Converging the Hamiltonian in Atomic Units?

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SUMMARY

The discussion centers on approximating the energy of a hydrogen atom using a trial function of the form \(\sum c_if_i\) with \(f_i = e^{-ar}\) in atomic units. The Hamiltonian in atomic units is defined as \(H = -\frac{1}{2}\nabla^2 + \frac{1}{r}\). The user encounters convergence issues when computing \(H_{ij} = \int_0^\infty f_iHf_j\) due to the integral \(\int_0^\infty f_1\frac{1}{r}f_2dr\) not converging. A suggested solution involves modifying the integral to \(\int_0^\infty f_1\frac{1}{(r+\epsilon)}f_2dr\) to achieve convergence.

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  • Understanding of quantum mechanics and the hydrogen atom model
  • Familiarity with Hamiltonian operators in quantum mechanics
  • Knowledge of trial wave functions and variational methods
  • Basic calculus, particularly integration techniques
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  • Study the variational principle in quantum mechanics
  • Learn about the properties of the hydrogen atom's wave functions
  • Research convergence issues in quantum mechanical integrals
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Students and researchers in quantum mechanics, particularly those focusing on atomic physics and computational methods for energy approximation.

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Homework Statement


So the question is I have to use some trial function of the form \sum c_if_i to approximate the energy of hydrogen atom where f_i=e^{-ar} for some number a (positive real number). Note that r is in atomic unit.

Homework Equations


Because r is in atomic unit, I think I should use the Hamiltonian in atomic unit, that is
H = -\frac{1}{2}\nabla^2 + \frac{1}{r}
or should I use the spherical Hamiltonian?

I try to compute H_{ij} = \int_0^\infty f_iHf_j but there will be the term \int_0^\infty f_1\frac{1}{r}f_2dr which cannot be integrated (not converged). So what's wrong with the way I approach the problem?

Thank you,
 
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might your coefficients be r dependent since they are in the full solution of the hydrogen like atom
 
<br /> \int_0^\infty f_1\frac{1}{r}f_2dr<br />
will converge if you change 1/r to 1/(r+eps)
 

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