see, $H$ must be hermitian, and so because of the factor of $i$ outside the sum, the sum itself must be antihermitian. This (I think) would lead to factors of $\sin k$, rather than $\cos k$, in agreement with you. But maybe we are missing something obvious. Have you considered asking the authors of the paper directly? $\endgroup$ – Clara Diaz Sanchez Jan 11 at 8:54

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.

enter image description here

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 ~$\cos(k_x)$.

| cite | improve this answer | |
  • 1
    $\begingroup$ This version of the answer is a revised version. The previous version was too compact and the arguments were not explained. In this version, I tried to put all the detail so that the arguments are transparent. $\endgroup$ – Mehedi Hasan Feb 4 at 17:11

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:

Kagome lattice, with electric fields

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?

| cite | improve this answer | |
  • $\begingroup$ Do you still have the same opinion? $\endgroup$ – Mehedi Hasan Feb 26 at 5:53
  • $\begingroup$ I'm more or less convinced by your answer (say 99%), but I still find it a little strange. I think we can agree that the authors of the paper made a poor job of explaining the reasoning. $\endgroup$ – Clara Diaz Sanchez Feb 26 at 17:38
  • $\begingroup$ I am not educated enough to make such a comment on a paper of X.G. Wen. $\endgroup$ – Mehedi Hasan Feb 28 at 14:26

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.