modifs
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@ -55,6 +55,8 @@
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\newcommand{\trashant}[1]{\textcolor{teal}{\sout{#1}}}
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\newcommand{\trashant}[1]{\textcolor{teal}{\sout{#1}}}
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\newcommand{\ANT}[1]{\ant{(\underline{\bf ANT}: #1)}}
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\newcommand{\ANT}[1]{\ant{(\underline{\bf ANT}: #1)}}
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\DeclareMathOperator{\sgn}{sgn}
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% addresses
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% addresses
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\newcommand{\LCPQ}{Laboratoire de Chimie et Physique Quantiques (UMR 5626), Universit\'e de Toulouse, CNRS, UPS, France}
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\newcommand{\LCPQ}{Laboratoire de Chimie et Physique Quantiques (UMR 5626), Universit\'e de Toulouse, CNRS, UPS, France}
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@ -363,7 +365,7 @@ For finite values of $s$, we have the following perturbation expansion of the Ha
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\label{eq:perturbation_expansionH}
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\label{eq:perturbation_expansionH}
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\bH(s) = \bH^{(0)}(s) + \lambda ~ \bH^{(1)}(s) + \lambda^2 \bH^{(2)}(s) + \cdots.
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\bH(s) = \bH^{(0)}(s) + \lambda ~ \bH^{(1)}(s) + \lambda^2 \bH^{(2)}(s) + \cdots.
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\end{equation}
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\end{equation}
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Hence, the generator $\boldsymbol{\eta}(s)$ admits a perturbation expansion of the same form.
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Hence, the generator $\boldsymbol{\eta}(s)$ admits a similar perturbation expansion.
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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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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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%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%
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@ -442,7 +444,7 @@ where $\bW^{(0)}(s)= \bV^{(0)}(s) \bU$.
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The matrix elements of $\bU$ and $\bD^{(0)}$ are
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The matrix elements of $\bU$ and $\bD^{(0)}$ are
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\begin{align}
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\begin{align}
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U_{p\nu,q\mu} &= \delta_{pq} \bX_{\nu\mu} \\
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U_{p\nu,q\mu} &= \delta_{pq} \bX_{\nu\mu} \\
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D_{p\nu,q\mu}^{(0)} &= \left(\epsilon_p + \text{sign}(\epsilon_p-\epsilon_\text{F})\Omega_\nu\right)\delta_{pq}\delta_{\nu\mu}
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D_{p\nu,q\mu}^{(0)} &= \left(\epsilon_p + \sgn(\epsilon_p-\epsilon_\text{F})\Omega_\nu\right)\delta_{pq}\delta_{\nu\mu}
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\end{align}
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\end{align}
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where $\epsilon_\text{F}$ is the Fermi level.
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where $\epsilon_\text{F}$ is the Fermi level.
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Note that the matrix $\bU$ is also used in the downfolding process of Eq.~\eqref{eq:GWlin}. \cite{Bintrim_2021}
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Note that the matrix $\bU$ is also used in the downfolding process of Eq.~\eqref{eq:GWlin}. \cite{Bintrim_2021}
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@ -475,7 +477,7 @@ Once again the two first equations are easily solved
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and the first order coupling elements are given by (up to a multiplication by $\bU^{-1}$)
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and the first order coupling elements are given by (up to a multiplication by $\bU^{-1}$)
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\begin{align}
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\begin{align}
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W_{p,q\nu}^{(1)}(s) &= W_{p,q\nu}^{(1)}(0) e^{- (F_{pp}^{(0)} - D_{q\nu,q\nu}^{(0)})^2 s} \\
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W_{p,q\nu}^{(1)}(s) &= W_{p,q\nu}^{(1)}(0) e^{- (F_{pp}^{(0)} - D_{q\nu,q\nu}^{(0)})^2 s} \\
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&= W_{p,q\nu}^{(1)}(0) e^{- (\epsilon_p - \epsilon_q - \text{sign}(\epsilon_q-\epsilon_F)\Omega_\nu)^2 s} \notag
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&= W_{p,q\nu}^{(1)}(0) e^{- [\epsilon_p - \epsilon_q - \sgn(\epsilon_q-\epsilon_F)\Omega_\nu]^2 s} \notag
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\end{align}
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\end{align}
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At $s=0$ the elements $W_{p,q\nu}^{(1)}(0)$ are equal to the two-electron screened integrals defined in Eq.~\eqref{eq:GW_sERI} while for $s\to\infty$ they go to zero.
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At $s=0$ the elements $W_{p,q\nu}^{(1)}(0)$ are equal to the two-electron screened integrals defined in Eq.~\eqref{eq:GW_sERI} while for $s\to\infty$ they go to zero.
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Therefore, $W_{p,q\nu}^{(1)}(s)$ are renormalized two-electrons screened integrals.
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Therefore, $W_{p,q\nu}^{(1)}(s)$ are renormalized two-electrons screened integrals.
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@ -508,12 +510,12 @@ This can be solved by simple integration along with the initial condition $\bF^{
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F_{pq}^{(2)}(s) = \sum_{r\mu} \frac{\Delta_{pr\mu}+ \Delta_{qr\mu}}{\Delta_{pr\mu}^2 + \Delta_{qr\mu}^2} W_{p,r\mu} W^{\dagger}_{r\mu,q} \times \\
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F_{pq}^{(2)}(s) = \sum_{r\mu} \frac{\Delta_{pr\mu}+ \Delta_{qr\mu}}{\Delta_{pr\mu}^2 + \Delta_{qr\mu}^2} W_{p,r\mu} W^{\dagger}_{r\mu,q} \times \\
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\left(1 - e^{-(\Delta_{pr\mu}^2 + \Delta_{qr\mu}^2) s}\right),
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\left(1 - e^{-(\Delta_{pr\mu}^2 + \Delta_{qr\mu}^2) s}\right),
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\end{multline}
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\end{multline}
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with $\Delta_{prv} = \epsilon_p - \epsilon_r - \text{sign}(\epsilon_r-\epsilon_F)\Omega_v$.
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with $\Delta_{prv} = \epsilon_p - \epsilon_r - \sgn(\epsilon_r-\epsilon_F)\Omega_v$.
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At $s=0$, this second-order correction is null while for $s\to\infty$ it tends towards the following static limit
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At $s=0$, this second-order correction is null while for $s\to\infty$ it tends towards the following static limit
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\begin{equation}
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\begin{equation}
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\label{eq:static_F2}
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\label{eq:static_F2}
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F_{pq}^{(2)}(\infty) = \sum_{r,v} \frac{\Delta_{prv}+ \Delta_{qrv}}{\Delta_{prv}^2 + \Delta_{qrv}^2} W_{p,r\mu} W^{\dagger}_{r\mu,q}.
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F_{pq}^{(2)}(\infty) = \sum_{r\nu} \frac{\Delta_{pr\nu}+ \Delta_{qr\nu}}{\Delta_{pr\nu}^2 + \Delta_{qr\nu}^2} W_{p,r\nu} W^{\dagger}_{r\nu,q}.
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\end{equation}
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\end{equation}
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Note that in the $s\to\infty$ limit the dynamic part of the self-energy [see Eq.~\eqref{eq:srg_sigma}] tends to zero.
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Note that in the $s\to\infty$ limit the dynamic part of the self-energy [see Eq.~\eqref{eq:srg_sigma}] tends to zero.
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Therefore, the SRG flow transforms the dynamic part of $\bSig(\omega)$ into a static correction.
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Therefore, the SRG flow transforms the dynamic part of $\bSig(\omega)$ into a static correction.
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