Hydrogen Atom Wavefunction Boundary conditions

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SUMMARY

The discussion focuses on solving the differential equation for the Hydrogen atom wavefunction in the ground state using Euler's method. The equation presented is d²u_nl/dr² - (l(l+1)/r²)*u_nl + 2k*(E_nl-V(r))*u_nl = 0, with the potential V(r) defined as -a/r, where a = 1/137.04. Initial conditions provided are u_nl(0) = 0 and du_nl(0)/dr = 1. The key insight is that the initial value for the second derivative is not necessary; instead, adjusting the energy value, referred to as the shooting parameter, will yield the correct wavefunction that decays to zero at infinity.

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  • Understanding of differential equations, specifically second-order linear equations.
  • Familiarity with numerical methods, particularly Euler's method for solving differential equations.
  • Knowledge of quantum mechanics concepts, especially the Hydrogen atom and wavefunctions.
  • Basic grasp of boundary conditions and their implications in physical systems.
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  • Research the application of Euler's method in solving differential equations in quantum mechanics.
  • Explore the concept of shooting methods for finding wavefunctions in quantum systems.
  • Study the implications of boundary conditions in quantum mechanics, particularly for the Hydrogen atom.
  • Learn about energy parameters and their role in determining wavefunction behavior at infinity.
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Chris333
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Hi, I have been given a differential equation to use in order to solve for the Hydrogen wavefunction in the ground state using Euler's method.

d^2u_nl/dr^2 -(l(l+1)/r^2)*u_nl + 2k*(E_nl-V(r))*u_nl = 0

V(r) = -a/r where a = 1/137.04

I have been given initial conditions u_nl(0) = 0 an du_nl(0)/dr = 1
However I need an initial value for the second derivative in order to proceed.

At r = 0, l(l+1/r^2 goes to infinity as does V(r) so I'm not sure what initial condition to use. Any help would be much appreciated.
 
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You already have two initial conditions for u_{nl} and there is no need for a third one.
You will find the right wavefunction by adjusting the value of the energy, which in this case is called the shooting parameter. Varying the value of the energy you will find the wavefunction decay at zero at infinity.
 

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