From 22afac608bef92c62d2ed407eae61b3275118f3a Mon Sep 17 00:00:00 2001 From: Pierre-Francois Loos Date: Thu, 30 Jul 2020 16:08:43 +0200 Subject: [PATCH] HF part for spherium --- RapportStage/Rapport.tex | 10 ++++------ 1 file changed, 4 insertions(+), 6 deletions(-) diff --git a/RapportStage/Rapport.tex b/RapportStage/Rapport.tex index bde598d..325a2ec 100644 --- a/RapportStage/Rapport.tex +++ b/RapportStage/Rapport.tex @@ -650,16 +650,14 @@ Here, we work in a minimal basis, composed of $Y_{0}$ and $Y_{1}$, or equivalent \end{equation} using a mixing angle between the two basis functions for each spin manifold. Hence we just minimize the energy with respect to the two mixing angles $\chi_\alpha$ and $\chi_\beta$. -\titou{STOPPED HERE.} -The minimization gives the three following solutions valid for all value of $R$: +This process provides the three following solutions valid for all value of $R$, which are respectively a minimum, a maximum and a saddle point of the HF equations: \begin{itemize} \item The two electrons are in the s orbital which is a RHF solution. This solution is associated with the energy $1/R^{2}$. -\item The two electrons are in the p\textsubscript{z} orbital which is a RHF solution. This solution is associated with the energy $1/R^{2} + 1/R$ -\item One electron is in the s orbital and the other is in the p\textsubscript{z} orbital which is a UHF solution. This solution is associated with the energy $2/R^{2}+ 29/(25R)$ +\item The two electrons are in the p\textsubscript{z} orbital which is a RHF solution. This solution is associated with the energy $1/R^{2} + 1/R$. +\item One electron is in the s orbital and the other is in the p\textsubscript{z} orbital which is a UHF solution. This solution is associated with the energy $2/R^{2}+ 29/(25R)$. \end{itemize} - -Those solutions are respectively a minimum, a maximum and a saddle point of the HF equations.\\ +\titou{STOPPED HERE.} In addition, there is also the well-known symmetry-broken UHF (sb-UHF) solution. For $R>3/2$ an other stationary UHF solution appears, this solution is a minimum of the HF equations. This solution corresponds to the configuration with the electron $\alpha$ on one side of the sphere and the electron $\beta$ on the opposite side and the configuration the other way round. The electrons can be on opposite sides of the sphere because the choice of p\textsubscript{z} as a basis function induced a privileged axis on the sphere for the electrons. This solution has the energy \eqref{eq:EsbUHF} for $R>3/2$.