creating the note on perturbative analysis and saving work
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Notes/PerturbativeAnalysis.tex
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Notes/PerturbativeAnalysis.tex
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\documentclass[aip,jcp,reprint,noshowkeys,superscriptaddress]{revtex4-1}
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\usepackage{graphicx,dcolumn,bm,xcolor,microtype,multirow,amscd,amsmath,amssymb,amsfonts,physics,longtable,wrapfig,txfonts,mleftright}
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\usepackage[version=4]{mhchem}
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\usepackage[utf8]{inputenc}
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\usepackage[T1]{fontenc}
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\usepackage{txfonts}
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\usepackage[
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colorlinks=true,
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citecolor=blue,
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breaklinks=true
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]{hyperref}
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\urlstyle{same}
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\newcommand{\ie}{\textit{i.e.}}
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\newcommand{\eg}{\textit{e.g.}}
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\newcommand{\alert}[1]{\textcolor{red}{#1}}
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\usepackage[normalem]{ulem}
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\newcommand{\titou}[1]{\textcolor{red}{#1}}
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\newcommand{\trashPFL}[1]{\textcolor{r\ed}{\sout{#1}}}
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\newcommand{\PFL}[1]{\titou{(\underline{\bf PFL}: #1)}}
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\newcommand{\mc}{\multicolumn}
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\newcommand{\fnm}{\footnotemark}
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\newcommand{\fnt}{\footnotetext}
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\newcommand{\tabc}[1]{\multicolumn{1}{c}{#1}}
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\newcommand{\QP}{\textsc{quantum package}}
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\newcommand{\T}[1]{#1^{\intercal}}
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% coordinates
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\newcommand{\br}{\boldsymbol{r}}
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\newcommand{\bx}{\boldsymbol{x}}
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\newcommand{\dbr}{d\br}
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\newcommand{\dbx}{d\bx}
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% methods
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\newcommand{\GW}{\text{$GW$}}
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\newcommand{\evGW}{ev$GW$}
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\newcommand{\qsGW}{qs$GW$}
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\newcommand{\GOWO}{$G_0W_0$}
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\newcommand{\Hxc}{\text{Hxc}}
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\newcommand{\xc}{\text{xc}}
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\newcommand{\Ha}{\text{H}}
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\newcommand{\co}{\text{c}}
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\newcommand{\x}{\text{x}}
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\newcommand{\KS}{\text{KS}}
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\newcommand{\HF}{\text{HF}}
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\newcommand{\RPA}{\text{RPA}}
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%
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\newcommand{\Ne}{N}
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\newcommand{\Norb}{K}
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\newcommand{\Nocc}{O}
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\newcommand{\Nvir}{V}
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% operators
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\newcommand{\hH}{\Hat{H}}
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\newcommand{\hS}{\Hat{S}}
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\newcommand{\ani}[1]{\hat{a}_{#1}}
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\newcommand{\cre}[1]{\hat{a}_{#1}^\dagger}
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\newcommand{\no}[2]{\mleft\{ \hat{a}_{#1}^{#2}\mright\} }
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% energies
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\newcommand{\Enuc}{E^\text{nuc}}
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\newcommand{\Ec}[1]{E_\text{c}^{#1}}
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\newcommand{\EHF}{E^\text{HF}}
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% orbital energies
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\newcommand{\eps}{\epsilon}
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\newcommand{\reps}{\Tilde{\epsilon}}
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\newcommand{\Om}{\Omega}
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% Matrix elements
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\newcommand{\SigC}{\Sigma^\text{c}}
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\newcommand{\rSigC}{\Tilde{\Sigma}^\text{c}}
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\newcommand{\MO}[1]{\phi_{#1}}
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\newcommand{\SO}[1]{\psi_{#1}}
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\newcommand{\eri}[2]{\braket{#1}{#2}}
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\newcommand{\aeri}[2]{\mel{#1}{}{#2}}
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\newcommand{\ERI}[2]{(#1|#2)}
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\newcommand{\rbra}[1]{(#1|}
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\newcommand{\rket}[1]{|#1)}
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% Matrices
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\newcommand{\bO}{\boldsymbol{0}}
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\newcommand{\bI}{\boldsymbol{1}}
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\newcommand{\bH}{\boldsymbol{H}}
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\newcommand{\bSigC}{\boldsymbol{\Sigma}^{\text{c}}}
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\newcommand{\be}{\boldsymbol{\epsilon}}
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\newcommand{\bOm}{\boldsymbol{\Omega}}
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\newcommand{\bA}{\boldsymbol{A}}
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\newcommand{\bB}{\boldsymbol{B}}
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\newcommand{\bC}{\boldsymbol{C}}
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\newcommand{\bD}{\boldsymbol{D}}
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\newcommand{\bF}{\boldsymbol{F}}
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\newcommand{\bU}{\boldsymbol{U}}
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\newcommand{\bV}{\boldsymbol{V}}
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\newcommand{\bW}{\boldsymbol{W}}
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\newcommand{\bX}{\boldsymbol{X}}
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\newcommand{\bY}{\boldsymbol{Y}}
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\newcommand{\bZ}{\boldsymbol{Z}}
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\newcommand{\bc}{\boldsymbol{c}}
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% orbitals, gaps, etc
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\newcommand{\IP}{I}
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\newcommand{\EA}{A}
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\newcommand{\HOMO}{\text{HOMO}}
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\newcommand{\LUMO}{\text{LUMO}}
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\newcommand{\Eg}{E_\text{g}}
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\newcommand{\EgFun}{\Eg^\text{fund}}
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\newcommand{\EgOpt}{\Eg^\text{opt}}
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\newcommand{\EB}{E_B}
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% shortcuts for greek letters
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\newcommand{\si}{\sigma}
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\newcommand{\la}{\lambda}
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\newcommand{\RHH}{R_{\ce{H-H}}}
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\newcommand{\ii}{\mathrm{i}}
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% addresses
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\newcommand{\LCPQ}{Laboratoire de Chimie et Physique Quantiques (UMR 5626), Universit\'e de Toulouse, CNRS, UPS, France}
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\begin{document}
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\title{Perturbative Analysis of the Similarity Renormalisation Group}
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\author{Antoine \surname{Marie}}
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\email{amarie@irsamc.ups-tlse.fr}
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\affiliation{\LCPQ}
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\author{Pierre-Fran\c{c}ois \surname{Loos}}
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\email{loos@irsamc.ups-tlse.fr}
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\affiliation{\LCPQ}
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%\begin{abstract}
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%Here comes the abstract.
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%\bigskip
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%\begin{center}
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% \boxed{\includegraphics[width=0.5\linewidth]{TOC}}
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%\end{center}
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%\bigskip
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%\end{abstract}
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\maketitle
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%%%%%%%%%%%%%%%%%%%%%%
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\section{Introduction}
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%%%%%%%%%%%%%%%%%%%%%%
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The aim of this document is two-fold.
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First, we want to re-derive the perturbative analysis of the similarity renormalisation group (SRG) formalism applied to the non-relativistic electronic Hamiltonian.
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In a second time, we want to apply the same formalism to the unfolded GW Hamiltonian.
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Before jumping into these analysis, we do a brief presentation of the SRG formalism.
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%%%%%%%%%%%%%%%%%%%%%%
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\section{The similarity renormalisation group}
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%%%%%%%%%%%%%%%%%%%%%%
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The similarity renormalization group aims at continuously transforming an Hamiltonian to a diagonal form, or more often to a block-diagonal form.
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Therefore, the transformed Hamiltonian
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\begin{equation}
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\bH(s) = \bU(s) \, \bH \, \bU^\dag(s)
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\end{equation}
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depends on a flow parameter $s$.
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The resulting Hamiltonian possess up to $N$-body operators with $N$ the number of particle.
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\begin{equation}
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\bH(s) = E_0(s) + \bF(s) + \bV(s) + \bW(s) + \dots
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\end{equation}
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In the following, we will truncate every contribution superior to two-body operators.
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We can easily derive an evolution equation for this Hamiltonian by taking the derivative of $\bH(s)$. This gives
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\begin{equation}
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\label{eq:flowEquation}
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\dv{\bH(s)}{s} = \comm{\boldsymbol{\eta}(s)}{\bH(s)}
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\end{equation}
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where $\boldsymbol{\eta}(s)$, the flow generator, is defined as
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\begin{equation}
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\boldsymbol{\eta}(s) = \dv{\bU(s)}{s} \bU^\dag(s) = - \boldsymbol{\eta}^\dag(s) .
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\end{equation}
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To solve this equation at a cost inferior to the one of diagonalizing the initial Hamiltonian, one needs to introduce approximation for $\boldsymbol{\eta}(s)$.
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Before doing so, we need to define what is the blocks to suppress in order to obtain a block-diagonal Hamiltonian.
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Therefore, the Hamiltonian is separated in two parts as
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\begin{equation}
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\bH(s) = \underbrace{\bH^\text{d}(s)}_{\text{diagonal}} + \underbrace{\bH^\text{od}(s)}_{\text{off-diagonal}}.
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\end{equation}
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By definition, we have the following condition on $\bH^\text{od}$
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\begin{equation}
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\bH^\text{od}(\infty) = \boldsymbol{0}.
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\end{equation}
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In this work, we will use Wegner's canonical generator which is defined as
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\begin{equation}
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\boldsymbol{\eta}^\text{W}(s) = \comm{\bH^\text{d}(s)}{\bH(s)} = \comm{\bH^\text{d}(s)}{\bH^\text{od}(s)}.
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\end{equation}
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This generator has the advantage of defining a true renormalisation scheme, \ie the coupling coefficients with the highest energy determinants are removed first.
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One of the flaws of this generator is that it generates a stiff set of ODE which is difficult to solve numerically.
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However, here we consider analytical perturbative expressions so we will not be affected by this problem.
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%%%%%%%%%%%%%%%%%%%%%%
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\section{The electronic Hamiltonian}
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%%%%%%%%%%%%%%%%%%%%%%
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In this part, we derive the perturbative expression for the SRG applied to the non-relativistic electronic Hamiltonian
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\begin{equation}
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\label{eq:hamiltonianSecondQuant}
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\hH = \sum_{pq} f_{pq} \cre{p}\ani{q} + \frac{1}{4} \sum_{pqrs} \aeri{pq}{rs} \cre{p}\cre{q}\ani{r}\ani{s}
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\end{equation}
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which can also be written in normal order wrt a reference determinant as
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\begin{equation}
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\label{eq:hamiltonianNormalOrder}
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\hH = E_0 + \sum_{pq} + f_p^q\no{q}{p} + \frac{1}{4} \sum_{pqrs}v_{pq}^{rs}\no{rs}{pq}.
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\end{equation}
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In this case, we want to decouple the reference determinant from every singly and doubly excited determinants.
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Hence, we define the off-diagonal Hamiltonian as
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\begin{equation}
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\label{eq:hamiltonianOffDiagonal}
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\hH^{\text{od}}(s) = \sum_{ia} + f_i^a(s)\no{a}{i} + \frac{1}{4} \sum_{ijab}v(s)_{ij}^{ab}\no{ab}{ij}.
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\end{equation}
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Note that each coefficients depend on $s$.
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The perturbative parameter $\la$ is such that
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\begin{equation}
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\bH(0) = E_0(0) + F(0) + \la V(0)
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\end{equation}
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In addition, we know the following initial conditions.
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We use the HF basis set of the reference such that $F^{\text{od}}(0) = 0$ and $F^{\mathrm{d}}(0)=\delta_{pq}\epsilon_p$
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Therefore, we have
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\begin{align}
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\bH^\text{d}(0)&=E_0(0) + F^{\mathrm{d}}(0) + \la V^{\mathrm{d}}(0) & \bH^\text{od}(0)&= \la V^{\mathrm{od}}(0)
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\end{align}
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Now, we want to compute the terms at each order of the following development
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\begin{equation}
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\label{eq:hamiltonianPTExpansion}
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\bH(s) = \bH^{(0)}(s) + \bH^{(1)}(s) + \bH^{(2)}(s) + \bH^{(3)}(s) + \dots
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\end{equation}
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by integrating Eq.~\eqref{eq:flowEquation}.
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First, we start by showing that the zeroth order Hamiltonian is independant of $s$ and therefore equal to $\bH^{(0)}(0)$
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\begin{align}
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\bH(\delta s) &= \bH(0) + \delta s \dv{\bH(s)}{s}\bigg|_{s=0} + O(\delta s^2) \\
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\dv{\bH(s)}{s}\bigg|_{s=0} &= [ \boldsymbol{\eta}(0), \bH(0)] \\
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\end{align}
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However, we have seen that $\bH^\text{od}(0)$ is of order 1.
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Hence, $\boldsymbol{\eta}(0)$ does not have a zero order contribution.
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Which gives us the following equality
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\begin{equation}
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\bH^{(0)}(\delta s) = \bH^{(0)}(0)
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\end{equation}
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meaning that the zeroth order Hamiltonian is independant of $s$.
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The right-hand side of \eqref{eq:flowEquation} has no zeroth order, so we want to compute its first order term.
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To do so, we first need to compute the first order term of $\boldsymbol{\eta}(s)$.
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We have seen that $\bH^\text{od}(0)$ has no zeroth order contribution so we have
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\begin{equation}
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\boldsymbol{\eta}^{(1)}(s) = \comm{\bH^{\text{d},(0)}(0)}{\bH^{\text{od},(1)}(s)}
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\end{equation}
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According to the appendix the one-body part of $\boldsymbol{\eta}^{(1)}(s) $ has four contributions. However, two of them involve the two-body part of $\bH^{\text{d},(0)}(0)$ which is equal to zero.
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In addition, the term $\left[A_{1}, B_{2}\right]_{p}^{q}$ involves the coefficients $A_i^a = f_i^a(0) = 0$.
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So finally we only have one term
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\begin{align}
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\eta_a^{i,(1)}(s) &= \comm{\bH_1^{\text{d},(0)}(0)}{\bH_1^{\text{od},(1)}(s)} \\
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&= \sum_r \left( f_a^r(0) f_r^{i,(1)}(s) - f_r^i(0) f_a^{r,(1)}(s) \right) \\
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&= (\epsilon_a - \epsilon_i)f_a^{i,(1)}(s)
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\end{align}
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Now turning to the two-body part of $\boldsymbol{\eta}^{(1)}(s) $ and once again two terms are zero because the two-body part of $\bH^{\text{d},(0)}(0)$ is equal to zero.
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So we have
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\begin{align}
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\eta_{ab}^{ij,(1)}(s) &= \comm{\bH_1^{\text{d},(0)}(0)}{\bH_2^{\text{od},(1)}(s)} \\
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&= \sum_t [P(ab) f_a^t(0) v_{tb}^{ij,(1)}(s) - P(ij) f_t^i(0) v_{ab}^{tj,(1)}(s) ] \\
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&= \sum_t [ P(ab) \epsilon_a \delta_{at}v_{tb}^{ij,(1)}(s) - P(ij) \epsilon_i \delta_{it} v_{ab}^{tj,(1)}(s) ] \\
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&= P(ab) \epsilon_a v_{ab}^{ij,(1)}(s) - P(ij) \epsilon_i v_{ab}^{ij,(1)}(s) \\
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&= \left( \epsilon_a + \epsilon_b - \epsilon_i - \epsilon_j \right) v_{ab}^{ij,(1)}
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\end{align}
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WIP...
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%%%%%%%%%%%%%%%%%%%%%%
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\section{The unfolded GW Hamiltonian}
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%%%%%%%%%%%%%%%%%%%%%%
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\appendix
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%%%%%%%%%%%%%%%%%%%%%%
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\section{Matrix elements of $C=[A, B]_{1,2}$}
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%%%%%%%%%%%%%%%%%%%%%%
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An operator $A$ containing at most two-body terms may be written in normal ordered form with respect to the reference $\Phi$ as
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$$
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A=A_{0}+A_{1}+A_{2},
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$$
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where $A_{0}$ is a scalar, and
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$$
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\begin{gathered}
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A_{1}=\sum_{p q} A_{p}^{q}\left\{\hat{a}_{q}^{p}\right\}, \\
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A_{2}=\frac{1}{4} \sum_{p q r s} A_{p q}^{r s}\left\{\hat{a}_{r s}^{p q}\right\},
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\end{gathered}
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$$
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with the second quantization operator written compactly as $\hat{a}_{q}^{p}=\hat{a}_{p}^{\dagger} \hat{a}_{q}$ and $\hat{a}_{r s}^{p q}=\hat{a}_{p}^{\dagger} \hat{a}_{q}^{\dagger} \hat{a}_{s} \hat{a}_{r}$. The commutator $C=[A, B]_{1,2}$ contains contributions from the following terms:
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$$
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\begin{gathered}
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C_{0}=\left\langle\Phi\left|\left[A_{1}, B_{1}\right]\right| \Phi\right\rangle+\left\langle\Phi\left|\left[A_{2}, B_{2}\right]\right| \Phi\right\rangle, \\
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C_{p}^{q}=\left[A_{1}, B_{1}\right]_{p}^{q}+\left[A_{1}, B_{2}\right]_{p}^{q}-\left[B_{1}, A_{2}\right]_{p}^{q}+\left[A_{2}, B_{2}\right]_{p}^{q}, \\
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C_{p q}^{r s}=\left[A_{1}, B_{2}\right]_{p q}^{r s}-\left[B_{1}, A_{2}\right]_{p q}^{r s}+\left[A_{2}, B_{2}\right]_{p q}^{r s},
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\end{gathered}
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$$
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where the unique contributions to the matrix elements are
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$$
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\begin{gathered}
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\left\langle\Phi\left|\left[A_{1}, B_{1}\right]\right| \Phi\right\rangle=\sum_{p} \sum_{i}\left(A_{i}^{p} B_{p}^{i}-B_{i}^{p} A_{p}^{i}\right), \\
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\left\langle\Phi\left|\left[A_{2}, B_{2}\right]\right| \Phi\right\rangle=\frac{1}{4} \sum_{i j} \sum_{a b}\left(A_{i j}^{a b} B_{a b}^{i j}-B_{i j}^{a b} A_{a b}^{i j}\right), \\
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{\left[A_{1}, B_{1}\right]_{p}^{q}=\sum_{r}\left(A_{p}^{r} B_{r}^{q}-B_{p}^{r} A_{r}^{q}\right),} \\
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{\left[A_{1}, B_{2}\right]_{p}^{q}=\sum_{i} \sum_{a} A_{i}^{a} B_{p a}^{q i}-A_{a}^{i} B_{p i}^{q a},} \\
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{\left[A_{2}, B_{2}\right]_{p}^{q}=\frac{1}{2} \sum_{i j} \sum_{a}\left(A_{a p}^{i j} B_{i j}^{a q}-A_{i j}^{a q} B_{a p}^{i j}\right)} \\
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+\frac{1}{2} \sum_{i} \sum_{a b}\left(A_{i p}^{a b} B_{a b}^{i q}-A_{a b}^{i q} B_{i p}^{a b}\right), \\
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{\left[A_{1}, B_{2}\right]_{p q}^{r s}=\sum_{t}\left[P(p q) A_{p}^{t} B_{t q}^{r s}-P(r s) A_{t}^{r} B_{p q}^{t s}\right],} \\
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{\left[A_{2}, B_{2}\right]_{p q}^{r s}=\frac{1}{2} \sum_{a b}\left(A_{p q}^{a b} B_{a b}^{r s}-A_{a b}^{r s} B_{p q}^{a b}\right)} \\
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-\frac{1}{2} \sum_{i j}\left(A_{p q}^{i j} B_{i j}^{r s}-A_{i j}^{r s} B_{p q}^{i j}\right) \\
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+\sum_{i} \sum_{a} P(p q) P(r s)\left[A_{p i}^{r a} B_{q a}^{s i}-A_{p a}^{r i} B_{q i}^{s a}\right] .
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\end{gathered}
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$$
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In these equations $P(r s)$ is the antisymmetric permutation operator.
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\end{document}
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