From 0272932350650df6b2f442c760043b328775021d Mon Sep 17 00:00:00 2001 From: AntoineMarie2 Date: Mon, 27 Jul 2020 17:11:31 +0200 Subject: [PATCH] few corrections sec 4 --- RapportStage/Rapport.tex | 34 +++++++++++++++++----------------- 1 file changed, 17 insertions(+), 17 deletions(-) diff --git a/RapportStage/Rapport.tex b/RapportStage/Rapport.tex index f747871..0060385 100644 --- a/RapportStage/Rapport.tex +++ b/RapportStage/Rapport.tex @@ -307,7 +307,7 @@ This is due to the following theorem \cite{Goodson_2012}: \begin{quote} \textit{``The Taylor series about a point $z_0$ of a function over the complex $z$ plane will converge at a value $z_1$ if the function is non-singular at all values of $z$ in the circular region centered at $z_0$ with radius $\abs{z_1 − z_0}$. If the function has a singular point $z_s$ such that $\abs{z_s − z_0} < \abs{z_1 − z_0}$, then the series will diverge when evaluated at $z_1$.''} \end{quote} -This theorem means that the radius of convergence of the perturbation series is equal to the distance to the origin of the closest singularity of $E(\lambda)$. To illustrate this result we consider the simple function \eqref{eq:DivExample}. This function is smooth for $x \in \mathbb{R}$ and infinitely differentiable in $\mathbb{R}$. One would expect that the Taylor series of such a function would be convergent for all $x \in \mathbb{R}$, however this series is divergent for $x\geq1$. This is because the function has four singularities in the complex plane ($x = e^{i\pi/4}, e^{-i\pi/4}, e^{i3\pi/4}, e^{-i3\pi/4}$) with a modulus equal to 1. This simple example shows the importance of the singularities in the complex plane to understand the convergence properties on the real axis. +This theorem means that the radius of convergence of the perturbation series is equal to the distance to the origin of the closest singularity of $E(\lambda)$. To illustrate this result we consider the simple function \eqref{eq:DivExample}. This function is smooth for $x \in \mathbb{R}$ and infinitely differentiable in $\mathbb{R}$. One would expect that the Taylor series of such a function would be convergent for all $x \in \mathbb{R}$, however this series is divergent for $x\geq1$. This is because the function has four singularities in the complex plane ($x = e^{i\pi/4}, e^{-i\pi/4}, e^{i3\pi/4}, e^{-i3\pi/4}$) with a modulus equal to 1. This simple example emphasizes the importance of the singularities in the complex plane to understand the convergence properties on the real axis. \begin{equation} \label{eq:DivExample} f(x)=\frac{1}{1+x^4} @@ -315,30 +315,30 @@ f(x)=\frac{1}{1+x^4} \subsection{The Hartree-Fock Hamiltonian} -In the Born-Oppenheimer approximation, the equation \eqref{eq:ExactHamiltonian} gives the exact electronic Hamiltonian for a chemical system with $n$ electrons and $N$ nuclei. The first term is the kinetic energy of the electrons, the two following terms account respectively for the electron-nuclei attraction and the electron-electron repulsion. +In the Born-Oppenheimer approximation, the equation \eqref{eq:ExactHamiltonian} gives the exact electronic Hamiltonian for a chemical system with $n$ electrons and $N$ nuclei with respective charge $Z_k$. The first term is the kinetic energy of the electrons, the two following terms account respectively for the electron-nuclei attraction and the electron-electron repulsion. \begin{equation}\label{eq:ExactHamiltonian} \bH=\sum\limits_{i=1}^{n}\left[ -\frac{1}{2}\grad_i^2 - \sum\limits_{k=1}^{N} \frac{Z_k}{|\vb{r}_i-\vb{R}_k|}+\sum\limits_{i3/2$ an other stationary UHF solution appear, this solution is a minimum of the Hartree-Fock 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 have the energy \eqref{eq:EsbUHF} for $R>3/2$. +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 have the energy \eqref{eq:EsbUHF} for $R>3/2$. \begin{equation}\label{eq:EsbUHF} E_{\text{sb-UHF}}=-\frac{75}{112R^3}+\frac{25}{28R^2}+\frac{59}{84R} @@ -581,13 +581,13 @@ Exact & 9.783874 & 0.852781 & 0.391959 & 0.247898 & 0.139471 & 0.064525 & 0.005 \label{tab:ERHFvsEUHF} \end{table} -There is also another symmetry-broken solution for $R>75/38$ but this one corresponds to a maximum of the HF equations. This solution is associated with another type of symmetry breaking somewhat less known. It corresponds to a configuration where both electrons are on the same side of the sphere, in the same spatial orbital. This solution is called symmetry-broken RHF. At the critical value of R, the repulsion of the two electrons on the same side of the sphere maximizes more the energy than the kinetic energy of the p\textsubscript{z} orbitals. This symmetry breaking is associated with a charge density wave: the system oscillates between the situations with the electrons on each side \cite{GiulianiBook}. +There is also another symmetry-broken solution for $R>75/38$ but this one corresponds to a maximum of the HF equations. This solution is associated with another type of symmetry breaking somewhat less known. It corresponds to a configuration where both electrons are on the same side of the sphere, in the same spatial orbital. This solution is called symmetry-broken RHF (sb-RHF). At the critical value of R, the repulsion of the two electrons on the same side of the sphere maximizes more the energy than the kinetic energy of the p\textsubscript{z} orbitals. This symmetry breaking is associated with a charge density wave, the system oscillates between the situations with the electrons on each side \cite{GiulianiBook}. \begin{equation} E_{\text{sb-RHF}}=\frac{75}{88R^3}+\frac{25}{22R^2}+\frac{91}{66R} \end{equation} -We can also consider negative value of R. This corresponds to the situation where one of the electrons is replaced by a positron. There are also a sb-RHF ($R<-3/2$) and a sb-UHF ($R<-75/38$) solution for negative values of R (see Fig.\ref{fig:SpheriumNrj} but in this case the sb-RHF solution is a minimum and the sb-UHF is a maximum. Indeed, the sb-RHF states minimize the energy by placing the electron and the positron on the same side of the sphere. And the sb-UHF states maximize the energy because the two particles are on opposite sides of the sphere. +We can also consider negative values of R. This corresponds to the situation where one of the electrons is replaced by a positron. There are also a sb-RHF ($R<-3/2$) and a sb-UHF ($R<-75/38$) solution for negative values of R (see Fig.\ref{fig:SpheriumNrj} but in this case the sb-RHF solution is a minimum and the sb-UHF is a maximum. Indeed, the sb-RHF states minimize the energy by placing the electron and the positron on the same side of the sphere. And the sb-UHF states maximize the energy because the two attracting particles are on opposite sides of the sphere. In addition, we can also consider the symmetry-broken solutions beyond their respective Coulson-Fischer points by analytically continuing their respective energies leading to the so-called holomorphic solutions \cite{Hiscock_2014, Burton_2019, Burton_2019a}. All those energies are plotted in \autoref{fig:SpheriumNrj}. The dotted curves corresponds to the holomorphic domain of the energies. @@ -748,6 +748,6 @@ For a non-Hermitian Hamiltonian the exceptional points can lie on the real axis. \newpage \appendix -\section{ERHF and EUHF} +\section{The unrestricted energy}\label{sec:UHF_NRJ} \end{document}