small corrections in Sec 4

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Antoine Marie 2023-03-10 11:06:33 +01:00
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@ -443,7 +443,7 @@ In particular, we focus here on the second-order renormalized quasiparticle equa
\subsection{Zeroth-order matrix elements}
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The choice of Wegner's generator in the flow equation [see Eq.~\eqref{eq:flowEquation}] implies that the off-diagonal correction is of order $\order*{\lambda}$ while the correction to the diagonal block is at least $\order*{\lambda^2}$.
The choice of Wegner's generator in the flow equation [see Eq.~\eqref{eq:flowEquation}] implies that the off-diagonal correction is of order $\order*{\lambda}$ while the correction to the diagonal block is at least $\order*{\lambda^2}$. \cite{Hergert_2016}
Therefore, the zeroth-order Hamiltonian is independent of $s$ and we have
\begin{equation}
\bH^{(0)}(s) = \bH^{(0)}(0).
@ -577,7 +577,7 @@ while the dynamic part of the self-energy [see Eq.~\eqref{eq:srg_sigma}] tends t
\lim_{s\to\infty} \widetilde{\bSig}(\omega; s) = \bO.
\end{equation}
Therefore, the SRG flow continuously transforms the dynamical self-energy $\widetilde{\bSig}(\omega; s)$ into a static correction $\widetilde{\bF}^{(2)}(s)$.
As illustrated in Fig.~\ref{fig:flow} (magenta curve), this transformation is done gradually starting from the states that have the largest denominators in Eq.~\eqref{eq:static_F2}.
As illustrated in Fig.~\ref{fig:flow} (magenta curve), this transformation is done gradually starting from the states that have \ant{the largest denominators} in Eq.~\eqref{eq:static_F2}.
For a fixed value of the energy cutoff $\Lambda$, if $\abs*{\Delta_{pr}^{\nu}} \gg \Lambda$, then $W_{pr}^{\nu} e^{-(\Delta_{pr}^{\nu})^2 s} \approx 0$, meaning that the state is decoupled from the 1h and 1p configurations, while, for $\abs*{\Delta_{pr}^{\nu}} \ll \Lambda$, we have $W_{pr}^{\nu}(s) \approx W_{pr}^{\nu}$, that is, the state remains coupled.