I Finite difference method for Schrödinger equation

aaaa202
Messages
1,144
Reaction score
2
Suppose I want to solve the time-independent Schrödinger equation
2/2m ∂2/∂x2 + V)ψ = Eψ
using a numerical approach. I then discretize the equation on a lattice of N points such that x=(x1,x2,...,xN) etc. Finally I approximate the second order derivative with the well known central difference formula:
2/∂x2 ≈ 1/Δx2i+1i-1-2ψi)
My question is now: How do you estimate the validity of this approximation? I have already talked to my teacher about it and he said the following:
The discrete approximation is a tight-binding model with dispersion:
E = ħ2/2m * 2/Δx2(1-cos(kΔx))
So for Δx<<1/k we can taylor expand this expression to give:
E ≈ ħ2/2m * 2/Δx2(1-(1-1/2(kΔx)2))=ħ2k2/2m
Which, according to my teacher, shows that the approximation holds provided that the lattice spacing is much shorter than the wavelength. What I don't get is how you can argue that because the dispersion is parabolic in k the finite difference approximation for the derivative ∂2/∂x2 is a good approximation. In short: What "connects" ħ2k2/2m with ħ2/2m ∂2/∂x2?
 
Physics news on Phys.org
aaaa202 said:
In short: What "connects" ħ2k2/2m with ħ2/2m ∂2/∂x2?
You mean apart from the fact that ##p = \hbar k## and ##\hat{p}^2= -\hbar^2 \frac{d^2}{dx^2}##?
 
But k is not an operator in this case. It is a wavenumber. What connects the number form with the operator form?
 
aaaa202 said:
But k is not an operator in this case. It is a wavenumber. What connects the number form with the operator form?
Plane waves. Eigenfunctions of the ##\hat{p}^2## operator are of the form ##\exp(i k x)## where ##k = p/\hbar##, with ##p## the momentum of the corresponding plane wave.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
Back
Top