Schrödinger eq. with 3D spherical potential

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SUMMARY

The discussion focuses on solving the radial Schrödinger equation for a 3D spherical symmetric potential, specifically in the context of quantum mechanics exam preparation. The equation is presented as: -h²/(2*m) * d²u/dr² + [V(r) + (l*(l+1) / r²)]*u = E*u, where u(r) = r*R(r). It is concluded that as r approaches infinity, the attractive potential V(r) vanishes exponentially, leading to the effective potential's second term diminishing rapidly, which provides insight into the asymptotic behavior of u(r) for bound states.

PREREQUISITES
  • Understanding of quantum mechanics principles
  • Familiarity with the Schrödinger equation
  • Knowledge of spherical coordinates in quantum systems
  • Basic concepts of bound states and potential energy
NEXT STEPS
  • Study the derivation of the radial Schrödinger equation in spherical coordinates
  • Learn about effective potential in quantum mechanics
  • Explore asymptotic analysis of wave functions in bound states
  • Investigate the behavior of wave functions at infinity in quantum systems
USEFUL FOR

Students preparing for quantum mechanics exams, particularly those focusing on the radial Schrödinger equation and bound state analysis in spherical potentials.

maethros
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Hello!

I'm trying to solve some old exam exercises to prepare for my qm exam next week.
Now I got a question I don't have any idea how to solve it. I hope somebody can help me:

"The radial Schrödinger equation in the case of a 3D spherical symmetric potential V(r) can be written in the form

-h^2/(2*m) * d^2u/dr^2 + [V(r) + (l*(l+1) / r^2)]*u = E*u

where u(r) = r*R(r). If V is attractive and vanishes exponentially at infinity, how does u(r) behave asymptotically for bound states?"



Thanks for helping!
 
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What can you say about the effective potential as [tex]r\rightarrow \infty[/tex]? That should give you a pretty good idea of where to start. Also, I think this belongs in Homework Help.
 
as r-> inf, the potential goes to zero and the second term in the effective potential goes rapidly to zero (as there is a r^2)
 

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