# coding: utf-8
# # Numerov's Algorithm & Shooting Method: Finite Square Well
#
# ### Time indepedent Schrodinger Equation
#
# In units where $\hbar = 1$, the 1D TISE can be expressed in the form:
#
# \begin{equation}
# \frac{d^2 \psi}{dx^2} = -2m\left(E - V(x) \right) \psi = -g(x) \psi
# \end{equation}
#
# This differential equation can be solved numerically via Numerov's method ([see page 10 - 11](http://www.fisica.uniud.it/~giannozz/Corsi/MQ/LectureNotes/mq-cap1.pdf)). For a 1D spatial grid, the wavefunction at the $n+1$th point along can be approximated by:
#
# \begin{equation}
# \psi_{n+1} = \frac{(12-10f_n) \ \psi_n-f_{n-1}\psi_{n-1}}{f_{n+1}}
# \end{equation}
#
# where:
# \begin{equation}
# f_n \equiv \left( 1 + \frac{\delta x^2}{12}g_n \right), \ \ \ \ \ \ \ g_n = 2m(E-V(x_n))
# \end{equation}
#
# Hence to start Numerov's method we require $\psi_0$ and $\psi_1$, in other words $\psi(x = x_{min})$ and $\psi(x = x_{min} + \delta x)$, where $\delta x$ is the step size.
# ### Shooting Method:
#
# When investigating bound states we require $\psi( x \to \pm \infty) = 0$. However we cannot consider a infinite domain, instead we must choose a big enough domain that setting $\psi( x = x_{min}) = 0$ is a good approximation (and similarly for $x_{max}$).
#
# With Numerov's Method in place, the shooting method can be used to find the energy eigenstates. It goes as follows:
#
# 1. Setting $\psi_0 = 0$ "satisfies" the boundary condition that the wavefunctions must vanish at the boundary
# 2. Since the Schrodinger Equation is linear and homogeneous we are free to set $\psi_1$ to any non-zero constant as multiplying by a constant does not affect the solution. In this case we shall set $\psi_1 = \delta x$.
# 3. Using the Numerov algorithm, $\psi(x)$ can be found. Exponential growth near $x_{max}$ can be observed if the input energy is not near a energy eigenvalue
# 4. A bisection search can be performed to calculate the energy eigenvalue and eigenstate
# #### Bisection Search:
#
# We wish to find a function $f(E) = 0$. First we must find values of $E$ which bracket the solution, that is:
# $f(E_1) > 0 \ $ and $\ f(E_2) > 0$
# Evaluating $f$ at the midpoint $E_3 = \frac{1}{2}(E_1+E_2)$, depending on the result we can rebracket our solution. Hence the solution will be converged about with many iterations. Solutions can converge from both above and below $0$ so your search algorithm should account for this.
#
# The search should be stopped when $\left| \ f(E)\right| < \epsilon$ where $\epsilon$ is a suitably small number.
#
#
#
# Dotted lines show the bracket solutions and the solid lines show the progression of the search to obtain the ground state of the infinite square well
# ## Exercises:
#
# 1. Write a Numerov routine for a finite square well potential of the form: $V = -V_0$ for $\left| \ x \ \right| < a$ elsewhere $V = 0$
# 2. Include a bisection search which calls the Numerov function until a solution is found. Choose large initial energy brackets to ensure a result is within range
# 3. Search for bound states and travelling wave states, by plotting them note their structure.
# 4. For $V_0 = 300$, $m = 1$ and $a = 0.15$, find and plot the **normalised** wavefunctions and energy eigenvalues
# In[2]:
#NAME: Shooting Method & Numerov's Algorithm
#DESCRIPTION: Solving the 1D time independent SchrÃ¶dinger equation using the shooting method and Numerov's algorithm.
import numpy as np
import matplotlib.pyplot as plt
from math import ceil
import pycav.quantum as pyq
get_ipython().run_line_magic('matplotlib', 'notebook')
def potential(params):
def V(x):
if abs(x) < params[1]:
return -params[2]
else:
return 0.0
return V
def normalise(psi,spacing):
area = np.trapz(psi**2,dx = spacing)
return psi / area**0.5
def squarewell_numbs(params):
u_0 = params[0]*params[1]**2*params[2]/2.
return 2*np.sqrt(u_0)/np.pi
domain = [-0.55,0.55]
steps = 500
L = domain[1]-domain[0]
dx = (domain[1]-domain[0])/steps
x = np.linspace(domain[0],domain[1],steps+1)
V_params = [1.0,0.15,300.]
V = potential(V_params)
fig = plt.figure(figsize = (12,6))
ax = plt.subplot(111)
bracket_E = [-150.,0.]
psi, E_bisect = pyq.bisection_search(x,dx,V,V_params,bracket_E,tolerance = 0.005)
print(E_bisect)
ax.plot(x,normalise(psi,dx)+0.05*E_bisect)
bracket_E = [-250,-35]
psi, E_bisect = pyq.bisection_search(x,dx,V,V_params,bracket_E,tolerance = 0.005)
print(E_bisect)
ax.plot(x,normalise(psi,dx)+0.05*E_bisect)
bracket_E = [-550,-170]
psi, E_bisect = pyq.bisection_search(x,dx,V,V_params,bracket_E,tolerance = 0.005)
print(E_bisect)
ax.plot(x,normalise(psi,dx)+0.05*E_bisect)
bracket_E = [0.,100.]
psi, E_bisect = pyq.bisection_search(x,dx,V,V_params,bracket_E,tolerance = 0.005)
print(E_bisect)
ax.plot(x,normalise(psi,dx)+0.05*E_bisect)
plot_V = [0.05*V(y) for y in x]
ax.plot(x,plot_V,'k')
ax.set_xlim([np.min(x),np.max(x)])