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Notes/PerturbativeAnalysis.tex

@@ -548,33 +548,44 @@ Now that we obtained the SRG-transformed Hamiltonian to a given order we can dow
  \right.
 \end{equation}
 
-\begin{align}
-  &(\bF^{(0)}(0) + \bF^{(2)}(s)) + (\bV{}{(1)}(s) + \bV{}{(2)}(s))(\omega \mathbb{1} - \bC{}{(0)}(0) +\bC{}{(2)}(s)  )^{-1}  \notag \\
-  &\dots  (\bV{}{(1),\dagger}(s) + \bV{}{(2),\dagger}(s)) \bR^{1h/1p} = \omega \bR^{1h/1p} 
-\end{align}
+\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}
+\end{widetext}
 
-\begin{align}
-  &(\omega \mathbb{1} - \bC{}{(0)}(0) - \bC{}{(1)}(s)  - \bC{}{(2)}(s)  )^{-1} = (\omega \mathbb{1} - \bC{}{(0)}(0))^{-1} \\
-  &+ (\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)
+  &\tilde{\bF} =\bF^{(0)}(0) + \bF^{(2)}(s) + \dots \\
+  &\tilde{\boldsymbol{\Sigma}}(\omega) = \tilde{\bV{}{}}(s) (\omega \mathbb{1} - \tilde{\bC{}{}}(s) )^{-1} \tilde{\bV{}{}}^\dagger(s) \\
+  &\tilde{\bC{}{}}(s) = \bC{\text{d}}{(0)}(0) + \bC{\text{od}}{(1)}(s) + \bC{}{(2)}(s)  + \dots \\
+  &\tilde{\bV{}{}}(s) = \bV{}{(1)}(s) + \bV{}{(2)}(s) + \dots
 \end{align}
 
+Note that the inverse of $\omega\mathbb{1} - \tilde{\bC{}{}}(s)$ does not need to be approximated using a Taylor expansion.
+Indeed, $\tilde{\bC{}{}}(s)$ can be used to define a renormalized RPA problem to diagonalize instead of the usual one.
+We need to find a way to truncate the above quasi-particle equation.
+For the moment, I'm not sure if we should truncate according to the quasi-particle equation or truncate directly the subblocks of the unfolded matrix\dots
+
+I'm not sure if this valid from a perturbation theory point of view but truncating each terms after their first non-zero correction could be handy.
+This would give first order for the coupling terms and second order for the diagonal ones.
+
 %%%%%%%%%%%%%%%%%%%%%%
 \subsection{The SRG(2) quasi-particle equations}
 %%%%%%%%%%%%%%%%%%%%%%
 
 In this section, we report the GF(2), GW and GT quasi-particle equations.
 
+We start with the GF(2) equation because it does not have non-diagonal $\bC{}{}$ contributions, this gives the following renormalized quasi-particle equation 
+\begin{widetext}
+  \begin{align}
+    \label{eq:selfenergies}
+    \tilde{f}_{pq} + \tilde{\Sigma}_{pq}^{GF(2)}(\omega) &= \delta_{pq}\eps_p + - \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]}  \\
+                                                         &+\sum_{klc} \frac{\aeri{pc}{kl}\aeri{qc}{kl}}{\omega + \eps _c -\eps_k -\eps_l - \ii \eta}e^{-s(\epsilon^{(0)}_p - \Delta\epsilon^{(0)}_{ckl} )^2} e^{-s(\epsilon^{(0)}_q - \Delta\epsilon^{(0)}_{ckl} )^2}  + \sum_{kcd} \frac{\aeri{pk}{cd}\aeri{qk}{cd}}{\omega + \eps _k -\eps_c -\eps_d + \ii \eta} e^{-s(\epsilon^{(0)}_p - \Delta\epsilon^{(0)}_{kcd} )^2} e^{-s(\epsilon^{(0)}_q - \Delta\epsilon^{(0)}_{kcd} )^2} \notag
+  \end{align}
+\end{widetext}
+
 \appendix
 
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