modificatons in Sec III

Manuscript/SRGGW.tex

@@ -310,11 +310,10 @@ The similarity renormalization group method aims at continuously transforming a
 Therefore, the transformed Hamiltonian
 \begin{equation}
   \label{eq:SRG_Ham}
-  \bH(s) = \bU(s) \, \bH \, \bU^\dag(s),
+  \bH(s) = \bU(s) \, \bH \, \bU^\dag(s)
 \end{equation}
-depends on a flow parameter $s$, such that $\bH(s=0)$ is the initial untransformed Hamiltonian and $\bH(s=\infty)$ is the (block)-diagonal Hamiltonian.
-An evolution equation for $\bH(s)$ can be easily obtained by deriving Eq.~\eqref{eq:SRG_Ham} with respect to $s$.
-This gives the flow equation
+depends on a flow parameter $s$, such that $\bH(s=0)$ is the initial untransformed Hamiltonian and $\bH(s=\infty)$ is the (block-)diagonal Hamiltonian.
+An evolution equation for $\bH(s)$ can be easily obtained by differentiating Eq.~\eqref{eq:SRG_Ham} with respect to $s$, yielding the flow equation
 \begin{equation}
   \label{eq:flowEquation}
   \dv{\bH(s)}{s} = \comm{\boldsymbol{\eta}(s)}{\bH(s)},
@@ -323,8 +322,8 @@ where $\boldsymbol{\eta}(s)$, the flow generator, is defined as
 \begin{equation}
 \boldsymbol{\eta}(s) = \dv{\bU(s)}{s}	\bU^\dag(s)	= - \boldsymbol{\eta}^\dag(s).
 \end{equation}
-To solve this equation at a cost inferior to the one of diagonalizing the initial Hamiltonian, one needs to introduce an approximation for $\boldsymbol{\eta}(s)$.
-Before defining such an approximation, we need to define what are the blocks to suppress to obtain a block-diagonal Hamiltonian.
+To solve this equation at a lower cost than the one of diagonalizing the initial Hamiltonian, one must introduce an approximate form for $\boldsymbol{\eta}(s)$.
+\titou{Before defining such an approximation, we need to define what are the blocks to suppress to obtain a block-diagonal Hamiltonian.
 The Hamiltonian is separated into two parts as
 \begin{equation}
   \bH(s) = \underbrace{\bH^\text{d}(s)}_{\text{diagonal}} + \underbrace{\bH^\text{od}(s)}_{\text{off-diagonal}},
@@ -332,9 +331,10 @@ The Hamiltonian is separated into two parts as
 and, by definition, we have the following condition on $\bH^\text{od}$
 \begin{equation}
   \bH^\text{od}(s=\infty) = \boldsymbol{0}.
-\end{equation}
+\end{equation}}
+\PFL{Move this part at the start of the section.}
 
-In this work, we will use Wegner's canonical generator which is defined as \cite{Wegner_1994}
+In this work, we consider Wegner's canonical generator which is defined as \cite{Wegner_1994}
 \begin{equation}
   \boldsymbol{\eta}^\text{W}(s) = \comm{\bH^\text{d}(s)}{\bH(s)} = \comm{\bH^\text{d}(s)}{\bH^\text{od}(s)}.
 \end{equation}
@@ -343,21 +343,21 @@ If this generator is used, the following condition is verified  \cite{Kehrein_20
   \label{eq:derivative_trace}
   \dv{s}\text{Tr}\left[ \bH^\text{od}(s)^2 \right] \leq 0
 \end{equation}
-which implies that the matrix elements of the off-diagonal part will decrease in a monotonic way.
-Even more, the coupling coefficients associated with the highest energy determinants are removed first as will be evidenced by the perturbative analysis after.
-The main flaw of this generator is that it generates a stiff set of ODE which is therefore difficult to solve numerically. \ant{ref}
-However, here we will not tackle the full SRG problem but only consider analytical low-order perturbative expressions so we will not be affected by this problem. \cite{Evangelista_2014, Hergert_2016}
+which implies that the matrix elements of the off-diagonal part decrease in a monotonic way throughout the transformation.
+Even more, the coupling coefficients associated with the highest-energy determinants are removed first as we shall evidence in the perturbative analysis below.
+The main drawback of this generator is that it generates a stiff set of ODE which is therefore difficult to solve numerically. \ant{ref}
+However, here we will not tackle the full SRG problem but only consider analytical low-order perturbative expressions so we will not be affected by this problem. \cite{Evangelista_2014,Hergert_2016}
 
 Let us now perform the perturbative analysis of the SRG equations.
 For $s=0$, the initial problem is
 \begin{equation}
-  \bH(0) = \bH^\text{d}(0) + \lambda ~ \bH^\text{od}(0)
+  \bH(0) = \bH^\text{d}(0) + \lambda \bH^\text{od}(0)
 \end{equation}
 where $\lambda$ is the usual perturbation parameter and the off-diagonal part of the Hamiltonian has been defined as the perturber.
 For finite values of $s$, we have the following perturbation expansion of the Hamiltonian
 \begin{equation}
   \label{eq:perturbation_expansionH}
-  \bH(s) = \bH^{(0)}(s) + \lambda ~ \bH^{(1)}(s) + \lambda^2 \bH^{(2)}(s) + \dots
+  \bH(s) = \bH^{(0)}(s) + \lambda ~ \bH^{(1)}(s) + \lambda^2 \bH^{(2)}(s) + \cdots
 \end{equation}
 Hence, the generator $\boldsymbol{\eta}(s)$ admits a perturbation expansion as well.
 Then, one can collect order by order the terms in Eq.~\eqref{eq:flowEquation} and solve analytically the low-order differential equations.
@@ -369,7 +369,7 @@ Then, one can collect order by order the terms in Eq.~\eqref{eq:flowEquation} an
 
 Finally, the SRG formalism exposed above will be applied to $GW$.
 The first step is to define the diagonal and off-diagonal parts of the $GW$ effective Hamiltonian.
-As hinted at the end of section~\ref{sec:gw}, the diagonal and off-diagonal parts will be defined as
+As hinted at the end of Sec.~\ref{sec:gw}, the diagonal and off-diagonal parts are defined as
 \begin{align}
   \label{eq:diag_and_offdiag}
   \bH^\text{d}(s) &=