Add numerov solve
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numerov_solve.py
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numerov_solve.py
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"""
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MO 26 April 2024
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modified from: plh 26 April 2024
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Template for Numerical integration using Numerov method.
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Ported from SHOsolve.py
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See https://physics.bu.edu/~py502/lectures4/schrod.pdf and
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https://mtdevans.com/2013/07/solving-the-schrodinger-equation-with-numerovs-algorithm/
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"""
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# 1-D Schrodinger wave equation
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# (-hbar^2/2m) d^2/dx^2 psi(x) + V(x) psi(x) = E psi(x)
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#
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# Spherically symmetric Psi(xvec)_L,Lz,n = R(r)_L,n Y(phi,theta)_L,Lz
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# [(-hbar^2/2mu)(1/r)d^2/dr^2 + L(L+1)hbar^2/2mu-r^2 + V(r)] R(r)_L,n = E_L,n R(r)_L,n
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# Define U(r) = r R(r)
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# [(-hbar^2/2mu) d^2/dr^2 + L(L+1)hbar^2/2mu-r^2 + V(r)] U(r)_L,n = E_L,n U(r)_L,n
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#
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from typing import Callable
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import numpy as np
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import pylab as pl
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# version 0.3
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# version history
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# 0.1 - ported from BohrSolve.py of 2018(ca), retains constants
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# 0.2 - simplified to calculating psi(0) for a list of energies
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# 0.3 - version used for 2024SP PS
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# constants
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hbar = 6.626070040e-34 / np.pi / 2 # J-s = Planck's constant
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light_speed = 299792458 # m s-1 = speed of light
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bohr_radius = 0.52917721067e-10 # m = Bohr radius
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mass_proton = 1.672621898e-27 # kg = mass of proton
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mass_electron = 9.10938356e-31 # kg = mass of electron
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mass_electron_reduced = mass_electron / (1.0 + mass_electron / mass_proton) # kg = reduced mass of electron in H atom
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electrical_constant = 9.0e9 # N-m^2/s^2 = electrical constant
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charge = 1.602176634e-19 # C = electronic charge
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potential_1 = -electrical_constant * charge * charge # = potential at r=1m
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potential_0 = potential_1 / bohr_radius # = potential at r=a0
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energy_1_ev = 13.6 # eV = ground state energy of H atom in eV
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energy_1_j = energy_1_ev * charge # J = ground state energy of H atom in J = 21.88E-19 J
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# initial values
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intervals = 10000 # number of intervals
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radius_max = 200.0 * bohr_radius # maximum radius = starting point
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radius_min = bohr_radius / 500000.0 # minimum radius = ending point
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# dr = a0/1000000.0 # radius step
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energy = -energy_1_j # energy, turning point is where E=V(x)
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energy_min = -1 * energy_1_j # min energy for scan
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energy_max = -0.005 * energy_1_j # max energy for scan
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energy_intervals = 1000
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def bohr_potential(r):
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"""Models the Bohr atom electrostatic potential"""
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return potential_1 / r
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def do_numerov(radial_vals: np.ndarray, energy: float, initial: tuple[float, float],
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potential_func: Callable[[float], float], angular_momentum=0.0) -> np.ndarray[float]:
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"""
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Solves the differential equation F''(r) + g(r) F(r) = 0 using the Numerov method.
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In the context of the Schrödinger equation, g(r) = k^2 = (E - V(r))*2/m
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Here, the argument 'potential' points to a function definition that calculates the potential V(r).
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"""
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hbar_sqrd_over_2m = hbar * hbar / 2 / mass_electron_reduced
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step_size = radial_vals[1] - radial_vals[0]
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solution = np.zeros(len(radial_vals))
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solution[0], solution[1] = initial
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for i in range(2, len(radial_vals)):
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g_factor = step_size * step_size * (((energy - potential_func(float(radial_vals[i]))) / hbar_sqrd_over_2m)
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- (angular_momentum * (angular_momentum + 1) / radial_vals[i] ** 2)) / 12
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solution[i] = ((solution[i - 1] * (2.0 - 10.0 * g_factor) - solution[i - 2] * (1.0 + g_factor))
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/ (1.0 + g_factor))
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# Normalize the solution so that the total probability equals 1 in the center
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normalization_factor = solution.max()
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solution = solution / normalization_factor
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return solution
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def opposite_signs(a, b):
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return a * b < 0
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def main():
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# Main program begins here
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x_vals = np.linspace(radius_max, radius_min, intervals) # generate x values
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dx = abs(x_vals[1] - x_vals[0])
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energies = np.linspace(energy_min, energy_max, energy_intervals).tolist()
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results = [] # these will be wave function(at r near 0) for each of the energies
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zeros = []
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print(f'\nTesting {len(energies)} energies from {energy_min} to {energy_max} for solutions...')
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for i, energy in enumerate(energies):
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if i % 100 == 0:
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print(f' - Testing energy: {i}/{len(energies)} = {energy}...')
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initial = (0.0, 1.00) # initial value of function and slope, at large r
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numerov = do_numerov(x_vals, energy, initial, bohr_potential) # does integration, returns wave function
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if len(results) != 0:
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if opposite_signs(results[-1], numerov[-1]):
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zero = (energy + energies[i - 1]) / 2
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print(f'\tFound solution at {zero}')
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zeros.append(zero)
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results.append(numerov[-1])
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print(f'Done testing energies\n')
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# guess at best solution E
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# first plot the results as a function of energy
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pl.figure(4)
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pl.xlabel('Energy')
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pl.ylabel('Wave function near r=0')
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pl.title('Trial results vs Energy')
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for zero in zeros:
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pl.axvline(x=zero, color='red')
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pl.plot(energies, results)
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for n, energy_solution in enumerate(zeros):
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print(f'Trying E = {energy_solution:.3e} J, {energy_solution / charge:.3} eV')
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initial = (0.0, 1.0)
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func_reduced = do_numerov(x_vals, energy_solution, initial, bohr_potential) # U(r) = reduced wave function
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# Figure 1 = reduced radial wave function
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pl.figure(1)
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pl.xlabel('r')
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pl.ylabel('U(r)')
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pl.title('Reduced U(r) = r R(r)')
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pl.plot(x_vals, func_reduced, label=f'n={n:d}')
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pl.legend()
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prob = dx * func_reduced.dot(func_reduced) # calculate total probability, should we x 4pi?
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func_reduced = func_reduced / np.sqrt(prob) # total probability should = 1.0!
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func_radial = func_reduced / x_vals # get the radial wave function R(r)
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mean_x = dx * (x_vals * func_reduced).dot(func_reduced)
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print(f'Expectation value <r> = {mean_x:.3e} m')
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# Figure 2 = actual radial wave function R(r)
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pl.figure(2)
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pl.title('Radial wave function R(r) = U(r)/r')
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pl.plot(x_vals, func_radial, label=f'n={n:d}')
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pl.xlabel('r')
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pl.ylabel('R(r)')
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# pl.ylim(-1,1.1*E)
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pl.legend()
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# Figure 3 = probability integrand r^2 R(r)^2 = U^2
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pl.figure(3)
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pl.title('Probability Integrand [r^2 R(r)^2]')
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pl.plot(x_vals, func_reduced * func_reduced, label=f'n={n:d}')
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pl.xlabel('r')
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pl.ylabel('P(r)')
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# pl.ylim(-1,1.1*E)
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pl.legend()
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pl.show() # show graph window until closed
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main()
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