Núñez Marco A
Departamento de Física, Universidad Autónoma Metropolitana, Iztapalapa, Apartado Postal 55-534, CP 09340 México, Distrito Federal, Mexico.
Phys Rev E Stat Nonlin Soft Matter Phys. 2003 Jul;68(1 Pt 2):016703. doi: 10.1103/PhysRevE.68.016703. Epub 2003 Jul 25.
A perturbation approach to compute the bound states of the Schrödinger equation HPsi=EPsi with H0+lambdaV and Psi(x=+/- infinity)=0 is studied. The approach involves solving the corresponding Dirichlet problem H(R)Psi(R)=E(R)Psi(R) on a finite interval [-R,R] by the Rayleigh-Schrödinger perturbation theory (RSPT). The method is based on the fact that E(R),Psi(R) converge to E,Psi as R--> infinity. The model problems to study the summability properties of the RSPT series E(R)= sum(infinity)(k=0)E((k))(R)lambda(k) are the anharmonic oscillators H=p(2)+x(2)+lambda(x)(2M), with M=2,3,4 for which the RSPT produces strongly divergent series E= sum(infinity)(k=0)E((k))lambda(k). The summation of the latter series with large lambda for the octic case is considered as an extremely challenging summation problem, in part, since it was rigorously proven that the Padé approximants cannot converge and the two-point Padé approximants, which combine information of the renormalized weak coupling and strong coupling expansions, give relatively good results. The calculations of this work show that the ordinary Padé approximants from the sole un-normalized E(R) series for the octic oscillator give accurate results with small or large lambda. The coefficients E((k))(R) are calculated with the eigenvalue series of an operator H(Rn), whose resolvent converges to that of H(R) as n--> infinity. The Padé approximants of the RSPT eigenfunction series Psi(R)= sum(infinity)(k=0)psi((k))(R)lambda(k) also provide accurate results for the octic oscillator.
研究了一种微扰方法，用于计算薛定谔方程$H\Psi = E\Psi$（其中$H = H_0+\lambda V$且$\Psi(x = \pm\infty)=0$）的束缚态。该方法涉及通过瑞利 - 薛定谔微扰理论（RSPT）在有限区间$[-R,R]$上求解相应的狄利克雷问题$H(R)\Psi(R)=E(R)\Psi(R)$。此方法基于这样一个事实，即当$R\to\infty$时，$E(R)$、$\Psi(R)$收敛于$E$、$\Psi$。用于研究RSPT级数$E(R)=\sum_{k = 0}^{\infty}E^{(k)}(R)\lambda^k$可和性性质的模型问题是非谐振子$H = p^2 + x^2+\lambda x^{2M}$，对于$M = 2,3,4$的情况，RSPT会产生强烈发散的级数$E=\sum_{k = 0}^{\infty}E^{(k)}\lambda^k$。对于八次情况，当$\lambda$较大时对后一级数求和被认为是一个极具挑战性的求和问题，部分原因在于已严格证明帕德逼近不能收敛，而结合了重整化弱耦合和强耦合展开信息的两点帕德逼近给出了相对较好的结果。这项工作的计算表明，对于八次振子，仅来自未归一化$E(R)$级数的普通帕德逼近在$\lambda$较小或较大时都能给出准确结果。系数$E^{(k)}(R)$是通过算子$H(R_n)$的特征值级数计算得到的，当$n\to\infty$时，其预解式收敛于$H(R)$的预解式。RSPT本征函数级数$\Psi(R)=\sum_{k = 0}^{\infty}\psi^{(k)}(R)\lambda^k$的帕德逼近对于八次振子也能提供准确结果。
