saving work in matrix perturbation theory
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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{graphicx,dcolumn,bm,xcolor,microtype,multirow,amscd,amsmath,amssymb,amsfonts,physics,longtable,wrapfig,txfonts,mleftright,bbold}
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\usepackage[version=4]{mhchem}
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\usepackage[utf8]{inputenc}
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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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\newcommand{\Sig}[2]{\Sigma_{#1}^{#2}}
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\newcommand{\dRPA}{\text{dRPA}}
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% coordinates
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\newcommand{\br}{\boldsymbol{r}}
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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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\newcommand{\Om}[2]{\Omega_{#1}^{#2}}
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\newcommand{\sERI}[2]{(#1|#2)}
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\newcommand{\e}[2]{\epsilon_{#1}^{#2}}
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%
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\newcommand{\Ne}{N}
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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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@ -92,11 +96,11 @@
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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{\bC}[2]{\boldsymbol{C}_{#1}^{#2}}
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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{\bV}[2]{\boldsymbol{V}_{#1}^{#2}}
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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{\RHH}{R_{\ce{H-H}}}
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\newcommand{\ii}{\mathrm{i}}
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\newcommand{\bEta}[1]{\boldsymbol{\eta}^{(#1)}(s)}
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\newcommand{\bHd}[1]{\bH_\text{d}^{(#1)}}
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\newcommand{\bHod}[1]{\bH_\text{od}^{(#1)}}
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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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\title{Notes on the project: Similarity Renormalization Group formalism applied to Green's function theory}
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\author{Antoine \surname{Marie}}
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\email{amarie@irsamc.ups-tlse.fr}
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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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First, we want to re-derive (in details) 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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To do so, we first need to find a second quantization effective Hamiltonian for Green function theory.
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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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@ -360,8 +369,185 @@ After integration, using the initial condition $E_0^{(2)}(0)=0$, we obtain
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\end{equation}
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%=================================================================%
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\section{The unfolded GW Hamiltonian}
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%=================================================================%
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\section{The unfolded Green's function}
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% =================================================================%
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%%%%%%%%%%%%%%%%%%%%%%
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\subsection{Initial conditions}
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%%%%%%%%%%%%%%%%%%%%%%
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Finding a second quantized effective Hamiltonian for MBPT is far from being trivial so we start the project with matrix perturbation theory.
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A general upfolded MBPT matrix can be written as
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\begin{equation}
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\label{eq:H_MBPT}
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H =
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\begin{pmatrix}
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\bF & \bV{}{} \\
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\bV{}{\dagger} & \bC{}{}
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\end{pmatrix}
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\end{equation}
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Using SRG language, we define the diagonal and off-diagonal parts as
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\begin{equation}
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\label{eq:H_MBPT_partitioning}
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H(0) =
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\begin{pmatrix}
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\bF & \bO \\
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\bO & \bC{}{}
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\end{pmatrix}
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+ \lambda
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\begin{pmatrix}
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\bO & \bV{}{} \\
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\bV{}{\dagger} & \bO
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\end{pmatrix}
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\end{equation}
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which gives the following conditions
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\begin{align}
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\bHd{0}(0) &= \begin{pmatrix}
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\bF & \bO \\
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\bO & \bC{}{}
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\end{pmatrix} & \bHod{0}(0) &= \bO \\
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\bHd{1}(0) &= \bO & \bHod{1}(0) &= \begin{pmatrix}
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\bO & \bV{}{} \\
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\bV{}{\dagger} & \bO
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\end{pmatrix}
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\end{align}
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%%%%%%%%%%%%%%%%%%%%%%
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\subsection{Zeroth order Hamiltonian}
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%%%%%%%%%%%%%%%%%%%%%%
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The zero-th order commutator of the Wegner generator therefore gives
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\begin{equation}
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\bEta{0} = \comm{\bHd{0}}{\bHod{0}} = \bO
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\end{equation}
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and similarly
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\begin{equation}
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\dv{\bH^{(0)}}{s} = \comm{\bEta{0}}{\bH^{(0)}} = \bO
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\end{equation}
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Finally, we have
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\begin{equation}
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\color{red}{\boxed{
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\color{black}{\bH^{(0)}(s) = \bH^{(0)}(0)}
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}}
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\end{equation}
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%%%%%%%%%%%%%%%%%%%%%%
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\subsection{First order Hamiltonian}
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%%%%%%%%%%%%%%%%%%%%%%
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Now turning to the first-order contribution to the MBPT matrix, we start by computing the first order part of the Wegner generator.
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\begin{align}
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&\bEta{1} = \comm{\bHd{0}}{\bHod{1}} \\
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&= \begin{pmatrix}
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\bO & \bF^{(0)}\bV{}{(1)} - \bV{}{(1)}\bF^{(0)}\\
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\bC{}{(0)}\bV{}{(1),\dagger} - \bV{}{(1),\dagger}\bC{}{(0)} & \bO
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\end{pmatrix}
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\end{align}
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\begin{align}
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\dv{\bH^{(1)}}{s} &= \comm{\bEta{1}}{\bHd{0}} = \begin{pmatrix}
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\dv{\bF^{(1)}}{s} & \dv{\bV{}{(1)}}{s} \\
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\dv{\bV{}{(1),\dagger}}{s} & \dv{\bC{}{(1)}}{s}
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\end{pmatrix} \\
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\dv{\bF^{(1)}}{s} &= \bO \Longleftrightarrow \color{red}{\boxed{\color{black}{\bF^{(1)}= \bO}}} \\
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\dv{\bC{}{(1)}}{s} &= \bO \Longleftrightarrow \color{red}{\boxed{\color{black}{\bC{}{(1)}= \bO}}} \\
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\dv{\bV{}{(1)}}{s} &= 2 \bF^{(0)}\bV{}{(1)}\bC{}{(0)} - (\bF^{(0)})^2\bV{}{(1)} - \bV{}{(1)}(\bC{}{(0)})^2 \\
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\dv{\bV{}{(1),\dagger}}{s} &= 2 \bC{}{(0)}\bV{}{(1),\dagger}\bF^{(0)} - \bV{}{(1),\dagger}(\bF^{(0)})^2 - (\bC{}{(0)})^2\bV{}{(1),\dagger}
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\end{align}
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The two last equations can be solved differently depending on the form of $\bF$ and $\bC{}{}$.
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\subsubsection*{Diagonal $\bC{}{(0)}$}
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In the following, upper case indices correspond to the 2h1p and 2p1h sectors while lower case indices correspond to the 1h and 1p sectors. Also the $\Delta\eps_R$ corresponds to the diagonal elements of the 2h1p and 2p1h sectors.
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\begin{align}
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(\dv{\bV{}{(1)}}{s})_{pQ} &= (2 \bF^{(0)}\bV{}{(1)}\bC{}{(0)} - (\bF^{(0)})^2\bV{}{(1)} - \bV{}{(1)}(\bC{}{(0)})^2 )_{pQ}\\
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&= \sum_{rS} 2 f^{(0)}_{pr} v^{(1)}_{rS}c^{(0)}_{SQ} - \sum_{rs} f^{(0)}_{pr} f^{(0)}_{rs} v^{(1)}_{sQ} - \sum_{RS} v^{(1)}_{pR} c^{(0)}_{RS}c^{(0)}_{SQ} \\
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&= \sum_{rS} 2 \epsilon^{(0)}_p\delta_{pr} v^{(1)}_{rS}\Delta\epsilon^{(0)}_Q\delta_{SQ} \\
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&- \sum_{rs} \epsilon^{(0)}_p\delta_{pr} \epsilon^{(0)}_r\delta_{rs} v^{(1)}_{sQ} \\
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&- \sum_{RS} v^{(1)}_{pR} \Delta\epsilon^{(0)}_R\delta_{RS} \Delta\epsilon^{(0)}_Q\delta_{SQ} \\
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&= (2 \epsilon^{(0)}_p\Delta\epsilon^{(0)}_Q - (\epsilon^{(0)}_p)^2 - (\Delta\epsilon^{(0)}_Q )^2) v^{(1)}_{pQ} \\
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\dv{v^{(1)}_{pQ}}{s} &= - (\epsilon^{(0)}_p - \Delta\epsilon^{(0)}_Q )^2 v^{(1)}_{pQ} \\
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&\color{red}{\boxed{\color{black}{v^{(1)}_{pQ}(s) = v^{(1)}_{pQ}(0) e^{-s(\epsilon^{(0)}_p - \Delta\epsilon^{(0)}_Q )^2} }}}
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\end{align}
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Note the close similarity with Evangelista's expressions for the off-diagonal part at first order!
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\subsubsection*{Non-diagonal $\bC{}{(0)}$}
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We follow the same development as before
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\begin{align}
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(\dv{\bV{}{(1)}}{s})_{pQ} &= (2 \bF^{(0)}\bV{}{(1)}\bC{}{(0)} - (\bF^{(0)})^2\bV{}{(1)} - \bV{}{(1)}(\bC{}{(0)})^2 )_{pQ}\\
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&= \sum_{rS} 2 f^{(0)}_{pr} v^{(1)}_{rS}c^{(0)}_{SQ} - \sum_{rs} f^{(0)}_{pr} f^{(0)}_{rs} v^{(1)}_{sQ} - \sum_{RS} v^{(1)}_{pR} c^{(0)}_{RS}c^{(0)}_{SQ} \\
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&= \sum_{rS} 2 \epsilon^{(0)}_p\delta_{pr} v^{(1)}_{rS} c^{(0)}_{SQ} \\
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&- \sum_{rs} \epsilon^{(0)}_p\delta_{pr} \epsilon^{(0)}_r\delta_{rs} v^{(1)}_{sQ} \\
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&- \sum_{RS} v^{(1)}_{pR} c^{(0)}_{RS} c^{(0)}_{SQ} \\
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&= - (\epsilon^{(0)}_p)^2v^{(1)}_{pQ}+ \sum_{S} 2 \epsilon^{(0)}_p v^{(1)}_{pS} c^{(0)}_{SQ} - \sum_{RS} v^{(1)}_{pR} c^{(0)}_{RS} c^{(0)}_{SQ}
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\end{align}
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We obtain a set of coupled differential equations which seems far from being trivial to solve.
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In order to simplify the problem we consider the case when $\bF = \eps_p$.
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\begin{align}
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\dv{\bV{}{(1)}}{s} &= 2 \bF^{(0)}\bV{}{(1)}\bC{}{(0)} - (\bF^{(0)})^2\bV{}{(1)} - \bV{}{(1)}(\bC{}{(0)})^2 \\
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&= 2 \eps_p\bV{}{(1)}\bC{}{(0)} - (\eps_p)^2\bV{}{(1)} - \bV{}{(1)}(\bC{}{(0)})^2 \\
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&= \bV{}{(1)} (\eps_p\mathbb{1} - \bC{}{(0)})^2
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\end{align}
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Now to solve this matrix differential equation, we just need to diagonalize $(\eps_p \mathbb{1} - \bC{}{(0)})^2$.
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Fortunately, this can be easily done because the eigenvalues of $\bC{}{(0)}$ are known to be the shifted RPA eigenvalues and the eigenvectors are given in Bintrim 2021.
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\textbf{\color{red}{IDEA: Can we put the non-diagonal part of C in the off-diag H?}}
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%%%%%%%%%%%%%%%%%%%%%%
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\subsection{Second order Hamiltonian}
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%%%%%%%%%%%%%%%%%%%%%%
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Recalling that $\bHod{0} = \bO$ and $\bHd{1} = \bO$, we derive
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\begin{align}
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&\bEta{2} = \comm{\bHd{0}}{\bHod{2}} + \comm{\bHd{1}}{\bHod{1}} \\
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&= \comm{\bHd{0}}{\bHod{2}} \\
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&= \begin{pmatrix}
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\bO & \bF^{(0)}\bV{}{(2)} - \bV{}{(2)}\bF^{(0)}\\
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\bC{}{(0)}\bV{}{(2),\dagger} - \bV{}{(2),\dagger}\bC{}{(0)} & \bO
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\end{pmatrix}
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\end{align}
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\begin{align}
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&\dv{\bH^{(2)}}{s} = \comm{\bEta{2}}{\bHd{0}} + \comm{\bEta{1}}{\bHd{1}} \\
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&= \begin{pmatrix}
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\dv{\bF^{(2)}}{s} & \dv{\bV{}{(2)}}{s} \\
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\dv{\bV{}{(2),\dagger}}{s} & \dv{\bC{}{(2)}}{s}
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\end{pmatrix} \\
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\dv{\bF^{(2)}}{s} &= \bF^{(0)}\bV{}{(1)}\bV{}{(1),\dagger} + \bV{}{(1)}\bV{}{(1),\dagger}\bF^{(0)} - 2 \bV{}{(1)}\bC{}{(0)}\bV{}{(1),\dagger}\\
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\dv{\bC{}{(2)}}{s} &= \bC{}{(0)}\bV{}{(1),\dagger }\bV{}{(1)} + \bV{}{(1),\dagger }\bV{}{(1)}\bC{}{(0)} - 2 \bV{}{(1)}\bF^{(0)}\bV{}{(1),\dagger}\\
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\dv{\bV{}{(2)}}{s} &= 2 \bF^{(0)}\bV{}{(2)}\bC{}{(0)} - (\bF^{(0)})^2\bV{}{(2)} - \bV{}{(2)}(\bC{}{(0)})^2 \\
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\dv{\bV{}{(2),\dagger}}{s} &= 2 \bC{}{(0)}\bV{}{(2),\dagger}\bF^{(0)} - \bV{}{(2),\dagger}(\bF^{(0)})^2 - (\bC{}{(0)})^2\bV{}{(2),\dagger}
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\end{align}
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Once again the integration of these equations is much simpler if $\bC{}{(0)}$ is diagonal.
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\subsubsection*{Diagonal $\bC{}{(0)}$}
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\begin{align}
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&(\dv{\bF^{(2)}}{s})_{pq} = (\bF^{(0)}\bV{}{(1)}\bV{}{(1),\dagger} + \bV{}{(1)}\bV{}{(1),\dagger}\bF^{(0)} - 2 \bV{}{(1)}\bC{}{(0)}\bV{}{(1),\dagger})_{pq} \notag \\
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&= \sum_{rS} f^{(0)}_{pr} v^{(1)}_{rS} v^{(1),\dagger}_{Sq} + \sum_{Rs} v^{(1)}_{pR} v^{(1),\dagger}_{Rs} f^{(0)}_{sq} - 2\sum_{RS} v^{(1)}_{pR} c^{(0)}_{RS} v^{(1),\dagger}_{Sq} \notag \\
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&= \sum_{S} \eps^{(0)}_{p} v^{(1)}_{pS} v^{(1)}_{qS} + \sum_{R} \eps^{(0)}_{q} v^{(1)}_{pR} v^{(1)}_{qR} - 2\sum_{R} \Delta\eps^{(0)}_R v^{(1)}_{pR} v^{(1)}_{qR} \notag \\
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&= \sum_R (\eps^{(0)}_{p} + \eps^{(0)}_{q} - 2 \Delta\eps^{(0)}_R) v^{(1)}_{pR} v^{(1)}_{qR} \notag \\
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&= \sum_R (\eps^{(0)}_{p} + \eps^{(0)}_{q} - 2 \Delta\eps^{(0)}_R) v^{(1)}_{pR}(0) v^{(1)}_{qR}(0) e^{-s [ (\eps^{(0)}_p - \Delta\eps^{(0)}_R)^2+ (\eps^{(0)}_q - \Delta\eps^{(0)}_R)^2]} \notag \\
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&f^{(2)}_{pq}(s) = \notag \\
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&\color{red}{\boxed{\color{black}{- \sum_R \frac{\eps^{(0)}_{p} + \eps^{(0)}_{q} - 2 \Delta\eps^{(0)}_R}{(\eps^{(0)}_p - \Delta\eps^{(0)}_R)^2+ (\eps^{(0)}_q - \Delta\eps^{(0)}_R)^2}(1 - e^{-s [ (\eps^{(0)}_p - \Delta\eps^{(0)}_R)^2+ (\eps^{(0)}_q - \Delta\eps^{(0)}_R)^2]})}}} \notag
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\end{align}
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A similar derivation should give (\textbf{\textcolor{red}{TO CHECK}})
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\begin{align}
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&c^{(2)}_{PQ}(s) = \notag \\
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&\color{red}{\boxed{\color{black}{- \sum_r \frac{\Delta\eps^{(0)}_{P} + \Delta\eps^{(0)}_{Q} - 2 \eps^{(0)}_r}{(\Delta\eps^{(0)}_P - \eps^{(0)}_r)^2+ (\Delta\eps^{(0)}_Q - \eps^{(0)}_r)^2}(1 - e^{-s [ (\Delta\eps^{(0)}_P - \eps^{(0)}_r)^2+ (\Delta\eps^{(0)}_P - \eps^{(0)}_r)^2]})}}} \notag
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\end{align}
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\begin{align}
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&\dv{v^{(2)}_{pQ}}{s} = - (\epsilon^{(0)}_p - \Delta\epsilon^{(0)}_Q )^2 v^{(2)}_{pQ} \\
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&\color{red}{\boxed{\color{black}{v^{(2)}_{pQ}(s) = v^{(2)}_{pQ}(0) e^{-s(\epsilon^{(0)}_p - \Delta\epsilon^{(0)}_Q )^2} = 0 }}}
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\end{align}
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\subsubsection*{Non-diagonal $\bC{}{(0)}$}
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\appendix
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