loos/SRGGW
2
0
Fork 0
Code Issues Pull Requests Projects Releases Wiki Activity

Equations for third order

Browse Source
This commit is contained in:
Antoine Marie 2022-11-09 15:39:32 +01:00
parent f54ee2b4c1
commit 003b2a18dd
1 changed files with 87 additions and 47 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

130
Notes/Notes.tex
View File

@ -415,7 +415,7 @@ A general upfolded MBPT matrix can be written as
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} \begin{equation}
\label{eq:H_MBPT_partitioning} \label{eq:H_MBPT_partitioning}
H(0) = \bH(0) =
\begin{pmatrix} \begin{pmatrix}
\bF{}{} & \bO \\ \bF{}{} & \bO \\
\bO & \bC{}{} \bO & \bC{}{}
@ -440,7 +440,7 @@ we could also have defined it like this
\bV{}{\dagger} & \bC{\text{od}}{} \bV{}{\dagger} & \bC{\text{od}}{}
\end{pmatrix} \end{pmatrix}
\end{equation} \end{equation}
However, in the end this choice was not wise and the reason is explained why in Appendix~\ref{sec:partitioning}. 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 The initial conditions corresponding to the first partitioning are
\begin{align} \begin{align}
@ -521,8 +521,8 @@ Recalling that $\bHod{0} = \bO$ and $\bHd{1} = \bO$, we derive
\end{align} \end{align}
\begin{align} \begin{align}
\dv{\bH^{(2)}}{s} &= \comm{\bEta{2}}{\bHd{0}} + \comm{\bEta{1}}{\bHd{1}} + \comm{\bEta{0}}{\bHd{2}} \\ \dv{\bH^{(2)}}{s} &= \comm{\bEta{2}}{\bH^{(0)}} + \comm{\bEta{1}}{\bH^{(1)}} + \comm{\bEta{0}}{\bH^{(2)}} \\
&= \comm{\bEta{2}}{\bHd{0}} \notag &= \comm{\bEta{2}}{\bHd{0}} + \comm{\bEta{1}}{\bHod{1}} \notag
\end{align} \end{align}
\begin{align} \begin{align}
@ -540,6 +540,25 @@ Recalling that $\bHod{0} = \bO$ and $\bHd{1} = \bO$, we derive
\bEta{3} = \comm{\bHd{0}}{\bHod{3}} + \comm{\bHd{2}}{\bHod{1}} \bEta{3} = \comm{\bHd{0}}{\bHod{3}} + \comm{\bHd{2}}{\bHod{1}}
\end{equation} \end{equation}
We will show after that $\bV{}{(2)} = \bO$, and hence $\bEta{2} = 0$, but we already use this result to simplify derivations
\begin{equation}
\dv{\bH^{(3)}}{s} = \comm{\bEta{3}}{\bHd{0}} + \comm{\bEta{1}}{\bHd{2}}
\end{equation}
\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} \subsubsection{Forth order differential equations}
% ///////////////////////////% % ///////////////////////////%
@ -548,13 +567,17 @@ Recalling that $\bHod{0} = \bO$ and $\bHd{1} = \bO$, we derive
\bEta{4} = \comm{\bHd{0}}{\bHod{4}} + \comm{\bHd{2}}{\bHod{2}} + \comm{\bHd{3}}{\bHod{1}} \bEta{4} = \comm{\bHd{0}}{\bHod{4}} + \comm{\bHd{2}}{\bHod{2}} + \comm{\bHd{3}}{\bHod{1}}
\end{equation} \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} \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 Hamiltonian elements} \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}}} \\
@ -562,16 +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{\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}}} \dv{\bC{}{(1)}}{s} &= \bO \Longleftrightarrow \bC{}{(1)}(s) = \bC{}{(1)}(0) \Longleftrightarrow \color{red}{\boxed{\color{black}{\bC{}{(1)}(s)= \bO}}}
\end{align} \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}). The differential equation for the coupling blocks can be solved in the GF(2) case because $\bC{}{\GF}(0)$ is diagonal.
Inspired by this remark we transform $\bC{}{(0)}$ to its diagonal representation using 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 $\bV{}{(1)}$ $\bC{}{(0)} = \bU \bD^{(0)} \bU^{-1}$ and insert it in the differential equation for $\bV{}{(1)}$
\begin{align} \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} &= 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{\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 \dv{\bW^{(1)}}{s} &= 2 \bF{}{(0)}\bW^{(1)} \bD^{(0)} - (\bF{}{(0)})^2\bW^{(1)} - \bW^{(1)} (\bD^{(0)})^2
\end{align} \end{align}
where in the last line we have defined the matrix of screened integral. 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 $\bW = \bV{}{}$. Note that in the GF(2) case $\bU = \mathbb{1}$ and thus $\bW^{(1)}=\bV{}{(1)}$.
\begin{align} \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 \\ &(\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 \\ &= \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 \\
@ -581,10 +604,12 @@ Note that in the GF(2) case $\bU = \mathbb{1}$ and thus $\bW = \bV{}{}$.
\end{align} \end{align}
This equation can be integrated to give This equation can be integrated to give
\begin{equation} \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} \end{equation}
%///////////////////////////%
\subsubsection{Second order Hamiltonian elements} \subsubsection{Second order Hamiltonian elements}
%///////////////////////////%
\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} \\
@ -646,6 +671,43 @@ Which finally gives
\end{itemize} \end{itemize}
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} \subsection{Downfolding the SRG-transformed matrix}
%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%
@ -688,7 +750,7 @@ To zeroth order in the coupling we recover the HF quasiparticle energies which m
Now turning to the first non-zero correction to the quasiparticle equation which is the second order equation Now turning to the first non-zero correction to the quasiparticle equation which is the second order equation
\begin{align} \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 \\ \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 \\ &+ \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^{\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 &+ \bW^{\pph,(1)} \text{diag}(\frac{1}{\omega - D_{(a,v),(a,v)}}) (\bW^{\pph,(1)})^{\mathsf{T}} \notag
@ -700,12 +762,12 @@ Now turning to the first non-zero correction to the quasiparticle equation which
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}). 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} \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 \\ (\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_{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_{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_{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_{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}^{(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 &+ \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} \end{align}
\item \textbf{GW} \item \textbf{GW}
@ -714,8 +776,8 @@ Now turning to the first non-zero correction to the quasiparticle equation which
\begin{align} \begin{align}
\label{eq:SRGGW_selfenergy} \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 \\ (\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}^{(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 &+ \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} \end{align}
\item \textbf{GT} \item \textbf{GT}
@ -724,17 +786,18 @@ Now turning to the first non-zero correction to the quasiparticle equation which
\end{itemize} \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 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. 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. 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}). In addition, this gives us an unambiguous way to define a non-diagonal static approximation.
The usual non-diagonal static approximation used in qsGW is On the other hand the usual non-diagonal static self-energy used in qsGW,
\begin{equation} \begin{equation}
(\Sigma_c^{qsGW})_{pq} = \frac{\Sigma_c(\epsilon_p)_{pq} + \Sigma_c(\epsilon_q)_{qp}}{2} (\Sigma_c^{qsGW})_{pq} = \frac{\Sigma_c(\epsilon_p)_{pq} + \Sigma_c(\epsilon_q)_{qp}}{2}
\end{equation} \end{equation}
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 $(x+y)/2xy$. 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. 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.
@ -929,8 +992,7 @@ We have
\label{sec:partitioning} \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 this section we discuss the alternative partitioning of the Hamiltonian mentioned previously.
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} \subsubsection{First order Hamiltonian}
@ -956,7 +1018,6 @@ Now turning to the first-order contribution to the MBPT matrix, we start by comp
\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{\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{\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 last two equations can be solved differently depending on the form of $\bF{}{}$ and $\bC{}{}$.
%///////////////////////////% %///////////////////////////%
\subsubsection{Second order Hamiltonian} \subsubsection{Second order Hamiltonian}
@ -994,24 +1055,7 @@ Recalling that $\bHod{0} = \bO$ and $\bHd{1} = \bO$, we derive
In order to choose what to do with $\bC{\text{od}}{}$ we look at the downfolded SRG quasiparticle equation. 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{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} \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} (\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}
@ -1023,10 +1067,6 @@ In order to choose what to do with $\bC{\text{od}}{}$ we look at the downfolded
(\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} (\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{equation}
\end{widetext} \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}