HF part for spherium
This commit is contained in:
parent
418e648d2c
commit
22afac608b
@ -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 $R>3/2$.
|
||||
|
||||
|
Loading…
Reference in New Issue
Block a user