why the simulation period is wrong about schrodinger equation in a harmonic potential
2 views (last 30 days)
Show older comments
Daniel Niu
on 22 Jun 2023
Commented: Daniel Niu
on 25 Jun 2023
clear; % Clear workspace variables
N = 1024;
x = linspace(-64, 64, N);
dx = x(2) - x(1);
psi = gaussian_wavepacket(x, -32.0, 3.0, 0.0);
V = (x / 32.0).^2 / 2;
H = hamiltonian(N, dx, V);
simulate = create_simulator(H, 1.0); % Create a new simulator function each time the script runs
figure;
fill([x, fliplr(x)], [V/10.0, zeros(1, numel(V))], 'r', 'FaceAlpha', 0.1, 'EdgeColor', 'none')
hold on;
h_plot = plot(x, probability_density(psi));
hold on;
title('Time Evolution of a Gaussian Wavepacket');
xlabel('Position');
ylabel('Probability Density');
axis([-64 64 0 0.15])
h_text = text(min(x), max(probability_density(psi)), sprintf('t = %.2f', 0), 'VerticalAlignment', 'middle');
M=500;
for k = 1:M
[psi, time] = simulate(psi);
set(h_plot, 'YData', probability_density(psi));
set(h_text, 'String', sprintf('t = %.2f', time));
movieVector(M) = getframe;
pause(0.1);
end
function H = hamiltonian(N, dx, V)
e = ones(N, 1);
A = spdiags([e -2 * e e], -1:1, N, N);
L = A / (dx^2);
H = -L + spdiags(V', 0, N, N);
H = sparse(H);
end
function psi = gaussian_wavepacket(x, x0, sigma0, p0)
A = (2 * pi * sigma0^2)^(-0.25);
psi = A * exp(1i * p0 * x - ((x - x0) / (2 * sigma0)).^2);
psi = psi.'; % Return as a column vector
end
function U = time_evolution_operator(H, dt)
U = expm(-1i * H * dt);
U(abs(U) < 1E-12) = 0;
U = sparse(U);
end
function prob_density = probability_density(psi)
prob_density = abs(psi).^2;
end
function simulate = create_simulator(H, dt)
U = time_evolution_operator(H, dt);
t = 0;
simulate = @simulate_func; % Return handle to nested function
function [psi, time] = simulate_func(psi)
time = t * dt;
psi = U * psi;
t = t + 1;
end
end
%my potential as V = (1/2) * (x/32)^2. I create a harmonic potential of the form (1/2) * m * ω^2 * x^2,
%and assuming m = 1 for simplicity, then ω^2 = 1/32^2, and ω = 1/32.
%This means my classical period T should be T = 2π * 32 = ~200.6
%but in the simulation, the period is about 142.
%I don't know why?
%Your help would be highly appreciated.
1 Comment
Florian Rössing
on 22 Jun 2023
I have studied physics, so I have seen the problem already. But: This is a Matlab Forum. Could you please provide the Math that yields your expected result so we can compare that against the code.
Accepted Answer
David Goodmanson
on 23 Jun 2023
Edited: David Goodmanson
on 23 Jun 2023
Hi Daniel,
Nice animation. Looks like in your 'hamiltonian' function, the kinetic energy is missing a factor of 1/2. Changing the line to
L = (1/2)*A / (dx^2)
gives a period right around 200.
More Answers (0)
See Also
Categories
Find more on Quantum Mechanics in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!