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small corrections in Sec III

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Pierre-Francois Loos 2023-02-01 22:39:53 +01:00
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Manuscript/SRGGW.tex
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@ -316,24 +316,20 @@ Therefore, these blocks will be the target of the SRG transformation but before
\label{sec:srg}
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The SRG method aims at continuously transforming a Hamiltonian to a diagonal form, or more often to a block-diagonal form.
\ant{Hence, the first step is to decompose the Hamiltonian into an off-diagonal part which will be suppressed and the remaining diagonal part
The SRG method aims at continuously transforming a general Hamiltonian matrix to its diagonal form, or more often, to a block-diagonal form.
Hence, the first step is to decompose this Hamiltonian matrix
\begin{equation}
\bH = \underbrace{\bH^\text{d}}_{\text{diagonal}} + \underbrace{\bH^\text{od}}_{\text{off-diagonal}}.
\bH = \bH^\text{d} + \bH^\text{od}.
\end{equation}
The SRG transformed Hamiltonian
into an off-diagonal part, $\bH^\text{od}$, that we aim at removing and the remaining diagonal part, $\bH^\text{d}$.
This transformation can be performed continuously via a unitary matrix $\bU(s)$, as follows:
\begin{equation}
\label{eq:SRG_Ham}
\bH(s) = \bU(s) \, \bH \, \bU^\dag(s),
\end{equation}
and the unitary matrix $\bU(s)$ both depend on a flow parameter $s$, such that $\bH(s=0)$ is the initial untransformed Hamiltonian and $\bU(s=0)$ is the identity.
By definition, we also have the following condition on the off-diagonal part
\begin{equation}
\bH^\text{od}(s=\infty) = \boldsymbol{0}.
\end{equation}
Therefore, the flow parameter controls the extent of the decoupling.
}
where $s$ is the so-called flow parameter that controls the extent of the decoupling.
By definition, we have $\bH(s=0) = \bH$ [or $\bU(s=0) = \bI$] and $\bH^\text{od}(s=\infty) = \bO$.
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}
@ -344,36 +340,35 @@ 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 lower cost than the one of diagonalizing the initial Hamiltonian, one must introduce an approximate form for $\boldsymbol{\eta}(s)$.
In this work, we consider Wegner's canonical generator which is defined as \cite{Wegner_1994}
To solve \titou{this equation} at a lower cost than the one associated with the diagonalization of the initial Hamiltonian, one must introduce an approximate form for $\boldsymbol{\eta}(s)$.
In this work, we consider Wegner's canonical generator \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)}.
\boldsymbol{\eta}^\text{W}(s) = \comm{\bH^\text{d}(s)}{\bH(s)} = \comm{\bH^\text{d}(s)}{\bH^\text{od}(s)},
\end{equation}
If this generator is used, the following condition is verified \cite{Kehrein_2006}
which satisfied the following condition \cite{Kehrein_2006}
\begin{equation}
\label{eq:derivative_trace}
\dv{s}\text{Tr}\left[ \bH^\text{od}(s)^2 \right] \leq 0
\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 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.
This implies that the matrix elements of the off-diagonal part decrease in a monotonic way throughout the transformation.
Moreover, 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.
where $\lambda$ is the usual perturbation parameter and the off-diagonal part of the Hamiltonian has been defined as the pertrubation.
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) + \cdots
\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.
Hence, the generator $\boldsymbol{\eta}(s)$ admits a perturbation expansion of the same form.
Then, as performed in Sec.~\ref{sec:srggw}, one can collect order by order the terms in Eq.~\eqref{eq:flowEquation} and solve analytically the low-order differential equations.
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\section{Regularized $GW$ approximation}