The total potential energy in the spherical symmetry

The energy related to one term in the energy density (38) reads

$${\cal E}^{n'L'v'J'}_{mI,nLvJ}=\int {\rm d}^{3}\vec{r}\, C_{m... ...}\right) \left(\rho_{mI,nLvJ,J',-M'}^{J_{1}''M_{1}''}\right) .$$ (110)

We carry out the derivation for arbitrary values of multipole components, that is for $J_{2}''M_{2}''\neq J_{1}''M_{1}''\neq00$, while the standard HF total energy in spherical symmetry corresponds to the particular case of $J_{2}''M_{2}'' = J_{1}''M_{1}'' = 00$. In the general case, the result corresponds to the energy matrix element between two multipole components, and allows one to calculated the total energy for the density matrix being a linear combination of multipole components, that is, for a deformed state.

After using the definition of reduced density in Eq. (102), the energy (116) becomes equal to:

$$\begin{eqnarray*} {\cal E}^{n'L'v'J'}_{mI,nLvJ}=\int {\rm d}^{3}\vec{r}\, C_{mI,... ...,-M',J_{1}''M_{1}''}^{TM_{T}}Y_{TM_{T}}\left(\theta,\phi\right). \end{eqnarray*}$$

By using the multiplication theorem for the spherical harmonics, one obtains the expression

$${\cal E}^{n'L'v'J'}_{mI,nLvJ} = \int {\rm d}^{3}\vec{r}\, C_{mI,nLvJ}^{n'L'v'J'}e^{-2\left(br\ri... ...{T'J_{2}''}\left(br\right) \tilde{\rho}_{mI,nLvJ,J'}^{TJ_{1}''}\left(br\right)$$

$$\times C_{T0,T'0}^{Q0}\sqrt{\frac{\left(2T+1\right)\left(2T'+1\right)}{4... ...extstyle{\frac{(-1)^{J'-M'}}{\sqrt{2J'+1}}}} C_{J'M',J_{2}''M_{2}''}^{T'M_{T}'}$$

$$\times C_{J',-M',J_{1}''M_{1}''}^{TM_{T}}C_{TM_{T},T'M_{T}'}^{QM_{Q}}Y_{QM_{Q}}\left(\theta,\phi\right) ,$$ (111)

which, for density-independent coupling constants, can be integrated over $\theta$ and $\phi$. After summing up the Clebsh-Gordan coefficients, one obtains:

$${\cal E}^{n'L'v'J'}_{mI,nLvJ} = {\textstyle{\frac{(-1)^{J_1''-M_1''}}{\sqrt{2J'+1}(2J_1''+1)}}} \... ...-M_1''} \int r^2{\rm d}\,{r}\, C_{mI,nLvJ}^{n'L'v'J'}e^{-2\left(br\right)^{2}}$$

$$\times \sum_{T}(-1)^T(2T+1)\tilde{\rho}_{00,n'L'v'J',J'}^{TJ_{2}''}\left(br\right) \tilde{\rho}_{mI,nLvJ,J'}^{TJ_{1}''}\left(br\right) .$$ (112)

One thus obtains the correct result that for density-independent coupling constants, the potential energy is diagonal in different multipoles.

When the coupling constants do depend on density, the angular integral cannot be performed, and a similar derivation gives the general result:

$${\cal E}^{n'L'v'J'}_{mI,nLvJ} = \int {\rm d}^{3}\vec{r}\, C_{mI,nLvJ}^{n'L'v'J'}e^{-2\left(br\ri... ...{T'J_{2}''}\left(br\right) \tilde{\rho}_{mI,nLvJ,J'}^{TJ_{1}''}\left(br\right)$$

$$\times C_{T0,T'0}^{Q0}\frac{\left(2T+1\right)\left(2T'+1\right)}{\sqrt{4\pi}}$$

$$\times \sum_{M_{Q}}C_{J_{1}''M_{1}'',J_{2}''M_{2}''}^{QM_{Q}}\left\{ \be... ... J_{2}'' & J_{1}'' & Q\end{array}\right\} Y_{QM_{Q}}\left(\theta,\phi\right).$$ (113)

For the spherical case, one has $J_{1}''=J_{2}''=0$, which implies $T=J'$, and both expressions (118) and (119) reduce to:

$${\cal E}^{n'L'v'J'}_{mI,nLvJ} = (-1)^{J'}\sqrt{2J'+1} \int r^2{\rm d}\,{r}\, C_{mI,nLvJ}^{n'L'v'J'}e^{-2\left(br\right)^{2}}$$

$$\times \tilde{\rho}_{00,n'L'v'J',J'}^{J'0}\left(br\right) \tilde{\rho}_{mI,nLvJ,J'}^{J'0}\left(br\right) .$$ (114)

For density-dependent terms, such as those in Eqs. (62) and (63), the total energy reads

$${\cal E}^{\alpha,n'L'v'J'}_{mI,nLvJ} = (-1)^{J'}\sqrt{2J'+1} C_{mI,nLvJ}^{\alpha,n'L'v'J'} \int r^2{\rm d}\,{r}\, e^{-2\left(br\right)^{2}} \rho_0^\alpha \left(br\right)$$

$$\times \tilde{\rho}_{00,n'L'v'J',J'}^{J'0}\left(br\right) \tilde{\rho}_{mI,nLvJ,J'}^{J'0}\left(br\right) ,$$ (115)

where the density $\rho_0\left(br\right)$ depends on the reduced density as:

$$\rho_0 \left(br\right)=\tilde{\rho}_{00,0000,00}^{00}\left(br\right)\frac{e^{-(br)^{2}}}{\sqrt{4\pi}} .$$ (116)