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did some changes to include offdiag C in the perturbative part

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Antoine Marie 2022-10-21 18:59:25 +02:00
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Notes/PerturbativeAnalysis.tex
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@ -100,6 +100,7 @@
\newcommand{\bD}{\boldsymbol{D}} \newcommand{\bD}{\boldsymbol{D}}
\newcommand{\bF}{\boldsymbol{F}} \newcommand{\bF}{\boldsymbol{F}}
\newcommand{\bU}{\boldsymbol{U}} \newcommand{\bU}{\boldsymbol{U}}
\newcommand{\bR}{\boldsymbol{R}}
\newcommand{\bV}[2]{\boldsymbol{V}_{#1}^{#2}} \newcommand{\bV}[2]{\boldsymbol{V}_{#1}^{#2}}
\newcommand{\bW}{\boldsymbol{W}} \newcommand{\bW}{\boldsymbol{W}}
\newcommand{\bX}{\boldsymbol{X}} \newcommand{\bX}{\boldsymbol{X}}
@ -134,7 +135,7 @@
\begin{document} \begin{document}
\title{Notes on the project: Similarity Renormalization Group formalism applied to Green's function theory} \title{Notes on the project: Perturbative Analysis of the Similarity Renormalization Group formalism applied to the electronic Hamiltonian and Green's function theory}
\author{Antoine \surname{Marie}} \author{Antoine \surname{Marie}}
\email{amarie@irsamc.ups-tlse.fr} \email{amarie@irsamc.ups-tlse.fr}
@ -392,23 +393,23 @@ Using SRG language, we define the diagonal and off-diagonal parts as
H(0) = H(0) =
\begin{pmatrix} \begin{pmatrix}
\bF & \bO \\ \bF & \bO \\
\bO & \bC{}{} \bO & \bC{\text{d}}{}
\end{pmatrix} \end{pmatrix}
+ \lambda + \lambda
\begin{pmatrix} \begin{pmatrix}
\bO & \bV{}{} \\ \bO & \bV{}{} \\
\bV{}{\dagger} & \bO \bV{}{\dagger} & \bC{\text{od}}{}
\end{pmatrix} \end{pmatrix}
\end{equation} \end{equation}
which gives the following conditions which gives the following conditions
\begin{align} \begin{align}
\bHd{0}(0) &= \begin{pmatrix} \bHd{0}(0) &= \begin{pmatrix}
\bF & \bO \\ \bF & \bO \\
\bO & \bC{}{} \bO & \bC{\text{d}}{}
\end{pmatrix} & \bHod{0}(0) &= \bO \\ \end{pmatrix} & \bHod{0}(0) &= \bO \\
\bHd{1}(0) &= \bO & \bHod{1}(0) &= \begin{pmatrix} \bHd{1}(0) &= \bO & \bHod{1}(0) &= \begin{pmatrix}
\bO & \bV{}{} \\ \bO & \bV{}{} \\
\bV{}{\dagger} & \bO \bV{}{\dagger} & \bC{\text{od}}{}
\end{pmatrix} \end{pmatrix}
\end{align} \end{align}
@ -440,8 +441,8 @@ Now turning to the first-order contribution to the MBPT matrix, we start by comp
\begin{align} \begin{align}
&\bEta{1} = \comm{\bHd{0}}{\bHod{1}} \\ &\bEta{1} = \comm{\bHd{0}}{\bHod{1}} \\
&= \begin{pmatrix} &= \begin{pmatrix}
\bO & \bF^{(0)}\bV{}{(1)} - \bV{}{(1)}\bF^{(0)}\\ \bO & \bF^{(0)}\bV{}{(1)} - \bV{}{(1)}\bC{\text{d}}{(0)}\\
\bC{}{(0)}\bV{}{(1),\dagger} - \bV{}{(1),\dagger}\bC{}{(0)} & \bO \bC{\text{d}}{(0)}\bV{}{(1),\dagger} - \bV{}{(1),\dagger} \bF^{(0)} & \bC{\text{d}}{(0)} \bC{\text{od}}{(1)} - \bC{\text{od}}{(1)} \bC{\text{d}}{(0)}
\end{pmatrix} \end{pmatrix}
\end{align} \end{align}
@ -451,16 +452,16 @@ Now turning to the first-order contribution to the MBPT matrix, we start by comp
\dv{\bV{}{(1),\dagger}}{s} & \dv{\bC{}{(1)}}{s} \dv{\bV{}{(1),\dagger}}{s} & \dv{\bC{}{(1)}}{s}
\end{pmatrix} \\ \end{pmatrix} \\
\dv{\bF^{(1)}}{s} &= \bO \Longleftrightarrow \color{red}{\boxed{\color{black}{\bF^{(1)}= \bO}}} \\ \dv{\bF^{(1)}}{s} &= \bO \Longleftrightarrow \color{red}{\boxed{\color{black}{\bF^{(1)}= \bO}}} \\
\dv{\bC{}{(1)}}{s} &= \bO \Longleftrightarrow \color{red}{\boxed{\color{black}{\bC{}{(1)}= \bO}}} \\ \dv{\bV{}{(1)}}{s} &= 2 \bF^{(0)}\bV{}{(1)}\bC{\text{d}}{(0)} - (\bF^{(0)})^2\bV{}{(1)} - \bV{}{(1)}(\bC{\text{d}}{(0)})^2 \\
\dv{\bV{}{(1)}}{s} &= 2 \bF^{(0)}\bV{}{(1)}\bC{}{(0)} - (\bF^{(0)})^2\bV{}{(1)} - \bV{}{(1)}(\bC{}{(0)})^2 \\ \dv{\bV{}{(1),\dagger}}{s} &= 2 \bC{\text{d}}{(0)}\bV{}{(1),\dagger}\bF^{(0)} - \bV{}{(1),\dagger}(\bF^{(0)})^2 - (\bC{\text{d}}{(0)})^2\bV{}{(1),\dagger} \\
\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} \dv{\bC{}{(1)}}{s} &= 2 \bC{\text{d}}{(0)}\bC{\text{od}}{(1)}\bC{\text{d}}{(0)}- (\bC{\text{d}}{(0)})^2\bC{\text{od}}{(1)} - \bC{\text{od}}{(1)}(\bC{\text{d}}{(0)})^2
\end{align} \end{align}
The two last equations can be solved differently depending on the form of $\bF$ and $\bC{}{}$. The two last equations can be solved differently depending on the form of $\bF$ and $\bC{}{}$.
\subsubsection*{Diagonal $\bC{}{(0)}$} \subsubsection*{Diagonal $\bC{}{(0)}$}
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. 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.
\begin{align} \begin{align}
(\dv{\bV{}{(1)}}{s})_{pQ} &= (2 \bF^{(0)}\bV{}{(1)}\bC{}{(0)} - (\bF^{(0)})^2\bV{}{(1)} - \bV{}{(1)}(\bC{}{(0)})^2 )_{pQ}\\ (\dv{\bV{}{(1)}}{s})_{pQ} &= (2 \bF^{(0)}\bV{}{(1)}\bC{\text{d}}{(0)} - (\bF^{(0)})^2\bV{}{(1)} - \bV{}{(1)}(\bC{\text{d}}{(0)})^2 )_{pQ}\\
&= \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} \\ &= \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} \\
&= \sum_{rS} 2 \epsilon^{(0)}_p\delta_{pr} v^{(1)}_{rS}\Delta\epsilon^{(0)}_Q\delta_{SQ} \\ &= \sum_{rS} 2 \epsilon^{(0)}_p\delta_{pr} v^{(1)}_{rS}\Delta\epsilon^{(0)}_Q\delta_{SQ} \\
&- \sum_{rs} \epsilon^{(0)}_p\delta_{pr} \epsilon^{(0)}_r\delta_{rs} v^{(1)}_{sQ} \\ &- \sum_{rs} \epsilon^{(0)}_p\delta_{pr} \epsilon^{(0)}_r\delta_{rs} v^{(1)}_{sQ} \\
@ -470,28 +471,13 @@ In the following, upper case indices correspond to the 2h1p and 2p1h sectors whi
&\color{red}{\boxed{\color{black}{v^{(1)}_{pQ}(s) = v^{(1)}_{pQ}(0) e^{-s(\epsilon^{(0)}_p - \Delta\epsilon^{(0)}_Q )^2} }}} &\color{red}{\boxed{\color{black}{v^{(1)}_{pQ}(s) = v^{(1)}_{pQ}(0) e^{-s(\epsilon^{(0)}_p - \Delta\epsilon^{(0)}_Q )^2} }}}
\end{align} \end{align}
Note the close similarity with Evangelista's expressions for the off-diagonal part at first order! Note the close similarity with Evangelista's expressions for the off-diagonal part at first order!
\subsubsection*{Non-diagonal $\bC{}{(0)}$}
We follow the same development as before
\begin{align} \begin{align}
(\dv{\bV{}{(1)}}{s})_{pQ} &= (2 \bF^{(0)}\bV{}{(1)}\bC{}{(0)} - (\bF^{(0)})^2\bV{}{(1)} - \bV{}{(1)}(\bC{}{(0)})^2 )_{pQ}\\ (\dv{\bC{}{(1)}}{s})_{PQ} &= (2 \bC{\text{d}}{(0)}\bC{\text{od}}{(1)}\bC{\text{d}}{(0)}- (\bC{\text{d}}{(0)})^2\bC{\text{od}}{(1)} - \bC{\text{od}}{(1)}(\bC{\text{d}}{(0)})^2)_{PQ} \\
&= \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} \\ &= \sum_{RS} 2 c^{(0)}_{PR} c^{(1)}_{RS} c^{(0)}_{SQ} - c^{(0)}_{PR} c^{(0)}_{RS} c^{(1)}_{SQ} - c^{(1)}_{PR} c^{(0)}_{RS} c^{(0)}_{SQ} \\
&= \sum_{rS} 2 \epsilon^{(0)}_p\delta_{pr} v^{(1)}_{rS} c^{(0)}_{SQ} \\ &= 2 \Delta\epsilon^{(0)}_Pc^{(1)}_{PQ}\Delta\epsilon^{(0)}_Q - (\Delta\epsilon^{(0)}_P)^2 c^{(1)}_{PQ} - c^{(1)}_{PQ} (\Delta\epsilon^{(0)}_Q)^2 \\
&- \sum_{rs} \epsilon^{(0)}_p\delta_{pr} \epsilon^{(0)}_r\delta_{rs} v^{(1)}_{sQ} \\ &= - (\Delta\epsilon^{(0)}_P - \Delta\epsilon^{(0)}_Q )^2 c^{(1)}_{PQ} \\
&- \sum_{RS} v^{(1)}_{pR} c^{(0)}_{RS} c^{(0)}_{SQ} \\ &\color{red}{\boxed{\color{black}{c^{(1)}_{PQ}(s) = c^{(1)}_{PQ}(0) e^{-s(\Delta\epsilon^{(0)}_P - \Delta\epsilon^{(0)}_Q )^2} }}}
&= - (\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}
\end{align} \end{align}
We obtain a set of coupled differential equations which seems far from being trivial to solve.
In order to simplify the problem we consider the case when $\bF = \eps_p$.
\begin{align}
\dv{\bV{}{(1)}}{s} &= 2 \bF^{(0)}\bV{}{(1)}\bC{}{(0)} - (\bF^{(0)})^2\bV{}{(1)} - \bV{}{(1)}(\bC{}{(0)})^2 \\
&= 2 \eps_p\bV{}{(1)}\bC{}{(0)} - (\eps_p)^2\bV{}{(1)} - \bV{}{(1)}(\bC{}{(0)})^2 \\
&= \bV{}{(1)} (\eps_p\mathbb{1} - \bC{}{(0)})^2
\end{align}
Now to solve this matrix differential equation, we just need to diagonalize $(\eps_p \mathbb{1} - \bC{}{(0)})^2$.
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.
\textbf{\color{red}{IDEA: Can we put the non-diagonal part of C in the off-diag H?}}
%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%
\subsection{Second order Hamiltonian} \subsection{Second order Hamiltonian}
@ -502,8 +488,8 @@ Recalling that $\bHod{0} = \bO$ and $\bHd{1} = \bO$, we derive
&\bEta{2} = \comm{\bHd{0}}{\bHod{2}} + \comm{\bHd{1}}{\bHod{1}} \\ &\bEta{2} = \comm{\bHd{0}}{\bHod{2}} + \comm{\bHd{1}}{\bHod{1}} \\
&= \comm{\bHd{0}}{\bHod{2}} \\ &= \comm{\bHd{0}}{\bHod{2}} \\
&= \begin{pmatrix} &= \begin{pmatrix}
\bO & \bF^{(0)}\bV{}{(2)} - \bV{}{(2)}\bF^{(0)}\\ \bO & \bF^{(0)}\bV{}{(2)} - \bV{}{(2)}\bC{\text{d}}{(0)}\\
\bC{}{(0)}\bV{}{(2),\dagger} - \bV{}{(2),\dagger}\bC{}{(0)} & \bO \bC{\text{d}}{(0)}\bV{}{(2),\dagger} - \bV{}{(2),\dagger}\bF^{(0)} & \bC{\text{d}}{(0)} \bC{\text{od}}{(2)} - \bC{\text{od}}{(2)} \bC{\text{d}}{(0)}
\end{pmatrix} \end{pmatrix}
\end{align} \end{align}
@ -513,18 +499,15 @@ Recalling that $\bHod{0} = \bO$ and $\bHd{1} = \bO$, we derive
\dv{\bF^{(2)}}{s} & \dv{\bV{}{(2)}}{s} \\ \dv{\bF^{(2)}}{s} & \dv{\bV{}{(2)}}{s} \\
\dv{\bV{}{(2),\dagger}}{s} & \dv{\bC{}{(2)}}{s} \dv{\bV{}{(2),\dagger}}{s} & \dv{\bC{}{(2)}}{s}
\end{pmatrix} \\ \end{pmatrix} \\
\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}\\ \dv{\bF^{(2)}}{s} &= \bF^{(0)}\bV{}{(1)}\bV{}{(1),\dagger} + \bV{}{(1)}\bV{}{(1),\dagger}\bF^{(0)} - 2 \bV{}{(1)}\bC{\text{d}}{(0)}\bV{}{(1),\dagger}\\
\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}\\ \dv{\bC{}{(2)}}{s} &= \\
\dv{\bV{}{(2)}}{s} &= 2 \bF^{(0)}\bV{}{(2)}\bC{}{(0)} - (\bF^{(0)})^2\bV{}{(2)} - \bV{}{(2)}(\bC{}{(0)})^2 \\ \dv{\bV{}{(2)}}{s} &= 2 \bF^{(0)}\bV{}{(2)}\bC{\text{d}}{(0)} - (\bF^{(0)})^2\bV{}{(2)} - \bV{}{(2)}(\bC{\text{d}}{(0)})^2 \\
\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} &- 2 \bV{}{(1)} \bC{\text{d}}{(0)} \bC{\text{od}}{(1)} + \bF^{(0)} \bV{}{(1)} \bC{\text{od}}{(1)} + \bV{}{(1)} \bC{\text{od}}{(1)} \bC{\text{d}}{(0)} \\
\dv{\bV{}{(2),\dagger}}{s} &=
\end{align} \end{align}
Once again the integration of these equations is much simpler if $\bC{}{(0)}$ is diagonal.
\subsubsection*{Diagonal $\bC{}{(0)}$}
\begin{align} \begin{align}
&(\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 \\ &(\dv{\bF^{(2)}}{s})_{pq} = (\bF^{(0)}\bV{}{(1)}\bV{}{(1),\dagger} + \bV{}{(1)}\bV{}{(1),\dagger}\bF^{(0)} - 2 \bV{}{(1)}\bC{\text{d}}{(0)}\bV{}{(1),\dagger})_{pq} \notag \\
&= \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 \\ &= \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 \\
&= \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 \\ &= \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 \\
&= \sum_R (\eps^{(0)}_{p} + \eps^{(0)}_{q} - 2 \Delta\eps^{(0)}_R) v^{(1)}_{pR} v^{(1)}_{qR} \notag \\ &= \sum_R (\eps^{(0)}_{p} + \eps^{(0)}_{q} - 2 \Delta\eps^{(0)}_R) v^{(1)}_{pR} v^{(1)}_{qR} \notag \\
@ -533,21 +516,55 @@ Once again the integration of these equations is much simpler if $\bC{}{(0)}$ is
&\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 &\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
\end{align} \end{align}
A similar derivation should give (\textbf{\textcolor{red}{TO CHECK}}) %%%%%%%%%%%%%%%%%%%%%%
\subsection{Downfolding the SRG-transformed matrix}
%%%%%%%%%%%%%%%%%%%%%%
Now that we obtained the SRG-transformed Hamiltonian to a given order we can downfold it back to obtain a SRG-renormalized self-energy up to a given order.
\begin{equation}
\label{eq:H_SRGMBPT}
H(s) =
\begin{pmatrix}
\bF^{(0)}(0) + \bF^{(2)}(s) & \bV{}{(1)}(s) + \bV{}{(2)}(s) \\
\bV{}{(1),\dagger}(s) + \bV{}{(2),\dagger}(s) & \bC{}{(0)}(0) +\bC{}{(2)}(s)
\end{pmatrix}
\end{equation}
\begin{equation}
\left\{
\begin{aligned}
(\bF^{(0)}(0) + \bF^{(2)}(s)) \bR^{1h/1p} + \bV{}{(1)}(s) \bR^{2h1p/2p1h} &= \omega \bR^{1h/1p} \\
\bV{}{(1),\dagger}(s) \bR^{1h/1p} + (\bC{}{(0)}(0) +\bC{}{(2)}(s) ) \bR^{2h1p/2p1h}&= \omega \bR^{2h1p/2p1h}
\end{aligned}
\right.
\end{equation}
\begin{align} \begin{align}
&c^{(2)}_{PQ}(s) = \notag \\ &(\bF^{(0)}(0) + \bF^{(2)}(s)) + (\bV{}{(1)}(s) + \bV{}{(2)}(s))(\omega \mathbb{1} - \bC{}{(0)}(0) +\bC{}{(2)}(s) )^{-1} \notag \\
&\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 &\dots (\bV{}{(1),\dagger}(s) + \bV{}{(2),\dagger}(s)) \bR^{1h/1p} = \omega \bR^{1h/1p}
\end{align} \end{align}
\begin{align} \begin{align}
&\dv{v^{(2)}_{pQ}}{s} = - (\epsilon^{(0)}_p - \Delta\epsilon^{(0)}_Q )^2 v^{(2)}_{pQ} \\ &(\omega \mathbb{1} - \bC{}{(0)}(0) - \bC{}{(1)}(s) - \bC{}{(2)}(s) )^{-1} = (\omega \mathbb{1} - \bC{}{(0)}(0))^{-1} \\
&\color{red}{\boxed{\color{black}{v^{(2)}_{pQ}(s) = v^{(2)}_{pQ}(0) e^{-s(\epsilon^{(0)}_p - \Delta\epsilon^{(0)}_Q )^2} = 0 }}} &+ (\omega \mathbb{1} - \bC{}{(0)}(0))^{-1}(\bC{}{(1)}(s) + \bC{}{(2)}(s) )(\omega \mathbb{1} - \bC{}{(0)}(0))^{-1} \notag \\
&+ \dots \notag
\end{align}
Using this taylor expansion we can see that only the first term will contribute to second order in the self energy. Hence we have
\begin{equation*}
(\bF^{(0)}(0) + \bF^{(2)}(s) + \bV{}{(1)}(s)(\omega \mathbb{1} - \bC{}{(0)}(0))^{-1} \bV{}{(1),\dagger}(s))\bR^{1h/1p} = \omega \bR^{1h/1p}
\end{equation*}
Therefore we have to solve the following equation
\begin{align}
&(\tilde{\bF} + \tilde{\boldsymbol{\Sigma}}(\omega)) \bX{}{} = \omega \bX \\
&\tilde{\bF} =\bF^{(0)}(0) + \bF^{(2)}(s) \\
&\tilde{\boldsymbol{\Sigma}}(\omega) = \bV{}{(1)}(s)(\omega \mathbb{1} - \bC{}{(0)}(0))^{-1} \bV{}{(1),\dagger}(s)
\end{align} \end{align}
%%%%%%%%%%%%%%%%%%%%%%
\subsection{The SRG(2) quasi-particle equations}
%%%%%%%%%%%%%%%%%%%%%%
\subsubsection*{Non-diagonal $\bC{}{(0)}$} In this section, we report the GF(2), GW and GT quasi-particle equations.
\appendix \appendix