I Finite difference method for Schrödinger equation

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The discussion focuses on using the finite difference method to solve the time-independent Schrödinger equation numerically. The central difference formula is employed to approximate the second derivative, leading to a tight-binding model with a specific dispersion relation. The validity of this approximation is contingent on the lattice spacing being much smaller than the wavelength, as explained by Taylor expansion. The connection between the energy expressions and the finite difference approximation is clarified through the relationship between momentum and the wave number, emphasizing that plane waves serve as eigenfunctions of the momentum operator. This highlights the underlying quantum mechanical principles that justify the numerical approach.
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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?
 
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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.
 
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