Help with definitions of SOC and ferromagnetic exchange terms in MoS2 Hamiltonian

Question

I am trying to write eq. 2 in this PDF into matrix form (reproduced below) for numerical purposes. There are some definitions that I am not too familiar with, and don't seem to be given in the paper.

$$H_0=\alpha k^2+\begin{pmatrix} 0 & \beta k_+^2 & 0 & 0 \\ \beta^* k_-^2 & 0 & 0 & 0 \\ 0 & 0 & 0 & \beta k_+^2 \\ 0 & 0 & \beta^* k_-^2 & 0 \end{pmatrix}$$ with $k_\pm=k_x\pm k_y$.

For the atomic SOC effect, $H_{so}=\lambda_{so}L\cdot S$ is diagonal in the selected basis. The ferromagnetic exchange term is added as $H_M=M_{\sigma z}\otimes1$. So the the total Hamiltonian can be written as: $$H=H_0+H_{so}+H_M$$

I figured that $\alpha k^2$ is added to the diagonal of eq. 1, that $k^2 = k_x^2 + k_y^2$, and that $H_M = M_{\sigma z} \otimes 1 = \sigma_z \otimes 1$ (where $\sigma_z$ is the Pauli z matrix, and $1$ is the $2\times 2$ identity matrix).

However, I am stumped on the definition of $H_{SO}=\lambda_{SO} L\cdot S$. What is $L\cdot S$? They mention that it is diagonal in the selected basis. Does this mean that it is simply $\lambda_{SO} 1$, where $1$ is now the $4\times 4$ identity matrix? But then, how would the spin-up $2\times 2$ block be differentiated from the spin-down $2\times 2$ block? Am I missing some eigenvalue spin index that changes sign (perhaps $H_{SO}$ is just the constant $\pm \lambda_{SO}$ added to the diagonal?)? Any corrections to my assumptions would help. Thanks.

Answer 1

"However, I am stumped on the definition of $H_{SO}=\lambda_{SO} L\cdot S$. What is $L\cdot S$?"

Spin-orbit coupling is typically written as a scalar coupling constant (in this case $\lambda_{\textrm{SO}}$) multiplied with $\hat{L}\cdot \hat{S}$ where $\hat{L}$ is an orbital angular momentum operator and $\hat{S}$ is a spin angular momentum operator (hence ther term "spin-orbit" coupling). This can be further turned into:

$$ \tag{1} \hat{L} \cdot \hat{S} = \frac{1}{2}\left(\hat{J}^2 - \hat{L}^2 - \hat{S}^2 \right), $$

$$\tag{2} \hat{J} \equiv \hat{L} + \hat{S}. $$

This is the origin of one of the most popular approximations for the SO energy shift $\Delta E_{\textrm{SO}}$ in hydrogenic atoms:

$$ \langle \hat{L} \cdot \hat{S} \rangle = \frac{\hbar^2}{2}\left( j(j+1) - l(l+1) - s(s+1) \right),\tag{3} $$

$$\tag{4} \Delta E_{\textrm{SO}} = \frac{\beta}{2}\left( j(j+1) - l(l+1) - s(s+1) \right), $$

$$\tag{5} \beta \equiv Z^4 \frac{\mu_0}{4\pi} g_{\textrm{s}} \mu_{\textrm{B}}^2 \frac{1}{n^3 a_0^3\;l(l+1/2)(l+1)}. $$

"They mention that it is diagonal in the selected basis. Does this mean that it is simply $\lambda_{SO} 1$, where $1$ is now the $4\times 4$ identity matrix?

It seems that the overall Hamiltonian (in your case, it's labelled as $H$) is being written in the basis of eigenstates of the $\hat{L}\cdot \hat{S}$ operator, which would indeed make $H_{\textrm{SO}}$ diagonal, but diagonal does not necessarily mean proportional to the identity matrix. If the eigenvalues of $\hat{L}\cdot \hat{S}$ are, for example: $\left(0,\frac{1}{2},2,3\right)$ then $H_{\textrm{SO}}$ could be $\textrm{diag}\left(0,\frac{\lambda_{\textrm{SO}}}{2},2\lambda_{\textrm{SO}},3\lambda_{\textrm{SO}}\right)$. This would happen if the second excited state under the $H_{\textrm{SO}}$ Hamiltonian was higher in energy than the ground state by 4x more than the first excited state is.

"But then, how would the spin-up $2\times 2$ block be differentiated from the spin-down $2\times 2$ block?"

The spin-up and spin-down blocks will be identifiable if you keep track of how the states are ordered in your basis. They don't even have to be "spin-up block" and "spin-down block". you could order the states with spin-up and spin-down alternating.

In thi case of this paper, they do seem to have ordered the basis with spin-up followed by spin-down, as you can see earlier in the paragraph that they have labeled the basis vectors in the following way:

$$ \tag{6} |d_{+2} \uparrow\rangle ,|d_{-2} \uparrow\rangle ,|d_{+2} \downarrow\rangle ,|d_{-2} \downarrow\rangle , |d_{\pm2}\rangle \equiv |d_{x^2 - y^2}\rangle \pm 2\textrm{i}|d_{xy}\rangle. $$