🫐 2D quantum particle

What are you looking at?

This demo visualizes the stationary eigenstates belonging to the Hamiltonian of a 2D quantum particle:

$$i\hbar \frac{\partial \psi (x,y,t)}{\partial t}=\left[ -\frac{\hbar ^{2}}{2m}\left(\frac{\partial ^{2}}{\partial x^{2}}+\frac{\partial ^{2}}{\partial y^{2}}\right)+V(x,y) \right]\psi (x,y,t)$$

Where:

• $\psi(x, y, t)$ is the wave function of the system.
• $\partial^2/\partial x^2 + \partial^2/\partial y^2 = \nabla^2$ is the 2D Laplacian operator.
• $\hbar$ is the reduced Planck’s constant, set to $1$ in the code.
• $m$ is the mass of the particle, set to $1$ in the code.
• $V(x, y)$ is the potential energy function, e.g. an isotropic harmonic oscillator potential:

$$V(x,y)=\dfrac{1}{2}k(x^2+y^2)$$

Demo

Select different eigenstates to explore how their amplitude and phase form distinct wave-like patterns, including the characteristic lobes and nodes of quantum wavefunctions.

Validation of solver

The solver implements the so-called Lanczos algorithm featuring full reorthogonalization, matrix-free Hamiltonian application, Ritz vectors, eigenvalue ordering, residual checks and QL-diagonalization.

Isotropic harmonic oscillator

For the isotropic harmonic oscillator, the analytic energies are given by $\sqrt{k}(n_x +n_y+1)$ with $k=0.05$:

$n_x+n_y$	degeneracy	analytic $E_n$	calulated $E_n$
0	1	0.22361	0.22354
1	2	0.44721	0.44701, 0.44701
2	3	0.67082	0.67043, 0.67043, 0.67049
3	4	0.89443	0.89391, 0.89391, 0.89433, 0.89433
4	5	1.11803	1.11733, 1.11780, 1.11780, 1.12074, 1.12074

The degeneracies arise because e.g. the second level can be formed in two ways ($n_x=0, n_y=1$ and $n_x=1, n_y=0$), the third level in three ways ($n_x=0, n_y=2$, $n_x=1, n_y=1$, $n_x=2, n_y=0$), etc.

The small differences between the energies of the degenerate shells are a consequence of the discrete numeric representation on a finite lattice. The differences with the analytic values increase with the energy levels, which is expected for a numeric solver like this. However, they are still extremely small, e.g. in the order of 0.06 % for the highest energy levels.

Infinite square well

The analytic expression for the energies are now given by:

$$E_{n_x,n_y} = \frac{\pi^2\hbar^2}{2mL^2}(n_x^2+n_y^2), \qquad n_x,n_y=1,2,\ldots$$

In the code we configured the constants to be:

hbar = 1
m = 1
N = 110
extent = 0.15 * (N - 1) = 16.35

So the fundamental unit of energy is:

$$E_0=\frac{\pi^2}{2L^2} \approx 0.01845$$

This leads to the following result:

$n_x^2+n_y^2$	degeneracy	analytic E	Our E
2	1	0.03690	0.03692
5	2	0.09225	0.09228
8	2	0.14760	0.14764
10	1	0.18450	0.18450
13	2	0.23985	0.23986
17	2	0.31365	0.31349
18	1	0.33210	0.33207
20	2	0.36900	0.36885
25	2	0.46125	0.46107

The degeneracies are in accordance with theory:

0.0369177                    1
0.0922788  0.0922788         2
0.1476400                    1
0.1844963  0.1844963         2
0.2398575  0.2398575         2
...

Also note that the InfiniteSquareWell in our code is realised by imposing a boundary condition on the Hamiltonian.apply():

static withoutParameters = () => /** @type {SingleParticle} */ particle => 0;

As such, it is not genuine potential barrier, but a boundary condition on the domain:

for (let y = 1; y < ny - 1; y++)
    for (let x = 1; x < nx - 1; x++)

This is the correct way to represent an infinite square well with Dirichlet boundary conditions and makes this test useful to test both our Lanczos-solver, as well as the implementation of the discrete Laplace-operator + boundary conditions.