1
$\begingroup$

I have been trying to derive the dispersion relation for the low energy Hamiltonian described in Ref. 1.

The relevant equations are (1a) through (1d). I will re-write the equations here to save time: $$H_0 = \nu_F\left(\tau\sigma_x p_x + \sigma_y p_y\right)+\frac{m}{2}\sigma_z$$ $$H_{SOC} = \tau s_z\left(\lambda_c\sigma_+ + \lambda_v \sigma_-\right)$$ $$H_{ex} = -s_z\left(B_c\sigma_+ + B_v\sigma_-\right)$$ $$H_R = \lambda_R\left(\tau s_y \sigma_x + s_x \sigma_y\right)$$

The authors then state "The Pauli matrices $s_\alpha$ and $\sigma_\alpha$ $(\alpha = 0,x,y,z)$ refer to the real spin and orbital pseudo-spins respectively, and $\sigma_\pm =\frac{1}{2}\left(\sigma_0 \pm \sigma_z\right).$" $p$ refers to momentum (relative to the K or K' valley I think?) and the other constants are fitting parameters.

My issue is that I don't understand the shorthand that they are using here. When they refer to the Pauli matrices of the real spin and orbital pseudo-spin am I supposed to assume that they have the following form: $$\sigma_0=s_0=\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix},\space\space \sigma_x=s_x=\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}, \space\space \sigma_y=s_y=\begin{pmatrix}0 & -i\\i & 0\end{pmatrix}, \space\space \sigma_z=s_z=\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}$$ Or are they different for $\sigma_{\alpha}$ and $s_{\alpha}$? My confusion stems from the following features of the Hamiltonian: (Note that $\tau = \pm 1$ to select between the K and K' valleys)

  • Am I supposed to use $s_z$'s eigenvalue or Pauli matrix? If I'm using the eigenvalue then $s_z = \pm 1$ for spin-up and spin-down electrons. However, what does that mean for $s_y$ and $s_x$?
  • Do I simply multiply these matrices together or do I take their direct product? If I multiply them like normal matrices then I'll end up with a $2X2$ square matrix, but if I take the Kronecker Product I end up with a $4X4$ square matrix.

It's worth noting that these authors seemed to have borrowed this particular low-energy effective Hamiltonian (LEH) from earlier works with Graphene. Specifically from Ref. 2.

In this paper the authors introduce a Rashba Hamiltonian to the Graphene Hamiltonian to take into account spin-orbit coupling. In the end they derive a dispersion relationship which is exactly what I am trying to do here. To summarize, my main question is:

  • What is the full matrix form of the Rashba Hamiltonian ($H_R$) for 2D materials?
  • How can this be used to derive the dispersion relationship like that shown in Ref. 2 (eq. 5).

References:

  1. J. Qi, X. Li, Q. Niu, and J. Feng, "Giant and tunable valley degeneracy splitting in $\require{mhchem}\ce{MoTe2}$", Phys. Rev. B 92, 121403 (2015), arXiv:1504.04434.
  2. M. Gmitra and J. Fabian, "Graphene on transition-metal dichalcogenides: A platform for proximity spin-orbit physics and optospintronics", Phys. Rev. B 92, 155403 (2015) , arXiv:1506.08954.
$\endgroup$
1
  • 1
    $\begingroup$ I don't know the answer to your questions but I'd be shocked if expressions like $s_x\sigma_y$ did not intend a tensor product/Kronecker product (ie you end up with 4x4 matrices). Otherwise why distinguish $s,\sigma$ at all? $\endgroup$
    – jacob1729
    Commented Oct 1, 2020 at 21:16

1 Answer 1

1
$\begingroup$

After discussing this problem with one of the original authors of Ref. 1 (Xiao Li), I can now answer the first part of this question. The Rashba Hamiltonian does indeed assume that we take the direct product of $s_y$ and $\sigma_x$ as well as $s_x$ and $\sigma_y$. Apparently, according to another author of a similar paper (Klaus Zollner Ref. 3) the direct product is assumed but in his own papers he explicitly writes the Hamiltonians using the Kroncker direct product symbol. Re-writing the Rashba Hamiltonian in this way we obtain the following:

$$H_R = \lambda_R \left( \tau s_y \otimes \sigma_x + s_x \otimes \sigma_y \right)$$ Therefore: $$H_R = \begin{pmatrix} 0 & 0 & 0 & (i-i\tau)\lambda_R\\ 0 & 0 & (-i-i\tau)\lambda_R & 0\\ 0 & (i+i\tau)\lambda_R & 0 & 0\\ (-i+i\tau)\lambda_R & 0 & 0 & 0\\ \end{pmatrix}$$

This answers the first question of what form the Rashba Hamiltonian takes. As far as the other Hamiltonians are concerned they need to have an identity matrix form a direct product with them to obtain a 4x4 matrix for each term thus allowing the Hamiltonians to be added together.

In order to solve the resulting Hamiltonian the authors Li and Zollner did not actually derive a dispersion relationship for the final Hamiltonian. Instead they plugged in numerical values for the various constants and then numerically calculated the eigenvalues of the resulting matrix. This is how the TB Hamiltonian curves were plotted in Ref. 1. In order to determine whether or not an eigenvalue corresponds a spin-up or spin-down electron Li suggests that one merely take the eigenvalue and its associated eigenvector and use the following equation: $$\pm1 = \langle \psi_\alpha \mid s_0 \otimes s_z \mid \psi_\alpha \rangle$$ Where $\psi_\alpha$ is the resulting eigenvector for a given eigenvalue calculated from the Hamiltonian. The resulting sign indicates whether or not the eigenvector is for a spin-up or spin-down electron.

Regarding the second part of the question, I have not yet found a solution.

References:

  1. Klaus Zollner, Paulo E. Faria Junior, and Jaroslav Fabian, "Proximity exchange effects in $MoSe_2$ and $WSe_2$ heterostructures with $CrI_3$: Twist angle, layer, and gate dependence", Phys. Rev. B 100, 085128 (201), arXiv:1902.01631.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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