1 Dimensional Schrödinger Equation Simulation

The time dependent Schrödinger equation is:

A naive implementation using Eulers method where ${\psi }_{t+\mathrm{\Delta }t}(x)={\psi }_t(x)+{\displaystyle \frac{\partial {\psi }_t(x)}{\partial t}}\mathrm{\Delta }t$ will usually cause the waveform to explode due to floating point and integration errors.

Because $\psi$ is usually derived from quantizing a classical theory, it's an infinite dimensional vector in a Hilbert space. So $\psi \in (ℝ\to ℂ)$ assigning a complex number to each possible position. For the simulation we will approximate it as a finite dimensional vector by discreetizing it over $N$ points so $\psi \in {ℂ}^N$. Similarily $\hat{H}\in {ℂ}^{N\times N}$ is now a finite (Hermetian) matrix. The equation stays the same, but it's now in a finite dimensional Hilbert space:

Assuming the Hamiltonian isn't time dependent, the differential equation has the closed form solution using matrix exponentiation.

We should now actually define the Hamiltonian operator $\hat{H}$. The original discreet version for a non-relativistic particle with some potential function $V$ is the linear operator:

Since we discreetized $\psi$ as a finite dimensional vector, we replace the second derivative with the finite difference method.

So for each position of $-{\displaystyle \frac{{\hslash }^2}{2m}}{\displaystyle \frac{{\partial }^2}{\partial x^2}}$ we can make a matrix below. We assume no repeating boundary conditions. The matrix is equivelent to the Laplacian of a path graph.

The potential term $V(x)$ is simply the value at each point:

$\hat{H}$ is the sum of those two matricies

Each discrete step of the simulation will go through a certain change in time using the matrix exponent of the Hamiltonian operator. It should be simple to put it into a for loop now.