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使用scipy.integrate.odeint在调用(可能是错误的Dfun类型)上完成的过多工作

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我'm writing a code in Python that predicts the energy levels of Hydrogen which I will use as a template for research into quarkonium energy levels. I' m使用 scipy.integrate.odeint() 函数来解决Shroedinger方程,它可以在低至1818992_的较低能量水平下正常工作 . 我不需要超越它,但odeint会返回 Excess work done on this call (perhaps wrong Dfun type). ,这只会鼓励我扩展我能预测的内容 .

我正在使用的Shroedinger方程替换是:

u'' - (l*(l+1)/r**2 - 2mu_e(E-V_emag(r))) * u = 0
=>
u' = v
v' = ((l*(l+1))/(r**2) - 2.0*mu_e*(E - V_emag(r)))*u

然后我在它上面使用 scipy.integrate.odeint() 并迭代能量并使用我定义的其他函数来评估结果中的转折点和节点 . 我找到能量水平的方法是找到尽可能低的E值,其中转折点和节点的数量与它应该的相匹配;然后将 L 递增1并找到新的地面能量,例如如果 L=0 我会找到 n=1 能量,如果 L=3 ,我会找到 n=2 能量 .

一旦代码增加到 L=7 ,它就不会返回任何有用的内容 . r 的范围已经延长,但我已经阅读了关于雅各比人的内容 . 我已经确定了它们是什么,或者我还需要它们 . 有任何想法吗?

def v_emag(r):
    v = -alpha/r
    return v

def s_e(y,r,l,E): #Shroedinger equation for electromagntism
    x = numpy.zeros_like(y)
    x[0] = y[1]
    x[1] = ((l*(l+1))/(r**2) - 2.0*mu_e*(E - V_emag(r)))*y[0]
    return x

def i_s_e(l,E,start=0.001,stop=None,step=(0.005*bohr)):
    if stop is None:
        stop = ((l+1)*30-10)*bohr
    r = numpy.arange(start,stop,step)
    y = odeint(s_e,y0,r,args=(l,E))
    return y

def inormalise_e(l,E,start=0.001,stop=None,step=(0.005*bohr)):
    if stop is None:
        stop = ((l+1)*30-10)*bohr
    r = numpy.arange(start,stop,step)
    f = i_s_e(l,E,start,stop,step)[:,0]
    f2 = f**2
    area = numpy.trapz(f2,x=r)
    return f/(numpy.sqrt(area))

def inodes_e(l,E,start=0.001,stop=None,step=(0.005*bohr)):
    if stop is None:
        stop = ((l+1)*30-10)*bohr
    x = i_s_e(l,E,start,stop,step)
    r = numpy.arange(start,stop,step)
    k=0
    for i in range(len(r)-1):
        if x[i,0]*x[i+1,0] < 0: #If u value times next u value <0,
            k+=1               #crossing of u=0 has occured, therefore count node
    return k

def iturns_e(l,E,start=0.001,stop=None,step=(0.005*bohr)):
    if stop is None:
        stop = ((l+1)*30-10)*bohr
    x = i_s_e(l,E,start,stop,step)
    r = numpy.arange(start,stop,step)
    k = 0
    for i in range(len(r)-1):
        if x[i,1]*x[i+1,1] < 0: #If du/dr value times next du/dr value <0,
            k=k+1               #crossing of du/dr=0, therefore a maximum/minimum
    return k

l = 0
while l < 10:    #The ground state for a system with a non-zero angular momentum will
    E1 = -1.5e-08    #be the energy level of principle quantum number l-1, therefore
    E3 = 0           #by changing l, we can find n by searching for the ground state
    E2 = 0.5*(E1+E3)
    i = 0
    while i < 40:
        N1 = inodes_e(l,E1)
        N2 = inodes_e(l,E2)
        N3 = inodes_e(l,E3)
        T1 = iturns_e(l,E1)
        T2 = iturns_e(l,E2)
        T3 = iturns_e(l,E3)
        if N1 != N2:# and T1 != T2: #Looks in lower half first, therefore will tend to ground state
            E3 = E2
            E2 = 0.5*(E1+E3)
        elif N2 != N3:# and T2 != T3:
            E1 = E2
            E2 = 0.5*(E1+E3)
        else:
            print "Can't find satisfactory E in range"
            break

        i += 1
    x = inormalise_e(l,E2)
    if x[((l+1)**2)/0.005] > (x[2*((l+1)**2)/0.005]) and iturns_e(l,E2+1e-20)==1:
        print 'Energy of state: n =',(l+1),'is: ',(E2*(10**9)),'eV'
        l += 1
    else:
        E1 = E2+10e-20

1 Answer

  • 0

    我不完全确定你的 while i<40: 循环是做什么的,所以如果我错了,也许你可以更正以下内容 .

    如果您希望此系统的波函数为某个 n, l ,则可以将能量计算为E = RH / n ^ 2,其中RH是里德堡常数,因此您无需计算节点数 . 如果确实需要对节点进行计数,那么对应于 (n,l) 的数字是 n-l-1 ,因此您可以改变E并观察固定l的节点数量变化 .

    主要的问题,在我看来,你的r范围不够大,无法包含所有节点(对于大n~l),并且 odeint 不知道远离其他(非物理)渐近解决方案,psi~exp(cr),所以在某些条件下将psi发送到大于r的±无穷大 .

    如果它有用的话,这就是我想出来找到SE方程的数值解:你需要根据 n,l 改变 r -range但是要避免上述问题(例如,如果你要求 n, l = 10, 9 ,看看会发生什么) .

    import numpy as np
    import scipy as sp
    from scipy.integrate import odeint
    
    m_e, m_p, hbar = sp.constants.m_e, sp.constants.m_p, sp.constants.hbar
    mu_e = m_e*m_p/(m_e + m_p)
    bohr = sp.constants.physical_constants['Bohr radius'][0]
    Rinfhc = sp.constants.physical_constants['Rydberg constant times hc in J'][0]
    RHhc = Rinfhc * mu_e / m_e
    
    fac = sp.constants.e**2/4/sp.pi/sp.constants.epsilon_0
    
    def V(r):
        return -fac/r
    
    def deriv(y, r, l, E):
        y1, y2 = y
        dy1dr = y2
        dy2dr = -2*y2/r - (2*mu_e/hbar**2*(E - V(r)) - l*(l+1)/r**2)*y1
        return dy1dr, dy2dr
    
    def solveSE(l, E, y0):
        rstep = 0.001 * bohr
        rmin = rstep
        rmax = 200*l * bohr         # 
        r = np.arange(rmin, rmax, rstep)
        y, dydt = odeint(deriv, y0, r, args=(l,E)).T
        return r, y, dydt
    
    n = 10
    l = 2
    y0 = (bohr, -bohr)
    E = -RHhc / n**2
    r, psi, dpsi_dr = solveSE(l, E, y0)
    
    import pylab
    pylab.plot(r, psi)
    pylab.show()
    

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