SOC Matrix Element in DFT
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SOC in DFT
We start from this formula of SOC expression in DFT: \(H_{SOC}=\sum_i\xi_i(r)\vec{L}_i\cdot\vec{S}_i\)
decomposition of operators: \(\vec{L}_i\cdot\vec{S}_i=L_i^zS_i^z+\frac{1}{2}\left[L^{+}_iS^{-}_i+L^{-}_iS^{+}_i\right]\)
Since most DFT in LCAO basis use real harmonics, while the spin and angular momentum operator act on spherical harmonics, we need a basis change when estimating matrix elements. We define spherical harmonics as $Y_{lm}$, real harmonics as $X_{lm}$, they are able to transform according to:
\[X_{lm}=\begin{cases} \frac{i}{\sqrt{2}}\left(Y_{l,-|m|}-(-1)^{m}Y_{l,|m|}\right) & m<0\\ Y_{l0} & m=0\\ \frac{1}{\sqrt{2}}\left(Y_{l,-|m|}+(-1)^{m}Y_{l,|m|}\right) & m>0 \end{cases}\]This can be written as a matrix operation: \(X_{lm}=\sum_{m,m'}M_{m,m'}^lY_{l,m'}\)
Understanding Angular momentum
First, the quantum angular momentum $\vec{L}$ is represent directly from the classical expression as $r\times p$. In a semi-classical picture, it is just a angular momentum of a electron in rounding around a central orbital.
For more general definition, accordingly to: https://en.wikipedia.org/wiki/Angular_momentum_operator#Angular_momentum_as_the_generator_of_rotations
The angular momentum is defined as the generator of the rotation group elements:
\[R(\hat{n},\phi)=\exp(-\frac{i}{\hbar}\phi J_{\hat{n}})\]Where $J_{\hat{n}}$ is the angular momentum along $\hat{n}$ direction.
Mathematically, $H_{\hat{n}}$ can be recovered by taking the first direvative of $R(\hat{n}, \phi)$ w.r.t. $\phi$, which results in a infinisimal rotation. Therefore, if a system is rotational invariant along $\hat{n}$, $J_{\hat{n}}$ should not change the states. Which means, the states should be a eigenstates of $J_{\hat{n}}$. Therefore it have a concerved quantity under action of $J_{\hat{n}}$, which we called the angular momentum.
For the $\vec{L}$ and $\vec{S}$, they are generator of rotation in position ($SO(3)$) and spin ($SU(2)$).
The anticommutation relation of the angular and spin momentum operators are tightly connected to the representation of Lie algebra (https://en.wikipedia.org/wiki/Lie_algebra_representation). (Don’t understand Lie group and Lie algebra now, so leave it for future.)
Algebra of angular momentum operators
\[\vec{L}^2Y_{lm}=\hbar^2l(l+1)Y_{lm}\\ L_zY_{lm}=\hbar mY_{lm}\\ \vec{L}_{+}Y_{lm}=\hbar\sqrt{l(l+1)-m(m+1)}Y_{l,m+1}\\ \vec{L}_{-}Y_{lm}=\hbar\sqrt{l(l+1)-m(m-1)}Y_{l,m-1}\]| For spin, basis is composed by $ | s,m_s\rangle$, where $s=\frac{1}{2}$ for all electrons, while, $m_s$ can only taking value from $(-s,+s)$. |
Compute SOC matrix element.
We need to evaluate: \(\langle R_iX_{lm}|H_{SOC}|R_jX_{l'm'}\rangle\) Where $R_{i/j}$ is the radial part of the real harmonics. First by change of basis, this equals to: \(\begin{align*} &\sum_{m'',m'''}M^l_{m'',m}M^{l'}_{m''',m'}\langle R_iY_{ilm''}|\sum_k \xi_k(r)\vec{L}_k\cdot\vec{S}_k|R_jY_{jl'm'''}\rangle\\ =&\sum_k\sum_{m'',m'''}M^l_{m'',m}M^{l'}_{m''',m'}\langle R_i|\xi_k(r)|R_j\rangle\left\langle Y_{ilm''}\left|L_k^zS_k^z+\frac{1}{2}\left[L^{+}_kS^{-}_k+L^{-}_kS^{+}_k\right]\right|Y_{jl'm'''}\right\rangle \end{align*}\)
When both basis have spin up, only $S^z_k$ term contribute, so \(\begin{align*} H_{SOC,ij}^{\uparrow\uparrow}&=\frac{\hbar}{2}\sum_{m'',m'''}M^l_{m'',m}M^{l'}_{m''',m'}\left[\langle R_i|\xi_i(r)|R_j\rangle\left\langle Y_{ilm''}\left|L_i^z\right|Y_{jl'm'''}\right\rangle+\langle R_i|\xi_j(r)|R_j\rangle\left\langle Y_{ilm''}\left|L_j^z\right|Y_{jl'm'''}\right\rangle\right]\\ &=\frac{\hbar}{2}\sum_{m'',m'''}M^l_{m'',m}M^{l'}_{m''',m'}\left[\hbar m''\langle R_i|\xi_i(r)|R_j\rangle\left\langle Y_{ilm''}|Y_{jl'm'''}\right\rangle+\hbar m'''\langle R_i|\xi_j(r)|R_j\rangle\left\langle Y_{ilm''}|Y_{jl'm'''}\right\rangle\right] \end{align*}\)
Here we should notice that:
\[\langle R_i|\xi_i|R_j\rangle=\int dr\xi_i(|r-r_i|)R_i(|r-r_i|)R_j(|r-r_j|) \\ \xi_i(|r-r_i|)=\frac{1}{2m^2c^2}\frac{\partial U_i(|r-r_i|)}{\partial |r-r_i|}\]| Since the $U_i( | r-r_i | )$ is the fully-relativistic pseudopotential, which is spherical, so $\xi_i$ will also be spherical. So, we can parameterize: |
\(\frac{\langle R_i|\xi_i|R_j\rangle}{\langle R_i|R_j\rangle}=\lambda^{(0)}_{i,j}\\ \frac{\langle R_i|\xi_j|R_j\rangle}{\langle R_i|R_j\rangle}=\lambda^{(1)}_{i,j}\) By chaning the matrix again to the original basis, we have:
\[H_{SOC,ij}^{\uparrow\uparrow}=\frac{\hbar}{2}\sum_{m'',m'''}\left[M^l_{m'',m}\lambda_{i,j}^{(0)}\hbar m''\langle R_i|R_j\rangle\left\langle Y_{ilm''}|X_{jl'm'}\right\rangle+M^{l'}_{m''',m'}\lambda_{i,j}^{(1)}\hbar m'''\langle R_i|R_j\rangle\left\langle X_{ilm}|Y_{jl'm'''}\right\rangle\right]\\ =\frac{\hbar^2}{2}\left[\lambda_{i,j}^{(0)}\sum_{m'',m''''}M^l_{m'',m}m''{(M^{l})}^{-1}_{m'''',m''}\langle R_i|R_j\rangle\left\langle X_{ilm''''}|X_{jl'm'}\right\rangle+\lambda_{i,j}^{(1)}\sum_{m''',m'''''}M^{l'}_{m''',m'}m'''{(M^{l})}^{-1}_{m''''',m'''}\langle R_i|R_j\rangle\left\langle X_{ilm}|Y_{jl'm'''}\right\rangle\right]\\ =\frac{\hbar^2}{2}\left[\lambda_{i,j}^{(0)}\sum_{m'',m'''}M^l_{m'',m}m''{(M^{l})}^{-1}_{m''',m''}S_{i,j}^{lm''',l'm'}+\lambda_{i,j}^{(1)}\sum_{m'',m'''}M^{l'}_{m'',m'}m''{(M^{l})}^{-1}_{m''',m''}S_{i,j}^{lm,l'm'''}\right]\]For spin up down:
since $S_z$ would not flip spin, the overlap of spin up and down orbitals are zero, so neglecting the z component of the operator.
The result is:
\[\begin{align*} H^{\uparrow\downarrow}_{SOC,ij}&=\frac{\hbar^2}{2}[\lambda_{i,j}^{(0)}\sum_{m'',m'''}\sqrt{l(l+1)-m''(m''-1)}M^l_{m'',m}{(M^{l-1})}^{-1}_{m''',m''}\langle X_{i(l-1)m'''}|X_{jl'm'}\rangle\\ +&\lambda_{i,j}^{(1)}\sum_{m'',m'''}\sqrt{l'(l'+1)-m''(m''-1)}M_{m'',m'}^{l'}(M^{l'-1})^{-1}_{m',m'''}\langle X_{ilm}|X_{j(l'-1)m'''}\rangle] \end{align*}\]