Question

我有一个Python代码，可以为不同的潜力绘制波函数和能量：

    # -*- coding: cp1250 -*-
from math import *
from scipy.special import *
from pylab import *
from scipy.linalg import *

firebrick=(178./255.,34./255.,34./255.)
indianred=(176./255.,23./255.,31./255.)
steelblue=(70./255.,130./255.,180./255.)
slategray1=(198./255.,226./255.,255./255.)
slategray4=(108./255.,123./255.,139./255.)
lavender=(230./255.,230./255.,230./255.)
cobalt=(61./255.,89./255.,171./255.)
midnightblue=(25./255.,25./255.,112./255.)
forestgreen=(34./255.,139./255.,34./255.)

#definiranje mreze
Nmesh=512
L=4.0
dx=L/Nmesh
Xmax=L
x=arange(-L,L+0.0001,dx)
Npts=len(x)
numwav=0 #redni broj valne funkcije koji se iscrtava

V=zeros([Npts],float)
for i in range(Npts):
    V[i]=x[i]**50

a=zeros([2,Npts-2],float)
wave=zeros([Npts],float)

wave1=zeros([Npts],float)
encor=3.0/4*(3.0/4)**(1.0/3)

#numericko rjesenje
for i in range(1,Npts-1,1):
    a[0,i-1]= 1.0/dx**2+V[i]     #dijagonalni elementi
    a[1,i-1]=-1.0/dx**2/2        #elementi ispod dijagonale
a[1,Npts-3]=-99.0                #element se ne koristi
eig,vec=eig_banded(a,lower=1)    #rutina koja dijagonalizira tridijagonalnu matricu

for i in range(1,Npts-1,1):
    wave[i]=vec[i-1,numwav]
wave[0]=0.0             #valna funkcija u prvoj tocki na mrezi ima vrijednost nula
wave[Npts-1]=0.0        #valna funkcija u zadnjoj tocki na mrezi ima vrijednost nula

for i in range(1,Npts-1,1):
    wave1[i]=(2.0/pi*(3.0/4)**(1.0/3))**0.25*exp(-(3.0/4)**(1.0/3)*x[i]**2)
wave1[0]=0.0             #valna funkcija u prvoj tocki na mrezi ima vrijednost nula
#wave1[Npts-1]=0.0        #valna funkcija u zadnjoj tocki na mrezi ima vrijednost nula


#wave1=omjer*150*wave1+encor
wave=150*wave+eig[numwav]

#graf potencijala
line=plt.plot(x,V)
plt.setp(line,color='firebrick',linewidth=2)

#crtanje odabranog nivoa i odgovarajuce valne funkcije
plt.axhline(y=eig[numwav],linewidth=2,color='steelblue')
#plt.axhline(y=encor,linewidth=2,color='midnightblue')


#crtanje tocaka valne funkcije
plt.plot(x,wave,"b-",linewidth=2,color='forestgreen')
#plt.plot(x,wave1,"-",linewidth=2,color='indianred')

plt.xlabel(r'$x$',size=14)
plt.ylabel(r'$V(x)$',size=14)
plt.title(r'Valna funkcija i energija 3. pobuđenog stanja za $V(x)=x^{50}$')
plt.axis([-3.0,3.0,-8.0,100.0]) #raspon x i y osi
plt.grid(True)
plt.legend((r'$V(x)$',r'$E_0$',r'$\psi_0$'))
plt.show()

忽略忽略的行，它们对于这种情况并不重要，语言：D

无论如何，我有一个问题。如果我绘制了潜力（V部分），让我们说x^20它会很好，就像这样x^6：

potential1

如果我把潜力说出来，说x^50就变成了这个：

potential2

那么似乎是什么问题？为什么他犯了这么大的错误？它应该是平滑的，从理论上来说，当我到达非常大V(x)=x^p（p→∞）的点p时，潜力应该转到着名的无限方形井，看起来像这样：

potential3

所以我怀疑对于更大的潜力我需要更多的点来在给定的范围内绘制它。那么我应该增加Nmesh（网格）的数量吗？因为他说Npts=len(x) - 他正在服用的点数。我对吗？这似乎合乎逻辑，但我想确定。

感谢您的任何建议和帮助

编辑：我尝试扩展Nmesh，但是在非常大的数字我得到的网格太大，或者存在内存问题。

如果我拍摄，比如2048，我会得到相同的照片，但只是稍微偏移了一点。

Answer 1

对eig_banded()使用select参数：

#!/usr/bin/env python
from __future__ import division
import functools
import math
import sys

from timeit import default_timer as timer

import numpy as np
import matplotlib.pyplot as plt
import scipy.linalg as linalg

def report_time(func):
    @functools.wraps(func)
    def wrapper(*args, **kwargs):
        start = timer()
        try: return func(*args, **kwargs)
        finally:
            print '%s takes %.2f seconds' % (func.__name__, timer()-start)
    return wrapper

@report_time
def calc(Nmesh,POWER,L ,numwav=0):
    #
    dx=L/Nmesh
    x = np.arange(-L,L+0.0001,dx)
    Npts=len(x)

    V = x**POWER

    #
    ai = np.empty((2,Npts))   # ai[:,i] = a[:,i-1]
    ai[0,:] = 1/dx**2 + V     #
    ai[1,:] = -1.0/dx**2/2    #
    ai[1,Npts-2] = -99.0      #
    a = ai[:,1:-1]
    f = report_time(linalg.eig_banded)
    eig, vec = f(a, lower=True,overwrite_a_band=True,
                 select='i',select_range=(0,numwav)
                 ) #

    wave = np.empty(Npts)
    wave[1:-1] = vec[:,numwav]
    wave[0] = 0             #
    wave[Npts-1] = 0        #
    wave = 150*wave + eig[numwav]
    return x, V, wave, eig[numwav]

def main():
    try: numwav = int(sys.argv[1])
    except (IndexError, ValueError):
        numwav = 0
    POWER=100
    L=4.0
    Nmesh = 512
    print 'Nmesh=%d' % Nmesh
    x, V, wave, y = calc(Nmesh, POWER, L,numwav)

    #
    line = plt.plot(x,V)
    plt.setp(line,color='firebrick',linewidth=2)

    #
    plt.plot(x,wave,"b-",linewidth=2,color='forestgreen')

    #
    plt.axhline(y=y,linewidth=2,color='steelblue')

    plt.xlabel(r'$x$',size=14)
    plt.ylabel(r'$V(x)$',size=14)
    plt.title(r'$V(x)=x^{%d}$, ' % POWER)
    plt.axis([-(abs(L)-1), abs(L)-1,min(min(wave), y, min(V))-1, max(max(wave), y)+1]) #
    plt.grid(True)
    plt.legend((r'$V(x)$',r'$E_%d$' % numwav,r'$\psi_%d$' % numwav))
    plt.savefig('V_%d_%d_%d.png' % (Nmesh, POWER, numwav))
    plt.show()

if __name__=="__main__":
    main()

输出

Nmesh=512
eig_banded takes 0.01 seconds
calc takes 0.01 seconds

V(x)=x**100

以下是numwav=4的变体：

numwav

差异的差异。方程给出了奇怪的结果

1 个答案:

输出