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infinite potential well python

Poised Parrot answered on October 17, 2022 Popularity 1/10 Helpfulness 1/10

answer infinite potential well python

infinite potential well python

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from pylab import *
from scipy.integrate import odeint #A differential equation solver
from scipy.optimize import brentq #A root finding algorithm
import matplotlib.pyplot as plt
import numpy as np #An immensely useful library for manipulating arrays and lists (i.e matrices and vectors)

def V(x):
    global a #Means variable 'a' is available to any part of the code, thus we can set 'a' later and our V(x) will be aware of it.
    if abs(x) > a:
        return 100000
    else:
        return 0
        

def D1D2(psi, x):
    D1 = psi[1] #First derivative
    D2 = 2.0*(V(x) - E)*psi[0] #Second derivative
    return array([D1, D2])

def Wave_function(energy):
    global psi
    global E
    E = energy                #We need this so as to pass it to DD
    psi = odeint(D1D2, psi0, x)
    return psi[-1,0]
    
N = 10000                   # number of discrete points on the x-axis
a = 2                       # Set the width of the well
E = 0                       # initialize the global variable E
psi = np.zeros([N,2])       # Wave function values and its derivative (psi and psi')
psi0 = array([0,1])         # Wave function initial states
x = linspace(-a, 0, N)      # the points on the x-axis between left wall of the well and x=0
Emax = 100.0 
Ebottom = 0.0                # let us only look for solutions between Ebottom =0 and Emax=100
en = linspace(Ebottom, Emax, 1000) # A number of discrete energies where we will calculate psi(x=0)
psi_b = []                  # vector of wave function at x = 0 for all of the energies in en
for e1 in en:
    psi_b.append(Wave_function(e1))     # Determine psi(x=0) for every e1 in en
    
plt.plot(en, psi_b, 'r-')
plt.xlabel(r'$E$',size=20)
plt.ylabel(r'$\psi(x=0)$',size=20)
plt.grid(True)
plt.show()

def find_all_zeroes(x,y): #Returns list of values of x for which y(x)=0
    all_zeroes = []
    s = sign(y)
    for i in range(len(y)-1):
        if s[i]+s[i+1] == 0:
            zero = brentq(Wave_function, x[i], x[i+1])
            all_zeroes.append(zero)
    return all_zeroes
    
E_zeroes = find_all_zeroes(en, psi_b)   # now find the precise energies where psi(x=0) = 0 
print( "Energies for the bound states are: ")
for E in E_zeroes:
        print( '%11.7f' % E)
        
x = linspace(-a, a, N)
h=2*a/(N-1)
for E in E_zeroes[0:4]:
        Wave_function(E)
        psi[:,0]=psi[:,0]/sqrt(h*np.sum(np.square(psi[:,0])))
        plt.plot(x, psi[:,0], label="E = %.2f"%E)
        
plt.xlabel(r'$x$',size=20)
plt.ylabel(r'$\psi(x)$',size=20)
plt.legend()
plt.grid(True)
plt.show()

Popularity 1/10 Helpfulness 1/10 Language python

Source: www.physicsforums.com

Tags: infinite python

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Contributed on Oct 17 2022

Poised Parrot

0 Answers Avg Quality 2/10

infinite potential well python

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