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@ -395,9 +395,13 @@ The downfolding procedure to obtain the $GW$ self-energy is derived in details i
%=================================================================%
%%%%%%%%%%%%%%%%%%%%%%
\subsection{Initial conditions}
\subsection{Order by order differential equations}
%%%%%%%%%%%%%%%%%%%%%%
%///////////////////////////%
\subsubsection{Initial conditions}
%///////////////////////////%
Finding a second quantized effective Hamiltonian for MBPT is far from being trivial so we start the project with matrix perturbation theory.
A general upfolded MBPT matrix can be written as
\begin{equation}
@ -408,7 +412,21 @@ A general upfolded MBPT matrix can be written as
\bV{}{\dagger} & \bC{}{}
\end{pmatrix}
\end{equation}
Using SRG language, we define the diagonal and off-diagonal parts as
Using SRG language, we define the diagonal and off-diagonal parts as
\begin{equation}
\label{eq:H_MBPT_partitioning}
\bH(0) =
\begin{pmatrix}
\bF{}{} & \bO \\
\bO & \bC{}{}
\end{pmatrix}
+ \lambda
\begin{pmatrix}
\bO & \bV{}{} \\
\bV{}{\dagger} & \bO
\end{pmatrix}
\end{equation}
we could also have defined it like this
\begin{equation}
\label{eq:H_MBPT_partitioning}
\bH(0) =
@ -422,23 +440,32 @@ Using SRG language, we define the diagonal and off-diagonal parts as
\bV{}{\dagger} & \bC{\text{od}}{}
\end{pmatrix}
\end{equation}
which gives the following conditions
However, in the end this alternative choice was not judicious and the reason is explained why in Appendix~\ref{sec:partitioning}.
The initial conditions corresponding to the first partitioning are
\begin{align}
\bHd{0}(0) &= \begin{pmatrix}
\bF{}{} & \bO \\
\bO & \bC{\text{d}}{}
\end{pmatrix} & \bHod{0}(0) &= \bO \notag \\
\bHd{1}(0) &= \bO & \bHod{1}(0) &= \begin{pmatrix}
\bO & \bV{}{} \\
\bV{}{\dagger} & \bC{\text{od}}{} \notag
\end{pmatrix}
\bF{}{} & \bO \\
\bO & \bC{}{}
\end{pmatrix} &
\bHod{0}(0) &=\begin{pmatrix}
\bO & \bO \\
\bO & \bO
\end{pmatrix} \notag \\
\bHd{1}(0) &=\begin{pmatrix}
\bO & \bO \\
\bO & \bO
\end{pmatrix} &
\bHod{1}(0) &= \begin{pmatrix}
\bO & \bV{}{} \\
\bV{}{\dagger} & \bO \notag
\end{pmatrix} \notag
\end{align}
At this point, we aren't sure if the off-diagonal part of $\bC{}{}$ should be included or not in the off-diagonal part of the Hamiltonian $\bH_\text{od}$.
In the following derivation, we choose to do so because the other case can be obtained simply by taking $\bC{\text{od}}{} = \boldsymbol{0}$ and $\bC{\text{d}}{} = \bC{}{}$.
%%%%%%%%%%%%%%%%%%%%%%
\subsection{Zeroth order Hamiltonian}
%%%%%%%%%%%%%%%%%%%%%%
%///////////////////////////%
\subsubsection{Zeroth order differential equations}
%///////////////////////////%
The zeroth-order commutator of the Wegner generator therefore gives
\begin{equation}
@ -455,109 +482,102 @@ Finally, we have
}}
\end{equation}
%%%%%%%%%%%%%%%%%%%%%%
\subsection{First order Hamiltonian}
%%%%%%%%%%%%%%%%%%%%%%
%///////////////////////////%
\subsubsection{First order differential equations}
%///////////////////////////%
Now turning to the first-order contribution to the MBPT matrix, we start by computing the first order part of the Wegner generator.
\begin{align}
&\bEta{1} = \comm{\bHd{0}}{\bHod{1}} \\
&= \begin{pmatrix}
\bO & \bF{}{(0)}\bV{}{(1)} - \bV{}{(1)}\bC{\text{d}}{(0)}\\
\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)} \notag
\bO & \bF{}{(0)}\bV{}{(1)} - \bV{}{(1)}\bC{}{(0)}\\
\bC{}{(0)}\bV{}{(1),\dagger} - \bV{}{(1),\dagger} \bF{}{(0)} & \bO \notag
\end{pmatrix}
\end{align}
\begin{align}
\dv{\bH^{(1)}}{s} &= \comm{\bEta{1}}{\bHd{0}} = \begin{pmatrix}
\dv{\bF{}{(1)}}{s} & \dv{\bV{}{(1)}}{s} \\
\dv{\bV{}{(1),\dagger}}{s} & \dv{\bC{}{(1)}}{s}
\end{pmatrix} \\
\dv{\bH^{(1)}}{s} &= \comm{\bEta{1}}{\bHd{0}} \\
\dv{\bF{}{(1)}}{s} &= \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),\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{\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
\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{}{(0)}\bV{}{(1),\dagger}\bF{}{(0)} - \bV{}{(1),\dagger}(\bF{}{(0)})^2 - (\bC{}{(0)})^2\bV{}{(1),\dagger} \\
\dv{\bC{}{(1)}}{s} &= \bO
\end{align}
The last two equations can be solved differently depending on the form of $\bF{}{}$ and $\bC{}{}$.
%%%%%%%%%%%%%%%%%%%%%%
\subsection{Second order Hamiltonian}
%%%%%%%%%%%%%%%%%%%%%%
%///////////////////////////%
\subsubsection{Second order differential equations}
%///////////////////////////%
Recalling that $\bHod{0} = \bO$ and $\bHd{1} = \bO$, we derive
\begin{align}
&\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{2}}{\bHod{0}} \\
&= \comm{\bHd{0}}{\bHod{2}} \notag \\
&= \begin{pmatrix}
\bO & \bF{}{(0)}\bV{}{(2)} - \bV{}{(2)}\bC{\text{d}}{(0)}\\
\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)} \notag
\bO & \bF{}{(0)}\bV{}{(2)} - \bV{}{(2)}\bC{}{(0)}\\
\bC{}{(0)}\bV{}{(2),\dagger} - \bV{}{(2),\dagger}\bF{}{(0)} & \bO \notag
\end{pmatrix}
\end{align}
\begin{align}
\dv{\bH^{(2)}}{s} &= \comm{\bEta{2}}{\bHd{0}} + \comm{\bEta{1}}{\bHd{1}} \\
&= \begin{pmatrix}
\dv{\bF{}{(2)}}{s} & \dv{\bV{}{(2)}}{s} \\
\dv{\bV{}{(2),\dagger}}{s} & \dv{\bC{}{(2)}}{s}
\end{pmatrix} \notag \\
\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} &= 2 \bC{\text{d}}{(0)}\bC{\text{od}}{(2)}\bC{\text{d}}{(0)}- (\bC{\text{d}}{(0)})^2\bC{\text{od}}{(2)} - \bC{\text{od}}{(2)}(\bC{\text{d}}{(0)})^2 \\
&-2 \bC{\text{d}}{(1)}\bC{\text{od}}{(0)}\bC{\text{d}}{(1)} + (\bC{\text{d}}{(1)})^2\bC{\text{od}}{(0)} + \bC{\text{od}}{(0)}(\bC{\text{d}}{(1)})^2 \notag \\
&+ \bC{\text{d}}{(0)}\bV{}{(1)}\bV{}{(1),\dagger} + \bV{}{(1)}\bV{}{(1),\dagger}\bC{\text{d}}{(0)} - 2 \bV{}{(1)}\bF{}{(0)}\bV{}{(1),\dagger} \notag \\
\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 \\
&- 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)} \notag \\
\dv{\bV{}{(2),\dagger}}{s} &= 2 \bC{\text{d}}{(0)}\bV{}{(2),\dagger}\bF{}{(0)} - \bV{}{(2),\dagger}(\bF{}{(0)})^2 - (\bC{\text{d}}{(0)})^2\bV{}{(2),\dagger} \\
&- 2 \bC{\text{od}}{(1)} \bC{\text{d}}{(0)} \bV{}{(1),\dagger} + \bC{\text{od}}{(1)} \bV{}{(1),\dagger} \bF{}{(0)} + \bC{\text{d}}{(0)} \bC{\text{od}}{(1)} \bV{}{(1),\dagger} \notag
\dv{\bH^{(2)}}{s} &= \comm{\bEta{2}}{\bH^{(0)}} + \comm{\bEta{1}}{\bH^{(1)}} + \comm{\bEta{0}}{\bH^{(2)}} \\
&= \comm{\bEta{2}}{\bHd{0}} + \comm{\bEta{1}}{\bHod{1}} \notag
\end{align}
%%%%%%%%%%%%%%%%%%%%%%
\subsection{Downfolding the SRG-transformed matrix}
%%%%%%%%%%%%%%%%%%%%%%
\begin{align}
\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{\bC{}{(2)}}{s} &= \bC{}{(0)}\bV{}{(1)}\bV{}{(1),\dagger} + \bV{}{(1)}\bV{}{(1),\dagger}\bC{}{(0)} - 2 \bV{}{(1)}\bF{}{(0)}\bV{}{(1),\dagger} \\
\dv{\bV{}{(2)}}{s} &= 2 \bF{}{(0)}\bV{}{(2)}\bC{}{(0)} - (\bF{}{(0)})^2\bV{}{(2)} - \bV{}{(2)}(\bC{}{(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}
\end{align}
In order to choose what to do with $\bC{\text{od}}{}$ we look at the downfolded SRG quasiparticle equation.
%///////////////////////////%
\subsubsection{Third order differential equations}
% ///////////////////////////%
\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}
\bEta{3} = \comm{\bHd{0}}{\bHod{3}} + \comm{\bHd{2}}{\bHod{1}}
\end{equation}
\begin{widetext}
\begin{equation}
\left\{
\begin{aligned}
(\bF{}{(0)}(0) + \bF{}{(2)}(s)) \bR^{1h/1p} + (\bV{}{(1)}(s) + \bV{}{(2)}(s)) \bR^{2h1p/2p1h} &= \omega \bR^{1h/1p} \\
(\bV{}{(1),\dagger}(s) + \bV{}{(2),\dagger}(s)) \bR^{1h/1p} + (\bC{}{(0)}(0) +\bC{}{(2)}(s) ) \bR^{2h1p/2p1h}&= \omega \bR^{2h1p/2p1h}
\end{aligned}
\right.
\end{equation}
\begin{equation}
(\bF{}{(0)}(0) + \bF{}{(2)}(s)) + (\bV{}{(1)}(s) + \bV{}{(2)}(s)) (\omega \mathbb{1} - \bC{\text{d}}{(0)}(0) - \bC{\text{od}}{(1)}(s) -\bC{}{(2)}(s) )^{-1} (\bV{}{(1),\dagger}(s) + \bV{}{(2),\dagger}(s)) \bR^{1h/1p} = \omega \bR^{1h/1p}
\end{equation}
We will show after that $\bV{}{(2)} = \bO$, and hence $\bEta{2} = 0$, but we already use this result to simplify derivations
If we want to truncate the quasiparticle equation to the second order we obtain
\begin{equation}
\dv{\bH^{(3)}}{s} = \comm{\bEta{3}}{\bHd{0}} + \comm{\bEta{1}}{\bHd{2}}
\end{equation}
\begin{equation}
(\bF{}{(0)}(0) + \bF{}{(2)}(s)) + \bV{}{(1)}(s)(\omega \mathbb{1} - \bC{\text{d}}{(0)}(0))^{-1} \bV{}{(1),\dagger}(s) \bR^{1h/1p} = \omega \bR^{1h/1p}
\end{equation}
\end{widetext}
So if we choose to put the off-diagonal part of $\bC{}{}$ in the off-diagonal $\bH{}{}$ we see that the off diagonal part of $\bC{}{}$ is not present in the second order quasi-particle equation.
We believe that this is not desirable.
In the following, we will integrate order by order the differential equations obtained above in the case $\bC{\text{od}}{} = \boldsymbol{0}$ and $\bC{\text{d}}{} = \bC{}{}$.
The expression in the other case are given in Appendix~\ref{sec:diagC}.
\begin{align}
\dv{\bF{}{(3)}}{s} &= \bO \\
\dv{\bC{}{(3)}}{s} &= \bO\\
\dv{\bV{}{(3)}}{s} &= 2 \bF{}{(0)}\bV{}{(3)}\bC{}{(0)} - (\bF{}{(0)})^2\bV{}{(3)} - \bV{}{(3)}(\bC{}{(0)})^2 \\
&+ 2 \bF{}{(0)}\bV{}{(1)}\bC{}{(2)} + 2 \bF{}{(2)}\bV{}{(1)}\bC{}{(0)} \notag \\
&- \left( \bF{}{(0)}\bF{}{(2)} + \bF{}{(2)}\bF{}{(0)} \right)\bV{}{(1)} \notag \\
&- \bV{}{(1)}\left( \bC{}{(0)}\bC{}{(2)} + \bC{}{(2)}\bC{}{(0)} \right) \notag \\
\dv{\bV{}{(3),\dagger}}{s} &= 2 \bC{}{(0)}\bV{}{(3),\dagger}\bF{}{(0)} - \bV{}{(3),\dagger}(\bF{}{(0)})^2 - (\bC{}{(0)})^2\bV{}{(3),\dagger} \\
&+ 2 \bC{}{(0)}\bV{}{(1),\dagger}\bF{}{(2)} + 2 \bC{}{(2)}\bV{}{(1),\dagger}\bF{}{(0)} \notag \\
&- \bV{}{(1),\dagger}\left( \bF{}{(0)}\bF{}{(2)} + \bF{}{(2)}\bF{}{(0)} \right) \notag \\
&-\left( \bC{}{(0)}\bC{}{(2)} + \bC{}{(2)}\bC{}{(0)} \right)\bV{}{(1),\dagger} \notag
\end{align}
%///////////////////////////%
\subsubsection{Forth order differential equations}
% ///////////////////////////%
\begin{equation}
\bEta{4} = \comm{\bHd{0}}{\bHod{4}} + \comm{\bHd{2}}{\bHod{2}} + \comm{\bHd{3}}{\bHod{1}}
\end{equation}
\begin{equation}
\dv{\bH^{(4)}}{s} = \comm{\bEta{4}}{\bHd{0}} + \comm{\bEta{2}}{\bHd{2}} + \comm{\bEta{1}}{\bHd{3}}
\end{equation}
%%%%%%%%%%%%%%%%%%%%%%
\subsection{Integrating order by order}
%%%%%%%%%%%%%%%%%%%%%%
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 $\Delta\epsilon_R$ corresponds to the diagonal elements of the 2h1p and 2p1h sectors.
\subsubsection{First order}
%///////////////////////////%
\subsubsection{First order Hamiltonian elements}
%///////////////////////////%
\begin{align}
\dv{\bF{}{(1)}}{s} &= \bO \Longleftrightarrow \bF{}{(1)}(s) = \bF{}{(1)}(0) \Longleftrightarrow \color{red}{\boxed{\color{black}{\bF{}{(1)}(s)= \bO}}} \\
@ -565,16 +585,16 @@ In the following, upper case indices correspond to the 2h1p and 2p1h sectors whi
\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} &= \bO \Longleftrightarrow \bC{}{(1)}(s) = \bC{}{(1)}(0) \Longleftrightarrow \color{red}{\boxed{\color{black}{\bC{}{(1)}(s)= \bO}}}
\end{align}
The differential equation for the coupling blocks can be solved in the GF(2) case because in this case $\bC{}{}(0)$ is diagonal (see Appendix~\ref{sec:diagC}).
Inspired by this remark we transform $\bC{}{(0)}$ to its diagonal representation using
The differential equation for the coupling blocks can be solved in the GF(2) case because $\bC{}{\GF}(0)$ is diagonal.
Inspired by this fact we transform $\bC{}{(0)}$ to its diagonal representation using
$\bC{}{(0)} = \bU \bD^{(0)} \bU^{-1}$ and insert it in the differential equation for $\bV{}{(1)}$
\begin{align}
\dv{\bV{}{(1)}}{s} &= 2 \bF{}{(0)}\bV{}{(1)}\bU \bD^{(0)} \bU^{-1}- (\bF{}{(0)})^2\bV{}{(1)} - \bV{}{(1)}\bU (\bD^{(0)})^2 \bU^{-1}\\
\dv{\bV{}{(1)}}{s}\bU &= 2 \bF{}{(0)}\bV{}{(1)}\bU \bD^{(0)} - (\bF{}{(0)})^2\bV{}{(1)} \bU - \bV{}{(1)}\bU (\bD^{(0)})^2 \\
\dv{\bW^{(1)}}{s} &= 2 \bF{}{(0)}\bW^{(1)} \bD^{(0)} - (\bF{}{(0)})^2\bW^{(1)} - \bW^{(1)} (\bD^{(0)})^2
\end{align}
where in the last line we have defined the matrix of screened integral.
Note that in the GF(2) case $\bU = \mathbb{1}$ and thus $\bW = \bV{}{}$.
where in the last line we have defined the matrix of screened integral $\bW^{(1)}=\bV{}{(1)} \bU$.
Note that in the GF(2) case $\bU = \mathbb{1}$ and thus $\bW^{(1)}=\bV{}{(1)}$.
\begin{align}
&(\dv{\bW^{(1)}}{s})_{p,(q,v)} = (2 \bF{}{(0)}\bW^{(1)} \bD^{(0)} - (\bF{}{(0)})^2\bW^{(1)} - \bW^{(1)} (\bD^{(0)})^2)_{p,(q,v)} \notag \\
&= \sum_{r,(s,x)} 2 F_{pr}^{(0)}W_{r,(s,x)}^{(1)}D_{(s,x),(q,v)}^{(0)} - \sum_{r,s} F_{pr}^{(0)} F_{rs}^{(0)}W_{s,(q,v)}^{(1)} \notag \\
@ -584,48 +604,12 @@ Note that in the GF(2) case $\bU = \mathbb{1}$ and thus $\bW = \bV{}{}$.
\end{align}
This equation can be integrated to give
\begin{equation}
W_{p,(q,v)}^{(1)}(s) = W_{p,(q,v)}^{(1)}(0) e^{- (F_{pp}^{(0)} - D_{(q,v),(q,v)}^{(0)})^2 s}
\color{red}{\boxed{\color{black}{W_{p,(q,v)}^{(1)}(s) = W_{p,(q,v)}^{(1)}(0) e^{- (F_{pp}^{(0)} - D_{(q,v),(q,v)}^{(0)})^2 s}}}}
\end{equation}
Using this first order analytical blocks we can now evaluate the second order downfolded correlation part of the self-energy as
\begin{align}
\bSig^{(2)} (\omega) &= \bV{}{\hhp,(1)} \bU^{\hhp} \text{diag}(\frac{1}{\omega - D_{(i,v),(i,v)}}) (\bU^{\hhp})^{-1} (\bV{}{\hhp,(1)})^{\mathsf{T}} \notag \\
&+ \bV{}{\pph,(1)} \bU^{\pph} \text{diag}(\frac{1}{\omega - D_{(a,v),(a,v)}}) (\bU^{\pph})^{-1} (\bV{}{\pph,(1)})^{\mathsf{T}} \notag \\
&= \bW^{\hhp,(1)} \text{diag}(\frac{1}{\omega - D_{(i,v),(i,v)}}) (\bW^{\hhp,(1)})^{\mathsf{T}} \notag \\
&+ \bW^{\pph,(1)} \text{diag}(\frac{1}{\omega - D_{(a,v),(a,v)}}) (\bW^{\pph,(1)})^{\mathsf{T}} \notag
\end{align}
\begin{itemize}
\item \textbf{GF(2)}
In the GF(2) case we have $D_{(i,v),(i,v)} = D_{ija,ija} = \epsilon_i + \epsilon_j - \epsilon_a$ and $D_{(a,v),(a,v)} = D_{iab,iab} = \epsilon_a + \epsilon_b - \epsilon_i $ and the $W$ matrix elements have been defined in Eq.~(\ref{eq:GF2_sERI}).
\begin{align}
(\bSig^{(2)} (\omega,s))_{pq} &= \sum_{ija} W_{p,i[ja]}^{(1)} \frac{1}{\omega - D_{ija,ija}}(W^{\mathsf{T}})_{i[ja],q}^{(1)} \notag \\
&+ \sum_{iab} W_{p,[ia]b}^{(1)} \frac{1}{\omega - D_{iab,iab}}(W^{\mathsf{T}})_{[ia]b,q}^{(1)} \notag \\
&= \sum_{ija} \frac{W_{pa,ij}^{(1)} W_{qa,ij}^{(1)} }{\omega - (\epsilon_i + \epsilon_j - \epsilon_a)} \notag \\
&+ \sum_{iab} \frac{W_{pi,ab}^{(1)} W_{qi,ab}^{(1)}}{\omega - (\epsilon_a + \epsilon_b - \epsilon_i)} \notag \\
&= \sum_{ija} \frac{W_{pa,ij}^{(0)} W_{qa,ij}^{(0)} }{\omega - (\epsilon_i + \epsilon_j - \epsilon_a)} e^{-(\epsilon_p - \Delta_{ij}^a)^2s}e^{-(\epsilon_q - \Delta_{ij}^a)^2s} \notag \\
&+ \sum_{iab} \frac{W_{pi,ab}^{(0)} W_{qi,ab}^{(0)} }{\omega - (\epsilon_a + \epsilon_b - \epsilon_i)} e^{-(\epsilon_p - \Delta_{i}^{ab})^2s}e^{-(\epsilon_q - \Delta_{i}^{ab})^2s} \notag
\end{align}
\item \textbf{GW}
A similar derivation gives
\begin{align}
\label{eq:SRGGW_selfenergy}
\Sigma_{pq}^{\GW}(\omega) &= \sum_{iv} \frac{W_{pi,v}^{(0)} W_{qi,v}^{(0)}}{\omega - \epsilon_i + \Omega_{v}^{\dRPA} - \ii \eta}e^{-(\epsilon_p - \epsilon_i + \Omega_v)^2s} e^{-(\epsilon_q - \epsilon_i + \Omega_v)^2s} \notag \\
&+ \sum_{av} \frac{W_{pa,v}^{(0)} W_{qa,v}^{(0)} }{\omega - \epsilon_a - \Omega_{v}^{\dRPA} + \ii \eta} e^{-(\epsilon_p - \epsilon_a - \Omega_v)^2s}e^{-(\epsilon_q - \epsilon_a - \Omega_v)^2s} \notag
\end{align}
\item \textbf{GT}
\textcolor{red}{\textbf{TODO Give analytical expression for the GT case.}}
\end{itemize}
\subsubsection{Second order}
%///////////////////////////%
\subsubsection{Second order Hamiltonian elements}
%///////////////////////////%
\begin{align}
\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} \\
@ -684,30 +668,147 @@ Which finally gives
& - \frac{1}{2} \sum_{iab} \frac{\epsilon_{p}^{(0)} + \epsilon_{q}^{(0)} - 2\Delta_{i}^{ab}}{(\epsilon_p^{(0)} - \Delta_{ij}^a)^2 + (\epsilon_q^{(0)} - \Delta_{i}^{ab})^2} \aeri{pi}{ab}\notag \\
&\times\aeri{qi}{ab} \left(1 - e^{-(\epsilon_p - \Delta_{i}^{ab})^2s} e^{-(\epsilon_q - \Delta_{i}^{ab})^2s}\right) \notag
\end{align}
\end{itemize}
Now that we have all first order blocks and $\bF{}{(2)}$ in analytical form we have every ingredients for the second order quasi-particle equation.
The equation for $\bV{}{(2)}$ and $\bV{}{(2),\dagger}$ are the same that for their first order counterpart.
So we define $\bW^{(2)} = \bV{}{(2)} \bU$ and we obtain
\begin{equation}
W_{p,(q,v)}^{(2)}(s) = W_{p,(q,v)}^{(2)}(0) e^{- (F_{pp}^{(0)} - D_{(q,v),(q,v)}^{(0)})^2 s} = 0
\end{equation}
Therefore, we have
\begin{align}
&\color{red}{\boxed{\color{black}{\bV{}{(2)}(s)= \bO}}} & &\color{red}{\boxed{\color{black}{\bV{}{(2),\dagger}(s)= \bO}}}
\end{align}
Finally, the elements of $\bC{}{(2)}$ can be obtained exactly like their $\bF{}{(2)}$ counterpart just by swapping the role of $\bF{}{(0)}$ and $\bC{}{(0)}$ in the formula.
%///////////////////////////%
\subsubsection{Third order Hamiltonian elements}
%///////////////////////////%
\begin{align}
\dv{\bF{}{(3)}}{s} &= \bO \Longleftrightarrow \bF{}{(3)}(s) = \bF{}{(3)}(0) \Longleftrightarrow \color{red}{\boxed{\color{black}{\bF{}{(3)}(s)= \bO}}} \\
\dv{\bC{}{(3)}}{s} &= \bO \Longleftrightarrow \bC{}{(1)}(s) = \bC{}{(1)}(0) \Longleftrightarrow \color{red}{\boxed{\color{black}{\bC{}{(3)}(s)= \bO}}}
\end{align}
The homogeneous equations for $\bV{}{(3)}$ and $\bV{}{(3), \dagger}$ give homogeneous solutions equal to zero for the same reasons as $\bV{}{(2)}$ and $\bV{}{(2), \dagger}$.
Therefore the differential equations can be simplified to
\begin{align}
\dv{\bV{}{(3)}}{s} &= 2 \bF{}{(0)}\bV{}{(1)}\bC{}{(2)} + 2 \bF{}{(2)}\bV{}{(1)}\bC{}{(0)} \notag \\
&- \left( \bF{}{(0)}\bF{}{(2)} + \bF{}{(2)}\bF{}{(0)} \right)\bV{}{(1)} \notag \\
&- \bV{}{(1)}\left( \bC{}{(0)}\bC{}{(2)} + \bC{}{(2)}\bC{}{(0)} \right) \notag \\
\dv{\bV{}{(3),\dagger}}{s} &= 2 \bC{}{(0)}\bV{}{(1),\dagger}\bF{}{(2)} + 2 \bC{}{(2)}\bV{}{(1),\dagger}\bF{}{(0)} \notag \\
&- \bV{}{(1),\dagger}\left( \bF{}{(0)}\bF{}{(2)} + \bF{}{(2)}\bF{}{(0)} \right) \notag \\
&-\left( \bC{}{(0)}\bC{}{(2)} + \bC{}{(2)}\bC{}{(0)} \right)\bV{}{(1),\dagger} \notag
\end{align}
and can be solved by integration of the right side using the previous analytic formula for Hamiltonian elements.
%///////////////////////////%
\subsubsection{Forth order Hamiltonian elements}
%///////////////////////////%
%%%%%%%%%%%%%%%%%%%%%%
\subsection{Downfolding the SRG-transformed matrix}
%%%%%%%%%%%%%%%%%%%%%%
In order to choose what to do with $\bC{\text{od}}{}$ we look at the downfolded SRG quasiparticle equation.
\begin{equation}
\label{eq:H_SRGMBPT}
H(s) =
\begin{pmatrix}
\bF{}{(0)} + \bF{}{(2)}(s) + \bF{}{(4)}(s) & \bV{}{(1)}(s) + \bV{}{(3)}(s) \\
\bV{}{(1),\dagger}(s) + \bV{}{(3),\dagger}(s) & \bC{}{(0)} +\bC{}{(2)}(s) + \bC{}{(4)}(s)
\end{pmatrix}
\end{equation}
\begin{widetext}
\begin{equation}
\left\{
\begin{aligned}
(\bF{}{(0)} + \bF{}{(2)}(s) + \bF{}{(4)}(s)) \bR^{1h/1p} + (\bV{}{(1)}(s) + \bV{}{(3)}(s)) \bR^{2h1p/2p1h} &= \omega \bR^{1h/1p} \\
(\bV{}{(1),\dagger}(s) + \bV{}{(3),\dagger}(s)) \bR^{1h/1p} + (\bC{}{(0)} + \bC{}{(2)}(s) +\bC{}{(4)}(s) ) \bR^{2h1p/2p1h}&= \omega \bR^{2h1p/2p1h}
\end{aligned}
\right.
\end{equation}
\begin{equation}
(\bF{}{(0)} + \bF{}{(2)}(s) + \bF{}{(4)}(s)) + (\bV{}{(1)}(s) + \bV{}{(3)}(s)) (\omega \mathbb{1} - \bC{}{(0)} - \bC{}{(2)}(s) - \bC{}{(4)}(s))^{-1} (\bV{}{(1),\dagger}(s) + \bV{}{(3),\dagger}(s)) \bR^{1h/1p} = \omega \bR^{1h/1p}
\end{equation}
Then the quasiparticle equation truncated to zeroth, second and forth order is
\begin{align}
\left[ \bF{}{(0)} \right] \bR^{1h/1p} &= \omega \bR^{1h/1p} \\
\left[ \bF{}{(0)} + \bF{}{(2)}(s) + \bV{}{(1)}(s) (\omega \mathbb{1} - \bC{}{(0)} )^{-1} \bV{}{(1),\dagger}(s) \right] \bR^{1h/1p} &= \omega \bR^{1h/1p} \\
\left[ (\bF{}{(0)} + \bF{}{(2)}(s) + \bF{}{(4)}(s)) + (\bV{}{(1)}(s) + \bV{}{(3)}(s)) (\omega \mathbb{1} - \bC{}{(0)} - \bC{}{(2)}(s))^{-1} (\bV{}{(1),\dagger}(s) + \bV{}{(3),\dagger}(s)) \right] \bR^{1h/1p} &= \omega \bR^{1h/1p}
\end{align}
\end{widetext}
To zeroth order in the coupling we recover the HF quasiparticle energies which makes sense.
Now turning to the first non-zero correction to the quasiparticle equation which is the second order equation
\begin{align}
\bSig_c^{(2)} (\omega) &= \bV{}{\hhp,(1)} \bU^{\hhp} \text{diag}(\frac{1}{\omega - D_{(i,v),(i,v)}}) (\bU^{\hhp})^{-1} (\bV{}{\hhp,(1)})^{\mathsf{T}} \notag \\
&+ \bV{}{\pph,(1)} \bU^{\pph} \text{diag}(\frac{1}{\omega - D_{(a,v),(a,v)}}) (\bU^{\pph})^{-1} (\bV{}{\pph,(1)})^{\mathsf{T}} \notag \\
&= \bW^{\hhp,(1)} \text{diag}(\frac{1}{\omega - D_{(i,v),(i,v)}}) (\bW^{\hhp,(1)})^{\mathsf{T}} \notag \\
&+ \bW^{\pph,(1)} \text{diag}(\frac{1}{\omega - D_{(a,v),(a,v)}}) (\bW^{\pph,(1)})^{\mathsf{T}} \notag
\end{align}
\begin{itemize}
\item \textbf{GF(2)}
In the GF(2) case we have $D_{(i,v),(i,v)} = D_{ija,ija} = \epsilon_i + \epsilon_j - \epsilon_a$ and $D_{(a,v),(a,v)} = D_{iab,iab} = \epsilon_a + \epsilon_b - \epsilon_i $ and the $W$ matrix elements have been defined in Eq.~(\ref{eq:GF2_sERI}).
\begin{align}
(\bSig_c^{(2)} (\omega,s))_{pq} &= \sum_{ija} W_{p,i[ja]}^{(1)} \frac{1}{\omega - D_{ija,ija}}(W^{\mathsf{T}})_{i[ja],q}^{(1)} \notag \\
&+ \sum_{iab} W_{p,[ia]b}^{(1)} \frac{1}{\omega - D_{iab,iab}}(W^{\mathsf{T}})_{[ia]b,q}^{(1)} \notag \\
&= \sum_{ija} \frac{W_{pa,ij}^{(1)} W_{qa,ij}^{(1)} }{\omega - (\epsilon_i + \epsilon_j - \epsilon_a)} \notag \\
&+ \sum_{iab} \frac{W_{pi,ab}^{(1)} W_{qi,ab}^{(1)}}{\omega - (\epsilon_a + \epsilon_b - \epsilon_i)} \notag \\
&= \sum_{ija} \frac{W_{pa,ij}^{(1)}(0) W_{qa,ij}^{(1)}(0) }{\omega - (\epsilon_i + \epsilon_j - \epsilon_a)} e^{-(\epsilon_p - \Delta_{ij}^a)^2s}e^{-(\epsilon_q - \Delta_{ij}^a)^2s} \notag \\
&+ \sum_{iab} \frac{W_{pi,ab}^{(1)}(0) W_{qi,ab}^{(1)}(0) }{\omega - (\epsilon_a + \epsilon_b - \epsilon_i)} e^{-(\epsilon_p - \Delta_{i}^{ab})^2s}e^{-(\epsilon_q - \Delta_{i}^{ab})^2s} \notag
\end{align}
\item \textbf{GW}
A similar derivation gives
\begin{align}
\label{eq:SRGGW_selfenergy}
(\Sigma_c^{\GW}(\omega,s))_{pq} &= \sum_{iv} \frac{W_{pi,v}^{(1)}(0) W_{qi,v}^{(1)}(0)}{\omega - \epsilon_i + \Omega_{v}^{\dRPA} - \ii \eta}e^{-(\epsilon_p - \epsilon_i + \Omega_v)^2s} e^{-(\epsilon_q - \epsilon_i + \Omega_v)^2s} \notag \\
&+ \sum_{av} \frac{W_{pa,v}^{(1)}(0) W_{qa,v}^{(1)}(0)}{\omega - \epsilon_a - \Omega_{v}^{\dRPA} + \ii \eta} e^{-(\epsilon_p - \epsilon_a - \Omega_v)^2s}e^{-(\epsilon_q - \epsilon_a - \Omega_v)^2s} \notag
\end{align}
\item \textbf{GT}
\textcolor{red}{\textbf{TODO Give analytical expression for the GT case.}}
\end{itemize}
Now that we have analytical expression for the second order correlation part of the self-energy as well as for $\bF{}{(2)}$, we have every ingredients for the second order quasi-particle equation.
In the previous formula we can see that the diagonal elements at $s \to \infty$ correspond to the same values as in the usual diagonal static approximation.
Therefore, the SRG as a renormalization group method removes the coupling $\bV{}{}$ while incorporating some of its physics in the non-coupled problem $\bF{}{}$.
Therefore, the SRG as a true renormalization group method removes the coupling $\bV{}{}$ while incorporating some of its physics in the non-coupled problem $\bF{}{}$.
This formalism gives us a rationalization of the diagonal static approximation from a RG perspective.
In addition, this gives us a way to define a non-diagonal static approximation which is not straightforward to define by simply looking at Eq.~(\ref{eq:GW_selfenergy}).
The usual non-diagonal static approximation used in qsGW is
In addition, this gives us an unambiguous way to define a non-diagonal static approximation.
On the other hand the usual non-diagonal static self-energy used in qsGW,
\begin{equation}
(\Sigma_c^{qsGW})_{pq} = \frac{\Sigma_c(\epsilon_p)_{pq} + \Sigma_c(\epsilon_q)_{qp}}{2}
\end{equation}
If we define $x=\epsilon_p^{(0)} - \epsilon_r^{(0)} \pm \Omega_v$ and $y=\epsilon_q^{(0)} - \epsilon_r^{(0)} \pm \Omega_v$ then the SRGqsGW is of the form $(x+y)/(x^2 + y^2)$ while the usual qsGW is of the form $(x+y)/2xy$.
is not unambiguously defined because this is not the only possible choice starting from the $GW$ self-energy.
If we define $x=\epsilon_p^{(0)} - \epsilon_r^{(0)} \pm \Omega_v$ and $y=\epsilon_q^{(0)} - \epsilon_r^{(0)} \pm \Omega_v$ then the SRG qsGW is of the form $(x+y)/(x^2 + y^2)$ while the usual qsGW is of the form $(x+y)/2xy$.
Note that both diagonal are of the form $1/x$ which is consistent with the fact that the SRG diagonal correspond to the usual static diagonal.
Even more, the SRG formalism defines a hierarchy of static approximation by considering higher and higher perturbation order for $\bSig$.
This hierarchy could be compared to another hierarchy of static approximation obtained by perturbing the static self-energy by its difference to its dynamic counterpart.y
This hierarchy could be compared to another hierarchy of static approximation obtained by perturbing the static self-energy by its difference to its dynamic counterpart.
One of the con of the static approximation is that we loose information about the satellites and this is true for the SRG also when the coupling has been totally removed.
One of the con of the static approximation is that we loose information about the satellites and this is also true for the SRG also when the coupling has been totally removed.
However, SRG allows us to stop at a finite value $s$ corresponding to a renormalized coupling but the coupling is still present.
Therefore the satellites can still be observed due to the non-linearity induced by the renormalized coupling.
% =================================================================%
\section{Towards second quantized effective Hamiltonians for MBPT?}
\label{sec:second_quant_mbpt}
@ -887,46 +988,86 @@ We have
\end{align}
%=================================================================%
\section{Perturbative matrix coefficients for $C^{(0)}$ diagonal}
\label{sec:diagC}
\section{Alternative partitioning}
\label{sec:partitioning}
%=================================================================%
In this section we discuss the alternative partitioning of the Hamiltonian mentioned previously.
%///////////////////////////%
\subsubsection{First order Hamiltonian}
%///////////////////////////%
Now turning to the first-order contribution to the MBPT matrix, we start by computing the first order part of the Wegner generator.
\begin{align}
(\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 \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} v^{(1)}_{pR} \Delta\epsilon^{(0)}_R\delta_{RS} \Delta\epsilon^{(0)}_Q\delta_{SQ} \\
&= (2 \epsilon^{(0)}_p\Delta\epsilon^{(0)}_Q - (\epsilon^{(0)}_p)^2 - (\Delta\epsilon^{(0)}_Q )^2) v^{(1)}_{pQ} \\
\dv{v^{(1)}_{pQ}}{s} &= - (\epsilon^{(0)}_p - \Delta\epsilon^{(0)}_Q )^2 v^{(1)}_{pQ} \\
&\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}
Note the close similarity with Evangelista's expressions for the off-diagonal part at first order!
\begin{align}
(\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 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} \\
&= 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 \\
&= - (\Delta\epsilon^{(0)}_P - \Delta\epsilon^{(0)}_Q )^2 c^{(1)}_{PQ} \\
&\color{red}{\boxed{\color{black}{c^{(1)}_{PQ}(s) = c^{(1)}_{PQ}(0) e^{-s(\Delta\epsilon^{(0)}_P - \Delta\epsilon^{(0)}_Q )^2} }}}
&\bEta{1} = \comm{\bHd{0}}{\bHod{1}} \\
&= \begin{pmatrix}
\bO & \bF{}{(0)}\bV{}{(1)} - \bV{}{(1)}\bC{\text{d}}{(0)}\\
\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)} \notag
\end{pmatrix}
\end{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{\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_{S} \epsilon^{(0)}_{p} v^{(1)}_{pS} v^{(1)}_{qS} + \sum_{R} \epsilon^{(0)}_{q} v^{(1)}_{pR} v^{(1)}_{qR} - 2\sum_{R} \Delta\epsilon^{(0)}_R v^{(1)}_{pR} v^{(1)}_{qR} \notag \\
&= \sum_R (\epsilon^{(0)}_{p} + \epsilon^{(0)}_{q} - 2 \Delta\epsilon^{(0)}_R) v^{(1)}_{pR} v^{(1)}_{qR} \notag \\
&= \sum_R (\epsilon^{(0)}_{p} + \epsilon^{(0)}_{q} - 2 \Delta\epsilon^{(0)}_R) v^{(1)}_{pR}(0) v^{(1)}_{qR}(0) e^{-s [ (\epsilon^{(0)}_p - \Delta\epsilon^{(0)}_R)^2+ (\epsilon^{(0)}_q - \Delta\epsilon^{(0)}_R)^2]} \notag \\
&f^{(2)}_{pq}(s) = \notag \\
&\color{red}{\boxed{\color{black}{- \sum_R\frac{ v^{(1)}_{pR}(0) v^{(1)}_{qR}(0) (\epsilon^{(0)}_{p} + \epsilon^{(0)}_{q} - 2 \Delta\epsilon^{(0)}_R)}{(\epsilon^{(0)}_p - \Delta\epsilon^{(0)}_R)^2+ (\epsilon^{(0)}_q - \Delta\epsilon^{(0)}_R)^2}(1 - e^{-s [ (\epsilon^{(0)}_p - \Delta\epsilon^{(0)}_R)^2+ (\epsilon^{(0)}_q - \Delta\epsilon^{(0)}_R)^2]})}}} \notag
\dv{\bH^{(1)}}{s} &= \comm{\bEta{1}}{\bHd{0}} = \begin{pmatrix}
\dv{\bF{}{(1)}}{s} & \dv{\bV{}{(1)}}{s} \\
\dv{\bV{}{(1),\dagger}}{s} & \dv{\bC{}{(1)}}{s}
\end{pmatrix} \\
\dv{\bF{}{(1)}}{s} &= \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),\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{\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}
%///////////////////////////%
\subsubsection{Second order Hamiltonian}
%///////////////////////////%
Recalling that $\bHod{0} = \bO$ and $\bHd{1} = \bO$, we derive
\begin{align}
&\bEta{2} = \comm{\bHd{0}}{\bHod{2}} + \comm{\bHd{1}}{\bHod{1}} \\
&= \comm{\bHd{0}}{\bHod{2}} \notag \\
&= \begin{pmatrix}
\bO & \bF{}{(0)}\bV{}{(2)} - \bV{}{(2)}\bC{\text{d}}{(0)}\\
\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)} \notag
\end{pmatrix}
\end{align}
\begin{align}
(\dv{\bV{}{(2)}}{s})_{pQ} &= (2 \bF{}{(0)}\bV{}{(2)}\bC{\text{d}}{(0)} - (\bF{}{(0)})^2\bV{}{(2)} - \bV{}{(2)}(\bC{\text{d}}{(0)})^2 \\
& - 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)})_{pQ} \notag \\
v^{(2)}_{pQ}(s) &= v^{(2)}_{pQ}(0) e^{-s(\epsilon^{(0)}_p - \Delta\epsilon^{(0)}_Q )^2} + \text{Non-homogeneous solution} \notag \\
v^{(2)}_{pQ}(s) &= \text{Non-homogeneous solution}
\dv{\bH^{(2)}}{s} &= \comm{\bEta{2}}{\bHd{0}} + \comm{\bEta{1}}{\bHd{1}} \\
&= \begin{pmatrix}
\dv{\bF{}{(2)}}{s} & \dv{\bV{}{(2)}}{s} \\
\dv{\bV{}{(2),\dagger}}{s} & \dv{\bC{}{(2)}}{s}
\end{pmatrix} \notag \\
\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} &= 2 \bC{\text{d}}{(0)}\bC{\text{od}}{(2)}\bC{\text{d}}{(0)}- (\bC{\text{d}}{(0)})^2\bC{\text{od}}{(2)} - \bC{\text{od}}{(2)}(\bC{\text{d}}{(0)})^2 \\
&-2 \bC{\text{d}}{(1)}\bC{\text{od}}{(0)}\bC{\text{d}}{(1)} + (\bC{\text{d}}{(1)})^2\bC{\text{od}}{(0)} + \bC{\text{od}}{(0)}(\bC{\text{d}}{(1)})^2 \notag \\
&+ \bC{\text{d}}{(0)}\bV{}{(1)}\bV{}{(1),\dagger} + \bV{}{(1)}\bV{}{(1),\dagger}\bC{\text{d}}{(0)} - 2 \bV{}{(1)}\bF{}{(0)}\bV{}{(1),\dagger} \notag \\
\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 \\
&- 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)} \notag \\
\dv{\bV{}{(2),\dagger}}{s} &= 2 \bC{\text{d}}{(0)}\bV{}{(2),\dagger}\bF{}{(0)} - \bV{}{(2),\dagger}(\bF{}{(0)})^2 - (\bC{\text{d}}{(0)})^2\bV{}{(2),\dagger} \\
&- 2 \bC{\text{od}}{(1)} \bC{\text{d}}{(0)} \bV{}{(1),\dagger} + \bC{\text{od}}{(1)} \bV{}{(1),\dagger} \bF{}{(0)} + \bC{\text{d}}{(0)} \bC{\text{od}}{(1)} \bV{}{(1),\dagger} \notag
\end{align}
%%%%%%%%%%%%%%%%%%%%%%
\subsection{Downfolding the SRG-transformed matrix}
%%%%%%%%%%%%%%%%%%%%%%
In order to choose what to do with $\bC{\text{od}}{}$ we look at the downfolded SRG quasiparticle equation.
\begin{widetext}
\begin{equation}
(\bF{}{(0)}(0) + \bF{}{(2)}(s)) + (\bV{}{(1)}(s) + \bV{}{(2)}(s)) (\omega \mathbb{1} - \bC{\text{d}}{(0)}(0) - \bC{\text{od}}{(1)}(s) -\bC{}{(2)}(s) )^{-1} (\bV{}{(1),\dagger}(s) + \bV{}{(2),\dagger}(s)) \bR^{1h/1p} = \omega \bR^{1h/1p}
\end{equation}
If we want to truncate the quasiparticle equation to the second order we obtain
\begin{equation}
(\bF{}{(0)}(0) + \bF{}{(2)}(s)) + \bV{}{(1)}(s)(\omega \mathbb{1} - \bC{\text{d}}{(0)}(0))^{-1} \bV{}{(1),\dagger}(s) \bR^{1h/1p} = \omega \bR^{1h/1p}
\end{equation}
\end{widetext}
% \section{Old stuff}
% However, in the general case this matrix differential equation is not trivial to solve.