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ISTPC/2022/postHF/ISTPC_Loos_postHF.tex
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\documentclass[aspectratio=169,9pt]{beamer}
% ***********
% * PACKAGE *
% ***********
\usepackage{amsmath,amssymb,amsfonts,pgfpages,graphicx,subfigure,xcolor,bm,multirow,microtype,wasysym,multimedia,hyperref,tabularx,amscd,pgfgantt,mhchem}
\usetikzlibrary{shapes.gates.logic.US,trees,positioning,arrows}
\usetheme{Pittsburgh}
\usecolortheme{seahorse}
\usepackage{mathpazo,libertine}
\usepackage[normalem]{ulem}
\usepackage{algorithmicx,algorithm,algpseudocode}
\algnewcommand\algorithmicassert{\texttt{assert}}
\algnewcommand\Assert[1]{\State \algorithmicassert(#1)}
%\algrenewcommand{\algorithmiccomment}[1]{$\triangleright$ #1}
%\usepackage[version=4]{mhchem}
\usepackage{amsmath,amsfonts,amssymb,bm,microtype,graphicx,wrapfig,geometry,physics,eurosym,multirow,pgfgantt}
\usepackage{hyperref}
\hypersetup{
colorlinks=true,
linkcolor=cyan,
filecolor=magenta,
urlcolor=blue,
citecolor=purple
}
% operators
\newcommand{\hI}{\Hat{1}}
\newcommand{\hH}{\Hat{H}}
\newcommand{\hO}{\Hat{\mathcal{O}}}
\newcommand{\hT}[2]{\Hat{T}_{#1}^{#2}}
\newcommand{\cre}[1]{\Hat{a}_{#1}^{\dag}}
\newcommand{\ani}[1]{\Hat{a}_{#1}}
\newcommand{\bH}{\mathbold{H}}
\newcommand{\br}{\mathbold{r}}
\newcommand{\la}{\lambda}
\newcommand{\si}{\sigma}
\newcommand{\cJ}{\mathcal{J}}
\newcommand{\cK}{\mathcal{K}}
% wave functions
\newcommand{\PsiO}{\Psi_0}
\newcommand{\PsiHF}{\Psi_\text{RHF}}
\newcommand{\PsiFCI}{\Psi_\text{FCI}}
\newcommand{\PsiCC}{\Psi_\text{CC}}
\newcommand{\PsiCCD}{\Psi_\text{CCD}}
\newcommand{\amp}[2]{t_{#1}^{#2}}
\newcommand{\Det}[2]{\Psi_{#1}^{#2}}
% energies
\newcommand{\EHF}{E_\text{HF}}
\newcommand{\EO}{E_\text{0}}
\newcommand{\ECC}{E_\text{CC}}
\newcommand{\ETCC}{E_\text{TCC}}
\newcommand{\EVCC}{E_\text{VCC}}
\newcommand{\EUCC}{E_\text{UCC}}
\newcommand{\ECCD}{E_\text{CCD}}
\newcommand{\nEl}{n}
\newcommand{\nBas}{N}
\newcommand{\ba}{\bm{a}}
\newcommand{\bb}{\bm{b}}
\newcommand{\bA}{\bm{A}}
\newcommand{\bB}{\bm{B}}
\newcommand{\bo}{\bm{0}}
\newcommand{\sbra}[1]{[ #1 |}
\newcommand{\sket}[1]{| #1 ]}
\newcommand{\sexpval}[1]{[ #1 ]}
\newcommand{\sbraket}[2]{[ #1 | #2 ]}
\newcommand{\smel}[3]{[ #1 | #2 | #3 ]}
\definecolor{darkgreen}{RGB}{0, 180, 0}
\definecolor{fooblue}{RGB}{0,153,255}
\definecolor{fooyellow}{RGB}{234,187,0}
\definecolor{lavender}{rgb}{0.71, 0.49, 0.86}
\definecolor{inchworm}{rgb}{0.7, 0.93, 0.36}
\newcommand{\violet}[1]{\textcolor{lavender}{#1}}
\newcommand{\orange}[1]{\textcolor{orange}{#1}}
\newcommand{\purple}[1]{\textcolor{purple}{#1}}
\newcommand{\blue}[1]{\textcolor{blue}{#1}}
\newcommand{\green}[1]{\textcolor{darkgreen}{#1}}
\newcommand{\yellow}[1]{\textcolor{fooyellow}{#1}}
\newcommand{\red}[1]{\textcolor{red}{#1}}
\newcommand{\highlight}[1]{\textcolor{fooblue}{#1}}
\newcommand{\pub}[1]{\small \textcolor{purple}{#1}}
\newcommand{\mc}{\multicolumn}
\newcommand{\mycirc}[1][black]{\Large\textcolor{#1}{\ensuremath\bullet}}
\usepackage{tikz}
\usetikzlibrary{arrows,positioning,shapes.geometric}
\usetikzlibrary{decorations.pathmorphing}
\tikzset{snake it/.style={
decoration={snake,
amplitude = .4mm,
segment length = 2mm},decorate}}
% *************
% * HEAD DATA *
% *************
\title[HF and post-HF methods]{
\purple{Hartree-Fock and post-Hartree-Fock methods: \\
Computational aspects}
}
\author[PF Loos]{Pierre-Fran\c{c}ois LOOS}
\date{2022 ISTPC --- June 23rd, 2022}
\institute[CNRS@LCPQ]{
Laboratoire de Chimie et Physique Quantiques (UMR 5626),\\
Universit\'e de Toulouse, CNRS, UPS, Toulouse, France.
}
\titlegraphic{
\includegraphics[width=0.3\textwidth]{fig/peppa}
\\
\vspace{0.05\textheight}
\includegraphics[height=0.05\textwidth]{fig/UPS}
\hspace{0.2\textwidth}
\includegraphics[height=0.05\textwidth]{fig/ERC}
\hspace{0.2\textwidth}
\includegraphics[height=0.05\textwidth]{fig/LCPQ}
\hspace{0.2\textwidth}
\includegraphics[height=0.05\textwidth]{fig/CNRS}
}
\begin{document}
%%% SLIDE 1 %%%
\begin{frame}
\titlepage
\end{frame}
%
%%% SLIDE 2 %%%
\begin{frame}{Today's program}
\begin{itemize}
\item How to perform a Hartree-Fock (HF) calculation in practice?
\begin{itemize}
\item Computation of integrals \pub{[Ahlrichs, PCCP 8 (2006) 3072]}
\item Orthogonalization matrix \pub{[Szabo \& Ostlund, Modern Quantum Chemistry]}
\item Construction of the Coulomb matrix \pub{[White \& Head-Gordon, JCP 104 (1996) 2620]}
\item Resolution of the identity \pub{[Weigend et al. JCP 130 (2009) 164106]}
\item DFT exchange via quadrature \pub{[Becke, JCP 88 (1988) 2547]}
\end{itemize}
\bigskip
\item Generalities on correlation methods
\begin{itemize}
\item Configuration Interaction (CI) \pub{[Szabo \& Ostlund, Modern Quantum Chemistry]}
\item Perturbation theory \pub{[Szabo \& Ostlund, Modern Quantum Chemistry]}
\item Coupled-cluster (CC) theory \pub{[Jensen, Introduction to Computational Chemistry]}
\end{itemize}
\bigskip
\item Computing the 2nd-order M{\o}ller-Plesset (MP2) correlation energy
\begin{itemize}
\item Atomic orbital (AO) to molecular orbital (MO) transformation \pub{[Frisch et al. CPL 166 (1990) 281]}
\item Laplace transform \pub{[Alml{\"o}f, CPL 181 (1991) 319]}
\end{itemize}
\bigskip
\item Coupled cluster with doubles (CCD)
\begin{itemize}
\item Introduction to CC methods \pub{[Shavitt \& Bartlett, \textit{``Many-Body Methods in Chemistry and Physics: MBPT and Coupled-Cluster Theory''}]}
\item Algorithm to compute the CCD energy \pub{[Pople et al. IJQC 14 (1978) 545]}
\end{itemize}
\end{itemize}
\end{frame}
%%% SLIDE X %%%
\begin{frame}{How to perform a HF calculation in practice?}
\begin{columns}
\begin{column}{0.7\textwidth}
\begin{block}{The SCF algorithm (p.~146)}
\begin{enumerate}
\item \orange{Specify molecule} $\{\br_A\}$ and $\{Z_A\}$ and \violet{basis set} $\{\phi_\mu\}$
\item Calculate integrals $S_{\mu \nu}$, $H_{\mu \nu}$ and $\langle \mu \nu | \lambda \sigma \rangle$
\item Diagonalize $\bm{S}$ and compute $\bm{X} = \bm{S}^{-1/2}$
\item Obtain \alert{guess density matrix} for $\bm{P}$
\begin{enumerate}
\item[1.] Calculate $\bm{J}$ and $\bm{K}$, then $\bm{F} = \bm{H} + \bm{J} + \bm{K}$
\item[2.] Compute $\bm{F}' = \bm{X}^\dag \cdot \bm{F} \cdot \bm{X}$
\item[3.] Diagonalize $\bm{F}'$ to obtain $\bm{C}'$ and $\bm{E}$
\item[4.] Calculate $\bm{C}= \bm{X} \cdot \bm{C}'$
\item[5.] Form a \blue{new density matrix} $\bm{P} = \bm{C} \cdot \bm{C}^\dag$
\item[6.] \alert{Am I converged?} If not go back to 1.
\end{enumerate}
\item Calculate stuff that you want, like $E_\text{HF}$ for example
\end{enumerate}
\end{block}
\end{column}
\begin{column}{0.3\textwidth}
\includegraphics[width=\textwidth]{fig/Szabo}
\end{column}
\end{columns}
\end{frame}
%
%-----------------------------------------------------
\begin{frame}{Assumptions \& Notations}
\begin{block}{Let's talk about notations}
\begin{itemize}
\bigskip
\item Number of \green{occupied orbitals} $O$
\item Number of \alert{vacant orbitals} $V$
\item \violet{Total number of orbitals} $N = O + V$
\bigskip
\item $i,j,k,l$ are \green{occupied orbitals}
\item $a,b,c,d$ are \alert{vacant orbitals}
\item $p,q,r,s$ are \violet{arbitrary (occupied or vacant) orbitals}
\item $\mu,\nu,\lambda,\sigma$ are \purple{basis function indexes}
\bigskip
\end{itemize}
\end{block}
\end{frame}
%-----------------------------------------------------
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{One- and two-electron integrals}
\begin{columns}
\begin{column}{0.7\textwidth}
\begin{block}{One-electron integrals: overlap \& core Hamiltonian (Appendix A)}
\begin{equation}
S_{\mu\nu}
= \braket{\mu}{\nu}
= \int \phi_\mu(\orange{\br}) \phi_\nu(\orange{\br}) d\orange{\br}
\end{equation}
\begin{equation}
H_{\mu\nu}
= \mel{\mu}{\hH^\text{c}}{\nu}
= \int \phi_\mu(\orange{\br}) \hH^\text{c}(\orange{\br}) \phi_\nu(\orange{\br}) d\orange{\br}
\end{equation}
\end{block}
\end{column}
\begin{column}{0.3\textwidth}
\includegraphics[width=\textwidth]{fig/SBG}
\end{column}
\end{columns} \begin{block}{Chemist/Mulliken notation for two-electron integrals (p.~68)}
\begin{equation}
( \mu \nu | \lambda \sigma )
= \iint \phi_\mu(\alert{\br_1}) \phi_\nu(\alert{\br_1}) \frac{1}{r_{12}} \phi_\lambda(\blue{\br_2}) \phi_\sigma(\blue{\br_2}) d\red{\br_1} d\blue{\br_2}
\end{equation}
% \begin{equation}
% ( \mu \blue{\nu} \orange{||} \lambda \red{\sigma} ) = ( \mu \blue{\nu} | \lambda \red{\sigma} ) - ( \mu \red{\sigma} | \lambda \blue{\nu} )
% \end{equation}
\end{block}
\begin{block}{Physicist/Dirac notation for two-electron integrals (p.~68)}
\begin{equation}
\langle \mu \nu | \lambda \sigma \rangle
= \iint \phi_\mu(\alert{\br_1}) \phi_\nu(\blue{\br_2}) \frac{1}{r_{12}} \phi_\lambda(\alert{\br_1}) \phi_\sigma(\blue{\br_2}) d\red{\br_1} d\blue{\br_2}
\end{equation}
% \begin{equation}
% \langle \mu \nu \orange{||} \blue{\lambda} \red{\sigma} \rangle = \langle \mu \nu | \blue{\lambda} \red{\sigma} \rangle - \langle \mu \nu | \red{\sigma} \blue{\lambda} \rangle
% \end{equation}
\end{block}
\end{frame}
\begin{frame}{Computing the electron repulsion integrals (ERIs)}
\begin{columns}
\begin{column}{0.7\textwidth}
\begin{block}{Four-center two-electron integrals}
\small
\begin{equation}
\begin{split}
\braket{\ba_1\ba_2}{\bb_1\bb_2}
& \equiv \mel{\ba_1\ba_2}{\alert{r_{12}^{-1}}}{\bb_1\bb_2}
\\
& = \iint \phi_{\ba_1}^{\bA_1}(\br_1) \phi_{\ba_2}^{\bA_2}(\br_2) \,\alert{\frac{1}{r_{12}}} \,
\phi_{\bb_1}^{\bB_1}(\br_1) \phi_{\bb_2}^{\bB_2}(\br_2) d\br_1 d\br_2
\end{split}
\end{equation}
\alert{Formally, one has to compute $\order{N^4}$ ERIs!}
\end{block}
\end{column}
\begin{column}{0.3\textwidth}
\includegraphics[width=\textwidth]{fig/STO}
\end{column}
\end{columns}
%
\begin{block}{Gaussian-type orbital (GTO)}
\small
\begin{align*}
\text{\violet{Contracted} GTO} & = \ket{\ba}
\equiv \phi_{\ba}^{\bA}(\br)
= \sum_k^K D_k \sket{\ba}_k
\\
\text{\blue{Primitive} GTO} & = \sket{\ba}
= (x-A_x)^{a_x} (y-A_y)^{a_y} (z-A_z)^{a_z} e^{-\alpha \abs{ \br -\bA }^2}
\end{align*}
\end{block}
\begin{itemize}
\item \textbf{\purple{Exponent:}} $\alpha$
\item \textbf{\purple{Center:}} $\bA = (A_x, A_y, A_z)$
\item \textbf{\purple{Angular momentum:}} $\ba = (a_x, a_y, a_z)$ and total angular momentum $a=a_x + a_y + a_z$
\end{itemize}
%
\end{frame}
\begin{frame}{The contraction problem}
\begin{columns}
\begin{column}{0.7\textwidth}
\begin{block}{Primitive vs Contracted}
\begin{itemize}
\item Same center $\bA$
\item Same angular momentum $\ba$
\item Different exponent $\violet{\alpha_k}$
\item Contraction coefficient $\blue{D_k}$ and degree $K$
\end{itemize}
\begin{equation}
\underbrace{\braket{\ba_1\ba_2}{\bb_1\bb_2}}_{\text{\green{contracted ERI}}}
= \sum_{k_1}^{K_1} \sum_{k_2}^{K_2} \sum_{k_3}^{K_3} \sum_{k_4}^{K_4}
\blue{D_{k_1} D_{k_2} D_{k_3} D_{k_4}}
\underbrace{\sbraket{\ba_{1,k_1}\ba_{2,k_2}}{\bb_{1,k_3}\bb_{2,k_4}}}_{\text{\red{primitive ERI}}}
\end{equation}
\centering
\green{One} contracted ERI required \red{$K_1 \times K_2 \times K_3 \times K_4$} primitive ERIs!
\end{block}
\begin{block}{Dunning's cc-pVTZ basis for the carbon atom}
\begin{equation}
\green{\braket{1s1s}{1s1s}}
= \sum_{k_1}^{10} \sum_{k_2}^{10} \sum_{k_3}^{10} \sum_{k_4}^{10}
\blue{D_{k_1} D_{k_2} D_{k_3} D_{k_4}}
\red{\sbraket{s_{k_1}^{\violet{\alpha_{k_1}}} s_{k_2}^{\violet{\alpha_{k_2}}}} {s_{k_3}^{\violet{\alpha_{k_3}}} s_{k_4}^{\violet{\alpha_{k_4}}} }}
\end{equation}
\centering
The $\green{\braket{1s1s}{1s1s}}$ integral requires $10^4$ \red{$s$-type integrals}!
\end{block}
\end{column}
\begin{column}{0.3\textwidth}
\begin{equation}
\boxed{\green{\ket{\ba}} = \sum_k^K \blue{D_k} \red{\sket{\ba_k}}}
\end{equation}
\\
\bigskip
\begin{block}{https://www.basissetexchange.org}
\bigskip
\centering
\includegraphics[width=\textwidth]{fig/C}
\end{block}
\end{column}
\end{columns}
\end{frame}
%%% SLIDE X %%%
\begin{frame}{Properties of Gaussian functions}
\begin{block}{Gaussian product rule: \textit{``The product of two gaussians is a gaussian''}}
\begin{equation}
G_{\red{\alpha},\red{\bm{A}}}(\br) = \exp(-\red{\alpha} \abs{\br - \red{\bA}}^2)
\qqtext{and}
G_{\blue{\beta},\blue{\bm{B}}}(\br) = \exp(-\blue{\beta} \abs{\br - \blue{\bB}}^2)
\qqtext{then}
\end{equation}
\begin{equation}
\boxed{G_{\red{\alpha},\red{\bm{A}}}(\br) G_{\blue{\beta},\blue{\bm{B}}}(\br) = \violet{K} \, G_{\violet{\zeta},\violet{\bm{P}}}(\br)}
\qqtext{with}
\violet{\zeta} = \red{\alpha} + \blue{\beta}
\qqtext{and}
\violet{\bm{P}} = \frac{\red{\alpha \bA} + \blue{\beta \bB}}{\red{\alpha} + \blue{\beta} }
\end{equation}
\begin{equation}
\violet{K} = \exp( -\frac{\red{\alpha} \blue{\beta}}{\red{\alpha} + \blue{\beta} } \abs{\red{\bA} - \blue{\bB}}^2)
\end{equation}
\end{block}
\begin{block}{Gaussian product rule for ERIs}
\begin{equation}
\begin{split}
(\bm{\red{a}} \bm{\blue{b}}|\bm{\orange{c}} \bm{\green{d}})
& = \iint G_{\red{\alpha},\red{\bm{A}}}(\br_1) G_{\blue{\beta},\blue{\bm{B}}}(\br_1) \frac{1}{r_{12}} G_{\orange{\gamma},\orange{\bm{C}}}(\br_2) G_{\green{\delta},\green{\bm{D}}}(\br_2) d\br_1 d\br_2
\\
& = \violet{K} \purple{K} \iint G_{\violet{\zeta},\violet{\bm{P}}}(\br_1) \frac{1}{r_{12}} G_{\purple{\eta},\purple{\bm{Q}}}(\br_2) d\br_1 d\br_2
\end{split}
\end{equation}
\alert{The number of ``significant'' ERIs in a large system is $\order{N^2}$!}
\end{block}
\end{frame}
%
\begin{frame}{Upper bounds for ERIs}
\begin{columns}
\begin{column}{0.35\textwidth}
\begin{block}{A ``good'' upper bound must be}
\begin{itemize}
\item tight (i.e., a good estimate)
\item simple (i.e, cheap to compute)
\end{itemize}
\end{block}
\end{column}
\begin{column}{0.65\textwidth}
\begin{equation}
\boxed{\abs{(\bm{\red{a}} \bm{\blue{b}}|\bm{\orange{c}} \bm{\green{d}})} \le B}
\end{equation}
\end{column}
\end{columns}
\bigskip
\begin{block}{Cauchy-Schwartz bound}
\begin{equation}
\abs{(\bm{\red{a}} \bm{\blue{b}}|\bm{\orange{c}} \bm{\green{d}})}
\le
\sqrt{(\bm{\red{a}} \bm{\blue{b}}|\bm{\red{a}} \bm{\blue{b}})}
\sqrt{(\bm{\orange{c}} \bm{\green{d}}|\bm{\orange{c}} \bm{\green{d}})}
\qqtext{or}
\abs{(\bm{\violet{P}}|\bm{\purple{Q}})}
\le
\sqrt{(\bm{\violet{P}}|\bm{\violet{P}})}
\sqrt{(\bm{\purple{Q}}|\bm{\purple{Q}})}
\end{equation}
\end{block}
\begin{block}{The family of generalized H\"older bounds}
\begin{equation}
\abs{(\bm{\red{a}} \bm{\blue{b}}|\bm{\orange{c}} \bm{\green{d}})}
\le
\qty[ (\bm{\red{a}} \bm{\blue{b}}|\bm{\red{a}} \bm{\blue{b}}) ]^{1/\purple{m}}
\qty[ (\bm{\orange{c}} \bm{\green{d}}|\bm{\orange{c}} \bm{\green{d}}) ]^{1/\violet{n}}
\qqtext{with}
\frac{1}{\purple{m}} + \frac{1}{\violet{n}} = 1
\qqtext{and}
\purple{m},\violet{n} > 1
\end{equation}
\end{block}
\end{frame}
\begin{frame}{Asymptotic scaling of two-electron integrals}
\begin{block}{Number of significant two-electron integrals}
\begin{equation}
(\bm{\red{a}} \bm{\blue{b}}|\bm{\orange{c}} \bm{\green{d}}) \equiv (\bm{\red{a}} \bm{\blue{b}}| \mathcal{O}_2 | \bm{\orange{c}} \bm{\green{d}})
\end{equation}
\end{block}
\bigskip
\begin{block}{Long-range vs short-range operators}
\begin{equation}
N_\text{sig} = c\,N^{\alpha}
\end{equation}
\center
\begin{tabular}{lcrccrc}
\hline
\hline
Molecule & $N$ & \mc{2}{c}{\red{$\hO = r_{12}^{-1}$}} && \mc{2}{c}{\orange{$\hO = e^{-r_{12}^2}$}} \\
\cline{3-4} \cline{6-7}
& & \mc{1}{c}{$N_\text{sig}$} & $\alpha$ && \mc{1}{c}{$N_\text{sig}$} & $\alpha$ \\
\hline
propene & 12 & 1\,625 & --- && 1\,650 & --- \\
butadiene & 16 & 5\,020 & 3.9 && 5\,020 & 3.9 \\
hexatriene & 24 & 24\,034 & 3.9 && 23\,670 & 3.8 \\
octatetraene & 32 & 63\,818 & 3.4 && 52\,808 & 2.8 \\
decapentaene & 40 & 119\,948 & 2.8 && 81\,404 & 1.9 \\
dodecaexaene & 48 & 192\,059 & 2.6 && 109\,965 & 1.6 \\
\hline
\hline
\end{tabular}
\bigskip
\end{block}
\end{frame}
\begin{frame}{Recipe for computing two-electron integrals}
\center
\begin{tikzpicture}
\begin{scope}[very thick,
node distance=4cm,on grid,>=stealth',
boxRR/.style={rectangle,draw,fill=green!40},
boxUB/.style={rectangle,draw,fill=orange!40},
boxFI/.style={rectangle,draw,fill=red!40},
integral/.style={rectangle,draw,fill=violet!40}],
\node [integral, align=center] (1) {\textbf{The cake:} \\ Two-electron integrals \\ $\braket{\ba_1 \ba_2}{ \bb_1 \bb_2}$};
\node [boxUB, align=center] (2A) [below=of 1] {\textbf{Ingredient number 2:} \\ Recurrence relations \\ $\expval*{\ba_1^+} = \expval*{\ba_1} + \expval*{\ba_1^-}$};
\node [boxRR, align=center] (2B) [right=of 2A] {\textbf{Ingredient number 3:} \\ Upper bounds \\ $\abs{\braket{\ba_1 \ba_2}{ \bb_1 \bb_2}} \le B$};
\node [boxFI, align=center] (2C) [left=of 2A] {\textbf{Ingredient number 1:} \\ Fundamental integrals \\ $\braket{\bo\bo}{\bo\bo}^{\bm{m}}$};
\path
(1) edge [<-] (2A)
(1) edge [<-,bend right] (2B)
(1) edge [<-,bend left] (2C)
;
\end{scope}
\end{tikzpicture}
\end{frame}
\begin{frame}{Late-contraction path algorithm (Head-Gordon-Pople \& PRISM inspired)}
\begin{tikzpicture}
\begin{scope}[
very thick,
node distance=1.5cm,on grid,>=stealth',
boxSP/.style={rectangle,draw,fill=purple!40},
box0m/.style={rectangle,draw,fill=red!40},
boxCm/.style={rectangle,draw,fill=gray!40},
boxA/.style={rectangle,draw,fill=red!40},
boxAA/.style={rectangle,draw,fill=red!40},
boxAAA/.style={rectangle,draw,fill=red!40},
boxC/.style={rectangle,draw,fill=gray!40},
boxCC/.style={rectangle,draw,fill=gray!40},
boxCCC/.style={rectangle,draw,fill=orange!40},
boxCCCCCC/.style={rectangle,draw,fill=green!40},
],
\node [boxSP, align=center] (SP) {Shell-pair \\ data};
\node [box0m, align=center] (0m) [right=of 1,xshift=1.25cm] {$\sbraket{00}{00}^{\bm{m}}$};
\node [boxCm, align=center] (Cm) [right=of 0m,xshift=1.75cm] {$\braket{00}{00}^{\bm{m}}$};
\node [boxA, align=center] (A) [below=of 0m] {$\sbraket{0 a_2}{00}^{\bm{m}}$};
\node [boxC, align=center] (C) [right=of A,xshift=1.75cm] {$\braket{0 a_2}{00}^{\bm{m}}$};
\node [boxAA, align=center] (AA) [below=of A] {$\sbraket{a_1 a_2}{00}$};
\node [boxCC, align=center] (CC) [right=of AA,xshift=1.75cm] {$\braket{a_1 a_2}{00}$};
\node [boxCCCCCC, align=center] (CCCC) [right=of CC,xshift=2cm] {$\braket{a_1 a_2}{b_1 b_2}$};
\path
(SP) edge[->] node[below,blue]{T$_0$} (0m)
(0m) edge[->] node[left,orange]{T$_1$} node [right,red]{VRR$_1$} (A)
(0m) edge[->,gray!70] (Cm)
(A) edge[->] node[left,orange]{T$_2$} node [right,red]{VRR$_2$} (AA)
(A) edge[->,gray!70] (C)
(AA) edge[->] node [below,blue]{CC} (CC)
(Cm) edge[->,gray!70] (C)
(C) edge[->,gray!70] (CC)
(CC) edge[->] node [above,orange]{T$_3$} node [below,red]{HRR} (CCCC)
;
\end{scope}
\end{tikzpicture}
\bigskip
\begin{itemize}
\item \red{HRR} = horizontal recurrence relation [Obara-Saika]
\item \red{VRR} = vertical recurrence relation
\item \blue{CC} = bra contraction
\end{itemize}
\end{frame}
%\begin{frame}{Screening algorithm for two-electron integrals}
%
%\resizebox{\textwidth}{!}{
%\begin{tikzpicture}
% \begin{scope}[very thick,
% node distance=2.5cm,on grid,>=stealth',
% bound2/.style={diamond,draw,fill=blue!40},
% bound4/.style={diamond,draw,fill=blue!40},
% bound6/.style={diamond,draw,fill=blue!40},
% shell/.style={circle,draw,fill=green!40},
% shellpair/.style={circle,draw,fill=green!40},
% shellquartet/.style={circle,draw,fill=green!40},
% shell1/.style={rectangle,draw,fill=yellow!40},
% shell2/.style={rectangle,draw,fill=orange!40},
% shell3/.style={rectangle,draw,fill=red!40},
% integral/.style={rectangle,draw,fill=violet!40}],
% \node [shell1, align=center] (1) {Primitive\\shells\\$\sket{a}$};
% \node [bound2, align=center] (B2) [right=of 1] {$\sexpval{B_2}$};
% \node [shell, align=center] (S1T) [above=of B2, yshift=-0.5cm] {$\sket{a}$};
% \node [shell, align=center] (S1B) [below=of B2, yshift=0.5cm] {$\sket{b}$};
% \node [shell2, align=center] (2) [right=of B2,xshift=0.75cm] {Contracted\\shell-pairs\\$\ket{ab}$};
% \node [bound4, align=center] (B4) [right=of 2] {$\expval{B_4}$} ;
% \node [shellpair, align=center] (S2T) [above=of B4, yshift=-0.5cm] {$\ket{a_1b_1}$};
% \node [shellpair, align=center] (S2B) [below=of B4, yshift=0.5cm] {$\ket{a_2b_2}$};
% \node [shell3, align=center] (3) [right=of B4] {Two-Electron\\integrals\\$\braket{a_1b_1}{a_2b_2}$};
% \path
% (1) edge [->,bend left] (S1T)
% (1) edge [->,bend right] (S1B)
% (S1T) edge [snake it] (B2)
% (S1B) edge [snake it] (B2)
% (B2) edge [->,color=red] node [below] {\small Contraction} (2)
% (2) edge [->,bend left] (S2T)
% (2) edge [->,bend right] (S2B)
% (S2T) edge [snake it] (B4)
% (S2B) edge [snake it] (B4)
% (B4) edge [->] (3)
% ;
% \end{scope}
%\end{tikzpicture}
%}
%\end{frame}
\begin{frame}{Orthogonalization matrix}
\red{\bf We are looking for a matrix in order to orthogonalize the AO basis, i.e.~$\bm{X}^\dag \cdot \bm{S} \cdot \bm{X} = \bm{1}$}
\\
\bigskip
\begin{block}{Symmetric (or L\"owdin) orthogonalization}
\begin{equation}
\text{$\bm{X} =\bm{S}^{-1/2} = \bm{U} \cdot \bm{s}^{-1/2} \cdot \bm{U}^\dag$ is one solution...}
\end{equation}
\purple{\bf Is it working?}
\begin{equation}
\bm{X}^\dag \cdot \bm{S} \cdot \bm{X}
= \bm{S}^{-1/2} \cdot \bm{S} \cdot \bm{S}^{-1/2}
= \bm{S}^{-1/2} \cdot \bm{S} \cdot \bm{S}^{-1/2}
= \bm{I} \quad \green{\checkmark}
\end{equation}
\end{block}
\begin{block}{Canonical orthogonalization}
\begin{equation}
\text{$\bm{X} =\bm{U} \cdot \bm{s}^{-1/2}$ is another solution (when you have linear dependencies)...}
\end{equation}
\purple{\bf Is it working?}
\begin{equation}
\bm{X}^\dag \cdot \bm{S} \cdot \bm{X}
= \bm{s}^{-1/2} \cdot \underbrace{\bm{U}^{\dag} \cdot \bm{S} \cdot \bm{U}}_{\bm{s}} \cdot \bm{s}^{-1/2}
= \bm{I} \quad \green{\checkmark}
\end{equation}
\end{block}
\end{frame}
\begin{frame}{Computation of the Fock matrix and energy}
\begin{block}{Density matrix (closed-shell system)}
\begin{equation}
P_{\red{\mu \nu}} = 2 \sum_{i}^\text{occ} C_{\red{\mu} i} C_{\red{\nu} i}
\qqtext{or}
\boxed{\bm{P} = \bm{C} \cdot \bm{C}^{\dag}}
\end{equation}
\end{block}
\begin{block}{Fock matrix in the AO basis (closed-shell system)}
\begin{equation}
F_{\red{\mu\nu}}
= H_{\red{\mu\nu}}
+ \underbrace{\sum_{\blue{\la \si}} P_{\blue{\la\si}} (\red{\mu\nu}|\blue{\la\si})}_{J_{\red{\mu \nu}} = \text{ Coulomb}}
\underbrace{ - \frac{1}{2} \sum_{\blue{\la \si}} P_{\blue{\la\si}} (\red{\mu}\blue{\si}|\blue{\la}\red{\nu})}_{K_{\red{\mu \nu}} = \text{ exchange}}
\end{equation}
\end{block}
\begin{block}{HF energy in the AO basis (closed-shell system)}
\begin{equation}
E_\text{HF} = \sum_{\red{\mu \nu}} P_{\red{\mu \nu}} H_{\red{\mu \nu}}
+ \frac{1}{2} \sum_{\red{\mu \nu} \blue{\la\si}} P_{\red{\mu \nu}} \qty[ (\red{\mu \nu} | \blue{\lambda \sigma}) - \frac{1}{2} (\red{\mu} \blue{\sigma} | \red{\lambda} \blue{\nu}) ] P_{\blue{\lambda\sigma}}
\qqtext{or}
\boxed{E_\text{HF} = \frac{1}{2} \text{Tr}{\qty[\bm{P} \cdot (\bm{H} + \bm{F})]}}
\end{equation}
\end{block}
\end{frame}
\begin{frame}{Computation of the Fock matrix and energy}
\begin{algorithmic}
\Procedure{Computing the Coulomb matrix}{}
\For{$\red{\mu}=1,N$}
\For{$\blue{\nu}=1,N$}
\State $J_{\red{\mu}\blue{\nu}} = 0$ \Comment{Initialization of the array}
\For{$\orange{\la}=1,N$}
\For{$\violet{\si}=1,N$}
\State $J_{\red{\mu}\blue{\nu}}
= J_{\red{\mu}\blue{\nu}}
+ P_{\orange{\la}\violet{\si}} (\red{\mu}\blue{\nu}|\orange{\la}\violet{\si})$
\Comment{Accumulation step}
\EndFor
\EndFor
\EndFor
\EndFor
\EndProcedure
\Comment{\bf \red{This is a $\order{N^4}$ algorithm as it involves four loops}}
\end{algorithmic}
\end{frame}
%%% SLIDE X %%%
\begin{frame}{Resolution of the identity}
\begin{block}{Resolution of the identity (RI)}
\begin{equation}
\sum_{\green{A}=1}^{\red{\infty}} \dyad{\green{A}} = \hI
\qq{with}
\braket{\green{A}}{\green{B}} = \delta_{AB}
\qq{$\Leftrightarrow$}
\sum_{\green{A}=1}^{\red{\infty}} \green{A}(\br_1) \green{A}(\br_2)
= \delta(\br_1 - \br_2)
\end{equation}
\end{block}
\begin{block}{Generalization to a two-body operator $\hO$}
\begin{equation}
\sum_{\green{\Tilde{A}}=1}^{\red{\infty}} \dyad{\green{\Tilde{A}}} = \hO
\qq{with}
\mel{\green{A}}{\hO}{\green{B}} = \delta_{AB}
\qq{and}
\hO \ket{\green{A}} = \ket{\green{\Tilde{A}}}
\qq{$\Leftrightarrow$}
\sum_{\green{\Tilde{A}}=1}^{\red{\infty}} \green{\Tilde{A}}(\br_1) \green{\Tilde{A}}(\br_2)
= \hO(\br_1,\br_2)
\end{equation}
\end{block}
\end{frame}
%
%%% SLIDE X %%%
\begin{frame}{Resolution of the Coulomb operator}
\begin{block}{RI in practice = RI \alert{approximation}}
\begin{equation}
\boxed{\sum_{\green{A}=1}^{\red{\infty}} \dyad{\green{A}} = \hI
\qqtext{and, in practice, }
\sum_{\green{A}=1}^{\red{K}} \dyad{\green{A}} \approx \hI}
\end{equation}
\end{block}
\begin{block}{Computing the Coulomb matrix within the RI approximation}
\begin{equation}
\begin{split}
J_{\red{\mu\nu}}
& = \sum_{\blue{\la \si}} P_{\blue{\la\si}} (\red{\mu\nu}|\blue{\la\si})
\\
& \stackrel{\text{\green{RI}}}{=} \sum_{\blue{\la \si}} P_{\blue{\la\si}} \sum_{\green{A}} (\red{\mu\nu}|\green{A}) (\green{A}|\blue{\la\si})
\\
& = \sum_{\green{A}} (\red{\mu\nu}|\green{A})
\underbrace{\sum_{\blue{\la \si}} P_{\blue{\la\si}} (\green{A}|\blue{\la\si})}_{\order{KN^2} \text{ and $K$ storage}}
= \underbrace{\sum_{\green{A}} (\red{\mu\nu}|\green{A}) \rho_{\green{A}}}_{\order{KN^2}}
\end{split}
\end{equation}
\\
Similar (more effective) approaches are named Cholesky decomposition, low-rank approximation, etc.
\end{block}
\end{frame}
%
\begin{frame}{Computation of exact exchange}
\begin{algorithmic}
\Procedure{Computing the exchange matrix}{}
\For{$\red{\mu}=1,N$}
\For{$\blue{\nu}=1,N$}
\State $K_{\red{\mu}\blue{\nu}} = 0$ \Comment{Initialization of the array}
\For{$\orange{\la}=1,N$}
\For{$\violet{\si}=1,N$}
\State $K_{\red{\mu}\blue{\nu}}
= K_{\red{\mu}\blue{\nu}}
+ P_{\orange{\la}\violet{\si}} (\red{\mu}\violet{\si}|\orange{\la}\blue{\nu})$
\Comment{Accumulation step}
\EndFor
\EndFor
\EndFor
\EndFor
\EndProcedure
\Comment{\bf \red{This is a $\order{N^4}$ algorithm and it's hard to play games...}}
\end{algorithmic}
\end{frame}
\begin{frame}{Computation of DFT exchange}
\begin{block}{LDA exchange (in theory) = cf \sout{Julien's} Manu's lectures}
\begin{gather}
K_{\mu\nu}^\text{LDA}
= \int \phi_{\mu}(\br) \violet{v_\text{x}^\text{LDA}}(\br) \phi_{\nu}(\br) d\br
= \frac{4}{3} C_\text{x} \overbrace{\int \phi_{\mu}(\br) \blue{\rho^{1/3}}(\br) \phi_{\nu}(\br) d\br}^{\text{\alert{no closed-form expression in general}}}
\\
\blue{\rho}(\br) = \sum_{\mu \nu} \phi_{\mu}(\br) \blue{P_{\mu \nu}} \phi_{\nu}(\br)
\end{gather}
\end{block}
\begin{block}{LDA exchange (in practice) = \alert{numerical integration via quadrature} = $\int f(x) dx \approx \sum_k w_k f(x_k)$}
\begin{gather}
\underbrace{K_{\mu\nu}^\text{LDA}}_{\green{\order{N_\text{grid} N^2}}}
\approx \sum_{k=1}^{\purple{N_\text{grid}}}
\underbrace{\orange{w_k}}_{\orange{\text{weights}}} \phi_{\mu}(\red{\br_k}) \violet{v_\text{x}^\text{LDA}}(\underbrace{\red{\br_k}}_{\text{\red{roots}}}) \phi_{\nu}(\red{\br_k})
= \frac{4}{3} C_\text{x} \sum_{k=1}^{\purple{N_\text{grid}}} \orange{w_k} \phi_{\mu}(\red{\br_k}) \blue{\rho^{1/3}}(\red{\br_k}) \phi_{\nu}(\red{\br_k})
\\
\underbrace{\blue{\rho}(\red{\br_k})}_{\green{\order{N_\text{grid} N^2}}} = \sum_{\mu \nu} \phi_{\mu}(\red{\br_k}) \blue{P_{\mu \nu}} \phi_{\nu}(\red{\br_k})
\end{gather}
\end{block}
\end{frame}
\begin{frame}{The correlation energy}
\begin{itemize}
\item HF replaces the e-e interaction by an \green{averaged interaction}
\bigskip
\item The error in the HF method is called the \purple{correlation energy}
$$\boxed{E_c = E - E_\text{HF}} $$
\item The correlation energy is small \orange{but cannot but neglected!}
\bigskip
\item HF energy \blue{roughly 99\%} of total but \blue{chemistry very sensitive to remaining 1\%}
\bigskip
\item The correlation energy is \alert{always negative}
\bigskip
\item Computing $E_c$ is one of the \violet{central problem of quantum chemistry}
\bigskip
\item In quantum chemistry, we usually \alert{``freeze'' the core electrons} for correlated calculations
\end{itemize}
\end{frame}
\begin{frame}{Most common correlation methods in quantum chemistry}
\begin{enumerate}
\item \alert{Configuration Interaction}: CID, CIS, CISD, CISDTQ, etc.
\bigskip
\item \alert{Coupled Cluster}: CCD, CCSD, CCSD(T), CCSDT, CCSDTQ, etc.
\bigskip
\item \alert{M{\o}ller-Plesset perturbation theory}: MP2, MP3, MP4, MP5, etc.
\bigskip
\item \alert{Multireference methods}: MCSCF, CASSCF, RASSCF, MRCI, MRCC, CASPT2, NEVPT2, etc. (C.~Angeli \& S. Knecht)
\bigskip
\item \alert{Density-functional theory}: DFT, TDDFT, etc. (J. Toulouse/E. Fromager, F. Sottile)
\bigskip
\item \alert{Quantum Monte Carlo}: VMC, DMC, FCIQMC, etc. (M.~Caffarel)
\end{enumerate}
\end{frame}
\begin{frame}{Configuration Interaction (CI)}
\begin{itemize}
\item This is the \blue{oldest} and perhaps the \blue{easiest method to understand}
\bigskip
\item CI is based on the \orange{variational principle} (like HF)
\bigskip
\item The CI wave function is a \blue{linear combination of determinants}
\bigskip
\item CI methods use \violet{excited determinants} to ``improve'' the reference (usually HF) wave function
\begin{equation}
\ket{\Phi_0}
= \underbrace{c_0 \ket*{\Psi_0}}_{\text{reference}}
+ \underbrace{\violet{\sum_{\substack{i \\ a}} c_i^a \ket*{\Psi_i^a}}}_{\text{singles}}
+ \underbrace{\purple{\sum_{\substack{i < j \\ a < b}} c_{ij}^{ab} \ket*{\Psi_{ij}^{ab}}}}_{\text{doubles}}
+ \underbrace{\orange{\sum_{\substack{i < j < k \\ a < b < c}} c_{ijk}^{abc} \ket*{\Psi_{ijk}^{abc}}}}_{\text{triples}}
+ \underbrace{\blue{\sum_{\substack{i < j < k < l \\ a < b < c < d}} c_{ijkl}^{abcd} \ket*{\Psi_{ijkl}^{abcd}}}}_{\text{quadruples}}
+ \ldots
\end{equation}
\end{itemize}
\end{frame}
\begin{frame}{CI method and Excited determinants}
\begin{block}{Excited determinants}
\center
\includegraphics[width=0.7\textwidth]{fig/det}
\end{block}
\begin{block}{CI wave function}
\begin{equation}
\boxed{
\ket{\Phi_0}
= c_0 \ket{\text{0}}
+ \violet{c_\text{S} \ket{\text{S}}}
+ \purple{c_\text{D} \ket{\text{D}}}
+ \orange{c_\text{T} \ket{\text{T}}}
+ \blue{c_\text{Q} \ket{\text{Q}}}
+ \ldots
}
\end{equation}
\end{block}
\end{frame}
\begin{frame}{Truncated CI}
\begin{itemize}
\item When $\ket{\text{S}}$ (\violet{singles}) are taken into account: \textbf{CIS}
\begin{equation}
\violet{\ket{\Phi_\text{CIS}} = c_0 \ket{\text{0}} + c_\text{S} \ket{\text{S}}}
\end{equation}
\textbf{NB:} CIS is an \violet{excited state method}
\item When $\ket{\text{D}}$ (\alert{doubles}) are taken into account: \textbf{CID}
\begin{equation}
\alert{\ket{\Phi_\text{CID}} = c_0 \ket{\text{0}} + c_\text{D} \ket{\text{D}}}
\end{equation}
\textbf{NB:} CID is the \alert{cheapest CI method}
\item When $\ket{\text{S}}$ and $\ket{\text{D}}$ are taken into account: \textbf{CISD}
\begin{equation}
\purple{\ket{\Phi_\text{CISD}} = c_0 \ket{\text{0}} + c_\text{S} \ket{\text{S}} + c_\text{D} \ket{\text{D}} }
\end{equation}
\textbf{NB:} CISD is the \purple{most commonly-used} CI method
\item When $\ket{\text{S}}$, $\ket{\text{D}}$ and $\ket{\text{T}}$ (\orange{triples}) are taken into account: \textbf{CISDT}
\begin{equation}
\orange{\ket{\Phi_\text{CISDT}} = c_0 \ket{\text{0}} + c_\text{S} \ket{\text{S}} + c_\text{D} \ket{\text{D}} + c_\text{T} \ket{\text{T}}}
\end{equation}
\item \textbf{CISDTQ}, etc.
\end{itemize}
\end{frame}
\begin{frame}{Full CI}
\begin{itemize}
\item When all possible excitations are taken into account,
\alert{this is called a Full CI calculation} (\textbf{FCI})
\begin{equation}
\alert{\ket{\Phi_\text{FCI}}
= c_0 \ket{\text{0}}
+ c_\text{S} \ket{\text{S}}
+ c_\text{D} \ket{\text{D}}
+ c_\text{T} \ket{\text{T}}
+ c_\text{Q} \ket{\text{Q}}
+ \ldots}
\end{equation}
\item FCI gives the \violet{exact solution of the Schr\"odinger equation within a given basis}
\bigskip
\item FCI is becoming more and more fashionable these days (e.g. \orange{FCIQMC and SCI methods})
\bigskip
\item \blue{So, why do we care about other methods?}
\bigskip
\item \alert{Because FCI is super computationally expensive!}
\bigskip
\end{itemize}
\end{frame}
\begin{frame}{Size of CI Matrix}
\violet{\textit{``Assume we have 10 electrons in 38 spin MOs: 10 are occupied and 28 are empty''}}
\bigskip
\begin{columns}
\begin{column}{0.65\textwidth}
\begin{itemize}
\item There is $C_{10}^k$ possible ways of selecting $k$ electrons out of the 10 occupied orbitals
$$ C_{n}^k = \frac{n!}{k!(n-k)!} $$
\item There is $C_{28}^k$ ways of distributing them out in the 28 virtual orbitals
\item For a given excitation level $k$, \alert{there is $C_{10}^k C_{28}^k$ excited determinants}
\item \violet{The total number of possible excited determinant} is
$$ \sum_{k=0}^{10}C_{10}^k C_{28}^k = C_{38}^{10} = 472,733,756$$
\item \alert{This is a lot...}
\end{itemize}
\end{column}
\begin{column}{0.35\textwidth}
\small
\orange{For $n = 10$ and $N = 38$:}
\\
\bigskip
\begin{tabular}{cr}
\hline \hline
$k$ & Num. of excitations \\
\hline
0 & 1 \\
1 & 280 \\
2 & 17,010 \\
3 & 393,120 \\
4 & 4,299,750 \\
5 & 24,766,560 \\
6 & 79,115,400 \\
7 & 142,084,800 \\
8 & 139,864,725 \\
9 & 69,069,000 \\
10 & 13,123,110 \\
\hline
Tot. & 472,733,756 \\
\hline \hline
\end{tabular}
\end{column}
\end{columns}
\end{frame}
\begin{frame}{The FCI matrix: \alert{before pruning}}
\begin{equation}
\boxed{
\ket{\Phi_0}
= c_0 \ket{\text{HF}}
+ c_\text{S} \ket{\text{S}}
+ c_\text{D} \ket{\text{D}}
+ c_\text{T} \ket{\text{T}}
+ c_\text{Q} \ket{\text{Q}}
+ \ldots
}
\end{equation}
\bigskip
\begin{equation}
\bH =
\begin{array}{ccccccc}
& | \text{HF} \rangle & | \text{S} \rangle & | \text{D} \rangle & | \text{T} \rangle & | \text{Q} \rangle & \cdots \\
\langle \text{HF} | & \langle \text{HF} | \hH | \text{HF} \rangle & \langle \text{HF} | \hH | \text{S} \rangle & \langle \text{HF} | \hH | \text{D} \rangle & \langle \text{HF} | \hH | \text{T} \rangle & \langle \text{HF} | \hH | \text{Q} \rangle & \cdots \\
\langle \text{S} | & \langle \text{S} | \hH | \text{HF} \rangle & \langle \text{S} | \hH | \text{S} \rangle & \langle \text{S} | \hH | \text{D} \rangle & \langle \text{S} | \hH | \text{T} \rangle & \langle \text{S} | \hH | \text{Q} \rangle & \cdots \\
\langle \text{D} | & \langle \text{D} | \hH |\text{HF} \rangle & \langle \text{D} | \hH | \text{S} \rangle & \langle \text{D} | \hH | \text{D} \rangle & \langle \text{D} | \hH| \text{T} \rangle & \langle \text{D} | \hH | \text{Q} \rangle & \cdots \\
\langle \text{T} | & \langle \text{T} | \hH |\text{HF} \rangle & \langle \text{T} | \hH | \text{S} \rangle & \langle \text{T} | \hH | \text{D} \rangle & \langle \text{T} |\hH | \text{T} \rangle & \langle \text{T} | \hH | \text{Q} \rangle & \cdots \\
\langle \text{Q} | & \langle \text{Q} | \hH | \text{HF} \rangle & \langle \text{Q} | \hH | \text{S} \rangle & \langle \text{Q} | \hH | \text{D} \rangle & \langle \text{Q} | \hH | \text{T} \rangle & \langle \text{Q} | \hH | \text{Q} \rangle & \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
\end{array}
\end{equation}
\end{frame}
\begin{frame}{The FCI matrix: \green{after pruning}}
\begin{equation}
\boxed{
\ket{\Phi_0}
= c_0 \ket{\text{HF}}
+ c_\text{S} \ket{\text{S}}
+ c_\text{D} \ket{\text{D}}
+ c_\text{T} \ket{\text{T}}
+ c_\text{Q} \ket{\text{Q}}
+ \ldots
}
\end{equation}
\bigskip
\begin{equation}
\bH =
\begin{array}{ccccccc}
& | \text{HF} \rangle & | \text{S} \rangle & | \text{D} \rangle & | \text{T} \rangle & | \text{Q} \rangle & \cdots \\
\langle \text{HF} | & \langle \text{HF} | \hH | \text{HF} \rangle & 0 & \langle \text{HF} | \hH | \text{D} \rangle & 0 & 0 & \cdots \\
\langle \text{S} | & 0 & \langle \text{S} | \hH | \text{S} \rangle & \langle \text{S} | \hH | \text{D} \rangle & \langle \text{S} | \hH | \text{T} \rangle & 0 & \cdots \\
\langle \text{D} | & \langle \text{D} | \hH | \text{HF} \rangle & \langle \text{D} | \hH | \text{S} \rangle & \langle \text{D} | \hH | \text{D} \rangle & \langle \text{D} | \hH | \text{T} \rangle & \langle \text{D} | \hH | \text{Q} \rangle & \cdots \\
\langle \text{T} | & 0 & \langle \text{T} | \hH | \text{S} \rangle & \langle \text{T} | \hH | \text{D} \rangle & \langle \text{T} | \hH | \text{T} \rangle & \langle \text{T} | \hH | \text{Q} \rangle & \cdots \\
\langle \text{Q} | & 0 & 0 & \langle \text{Q} | \hH | \text{D} \rangle & \langle \text{Q} | \hH | \text{T} \rangle & \langle \text{Q} | \hH | \text{Q} \rangle & \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
\end{array}
\end{equation}
\end{frame}
\begin{frame}{Rules \& Observations}
\begin{enumerate}
\item No coupling between HF ground state $\ket{ \text{HF} }$ and single excitations $ \ket{ \text{S} }$\\
\violet{$\Rightarrow$ Brillouin's theorem}
\begin{equation}
\violet{\mel{ \text{HF} }{ \hH }{ \text{S} } = 0}
\end{equation}
\item No coupling between $\ket{ \text{HF} }$ and triples $\ket{ \text{T} }$ , quadruples $ \ket{ \text{Q} }$ , etc. \\
\alert{$\Rightarrow$ Slater-Condon rules}
\begin{gather}
\alert{\mel{ \text{HF} }{ \hH }{ \text{T} } = \mel{ \text{HF} }{ \hH }{ \text{Q} } = \ldots = 0}
\\
\alert{\mel{ \text{S} }{ \hH }{ \text{Q} } = \ldots = 0}
\end{gather}
\item $ \ket{ \text{S} }$ have small effect but mix indirectly with $\ket{ \text{D} }$\\
\orange{$\Rightarrow$ CID $\neq$ CISD}
\begin{equation}
\orange{\mel{ \text{HF} }{ \hH }{ \text{S} } = 0 \qq{but} \mel{ \text{S} }{ \hH }{ \text{D} } \neq 0}
\end{equation}
\item $ \ket{ \text{D} }$ have large effect and $ \ket{ \text{Q} }$ more important than $ \ket{ \text{T} }$\\
\blue{$\Rightarrow$ CID gives most of the correlation energy}
\begin{equation}
\blue{\mel{ \text{HF} }{ \hH }{ \text{D} } \gg \mel{ \text{HF} }{ \hH }{ \text{Q} } \gg \mel{ \text{HF} }{ \hH }{ \text{T} }}
\end{equation}
\item \purple{Of course, this matrix is never explicitly built in practice (Davidson algorithm)...}
\end{enumerate}
\end{frame}
\begin{frame}{Example}
\begin{columns}
\begin{column}{0.5\textwidth}
\begin{block}{Weights of excited configurations for \ce{Ne}}
\center
\begin{tabular}{cc}
\hline \hline
Excit. level & Weight \\
\hline
0 & $9.6 \times 10^{-1}$ \\
1 & $9.8 \times 10^{-4}$ \\
2 & $3.4 \times 10^{-2}$ \\
3 & $3.7 \times 10^{-4}$ \\
4 & $4.5 \times 10^{-4}$ \\
5 & $1.9 \times 10^{-5}$ \\
6 & $1.7 \times 10^{-6}$ \\
7 & $1.4 \times 10^{-7}$ \\
8 & $1.1 \times 10^{-9}$ \\
\hline \hline
\end{tabular}
\end{block}
\end{column}
\begin{column}{0.5\textwidth}
\begin{block}{Correlation energy of \ce{Be} and Method scaling}
\center
\begin{table}
\begin{tabular}{lccc}
\hline \hline
Method & $\Delta E_c$ & \% & Scaling \\
\hline
HF & 0 & 0 & $N^4$ \\
CIS & 0 & 0 & $N^5$ \\
CISD & 0.075277 & 96.05 & $N^6$ \\
CISDT & 0.075465 & 96.29 & $N^8$ \\
CISDTQ & 0.078372 & 100 & $N^{10}$ \\
FCI & 0.078372 & 100 & $e^N$ \\
\hline \hline
\end{tabular}
\end{table}
\end{block}
\end{column}
\end{columns}
\end{frame}
\begin{frame}{Size consistency and size extensivity}
\begin{itemize}
\item Truncated CI methods are \alert{size inconsistent} i.e.
$$2E_c (\ce{H2}) \neq E_c (\ce{H2\bond{....}H2})$$
\item Size consistent defines for \purple{non-interacting fragment}
\bigskip
\item \violet{Size extensivity} refers to the scaling of $E_c$ with the number of electrons
\bigskip
\item \blue{NB:} FCI is size consistent and size extensive
\end{itemize}
\end{frame}
\begin{frame}{Rayleigh-Schr\"odinger perturbation theory}
Let's assume we want to find $\Psi_0$ and $E_0$, such as
\begin{equation}
(\hH^{(0)} + \alert{\la} \hH^{(1)}) \Psi_0 = E_0\,\Psi_0
\end{equation}
and that \blue{we know}
\begin{equation}
\boxed{ \hH^{(0)} \Psi^{(0)}_n = E^{(0)}_n \Psi_n^{(0)}, \quad n = 0,1,2,\ldots,\infty}
\end{equation}
Let's expand $\Psi_0$ and $E_0$ in term of $\la$:
\begin{equation}
E_0 = \orange{\la^0}\,E_0^{(0)} + \red{\la^1}\,E_0^{(1)} + \purple{\la^2}\,E_0^{(2)} + \violet{\la^3}\,E_0^{(3)} + \ldots
\end{equation}
\begin{equation}
\Psi_0 = \orange{\la^0}\,\Psi_0^{(0)} + \red{\la^1}\,\Psi_0^{(1)} + \purple{\la^2}\,\Psi_0^{(2)} + \violet{\la^3}\,\Psi_0^{(3)} + \ldots
\end{equation}
such as (\alert{intermediate normalization})
\begin{equation}
\braket{ \Psi_0^{(0)} }{ \Psi_0^{(0)} } = 1 \qquad \braket{ \Psi_0^{(0)} }{ \Psi_0^{(k)} } = 0, \quad k = 1,2,\ldots,\infty
\end{equation}
\end{frame}
\begin{frame}{Rayleigh-Schr\"odinger perturbation theory (Part 1)}
Gathering terms with respect to the power of $\la$:
\begin{align}
& \orange{\la^0:} \qquad \hH^{(0)}\Psi_0^{(0)} = E_0^{(0)} \Psi_0^{(0)}
\\
& \red{\la^1:} \qquad \hH^{(0)}\Psi_0^{(1)} + \hH^{(1)}\Psi_0^{(0)} = E_0^{(0)} \Psi_0^{(1)} + E_0^{(1)} \Psi_0^{(0)}
\\
& \purple{\la^2:} \qquad \hH^{(0)}\Psi_0^{(2)} + \hH^{(1)}\Psi_0^{(1)} = E_0^{(0)} \Psi_0^{(2)} + E_0^{(1)} \Psi_0^{(1)} + E_0^{(2)}
\\
& \violet{\la^3:} \qquad \hH^{(0)}\Psi_0^{(3)} + \hH^{(1)}\Psi_0^{(2)} = E_0^{(0)} \Psi_0^{(3)} + E_0^{(1)} \Psi_0^{(2)} + E_0^{(2)} \Psi_0^{(1)} + E_0^{(3)}
\end{align}
Using the intermediate normalization, we have
\begin{align}
& \orange{\la^0:} \qquad E_0^{(0)} = \mel{ \Psi_0^{(0)} }{ \hH^{(0)} }{ \Psi_0^{(0)} }
\\
& \red{\la^1:} \qquad E_0^{(1)} = \mel{ \Psi_0^{(0)} }{ \hH^{(1)} }{ \Psi_0^{(0)} }
\\
& \purple{\la^2:} \qquad E_0^{(2)} = \mel{ \Psi_0^{(0)} }{ \hH^{(1)} }{ \Psi_0^{(1)} } \qquad \blue{\text{Wigner's (2n+1) rule!}}
\\
& \violet{\la^3:} \qquad E_0^{(3)} = \mel{ \Psi_0^{(0)} }{ \hH^{(1)} }{ \Psi_0^{(2)} }
= \mel{ \Psi_0^{(1)} }{ \hH^{(1)} - E_0^{(1)} }{ \Psi_0^{(1)} }
\end{align}
\end{frame}
\begin{frame}{Rayleigh-Schr\"odinger perturbation theory (Part 2)}
Expanding $\Psi_0^{(1)}$ in the basis $\Psi_n^{(0)}$ with $n = 0,1,2,\ldots,\infty$
\begin{equation}
\ket{ \Psi_0^{(1)} } = \sum_n c_n^{(1)} \ket{ \Psi_n^{(0)} } \qq{$\Rightarrow$} c_n^{(1)} = \braket{ \Psi_n^{(0)} }{ \Psi_0^{(1)} }
\end{equation}
Therefore,
\begin{equation}
\ket{ \Psi_0^{(1)} } = \sum_{n \neq 0} \ket{ \Psi_n^{(0)} } \braket{ \Psi_n^{(0)} }{ \Psi_0^{(1)} }
\end{equation}
Using results from the previous slide, one can show that
\begin{equation}
\purple{\boxed{
E_0^{(2)} = \sum_{n \neq 0} \frac{\mel{ \Psi_0^{(0)} }{ \hH^{(1)} }{ \Psi_n^{(0)} }^2}{E_0^{(0)} - E_n^{(0)} }
}}
\end{equation}
\small
\begin{equation}
\violet{
\boxed{
E_0^{(3)} = \sum_{n,m \neq 0} \frac{\mel{ \Psi_0^{(0)} }{ \hH^{(1)} }{ \Psi_n^{(0)} } \mel{ \Psi_n^{(0)} }{ \hH^{(1)} }{ \Psi_m^{(0)} } \mel{ \Psi_m^{(0)} }{ \hH^{(1)} }{ \Psi_0^{(0)} }}{(E_0^{(0)} - E_n^{(0)})(E_0^{(0)} - E_m^{(0)})} - E_0^{(1)} \sum_{n \neq 0} \frac{\mel{ \Psi_0^{(0)} }{ \hH^{(1)} }{ \Psi_n^{(0)} }^2}{(E_0^{(0)} - E_n^{(0)})^2}
}}
\end{equation}
\end{frame}
\begin{frame}{M{\o}ller-Plesset (MP) perturbation theory}
In \alert{M{\o}ller-Plesset perturbation theory}, the partition is
\begin{equation}
\blue{\hH^{(0)} = \sum_{i=1}^N f(i) = \sum_{i=1}^N [h(i) + v^\text{HF}(i)]}, \qquad \green{\hH^{(1)} = \sum_{i<j} \frac{1}{r_{ij}} - \sum_i v^\text{HF}(i)}
\end{equation}
Therefore,
\begin{equation}
E_0^{(0)} = \sum_i^{\text{occ}} \varepsilon_i, \qquad E_0^{(1)} = - \frac{1}{2} \sum_{ij}^{\text{occ}} \langle ij || ij \rangle \quad \Rightarrow \quad \boxed{\orange{E_\text{HF} = E_0^{(0)} + E_0^{(1)}}}
\end{equation}
The first information about the correlation energy is given by the second-order correction
\begin{equation}
\boxed{
\blue{E_0^{(2)}} = \sum_{i<j}^{\text{occ}} \sum_{a<b}^{\text{virt}} \frac{\mel{ ij }{}{ ab}^2}{\varepsilon_i + \varepsilon_j - \varepsilon_a - \varepsilon_b}} \qq{\alert{This is the MP2 correlation energy!!}}
\end{equation}
\end{frame}
\begin{frame}{MP3 energy}
The third-order correction is a bit ugly...
\begin{equation}
\boxed{
\begin{split}
E_0^{(3)} \notag
& = \frac{1}{8} \sum_{ijkl}\sum_{ab} \frac{\langle ij || ab \rangle \langle kl || ij \rangle \langle ab || kl \rangle}{(\varepsilon_i + \varepsilon_j - \varepsilon_a - \varepsilon_b)(\varepsilon_k + \varepsilon_l - \varepsilon_a - \varepsilon_b)}
\\
& + \frac{1}{8} \sum_{ij}\sum_{abcd} \frac{\langle ij || ab \rangle \langle ab || cd \rangle \langle cd || ij \rangle}{(\varepsilon_i + \varepsilon_j - \varepsilon_a - \varepsilon_b)(\varepsilon_i + \varepsilon_j - \varepsilon_c - \varepsilon_d)}
\\
& + \sum_{ijk}\sum_{abc} \frac{\langle ij || ab \rangle \langle kb || cj \rangle \langle ac || ik \rangle}{(\varepsilon_i + \varepsilon_j - \varepsilon_a - \varepsilon_b)(\varepsilon_i + \varepsilon_k - \varepsilon_a - \varepsilon_c)}
\end{split}
}
\end{equation}
\begin{itemize}
\item \violet{MP2 and MP3 only requires only doubly excited determinants}\\
\item \purple{MP4 does need singly, doubly, triply and quadruply excited determinants!}
\end{itemize}
\end{frame}
\begin{frame}{Illustration for the \ce{Be} atom}
\begin{block}{Correlation energy of \ce{Be} in a 4s2p basis set}
\bigskip
\begin{table}
\begin{tabular}{llcclcc}
\hline \hline
Scaling & Level & $\Delta E_c$ & \% & Level & $\Delta E_c$ & \% \\
\hline
$N^5$ &MP2 & 0.053174 & 67.85 & & & \\
$N^6$ &MP3 & 0.067949 & 86.70 & CISD & 0.075277 & 96.05 \\
$N^7$ &MP4 & 0.074121 & 94.58 & & & \\
$N^8$ &MP5 & 0.076918 & 98.15 & CISDT & 0.075465 & 96.29 \\
$N^9$ &MP6 & 0.078090 & 99.64 & & & \\
$N^{10}$ &MP7 & 0.078493 & 100.15 & CISDTQ & 0.078372 & 100 \\
\hline \hline
\end{tabular}
\end{table}
\begin{itemize}
\item \alert{MP$n$ is not a variational method}, i.e. you can get \alert{an energy lower than the true ground state energy!}
\item MP$n$ \blue{fails} for systems with \blue{small HOMO-LUMO gap}
\item The MP$n$ series \orange{can oscillate} around the exact energy
\item \violet{MP$n$ is size-consistent!}
\end{itemize}
\end{block}
\end{frame}
\begin{frame}{MP2 correlation energy}
\begin{block}{MP2 is the simplest way of catching a good chunk of correlation:}
\begin{equation}
\begin{split}
\green{E_\text{c}^\text{(2)}}
&= \sum_{\red{ij}}^{\text{occ}}\sum_{\blue{ab}}^{\text{virt}}
\frac{ \braket{\red{ij}}{\blue{ab}} (2 \braket{\red{ij}}{\blue{ab}} - \braket{\red{ij}}{\blue{ba}})}
{\epsilon_{\red{i}} + \epsilon_{\red{j}} - \epsilon_{\blue{a}} - \epsilon_{\blue{b}}}
\\
& = \underbrace{
2 \sum_{\red{ij}}\sum_{\blue{ab}}
\frac{ \braket{\red{ij}}{\blue{ab}}^2}
{\epsilon_{\red{i}} + \epsilon_{\red{j}} - \epsilon_{\blue{a}} - \epsilon_{\blue{b}}}
}_{\text{direct part}}
- \underbrace{
\sum_{\red{ij}}\sum_{\blue{ab}}
\frac{ \braket{\red{ij}}{\blue{ab}} \braket{\red{ij}}{\blue{ba}}}
{\epsilon_{\red{i}} + \epsilon_{\red{j}} - \epsilon_{\blue{a}} - \epsilon_{\blue{b}}}
}_{\text{exchange part}}
\end{split}
\end{equation}
\centering
\includegraphics[width=0.5\textwidth]{fig/MP2}
\end{block}
\end{frame}
\begin{frame}{Computing the MP2 correlation energy}
\begin{block}{How much does it cost to compute the MP2 correlation energy?}
\begin{algorithmic}
\Procedure{MP2 correlation energy}{}
\State $\green{E_\text{c}^\text{(2)}} = 0$
\For{$\red{i}=1,O$}
\For{$\red{j}=1,O$}
\For{$\blue{a}=1,V$}
\For{$\blue{b}=1,V$}
\State $\purple{\Delta} = \epsilon_{\red{i}} + \epsilon_{\red{j}} - \epsilon_{\blue{a}} - \epsilon_{\blue{b}}$
\State $\green{E_\text{c}^\text{(2)}}
= \green{E_\text{c}^\text{(2)}}
+ (2 \braket{\red{ij}}{\blue{ab}}^2 - \braket{\red{ij}}{\blue{ab}}\braket{\red{ij}}{\blue{ba}})/\purple{\Delta} $
\EndFor
\EndFor
\EndFor
\EndFor
\EndProcedure
\Comment{\bf \red{$\order{N^4}$ because there are four loops!}}
\end{algorithmic}
\end{block}
\end{frame}
%
%%% SLIDE X %%%
\begin{frame}{AO to MO transformation (Take 1)}
\begin{block}{The naive way...}
\begin{equation}
\underbrace{\green{(pq|rs)}}_{\text{\purple{\bf MO integrals}}}
= \sum_{\red{\mu\nu\la\si}} c_{\red{\mu} \green{p}} c_{\red{\nu} \green{q}} c_{\red{\la} \green{r}} c_{\red{\si} \green{s}}
\underbrace{\red{(\mu\nu|\la\si)}}_{\text{\purple{\bf AO integrals}}}
\end{equation}
\end{block}
\begin{algorithmic}
\scriptsize
\Procedure{AO-to-MO Transformation}{}
\For{$\red{p}=1,N$}
\For{$\blue{q}=1,N$}
\For{$\orange{r}=1,N$}
\For{$\violet{s}=1,N$}
\State $(\red{p}\blue{q}|\orange{r}\violet{s}) = 0$
\Comment{Initialization of the array}
\For{$\red{\mu}=1,N$}
\For{$\blue{\nu}=1,N$}
\For{$\orange{\la}=1,N$}
\For{$\violet{\si}=1,N$}
\State
$(\red{p}\blue{q}|\orange{r}\violet{s}) = (\red{p}\blue{q}|\orange{r}\violet{s})
+ c_{\red{\mu p}} c_{\blue{\nu q}} c_{\orange{\la r}} c_{\violet{\si s}}
(\red{\mu}\blue{\nu}|\orange{\la}\violet{\si})$
\Comment{Accumulation step}
\EndFor
\EndFor
\EndFor
\EndFor
\EndFor
\EndFor
\EndFor
\EndFor
\Comment{\bf \red{This is a $\order{N^8}$ algorithm! You won't do much quantum chemistry with this...}}
\EndProcedure
\end{algorithmic}
\end{frame}
%
%%% SLIDE X %%%
\begin{frame}{AO to MO transformation (Take 2)}
\begin{block}{Semi-direct algorithm...}
\begin{equation}
(\red{p}\blue{q}|\orange{r}\violet{s})
=
\underbrace{
\sum_{\red{\mu p}} c_{\red{\mu p}}
\underbrace{
\qty{ \sum_{\blue{\nu q}} c_{\blue{\nu q}}
\underbrace{
\qty[ \sum_{\orange{\la r}} c_{\orange{\la r}}
\underbrace{
\qty( \sum_{\violet{\si s}} c_{\violet{\si s}}
(\red{\mu}\blue{\nu}|\orange{\la}\violet{\si})
)
}_{\text{\violet{Step \#1}}}
]
}_{\text{\orange{Step \#2}}}
}
}_{\text{\blue{Step \#3}}}
}_{\text{\red{Step \#4}}}
\end{equation}
\end{block}
\end{frame}
%%% SLIDE X %%%
\begin{frame}{Semi-direct algorithm}
\begin{block}{Semi-direct algorithm... \violet{Step \#1}}
\begin{algorithmic}
\Procedure{Semi-Direct Algorithm (\violet{Step \#1})}{}
\State Allocate temporary array $I$ of size $N^4$
\For{$\red{\mu}=1,N$}
\For{$\blue{\nu}=1,N$}
\For{$\orange{\la}=1,N$}
\For{$\violet{\si}=1,N$}
\For{$\violet{s}=1,N$}
\State
$I_{\red{\mu}\blue{\nu}\orange{\la}\violet{s}} = I_{\red{\mu}\blue{\nu}\orange{\la}\violet{s}}
+ c_{\violet{\si s}} (\red{\mu}\blue{\nu}|\orange{\la}\violet{\si})$
\EndFor
\EndFor
\EndFor
\EndFor
\EndFor
\Comment{\violet{Step \#1} costs $\order{N^5}$ and $\order{N^4}$ storage}
\EndProcedure
\end{algorithmic}
\end{block}
\end{frame}
%
\begin{frame}{Semi-direct algorithm}
\begin{block}{Semi-direct algorithm... \orange{Step \#2}}
\begin{algorithmic}
\Procedure{Semi-Direct Algorithm (\orange{Step \#2})}{}
\State Allocate temporary array $J$ of size $N^4$
\For{$\red{\mu}=1,N$}
\For{$\blue{\nu}=1,N$}
\For{$\orange{\la}=1,N$}
\For{$\orange{r}=1,N$}
\For{$\violet{s}=1,N$}
\State
$J_{\red{\mu}\blue{\nu}\orange{r}\violet{s}} = J_{\red{\mu}\blue{\nu}\orange{r}\violet{s}}
+ c_{\orange{\la r}} I_{\red{\mu}\blue{\nu}\orange{\la}\violet{s}}$
\EndFor
\EndFor
\EndFor
\EndFor
\EndFor
\Comment{\orange{Step \#2} costs $\order{N^5}$ and $\order{N^4}$ storage}
\EndProcedure
\end{algorithmic}
\end{block}
\end{frame}
%
\begin{frame}{Semi-direct algorithm}
\begin{block}{Semi-direct algorithm... \blue{Step \#3}}
\begin{algorithmic}
\Procedure{Semi-Direct Algorithm (\blue{Step \#3})}{}
\For{$\red{\mu}=1,N$}
\For{$\blue{\nu}=1,N$}
\For{$\blue{q}=1,N$}
\For{$\orange{r}=1,N$}
\For{$\violet{s}=1,N$}
\State
$I_{\red{\mu}\blue{q}\orange{r}\violet{s}} = I_{\red{\mu}\blue{q}\orange{r}\violet{s}}
+ c_{\blue{\nu q}} J_{\red{\mu}\blue{\nu}\orange{r}\violet{s}}$
\EndFor
\EndFor
\EndFor
\EndFor
\EndFor
\Comment{\blue{Step \#3} costs $\order{N^5}$ and no new storage}
\EndProcedure
\end{algorithmic}
\end{block}
\end{frame}
%
\begin{frame}{Semi-direct algorithm}
\begin{block}{Semi-direct algorithm... \red{Step \#4}}
\begin{algorithmic}
\Procedure{Semi-Direct Algorithm (\red{Step \#4})}{}
\For{$\red{\mu}=1,N$}
\For{$\red{p}=1,N$}
\For{$\blue{q}=1,N$}
\For{$\orange{r}=1,N$}
\For{$\violet{s}=1,N$}
\State
$(\red{p}\blue{q}|\orange{r}\violet{s}) = (\red{p}\blue{q}|\orange{r}\violet{s})
+ c_{\red{\mu p}} I_{\red{\mu}\blue{q}\orange{r}\violet{s}}$
\EndFor
\EndFor
\EndFor
\EndFor
\EndFor
\Comment{\red{Step \#4} costs $\order{N^5}$ and no new storage}
\EndProcedure
\end{algorithmic}
\end{block}
\end{frame}
\begin{frame}{Laplace transform}
\begin{block}{Alml{\"o}f's trick}
\begin{equation}
\boxed{\frac{1}{\purple{\Delta}} = \blue{\int_0^\infty} \exp(-\purple{\Delta} \blue{t}) d\blue{t}}
\end{equation}
\begin{equation}
\begin{split}
\green{E_\text{c}^\text{(2)}}
& = \frac{1}{4} \sum_{ij}\sum_{ab}
\frac{\mel{ij}{}{ab}^2}
{\purple{\epsilon_{i} + \epsilon_{j} - \epsilon_{a} - \epsilon_{b}}}
\\
& = \frac{1}{4} \blue{\int_0^{\infty}} \sum_{ij}\sum_{ab}
\mel{ij}{}{ab}^2 \exp[-(\purple{\epsilon_{i} + \epsilon_{j} - \epsilon_{a} - \epsilon_{b}}) \blue{t}] d\blue{t}
\\
& = \frac{1}{4} \blue{\int_0^{\infty}} \sum_{ij}\sum_{ab} \mel{i(\blue{t})j(\blue{t})}{}{a(\blue{t})b(\blue{t})}^2
\stackrel{\text{\blue{quad.}}}{\approx} \frac{1}{4} \blue{\sum_{k=1}^{N_\text{grid}}} \blue{w_k} \sum_{ij}\sum_{ab} \mel{i(\blue{t_k})j(\blue{t_k})}{}{a(\blue{t_k})b(\blue{t_k})}^2
\end{split}
\end{equation}
\begin{equation}
\ket*{p} \equiv \ket*{\varphi_p(0)}
\qqtext{and}
\ket*{p(\blue{t})} \equiv \ket*{\varphi_p(\blue{t})} = \ket*{\varphi_p(0) e^{\pm\frac{1}{2} \purple{\epsilon_p} \blue{t}}}
\end{equation}
\\
\bigskip
At this stage, one can play more games (e.g., localized orbitals, RI, stochastic sampling, quadrature, etc)
\end{block}
\end{frame}
%%% SLIDE 2 %%%
\begin{frame}{Coupled-Cluster Theory}
\begin{block}{A few random thoughts about coupled cluster (CC)}
\begin{itemize}
\bigskip
\item CC theory comes from \alert{nuclear physics}
\bigskip
\item The idea behind CC is to include \alert{all corrections} of a given type to \alert{infinite order}
\bigskip
\item The CC wave function is an \alert{exponential \textit{ansatz}}
\bigskip
\item The CC energy is \alert{size-extensive}, but \alert{non-variational}
\bigskip
\item CC is considered as the \alert{gold standard} for weakly correlated systems
\end{itemize}
\end{block}
\end{frame}
%%% SLIDE 3 %%%
\begin{frame}{Theory}
\begin{itemize}
\item CC wave function
\begin{equation}
\PsiCC = \alert{e^{\hT{}{}}} \PsiO
\qq{where $\PsiO$ is a reference wave function}
\end{equation}
\item Excitation operator
\begin{equation}
\hT{}{} = \hT{1}{} + \hT{2}{} + \ldots + \hT{\nEl}{}% \qquad \text{where $\nEl$ is the number of electrons}
\end{equation}
\item Exponential \textit{ansatz}
\begin{equation}
\begin{split}
e^{\hT{}{}}
& = \hI + \hT{}{} + \frac{1}{2!} \hT{}{2} + \frac{1}{3!} \hT{}{3} + \ldots
\\
& = \hI + \hT{1}{}
+ \qty( \underbrace{\hT{2}{}}_{\text{\green{connected}}} + \frac{1}{2} \underbrace{\hT{1}{2}}_{\text{\alert{disconnected}}} )
+ \qty( \hT{3}{} + \hT{2}{} \hT{1}{} + \frac{1}{6} \hT{1}{3} )
\\
& + \qty( \underbrace{\hT{4}{}}_{\text{
\purple{four electrons}}} + \hT{3}{} \hT{1}{} + \frac{1}{2} \underbrace{\hT{2}{2}}_{\text{\blue{two pairs of electrons}}} + \frac{1}{2} \hT{2}{} \hT{1}{2} + \frac{1}{24} \hT{1}{4} )
+ \ldots
\end{split}
\end{equation}
\end{itemize}
\end{frame}
%%% SLIDE 4 %%%
\begin{frame}{Excitation operators}
\begin{itemize}
\item Singles
\begin{equation}
\hT{1}{} = \sum_{ia} \underbrace{\amp{i}{a}}_{\text{\alert{amplitudes}}} \cre{a} \ani{i}
\qq{$\Rightarrow$}
\hT{1}{} \PsiO = \sum_{ia} \amp{i}{a} \Det{i}{a}
\end{equation}
\item Doubles
\begin{equation}
\hT{2}{} = \frac{1}{4} \sum_{ijab} \amp{ij}{ab} \cre{a} \cre{b} \ani{j} \ani{i}
\qq{$\Rightarrow$}
\hT{2}{} \PsiO = \frac{1}{4} \sum_{ijab} \amp{ij}{ab} \underbrace{\Det{ij}{ab}}_{\text{\alert{excited determinants}}}
\end{equation}
\item FCI wave function
\begin{equation}
\PsiFCI = ( \hI + \hT{}{}) \PsiO = ( \hI + \hT{1}{} + \hT{2}{} + \hT{3}{} + \ldots) \PsiO
\end{equation}
\item Anticommutation relation of the annihilation and creation operators
\begin{align}
\cre{p} \cre{q} + \cre{q} \cre{p} & = 0
&
\ani{p} \ani{q} + \ani{q} \ani{p} & = 0
&
\cre{p} \ani{q} + \ani{p} \cre{q} & = \delta_{pq}
\end{align}
\end{itemize}
\end{frame}
%%% SLIDE 5 %%%
\begin{frame}{CC energies}
\begin{itemize}
\item Schr\"odinger equation
\begin{equation}
\hH \ket{\PsiCC} = E \ket*{\PsiCC}
\, \Rightarrow \,
\hH e^{\hT{}{}} \ket*{\PsiO} = E e^{\hT{}{}} \ket*{\PsiO}
\, \Rightarrow \,
\underbrace{e^{-\hT{}{}} \hH e^{\hT{}{}}}_{\text{\green{$\Bar{H} = \text{similarity transform}$}}} \ket*{\PsiO} = E \ket*{\PsiO}
\end{equation}
\item Variational CC energy (\alert{factorial complexity})
\begin{equation}
\alert{\EVCC}
= \frac{\mel*{\PsiCC}{\hH}{\PsiCC}}{\braket*{\PsiCC}{\PsiCC}}
= \frac{\mel*{\PsiO (e^{\hT{}{}})^{\dag}}{\hH}{e^{\hT{}{}}\PsiO}}{\braket*{\PsiO (e^{\hT{}{}})^{\dag}}{e^{\hT{}{}}\PsiO}}
\ge E_\text{exact}
\end{equation}
\item (Traditional) projected CC energy (\blue{polynomial complexity})
\begin{equation}
\blue{\ETCC}
= \frac{\mel*{\PsiO}{\green{\Bar{H}}}{\PsiO}}{\braket*{\PsiO}{\PsiO}}
= \frac{\mel*{\PsiO e^{-\hT{}{}}}{\hH}{e^{\hT{}{}}\PsiO}}{\braket*{\PsiO e^{-\hT{}{}}}{e^{\hT{}{}}\PsiO}}
\end{equation}
\item Unitary CC energy (\purple{very expensive unless you have a quantum computer})
\begin{equation}
\purple{\EUCC}
= \frac{\mel*{\PsiO (e^{\Hat{\tau}})^{\dag}}{\hH}{e^{\Hat{\tau}}\PsiO}}{\braket*{\PsiO (e^{\Hat{\tau}})^{\dag}}{e^{\Hat{\tau}}\PsiO}}
= \frac{\mel*{\PsiO e^{-\Hat{\tau}}}{\hH}{e^{\Hat{\tau}}\PsiO}}{\braket*{\PsiO}{\PsiO}}
\qq{where
$\Hat{\tau} = \hT{}{} - \hT{}{\dag}$
is anti-Hermitian}
\end{equation}
\end{itemize}
\end{frame}
\begin{frame}{Similarity-transformed Hamiltonians \& Amplitude equations}
\begin{block}{Similarity-transformed Hamiltonians}
\begin{itemize}
\item The similarity-transformed Hamiltonian $\green{\Bar{H}} = e^{-\hT{}{}} \hH e^{\hT{}{}}$ is \alert{\bf not} Hermitian:
\begin{equation}
(e^{-\hT{}{}} \hH e^{\hT{}{}})^{\dag}
= (e^{\hT{}{}})^{\dag} \hH^{\dag} (e^{-\hT{}{}})^{\dag}
= e^{\hT{}{\dag}} \hH e^{-\hT{}{\dag}}
\neq e^{-\hT{}{}} \hH e^{\hT{}{}}
\end{equation}
\item The similarity-transformed Hamiltonian $e^{-\Hat{\tau}} \hH e^{\Hat{\tau}}$ is Hermitian:
\begin{equation}
(e^{-\Hat{\tau}} \hH e^{\Hat{\tau}})^{\dag}
= (e^{\Hat{\tau}})^{\dag} \hH^{\dag} (e^{-\Hat{\tau}})^{\dag}
= e^{\Hat{\tau}^{\dag}} \hH e^{-\Hat{\tau}^{\dag}}
= e^{-\Hat{\tau}} \hH e^{\Hat{\tau}}
\qq{because}
\Hat{\tau}^{\dag} = - \Hat{\tau}
\end{equation}
\end{itemize}
\end{block}
\begin{block}{The two most important equations in CC theory}
\begin{itemize}
\item \alert{The energy equation}
\begin{equation}
\boxed{\mel*{\PsiO }{e^{-\hT{}{}} \hH e^{\hT{}{}} }{\PsiO} = E}
\end{equation}
\item \alert{The amplitude equation}
\begin{equation}
\boxed{\mel*{\Psi_{ij\dots}^{ab\dots}} {e^{-\hT{}{}} \hH e^{\hT{}{}} }{\PsiO} = 0
\qq{$\Rightarrow$} t_{ij\dots}^{ab\dots}}
\end{equation}
\end{itemize}
\end{block}
\end{frame}
\begin{frame}{CISD vs CCSD}
Let's build the CISD and CCSD Hamiltonian matrix in the basis of $\ket{0}$, $\ket{\text{S}}$, and $\ket{\text{D}}$:
\bigskip
\begin{block}{CISD Hamiltonian}
\begin{equation}
\hH_\text{CISD} =
\begin{pmatrix}
E_\text{HF} & 0 & H_\text{0D} \\
0 & H_\text{SS} & H_\text{SD} \\
H_\text{D0} & H_\text{DS} & H_\text{DD} \\
\end{pmatrix}
\end{equation}
\end{block}
\begin{block}{CCSD Hamiltonian}
\begin{equation}
\Bar{H}_\text{CCSD} =
\begin{pmatrix}
E_\text{CC} & \Bar{H}_\text{0S} & \Bar{H}_\text{0D} \\
0 & \Bar{H}_\text{SS} & \Bar{H}_\text{SD} \\
0 & \Bar{H}_\text{DS} & \Bar{H}_\text{DD} \\
\end{pmatrix}
\end{equation}
\\
\bigskip
\textbf{NB:} This is the \alert{equation-of-motion} (EOM) CCSD Hamiltonian!
\end{block}
\end{frame}
%%% SLIDE 10 %%%
\begin{frame}{The Hausdorff expansion}
\begin{block}{Campbell-Baker-Hausdorff formula}
\begin{equation}
e^{-\hT{}{}} \hH e^{\hT{}{}}
= \hH + \comm{\hH}{\hT{}{}}
+ \frac{1}{2!} \comm{\comm{\hH}{\hT{}{}}}{\hT{}{}}
+ \frac{1}{3!} \comm{\comm{\comm{\hH}{\hT{}{}}}{\hT{}{}}}{\hT{}{}}
+ \frac{1}{4!} \comm{\comm{\comm{\hH}{\hT{}{}}}{\hT{}{}}}{\hT{}{}}
+ \dots
\end{equation}
\begin{equation}
\hH = \sum_{pq} h_{pq} \cre{p} \ani{q} + \frac{1}{4} \sum_{pqrs} \mel{pq}{}{rs} \cre{p} \cre{q} \ani{s} \ani{r}
\end{equation}
\begin{equation}
\comm{\cre{p} \ani{q}}{\cre{a} \ani{i}}
= \cre{p} \underbrace{\ani{q} \cre{a}}_{\delta_{qa} - \cre{a} \ani{q}} \ani{i} - \cre{a} \underbrace{\ani{i} \cre{p}}_{\delta_{ip} - \cre{p} \ani{i}} \ani{q}
= \cre{p} \delta_{qa} \ani{i} - \cre{a} \delta_{ip} \ani{q}
\end{equation}
\end{block}
\begin{itemize}
\item At the \blue{TCC} level, the BCH expansion \blue{truncates naturally after the first five terms}
\item At the \alert{VCC} level, the BCH expansion \alert{does not truncate but terminates}
\item At the \purple{UCC} level, the BCH expansion \purple{does not terminate}
\end{itemize}
\bigskip
For more details about normal-ordered operators, Wick's theorem, and diagrammatic techniques, see
\pub{Crawford \& Schaefer, Reviews in Computational Chemistry, Vol.~14, Chap.~2, 2000.}
\end{frame}
%%% SLIDE 6 %%%
\begin{frame}{Projected CC energy}
Assuming that $\braket*{\PsiO}{\PsiO} = 1$, we have
\begin{equation}
\begin{split}
\ECC
& = \mel*{\PsiO}{\hH e^{\hT{}{}}}{\PsiO}
\\
& = \mel*{\PsiO}{\hH \alert{(\hI + \hT{1}{} + \hT{2}{} + \frac{1}{2} \hT{1}{2} )}}{\PsiO}
\\
& = \mel*{\PsiO}{\hH}{\PsiO}
+ \mel*{\PsiO}{\hH \hT{1}{}}{\PsiO}
+ \mel*{\PsiO}{\hH \hT{2}{}}{\PsiO}
+ \frac{1}{2} \mel*{\PsiO}{\hH \hT{1}{2}}{\PsiO}
\\
& = \EO
+ \sum_{i} \sum_{a} \amp{i}{a} \green{\mel*{\PsiO}{\hH}{\Det{i}{a}}}
+ \frac{1}{4} \sum_{ij} \sum_{ab}(\amp{ij}{ab} + \amp{i}{a} \amp{j}{b} - \amp{i}{b} \amp{j}{a} ) \blue{\mel*{\PsiO}{\hH}{\Det{ij}{ab}}}
\end{split}
\end{equation}
\end{frame}
%%% SLIDE 7 %%%
\begin{frame}{CC energy with Hartree-Fock reference}
\begin{block}{HF reference wave function}
\begin{itemize}
\bigskip
\item $\orange{\EO = \EHF}$
\bigskip
\item $\green{\mel*{\PsiO}{\hH}{\Det{i}{a}} = \mel*{i}{f}{a} = 0} \quad \Leftrightarrow \quad \text{\alert{Brillouin's theorem}}$
\bigskip
\item $\blue{\mel*{\PsiO}{\hH}{\Det{ij}{ab}} = \mel*{ij}{}{ab} = \braket*{ij}{ab} - \braket*{ij}{ba}} \quad \Leftrightarrow \quad \text{\alert{Two-electron integrals}}$
\bigskip
\end{itemize}
\begin{equation}
\boxed{
\ECC = \orange{\EHF}
+ \frac{1}{4} \sum_{ij} \sum_{ab} (\amp{ij}{ab} + \amp{i}{a} \amp{j}{b} - \amp{i}{b} \amp{j}{a} ) \blue{\mel*{ij}{}{ab}}
}
\end{equation}
\end{block}
\end{frame}
%%% SLIDE 8 %%%
\begin{frame}{Truncated CC}
\begin{block}{CC with doubles (CCD)}
\begin{itemize}
\bigskip
\item Only doubles, doubles of doubles, etc $ \Rightarrow \alert{\hT{}{} = \hT{2}{}}$
\bigskip
\item Still an infinite series
\begin{equation}
\begin{split}
e^{\hT{2}{}}
= \hI + \hT{2}{} + \frac{1}{2} \hT{2}{2} + \frac{1}{6} \hT{2}{3} + \frac{1}{24} \hT{2}{4} + \ldots
\end{split}
\end{equation}
\bigskip
\item CCD energy
\begin{equation}
\ECCD = \EHF + \frac{1}{4} \sum_{ij} \sum_{ab} \amp{ij}{ab} \mel*{ij}{}{ab}
\end{equation}
\end{itemize}
\end{block}
\end{frame}
%%% SLIDE 9 %%%
\begin{frame}{CCD equations}
\begin{itemize}
\item Projection of similarity-transformed Hamiltonian onto doubles
\begin{equation}
\mel*{\alert{\Det{ij}{ab}}}{\green{\bar{H}}}{\PsiO} = \ECC \braket*{\alert{\Det{ij}{ab}}}{\PsiO} = 0
\quad \Rightarrow \quad
\mel*{\alert{\Det{ij}{ab}}}{e^{-\hT{}{}} \hH e^{\hT{}{}}}{\PsiO}= 0
\end{equation}
\item \alert{Residual equation}
\begin{equation}
\boxed{r_{ij}^{ab} = \mel*{ij}{}{ab} + \blue{\Delta_{ij}^{ab}} \amp{ij}{ab} + \green{u_{ij}^{ab}} + \alert{v_{ij}^{ab}} = 0}
\quad \Rightarrow \quad
\boxed{\amp{ij}{ab} = - \frac{\mel*{ij}{}{ab} + \green{u_{ij}^{ab}} + \alert{v_{ij}^{ab}}}{\blue{\Delta_{ij}^{ab}}}}
\end{equation}
\item Energy differences
\begin{equation}
\blue{\Delta_{ij}^{ab}} = \epsilon_a + \epsilon_b - \epsilon_i - \epsilon_j
\end{equation}
\item \green{Linear} array
\begin{equation}
\green{u_{ij}^{ab}} = f(\amp{ij}{ab}) = \order*{\green{\nBas^6}}
\end{equation}
\item \alert{Quadratic} array
\begin{equation}
\alert{v_{ij}^{ab}} = f(\amp{ij}{ab}) = \underbrace{\order*{\green{\nBas^6}}}_{\green{\text{smart}}} \text{ or } \underbrace{\order*{\alert{\nBas^8}}}_{\alert{\text{dumb}}}
\end{equation}
\end{itemize}
\end{frame}
%%% SLIDE 10 %%%
\begin{frame}{\green{Linear} array}
Each term of the linear term can be computed in $\order*{N^6}$:
\begin{equation}
\begin{split}
u_{ij}^{ab}
& = \frac{1}{2} \sum_{cd} \alert{\underbrace{\mel*{ab}{}{cd}}_{VVVV}} \amp{ij}{cd}
+ \frac{1}{2} \sum_{kl} \green{\underbrace{\mel*{kl}{}{ij}}_{OOOO}} \amp{kl}{ab}
\\
& + \sum_{kc} \qty[ - \orange{\underbrace{\mel*{kb}{}{jc}}_{OVOV}} \amp{ik}{ac} + \orange{\mel*{ka}{}{jc}} \amp{ik}{bc}
- \orange{\mel*{ka}{}{ic}} \amp{jk}{bc} + \orange{\mel*{kb}{}{ic}} \amp{jk}{ac} ]
\end{split}
\end{equation}
\alert{\bf NB:} CCD($v_{ij}^{ab} = 0$) $=$ linear CCD (LCCD)
\end{frame}
%%% SLIDE 11 %%%
\begin{frame}{\alert{Quadratic} array: the \alert{dumb} way}
The quadratic term is the computational bottleneck of CCD:
\begin{equation}
\begin{split}
v_{ij}^{ab}
= \frac{1}{4} \sum_{klcd} \purple{\underbrace{\mel*{kl}{}{cd}}_{OOVV}}
\bigg[
& \amp{ij}{cd} \amp{kl}{ab}
- 2(\amp{ij}{ac} \amp{kl}{bd} + \amp{ij}{bd} \amp{kl}{ac})
\\
& - 2 (\amp{ik}{ab} \amp{jl}{cd} + \amp{ik}{cd} \amp{jl}{ab})
+ 4 (\amp{ik}{ac} \amp{jl}{bd} + \amp{ik}{bd} \amp{jl}{ac})
\bigg]
\end{split}
\end{equation}
The ``formal'' scaling of the quadratic term is $\order*{N^8}$
\end{frame}
%%% SLIDE 12 %%%
\begin{frame}{\alert{Quadratic} array: the \green{smart} way}
One can ``sacrifice'' storage to gain in scaling:
\begin{align}
\underbrace{\alert{\mel*{kl}{X_1}{ij}}}_{\order*{N^6}} & = \sum_{cd} \mel*{kl}{}{cd} \amp{ij}{cd}
&
\underbrace{\orange{\mel*{b}{X_2}{c}}}_{\order*{N^5}} & = \sum_{kld} \mel*{kl}{}{cd} \amp{kl}{bd}
\\
\underbrace{\green{\mel*{k}{X_3}{j}}}_{\order*{N^5}} & = \sum_{lcd} \mel*{kl}{}{cd} \amp{jl}{cd}
&
\underbrace{\purple{\mel*{il}{X_4}{ad}}}_{\order*{N^6}} & = \sum_{kc} \mel*{kl}{}{cd} \amp{ik}{ac}
\end{align}
Now, the quadratic term can be computed in $\order*{N^6}$
\begin{equation}
\begin{split}
v_{ij}^{ab}
& = \frac{1}{4} \sum_{kl} \alert{\mel*{kl}{X_1}{cd}} \amp{kl}{ab}
- \frac{1}{2} \sum_{c} \qty[ \orange{\mel*{b}{X_2}{c}} \amp{ij}{ac} + \orange{\mel*{a}{X_2}{c}} \amp{ij}{cb} ]
\\
& - \frac{1}{2} \sum_{k} \qty[ \green{\mel*{k}{X_3}{j}} \amp{ik}{ab} + \green{\mel*{k}{X_3}{i}} \amp{kj}{ab} ]
+ \sum_{kc} \qty[ \purple{\mel*{ik}{X_4}{ac}} \amp{jk}{bc} + \purple{\mel*{ik}{X_4}{bc}} \amp{kj}{ac} ]
\end{split}
\end{equation}
\end{frame}
%%% SLIDE 12 %%%
\begin{frame}{CCD algorithm}
\begin{columns}
\begin{column}{0.7\textwidth}
\begin{block}{CCD subroutine}
\begin{algorithmic}
\Procedure{Iterative CCD algorithm}{}
\State Perform HF calculation to get $\epsilon_p$ and $\mel*{pq}{}{rs}$
\State Set $\green{u_{ij}^{ab}} = 0$, and $\orange{v_{ij}^{ab}} = 0$
\State Compute amplitudes $\red{t_{ij}^{ab}} = - \mel*{ij}{}{ab}/\Delta_{ij}^{ab}$ (MP2 guess)
\While{$\max{\abs*{\blue{r_{ij}^{ab}}}} > \tau$}
\State Form linear array $\green{u_{ij}^{ab}}$
\State Compute intermediate arrays $\mel*{kl}{X_1}{ij}$, $\mel*{b}{X_2}{c}$, $\mel*{k}{X_3}{j}$, and $\mel*{il}{X_4}{ad}$.
\State Form quadratic array $\orange{v_{ij}^{ab}}$
\State Compute residues: $\blue{r_{ij}^{ab}} = \mel*{ij}{}{ab} + \Delta_{ij}^{ab} \amp{ij}{ab} + \green{u_{ij}^{ab}} + \orange{v_{ij}^{ab}}$
\State Update amplitudes: $\red{t_{ij}^{ab}} \leftarrow \red{t_{ij}^{ab}} - \blue{r_{ij}^{ab}}/\Delta_{ij}^{ab}$
\EndWhile
\State Compute CCD energy: $\ECCD = \EHF + \frac{1}{4} \sum_{ij} \sum_{ab} \red{t_{ij}^{ab}} \mel*{ij}{}{ab}$
\EndProcedure
\end{algorithmic}
\end{block}
\end{column}
\begin{column}{0.3\textwidth}
\centering
\includegraphics[width=\textwidth]{fig/Diagrams-CCD}
\end{column}
\end{columns}
\end{frame}
\begin{frame}{Illustration for the \ce{Be} atom}
\begin{block}{Correlation energy of \ce{Be} in a 4s2p basis set}
\bigskip
\begin{table}
\small
\begin{tabular}{llcclcclcc}
\hline \hline
Scaling & Level & $\Delta E_c$ & \% & Level & $\Delta E_c$ & \% & Level & $\Delta E_c$ & \% \\
\hline
$N^5$ &MP2 & 0.053174 & 67.85 & & & & & & \\
$N^6$ &MP3 & 0.067949 & 86.70 & CISD & 0.075277 & 96.05 & CCSD & 0.078176 & 99.75 \\
$N^7$ &MP4 & 0.074121 & 94.58 & & & & CCSD(T) & 0.078361 & 99.99 \\
$N^8$ &MP5 & 0.076918 & 98.15 & CISDT & 0.075465 & 96.29 & CCSDT & 0.078364 & 99.99 \\
$N^9$ &MP6 & 0.078090 & 99.64 & & & & & & \\
$N^{10}$ &MP7 & 0.078493 & 100.15 & CISDTQ & 0.078372 & 100 & CCSDTQ & 0.078372 & 100 \\
\hline \hline
\end{tabular}
\end{table}
\bigskip
\alert{As a rule of thumb:}\\
HF $\ll$ MP2 $<$ CISD $<$ MP4(SDQ) $\sim$ CCSD $<$ MP4 $<$ CCSD(T)
\end{block}
\end{frame}
%%% FINAL SLIDE %%%
%-----------------------------------------------------
\begin{frame}{Good books}
\begin{columns}
\begin{column}{0.7\textwidth}
\begin{itemize}
\item Introduction to Computational Chemistry (Jensen)
\\
\vspace{1cm}
\item Essentials of Computational Chemistry (Cramer)
\\
\vspace{1cm}
\item Modern Quantum Chemistry (Szabo \& Ostlund)
\\
\vspace{1cm}
\item Molecular Electronic Structure Theory (Helgaker, Jorgensen \& Olsen)
\\
\vspace{1cm}
\end{itemize}
\end{column}
\begin{column}{0.3\textwidth}
\centering
\includegraphics[height=0.3\textwidth]{fig/Jensen}
\\
\bigskip
\includegraphics[height=0.3\textwidth]{fig/Cramer}
\\
\bigskip
\includegraphics[height=0.3\textwidth]{fig/Szabo}
\\
\bigskip
\includegraphics[height=0.3\textwidth]{fig/Helgaker}
\end{column}
\end{columns}
\end{frame}
\end{document}
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