割线法rungekutta schrodinger

Question

所以我试图使用正割方法找到一个基态能量，但是到目前为止我已经尝试了多种方法，所有这些都返回了错误。 到目前为止我的代码：

import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
###First importing the modules as always

Me = 9.1094*10**-31
hbar = 1.0546*10**-34
ec = 1.6022*10**-19
###Defining some constants we will need, electron mass, h bar and electron    charge

a = 5*10**-11    ###Halfwidth of the well
N = 1000
xsteps = (2*a)/1000
xpoints = np.arange(-a,a+xsteps,xsteps)   ###Array of calculation points inside the well
###using a+xsteps so that it ends at x=a, and not at x=a-step

def V(x):
    if abs(x)<a:
        V=0
    else:
        V=10**15
    return V     ###Can be modified to add other variables, for now should simply return V=0 for all steps

def s(r,x,E):      ###Where r is a vector comprised of phi and psi
    psi=r[0]
    phi=r[1]
    fpsi=phi      ###function of dpsi/dx
    fphi=((2*Me)/hbar**2)*(V(x)-E)*psi    ###function of dphi/dx
    return np.array([fpsi,fphi])
r=np.array([0,1])    ###Initial conditions

def RungeKuttaSchrod1(r,xpoints,E):
    psipoints = [] # getting empty arrays ready for the vector variables
    phipoints = []
    for x in xpoints:
        psipoints.append(r[0])
        phipoints.append(r[1])
        k1 = xsteps*s(r,x,E) 
        k2 = xsteps*s(r+0.5*k1, x+0.5*xsteps,E)
        k3 = xsteps*s(r+0.5*k2, x+0.5*xsteps,E)
        k4 = xsteps*s(r+k3, x+xsteps,E)
        r = r + (k1 + 2*k2 + 2*k3 + k4)/6
    return np.array([psipoints, phipoints])   

在尝试找到与边界条件匹配的E之前的所有原始代码（psi = 0，在x = -a和x = a）

到目前为止，我已尝试过这两种方法并且都失败了：

err = 1       ###error variable set
tolerance = ec/1000
E1=25*ec
E2=50*ec    ###intial guesses for the energy

#Secant method for this set of equations
while err > tolerance: 
        E3 = E2 - RungeKuttaSchrod1(r,xpoints,E2)*(E2-E1)/(RungeKuttaSchrod1(r,xpoints,E2)-RungeKuttaSchrod1(r,xpoints,E1))
        print "current best guess of ground state energy", E3
        err = abs(E2-E1)
        E1 = E2 #resetting E1 and E2 for the next iteration
        E2 = E3

还有这个：

def fphi2(x,psi,E):
    fphi2=((2*Me)/hbar**2)*(V(x)-E)*psi
    return fphi2

psi = 0 ###initial condition


err = 1.0  ###intialising error variable

#Secant method for this set of equations
while err > tolerance:
    for x in xpoints:
        E3 = E2 - RungeKuttaFphi(xpoints,psi,E2)*(E2-E1)/(RungeKuttaFphi(xpoints,psi,E2)-RungeKuttaFphi(xpoints,psi,E1))
        print "current best guess of ground state energy", E3
        err = abs(E2-E1)
        E1 = E2 #resetting E1 and E2 for the next iteration
        E2 = E3

这些都没有用，有人有什么想法吗？