2009 (Tkatchenko−Scheffler) TS
The Tkatchenko−Scheffler model for van der Waals interactions (vdW) defines the $C_6^{AB}$ parameters in an ab-initio fashion. In TS model the vdW energy $E_{vdw}$ is defined as,
\begin{equation}
E_{\text{vdW}} = -\frac{1}{2}\sum_{A,B}f_{\text{damp}}\left(R_{AB},R^{0}_{A},R^{0}_{B}\right)C_{6}^{AB}R^{-6}_{AB} \tag{1} \end{equation}
where $R^0_{A}$ and $R^0_{B}$ are the vdW radii. The $C_6^{AB}$ parameter can defined by the Casimir-Polder integral exactly:
$$
C_6^{AB}=\frac{3}{\pi}\int_{0}^{\infty}\alpha_{{A}}(i\omega)\alpha_{{B}}(i\omega)d\omega \tag{2} \label{eq:eq2}
$$
where $\alpha_{A/B}(i\omega)$ is the frequency-dependent polarizability of $A$ and $B$ evaluated at imaginary frequencies. The $\alpha_{A/B}(i\omega)$ can be replaced by an approximate $\alpha^1_{A/B}(i\omega)$, where $\alpha^1_{A}(i\omega)=\alpha^{0}_{A}/[1-(\omega/\eta_{A})^2]$. $\alpha^{0}_{A}$ is the static polarizability of $A$ and $\eta_{A}$ is an effective frequency. Simplifying \eqref{eq:eq2}, we get:
$$
C_6^{AB}=\frac{3}{2}[\eta_{A}\eta_{B}/(\eta_{A}+\eta_{B})]\alpha_{A}^0\alpha_{B}^0\tag{3}\label{eq:eq3}
$$
which after further simplification results in:
$$
C_6^{AB}=\frac{2C_6^{AA}C_6^{BB}}{[\frac{\alpha_{B}^0}{\alpha_{A}^0}C_6^{AA}+\frac{\alpha_{A}^0}{\alpha_{B}^0}C_6^{BB}]}\tag{4}
$$
$C_6^{AA}$ and $\alpha_{A}^0$ can be determined from highly accurate benchmark calculations.
Note: Here $C_6^{ij}\equiv C_6^{AB}$, $i\equiv A$ and $j\equiv B$