10
$\begingroup$

The orbital magnetization in periodic solids has been nicely described by the so-called modern theory of magnetization [1,2,3,4]. $$\tag{1} \mathcal{M}_{orb} = -\frac{1}{2} \Im \sum_{n,\mathbf{k}} f_{n\mathbf{k}} w_{\mathbf{k}} \langle \partial_\mathbf{k} u_{n\mathbf{k}} | \times (\mathcal{H}_{\mathbf{k}}+\epsilon_{\mathbf{k}}-2\mu)|\partial_\mathbf{k} u_{n\mathbf{k}} \rangle. $$ where $f_{n\mathbf{k}}$ is the Fermi occupation of band $n$ and wave vector $\mathbf{k}$. $w_{\mathbf{k}}$ is the k-point weight. $u_{n\mathbf{k}}$ is the cell periodic part of the Bloch function $\psi_{n\mathbf{k}}=e^{i\mathbf{k \cdot r}}u_{n\mathbf{k}}$ and $\mathcal{H}_{\mathbf{k}}u_{n\mathbf{k}}=\epsilon_{\mathbf{k}}u_{n\mathbf{k}}$. $\mu$ is the chemical potential. $\times$ is cross product.
I'm wondering if we could calculate $\mathcal{M}_{orb}$ in the following way (in atomic units): \begin{align} \mathcal{M}_{orb} &= - \frac{1}{2} \sum_{n,\mathbf{k}} f_{n\mathbf{k}} w_{\mathbf{k}} \langle \psi_{n\mathbf{k}} | \mathbf{r} \times \mathbf{v} | \psi_{n\mathbf{k}} \rangle \tag{2}\\ &= - \frac{1}{2} \sum_{n,\mathbf{k}} f_{n\mathbf{k}} w_{\mathbf{k}} \langle u_{n\mathbf{k}} | \mathbf{r} \times \mathbf{v_k} | u_{n\mathbf{k}} \rangle \tag{3}\\ &= - \frac{1}{2} \sum_{n,\mathbf{k}} f_{n\mathbf{k}} w_{\mathbf{k}} \sum_m \langle u_{n\mathbf{k}} | \mathbf{r} |u_{m\mathbf{k}}\rangle \times \langle u_{m\mathbf{k}} |\mathbf{v_k} | u_{n\mathbf{k}} \rangle \tag{4}\\ &= - \frac{1}{2} \sum_{n,\mathbf{k}} f_{n\mathbf{k}} w_{\mathbf{k}} \left( \mathcal{A}_{nn}(\mathbf{k}) \times \partial_\mathbf{k} \epsilon_\mathbf{k} + \frac{i}{\hbar}\sum_{m} \mathcal{A}_{nm}(\mathbf{k}) \times \mathcal{A}_{mn}(\mathbf{k})(\epsilon_{m\mathbf{k}}-\epsilon_{n\mathbf{k}}) \right) \tag{5} \end{align} where I have used in Eq. (4) the completeness relation:

$$\sum_m |u_{mk}\rangle \langle u_{mk}|=1\tag{6},$$ with $m$ running over all the states (which should be subjected to a convergence study for real calculations) and in Eq. (5) the relation: $$i\hbar \langle u_{m\mathbf{k}} |\mathbf{v_k} | u_{n\mathbf{k}} \rangle =\mathcal{A}_{mn}(\mathbf{k}) (\epsilon_{n\mathbf{k}}-\epsilon_{m\mathbf{k}})\tag{7}.$$ with $\mathcal{A}_{mn}(\mathbf{k})=i \langle u_{m\mathbf{k}} |\nabla_\mathbf{k} | u_{n\mathbf{k}} \rangle$. So the calculation of the orbital magnetization is reduced to evaluate the matrix elements of the Berry connection and band gradient.

$\endgroup$
5
  • 3
    $\begingroup$ +1. I just had to change your "split" environment to an "aligned" environment, because you can't label equations in the split environment. It's important to label equations even if you're not referring to them in the question, because someone might want to refer to them in the answer. Please see what I did! $\endgroup$ Oct 29 '20 at 22:12
  • 2
    $\begingroup$ @NikeDattani I did realize the numbering issue you mentioned, thanks for your help. $\endgroup$ Oct 29 '20 at 22:35
  • $\begingroup$ As it's now been 10 months since this question was asked, were you able to shed any light on it? $\endgroup$ Sep 5 at 22:31
  • $\begingroup$ As it's now been 6 months since my last comment, I'd like to point out some more things which may help get this question answered: (1) Defining the notation you're using. Sure the notations may have been defined in the 4 references you gave, but people would have to go and look for them. (2) Providing arXiv links for the 4 references, so that people can access them more easily. Not everyone has access to journals, and those that do, may have to go through a painstaking process to VPN into their institutional IP address to get it. (3) In Eq. 1, what does the $\times$ mean? $\endgroup$ Sep 21 at 22:26
  • $\begingroup$ Thanks @NikeDattani for your suggestions. I have modified the question accordingly. $\endgroup$ Sep 28 at 8:41

Your Answer

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

Browse other questions tagged or ask your own question.