diff --git a/Manuscript/SRGGW.tex b/Manuscript/SRGGW.tex index 4ae95be..54ef8c3 100644 --- a/Manuscript/SRGGW.tex +++ b/Manuscript/SRGGW.tex @@ -316,24 +316,20 @@ Therefore, these blocks will be the target of the SRG transformation but before \label{sec:srg} %%%%%%%%%%%%%%%%%%%%%% -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. %%%%%%%%%%%%%%%%%%%%%% \section{Regularized $GW$ approximation}