ISTPC/2021/Lecture_1/ISTPC_Loos_1.tex

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% ***********
% * PACKAGE *
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% operators
\newcommand{\hI}{\Hat{1}}
\newcommand{\hH}{\Hat{H}}
\newcommand{\hT}[2]{\Hat{T}_{#1}^{#2}}
\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{\EVCC}{E_\text{VCC}}
\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 ]}
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% *************
% * 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{Online ISTPC 2021 school --- April 27th, 2021}
\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 Computing the 2th-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}{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{$\mathcal{O}_2 = r_{12}^{-1}$}} && \mc{2}{c}{\orange{$\mathcal{O}_2 = e^{-\la r_{12}^2}$ ($\la = 1$)}} \\
\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) = a fancy way of writing 1}
\begin{equation}
\boxed{\sum_{\green{A}=1}^{\red{\infty}} |\green{A}) (\green{A}| = 1
\qqtext{and, in practice, }
\sum_{\green{A}=1}^{\red{K}} |\green{A}) (\green{A}| \approx 1}
\end{equation}
\end{block}
\begin{block}{Computing the Coulomb matrix with the RI}
\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}) I_{\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 Julien'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}{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} (Thank you 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% \qquad \text{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{3}{} )
\\
& + \qty( \hT{4}{} + \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} \underbrace{\hT{4}{}}_{\text{
\purple{four electrons}}} )
+ \ldots
\end{split}
\end{equation}
\end{itemize}
\end{frame}
%%% SLIDE 4 %%%
\begin{frame}{Excitation operators}
\begin{itemize}
\item Singles
\begin{equation}
\hT{1}{} \PsiO = \sum_{i} \sum_{a} \underbrace{\amp{i}{a}}_{\text{\alert{amplitudes}}} \Det{i}{a}
\end{equation}
\item Doubles
\begin{equation}
\hT{2}{} \PsiO = \sum_{i<j} \sum_{a<b} \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}
\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*{e^{\hT{}{}}\PsiO}{\hH}{e^{\hT{}{}}\PsiO}}{\braket*{e^{\hT{}{}}\PsiO}{e^{\hT{}{}}\PsiO}}
\end{equation}
\item Projected CC energy
\begin{equation}
\mel*{\PsiO}{\hH}{\PsiCC} = \ECC \braket*{\PsiO}{\PsiCC}
\quad \Rightarrow \quad
\boxed{\ECC = \frac{\mel*{\PsiO}{\alert{\hH e^{\hT{}{}}}}{\PsiO}}{\braket*{\PsiO}{\PsiO}}}
\end{equation}
\end{itemize}
\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}
%%%% FINAL SLIDE %%%
%\begin{frame}{Further reading for CC methods}
% \begin{itemize}
% \item Pople et al. IJQC 14 (1978) 545
% \bigskip
% \item Scuseria et al, JCP 86 (1987) 2881
% \bigskip
% \item Jensen, \textit{``Introduction to Computational Chemistry''} (Wiley)
% \bigskip
% \item Crawford \& Schaefer, \textit{``An Introduction to Coupled Cluster Theory for Computational Chemists''}, Reviews in Computational Chemistry, Vol.~14 (2000)
% \bigskip
% \item Shavitt \& Bartlett, \textit{``Many-Body Methods in Chemistry and Physics: MBPT and Coupled-Cluster Theory''} (Cambridge University Press)
% \end{itemize}
%\end{frame}
%%% FINAL SLIDE %%%
%-----------------------------------------------------
\section{Books}
%-----------------------------------------------------
\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}