From d964d2cd01147349b5d3ab7ac8f785f497b143de Mon Sep 17 00:00:00 2001 From: Pierre-Francois Loos Date: Tue, 1 Dec 2020 15:41:37 +0100 Subject: [PATCH] Done with IIC --- Manuscript/EPAWTFT.tex | 8 ++++---- 1 file changed, 4 insertions(+), 4 deletions(-) diff --git a/Manuscript/EPAWTFT.tex b/Manuscript/EPAWTFT.tex index 37a977b..984b0ce 100644 --- a/Manuscript/EPAWTFT.tex +++ b/Manuscript/EPAWTFT.tex @@ -388,8 +388,8 @@ setting $\lambda = 1$ then yields approximate solutions to Eq.~\eqref{eq:SchrEq} Mathematically, Eq.~\eqref{eq:E_expansion} corresponds to a Taylor series expansion of the exact energy around the reference system $\lambda = 0$. The energy of the target ``physical'' system is recovered at the point $\lambda = 1$. -However, like all series expansions, the Eq.~\eqref{eq:E_expansion} has a radius of convergence $\rc$. -When $\rc \le 1$, the Rayleigh--Sch\"{r}odinger expansion will diverge +However, like all series expansions, Eq.~\eqref{eq:E_expansion} has a radius of convergence $\rc$. +When $\rc < 1$, the Rayleigh--Schr\"{o}dinger expansion will diverge for the physical system. The value of $\rc$ can vary significantly between different systems and strongly depends on the particular decomposition of the reference and perturbation Hamiltonians in Eq.~\eqref{eq:SchrEq-PT}.\cite{Mihalka_2017b} @@ -418,9 +418,9 @@ This divergence occurs because $f(x)$ has four singularities in the complex ($\e^{\i\pi/4}$, $\e^{-\i\pi/4}$, $\e^{\i3\pi/4}$, and $\e^{-\i3\pi/4}$) with a modulus equal to $1$, demonstrating that complex singularities are essential to fully understand the series convergence on the real axis.\cite{BenderBook} -The radius of convergence for the perturbation series Eq.~\eqref{eq:E_expansion} is therefore dictated by the magnitude $\abs{\lambda_c}$ of the +The radius of convergence for the perturbation series Eq.~\eqref{eq:E_expansion} is therefore dictated by the magnitude $r_c = \abs{\lambda_c}$ of the singularity in $E(\lambda)$ that is closest to the origin. -Note that when $\lambda = \lambda_c$, one cannot \textit{a priori} predict if the series is convergent or not. +Note that when $\abs{\lambda} = r_c$, one cannot \textit{a priori} predict if the series is convergent or not. For example, the series $\sum_{k=1}^\infty \lambda^k/k$ diverges at $\lambda = 1$ but converges at $\lambda = -1$. Like the exact system in Sec.~\ref{sec:example}, the perturbation energy $E(\lambda)$ represents