Calculation of coefficients $D^{uU}_{rRpP}$ (72)

Coefficients $D^{uU}_{rRpP}$ (72) can easily be calculated by using the fact that they exactly correspond to the multiplication of tensors $X$ built from the position vector $x$.

$$X_{rRM_R}X_{pPM_P} = \sum_{uUM_U} C^{UM_U}_{RM_RPM_P} D^{uU}_{rRpP} X_{uUM_U} .$$ (119)

Tensors $X$ are simply related to spherical harmonics (see Eq. 5.2(40) in Ref. [6]):

$$X_{pPM_P} = \left(-{\textstyle{\frac{r^2}{\sqrt{3}}}}\right)^{{\textstyle{\fr... ...}}} r^{P} \sqrt{{\textstyle{\frac{4\pi P!}{(2P+1)!!}}}} Y_{PM_P}(\theta,\phi) .$$ (120)

Therefore, multiplication law of spherical harmonics (Eq. 5.6(8) in Ref. [6]) gives

$$X_{rRM_R}X_{pPM_P} = \left(-\sqrt{3}\right)^{{\textstyle{\frac{R-r}{2}}}}r^{r} \sqrt{{\textstyle{\frac{4\pi R!}{(2R+1)!!}}}}$$

$$\times \left(-\sqrt{3}\right)^{{\textstyle{\frac{P-p}{2}}}}r^{p} \sqrt{{... ...(2P+1)!!}}}} \sum_{UM_U} \sqrt{{\textstyle{\frac{(2R+1)(2P+1)}{4\pi (2U+1)}}}}$$

$$\times C^{U0}_{R0P0} C^{UM_U}_{RM_RPM_P} \left(-\sqrt{3}\right)^{{\text... ...\frac{u-U}{2}}}}r^{-u}\sqrt{{\textstyle{\frac{(2U+1)!!}{4\pi U!}}}} X_{uUM_U} ,$$ (121)

and thus the required coefficients read

$$D^{uU}_{rRpP} = \delta_{u,r+p}\left(-\sqrt{3}\right)^{{\textstyle{\frac{R+P-U}{2}... ...ac{R!P!}{U!}}}{\textstyle{\frac{(2U-1)!!}{(2R-1)!!(2P-1)!!}}}} C^{U0}_{R0P0} ,$$ (122)

and are independent of $urp$.