saving work in sec 4

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Pierre-Francois Loos 2020-07-30 10:28:08 +02:00
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@ -618,29 +618,29 @@ These one-electron orbitals are expanded in the basis of zonal spherical harmoni
It is possible to obtain the formula for the ground state UHF energy in this basis set \cite{Loos_2009}: It is possible to obtain the formula for the ground state UHF energy in this basis set \cite{Loos_2009}:
\begin{equation} \begin{equation}
E_{\text{UHF}} = E_{\text{c},\alpha} + E_{\text{c},\beta} + E_{\text{p}} E_{\text{UHF}} = T_{\alpha} + T_{\beta} + V
\end{equation}
\begin{equation}
E_{\text{c},\alpha} = \sum_{k=0}^{\infty} C_{\alpha,k}^2 \frac{k(k+1)}{R^2}
\end{equation}
\begin{equation}
E_{\text{p}} = \sum_{i,j,k,l=0}^{\infty}C_{\alpha,i}C_{\alpha,j}C_{\beta,k}C_{\beta,l} \frac{(-1)^{k+l}S_{i,j,k,l}}{R}\sum_{L=0}^{\infty} \begin{pmatrix}
i & j & L \\
0 & 0 & 0
\end{pmatrix}^2 \begin{pmatrix}
k & l & L \\
0 & 0 & 0
\end{pmatrix}^2
\label{eq:EUHF} \label{eq:EUHF}
\end{equation} \end{equation}
with
\begin{gather}
T_{\sigma} = \frac{1}{R^2} \sum_{\ell=0}^{\infty} C_{\sigma,\ell}^2 \, \ell(\ell+1)
\\
V = \frac{1}{R} \sum_{\ell_1,\ell_2,\ell_3,\ell_4=0}^{\infty} \sum_{L=0}^{\infty}
(-1)^{\ell_3+\ell_4} v^\alpha_{\ell_1,\ell_2,L} v^\beta_{\ell_3,\ell_4,L}
\end{gather}
and
\begin{equation}
v^\sigma_{\ell_1,\ell_2,L}
= \sqrt{(2\ell_1+1)(2\ell_2+1)} C_{\sigma,\ell_1}C_{\sigma,\ell_2}
\begin{pmatrix}
\ell_1 & \ell_2 & L
\\
0 & 0 & 0
\end{pmatrix}^2
\end{equation}
\titou{T2: please define Wigner 3j symbols.}
\begin{equation*} We obtained Eq.~\eqref{eq:EUHF} for the general form of the wave function \eqref{eq:UHF_WF}, but to be associated with a physical wave function the energy needs to be stationary with respect to the coefficients. The general method is to use the Hartree-Fock self-consistent field method \cite{SzaboBook} to get the coefficients of the wave functions corresponding to physical solutions. We will work in a minimal basis, composed of a $Y_{00}$ and a $Y_{10}$ spherical harmonic, or equivalently a s and a p\textsubscript{z} orbital, to illustrate the difference between the RHF and UHF solutions. In this basis there is a shortcut to find the stationary solutions. One can define the one-electron wave functions $\phi(\theta)$ using a mixing angle between the two basis functions. Hence we just need to minimize the energy with respect to the two mixing angles $\chi_\alpha$ and $\chi_\beta$.
S_{i,j,k,l}=\sqrt{(2i+1)(2j+1)(2k+1)(2l+1)}
\end{equation*}
We obtained the equation \eqref{eq:EUHF} for the general form of the wave function \eqref{eq:UHF_WF}, but to be associated with a physical wave function the energy needs to be stationary with respect to the coefficients. The general method is to use the Hartree-Fock self-consistent field method \cite{SzaboBook} to get the coefficients of the wave functions corresponding to physical solutions. We will work in a minimal basis, composed of a $Y_{00}$ and a $Y_{10}$ spherical harmonic, or equivalently a s and a p\textsubscript{z} orbital, to illustrate the difference between the RHF and UHF solutions. In this basis there is a shortcut to find the stationary solutions. One can define the one-electron wave functions $\phi(\theta)$ using a mixing angle between the two basis functions. Hence we just need to minimize the energy with respect to the two mixing angles $\chi_\alpha$ and $\chi_\beta$.
\begin{equation} \begin{equation}
\phi_\alpha(\theta_1)= \cos(\chi_\alpha)\frac{Y_{00}(\theta_1)}{R} + \sin(\chi_\alpha)\frac{Y_{10}(\theta_1)}{R} \phi_\alpha(\theta_1)= \cos(\chi_\alpha)\frac{Y_{00}(\theta_1)}{R} + \sin(\chi_\alpha)\frac{Y_{10}(\theta_1)}{R}