Using Matlab to solve 1D Schrödinger Equation (Strange Eigenfunctions)

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Niklas Kurz am 31 Mai 2023
Kommentiert: Torsten am 2 Jun. 2023
Following this splendid tutorial (in Python) I tried recreating the code in Matlab
First of, here the Python Code:
% import numpy as np
%
% import matplotlib.pyplot as plt
%
% from scipy.linalg import eigh_tridiagonal
%
% N = 2000
%
% dy = 1/N
%
% y = np.linspace(0,1,N+1)
%
% def mL2V(y):
% return 1000*(y-1/2)**2
%
% d = 1/dy**2 + mL2V(y)[1:-1]
%
% e = -1/(2*dy**2) * np.ones(len(d)-1)
%
% w, v = eigh_tridiagonal(d,e)
%
% plt.plot(v.T[0])
Now my attempt converting this to Matlab:
%% Initialise
% Number of Steps
N = 100;
dy = 1/N;
y = linspace(0,1,N+1);
% Potential
V = @(y,m,L) m*L*(y-1/2).^2;
Vy = V(y,1,1);
% Differential Matrix
d = 1/dy.^2 + Vy(1:end-1);
n = length(d);
e = -1/(2*dy.^2).*ones(n,1);
A = spdiags([e d' e],[-1 0 1],n,n);
%% Solve
[v,w] = eig(full(A));
%% Plot
plot(y(1:end-1),v(:,2),'-','LineWidth',2)
axis([0 1 -1 1])
However the solutions from Matlab are totally off also comparing them with textbook-ones
How is this to explain?
___________
Background:
the goal is to solve the 1D SG:
whose solution is mainly determined by the potential V
setting we get:
writing the derivatives in discrete form brings:
This can be written in matrix-shape:
so basically a Matrix with a main diagonal consisting of:
and two side diagonals with a constants entry of:
The initial conditions are implemented already: respectively
Now the function Ψ is directly given by determing the eigenfunctions of the Matrix
_________________________________________________________________
- This is where the magic happens and where Matlab somehow finds solutions that don't suite the physics content.
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Torsten am 31 Mai 2023
Most probably because you didn't transfer the Python code correctly.
But since I don't have experience with Python, it would be easier if you include the mathematical description of your problem instead of code in a different computer language.
Niklas Kurz am 1 Jun. 2023
I updated the physics. Me neither I'am not having many experiences in Python. I just thought it might help some of you.

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Torsten am 1 Jun. 2023
Bearbeitet: Torsten am 1 Jun. 2023
N = 2000;
dy = 1/N;
y = linspace(0,1,N+1).';
mL2V = @(y)1000*(y-0.5).^2;
e = -1/(2*dy^2)*ones(N-1,1);
d = 1/dy^2 + mL2V(y(2:end-1));
A = spdiags([e d e],-1:1,N-1,N-1);
A = full(A);
[v,w] = eig(A);
hold on
plot(v(:,1).^2)
plot(v(:,2).^2)
plot(v(:,3).^2)
hold off
grid on
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Niklas Kurz am 2 Jun. 2023
Awesome, thank you! Apparently my matrix was slightly different but my potential clearly too weak, causing the shape to differ strongly. Now it's all fine and my faith in Matlab still strong.
Torsten am 2 Jun. 2023
Especially the setting of the ylimits was not adequate.

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Weitere Antworten (1)

James Tursa am 1 Jun. 2023
Bearbeitet: James Tursa am 1 Jun. 2023
Python:
return is misspelled retrun
The mL2V( ) function returns 1000*(y-1/2)**2
I don't see where y is defined prior to the line d = 1/dy**2 + mL2V(y)[1:-1]
MATLAB:
The V( ) function returns 1*1*(y-1/2).^2
I stopped looking after that. Maybe post a working Python code that we can run on the side before comparing that to MATLAB.
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Niklas Kurz am 1 Jun. 2023
Bearbeitet: Niklas Kurz am 1 Jun. 2023
I also updated the Python Code. Works like a charm now.

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