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

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.

"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:

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

• Thank you Nike. This is great. Regarding the 'diagonal' SOC part, the authors give only one value for $\lambda_{SO}$ in their data. So, am I right in saying that they are effectively doing $H_{SO}=\lambda_{SO} diag(+1,+1,-1,-1)$, using the given basis? Jul 1 at 19:31
• That would imply that we have two pairs of degenerate energies. I've now updated my answer in several places, including the definition of $|d_{\pm 2}\rangle$ and the formula containing $l(l+1) + s(s+1)$. If the two $d$-like orbitals have the same energy, then we can imagine having their $l$ values set to 0 (as a reference) and getting $s(s+1) \in \{0,2\}$ which can be shifted by one unit downwards to give you $\pm 1$. Jul 1 at 22:54