modificatons in Sec III

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Pierre-Francois Loos 2023-01-31 16:11:36 +01:00
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@ -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}
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) &=