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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. 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 A general upfolded MBPT matrix can be written as
\begin{equation} \begin{equation}
@ -408,7 +412,21 @@ A general upfolded MBPT matrix can be written as
\bV{}{\dagger} & \bC{}{} \bV{}{\dagger} & \bC{}{}
\end{pmatrix} \end{pmatrix}
\end{equation} \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}
H(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} \begin{equation}
\label{eq:H_MBPT_partitioning} \label{eq:H_MBPT_partitioning}
H(0) = H(0) =
@ -422,23 +440,32 @@ Using SRG language, we define the diagonal and off-diagonal parts as
\bV{}{\dagger} & \bC{\text{od}}{} \bV{}{\dagger} & \bC{\text{od}}{}
\end{pmatrix} \end{pmatrix}
\end{equation} \end{equation}
which gives the following conditions However, in the end this choice was not wise and the reason is explained why in Appendix~\ref{sec:partitioning}.
The initial conditions corresponding to the first partitioning are
\begin{align} \begin{align}
\bHd{0}(0) &= \begin{pmatrix} \bHd{0}(0) &= \begin{pmatrix}
\bF{}{} & \bO \\ \bF{}{} & \bO \\
\bO & \bC{\text{d}}{} \bO & \bC{}{}
\end{pmatrix} & \bHod{0}(0) &= \bO \notag \\ \end{pmatrix} &
\bHd{1}(0) &= \bO & \bHod{1}(0) &= \begin{pmatrix} \bHod{0}(0) &=\begin{pmatrix}
\bO & \bV{}{} \\ \bO & \bO \\
\bV{}{\dagger} & \bC{\text{od}}{} \notag \bO & \bO
\end{pmatrix} \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} \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 The zeroth-order commutator of the Wegner generator therefore gives
\begin{equation} \begin{equation}
@ -455,101 +482,71 @@ Finally, we have
}} }}
\end{equation} \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. 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} \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)}\bC{\text{d}}{(0)}\\ \bO & \bF{}{(0)}\bV{}{(1)} - \bV{}{(1)}\bC{}{(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 \bC{}{(0)}\bV{}{(1),\dagger} - \bV{}{(1),\dagger} \bF{}{(0)} & \bO \notag
\end{pmatrix} \end{pmatrix}
\end{align} \end{align}
\begin{align} \begin{align}
\dv{\bH^{(1)}}{s} &= \comm{\bEta{1}}{\bHd{0}} = \begin{pmatrix} \dv{\bH^{(1)}}{s} &= \comm{\bEta{1}}{\bHd{0}} \\
\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{\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)}}{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 \dv{\bC{}{(1)}}{s} &= \bO
\end{align} \end{align}
The last two equations can be solved differently depending on the form of $\bF{}{}$ and $\bC{}{}$. 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 Recalling that $\bHod{0} = \bO$ and $\bHd{1} = \bO$, we derive
\begin{align} \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 \\ &= \comm{\bHd{0}}{\bHod{2}} \notag \\
&= \begin{pmatrix} &= \begin{pmatrix}
\bO & \bF{}{(0)}\bV{}{(2)} - \bV{}{(2)}\bC{\text{d}}{(0)}\\ \bO & \bF{}{(0)}\bV{}{(2)} - \bV{}{(2)}\bC{}{(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 \bC{}{(0)}\bV{}{(2),\dagger} - \bV{}{(2),\dagger}\bF{}{(0)} & \bO \notag
\end{pmatrix} \end{pmatrix}
\end{align} \end{align}
\begin{align} \begin{align}
\dv{\bH^{(2)}}{s} &= \comm{\bEta{2}}{\bHd{0}} + \comm{\bEta{1}}{\bHd{1}} \\ \dv{\bH^{(2)}}{s} &= \comm{\bEta{2}}{\bHd{0}} + \comm{\bEta{1}}{\bHd{1}} + \comm{\bEta{0}}{\bHd{2}} \\
&= \begin{pmatrix} &= \comm{\bEta{2}}{\bHd{0}} \notag
\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} \end{align}
%%%%%%%%%%%%%%%%%%%%%% \begin{align}
\subsection{Downfolding the SRG-transformed matrix} \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} \begin{equation}
\label{eq:H_SRGMBPT} \bEta{3} = \comm{\bHd{0}}{\bHod{3}} + \comm{\bHd{2}}{\bHod{1}}
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} \end{equation}
\begin{widetext} %///////////////////////////%
\begin{equation} \subsubsection{Forth order differential equations}
\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}
If we want to truncate the quasiparticle equation to the second order we obtain \begin{equation}
\bEta{4} = \comm{\bHd{0}}{\bHod{4}} + \comm{\bHd{2}}{\bHod{2}} + \comm{\bHd{3}}{\bHod{1}}
\begin{equation} \end{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}.
%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%
\subsection{Integrating order by order} \subsection{Integrating order by order}
@ -557,7 +554,7 @@ The expression in the other case are given in Appendix~\ref{sec:diagC}.
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. 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} \begin{align}
\dv{\bF{}{(1)}}{s} &= \bO \Longleftrightarrow \bF{}{(1)}(s) = \bF{}{(1)}(0) \Longleftrightarrow \color{red}{\boxed{\color{black}{\bF{}{(1)}(s)= \bO}}} \\ \dv{\bF{}{(1)}}{s} &= \bO \Longleftrightarrow \bF{}{(1)}(s) = \bF{}{(1)}(0) \Longleftrightarrow \color{red}{\boxed{\color{black}{\bF{}{(1)}(s)= \bO}}} \\
@ -587,45 +584,7 @@ This equation can be integrated to give
W_{p,(q,v)}^{(1)}(s) = W_{p,(q,v)}^{(1)}(0) e^{- (F_{pp}^{(0)} - D_{(q,v),(q,v)}^{(0)})^2 s} 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} \end{equation}
Using this first order analytical blocks we can now evaluate the second order downfolded correlation part of the self-energy as \subsubsection{Second order Hamiltonian elements}
\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}
\begin{align} \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{\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} \\
@ -687,6 +646,84 @@ Which finally gives
\end{itemize} \end{itemize}
%%%%%%%%%%%%%%%%%%%%%%
\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^{(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}
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. 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.
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. 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.
@ -702,12 +739,13 @@ If we define $x=\epsilon_p^{(0)} - \epsilon_r^{(0)} \pm \Omega_v$ and $y=\epsilo
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. 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 $\bSig$. 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. 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. 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?} \section{Towards second quantized effective Hamiltonians for MBPT?}
\label{sec:second_quant_mbpt} \label{sec:second_quant_mbpt}
@ -887,46 +925,109 @@ We have
\end{align} \end{align}
%=================================================================% %=================================================================%
\section{Perturbative matrix coefficients for $C^{(0)}$ diagonal} \section{Alternative partitioning}
\label{sec:diagC} \label{sec:partitioning}
%=================================================================% %=================================================================%
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 $\bC{\text{d}}{} = \bC{}{}$.
%///////////////////////////%
\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} \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}\\ &\bEta{1} = \comm{\bHd{0}}{\bHod{1}} \\
&= \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} \\ &= \begin{pmatrix}
&= \sum_{rS} 2 \epsilon^{(0)}_p\delta_{pr} v^{(1)}_{rS}\Delta\epsilon^{(0)}_Q\delta_{SQ} \\ \bO & \bF{}{(0)}\bV{}{(1)} - \bV{}{(1)}\bC{\text{d}}{(0)}\\
&- \sum_{rs} \epsilon^{(0)}_p\delta_{pr} \epsilon^{(0)}_r\delta_{rs} v^{(1)}_{sQ} \\ \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
&- \sum_{RS} v^{(1)}_{pR} \Delta\epsilon^{(0)}_R\delta_{RS} \Delta\epsilon^{(0)}_Q\delta_{SQ} \\ \end{pmatrix}
&= (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} }}}
\end{align} \end{align}
\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{\text{d}}{(0)}\bV{}{(1),\dagger})_{pq} \notag \\ \dv{\bH^{(1)}}{s} &= \comm{\bEta{1}}{\bHd{0}} = \begin{pmatrix}
&= \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 \\ \dv{\bF{}{(1)}}{s} & \dv{\bV{}{(1)}}{s} \\
&= \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 \\ \dv{\bV{}{(1),\dagger}}{s} & \dv{\bC{}{(1)}}{s}
&= \sum_R (\epsilon^{(0)}_{p} + \epsilon^{(0)}_{q} - 2 \Delta\epsilon^{(0)}_R) v^{(1)}_{pR} v^{(1)}_{qR} \notag \\ \end{pmatrix} \\
&= \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 \\ \dv{\bF{}{(1)}}{s} &= \bO \\
&f^{(2)}_{pq}(s) = \notag \\ \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 \\
&\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{\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}
The last two equations can be solved differently depending on the form of $\bF{}{}$ and $\bC{}{}$.
%///////////////////////////%
\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} \end{align}
\begin{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 \\ \dv{\bH^{(2)}}{s} &= \comm{\bEta{2}}{\bHd{0}} + \comm{\bEta{1}}{\bHd{1}} \\
& - 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 \\ &= \begin{pmatrix}
v^{(2)}_{pQ}(s) &= v^{(2)}_{pQ}(0) e^{-s(\epsilon^{(0)}_p - \Delta\epsilon^{(0)}_Q )^2} + \text{Non-homogeneous solution} \notag \\ \dv{\bF{}{(2)}}{s} & \dv{\bV{}{(2)}}{s} \\
v^{(2)}_{pQ}(s) &= \text{Non-homogeneous solution} \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} \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{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{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}
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}
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}.
% \section{Old stuff} % \section{Old stuff}
% However, in the general case this matrix differential equation is not trivial to solve. % However, in the general case this matrix differential equation is not trivial to solve.