5
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

"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:

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

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$ Jul 1 at 19:31
  • 1
    $\begingroup$ 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 $\pm 1$. $\endgroup$ Jul 1 at 22:54

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.