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Pierre-Francois Loos 2020-12-01 15:41:37 +01:00
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@ -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 Mathematically, Eq.~\eqref{eq:E_expansion} corresponds to a Taylor series expansion of the exact energy
around the reference system $\lambda = 0$. around the reference system $\lambda = 0$.
The energy of the target ``physical'' system is recovered at the point $\lambda = 1$. 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 $\rc$. However, like all series expansions, Eq.~\eqref{eq:E_expansion} has a radius of convergence $\rc$.
When $\rc \le 1$, the Rayleigh--Sch\"{r}odinger expansion will diverge When $\rc < 1$, the Rayleigh--Schr\"{o}dinger expansion will diverge
for the physical system. for the physical system.
The value of $\rc$ can vary significantly between different systems and strongly depends on the particular decomposition 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} 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 ($\e^{\i\pi/4}$, $\e^{-\i\pi/4}$, $\e^{\i3\pi/4}$, 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} 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. 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 $\lambda = -1$. 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 Like the exact system in Sec.~\ref{sec:example}, the perturbation energy $E(\lambda)$ represents