Coulomb potential removes the degnerecay of states

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Discussion Overview

The discussion revolves around the effect of adding a small term proportional to \(1/r^2\) to the Coulomb potential and how this addition removes the degeneracy of states with different angular momentum quantum numbers (L), specifically in the context of quantum defects.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant inquires about the mechanism by which a small \(1/r^2\) term in the Coulomb potential removes the degeneracy of states with different small L values.
  • Another participant asks whether a full mathematical calculation or a quantitative answer is desired.
  • A subsequent reply indicates that generally, a perturbation lifts the degeneracy of states.
  • It is noted that the \(1/r^2\) term is related to spin-orbit interaction, with the derivative of the Coulomb potential being \(1/r^2\) and angular momentum L being proportional to r.
  • A suggestion is made to refer to a textbook, specifically Griffiths chapter 6, for a more detailed explanation and to utilize perturbation theory to calculate approximate energy levels.

Areas of Agreement / Disagreement

Participants appear to agree on the general principle that perturbations can lift degeneracy, but there is no consensus on the specific calculations or implications of the \(1/r^2\) term in this context.

Contextual Notes

The discussion does not resolve the specific mathematical steps or assumptions involved in applying perturbation theory to this problem.

eman2009
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Hi every one this is the first time in this wonderful forum :)

and i have a question i hope i find an answer ?

how can the additiona of a smalll (c/r square)term to the coulomb potential removes the degnerecay of states with different (small) L. (quantum defect)?

:confused:

thanks
 
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Do you want a full mathematical calculation or a quantitative answer?
 


no just quantitative answer?
__________________
 


do you know perturbation theory?
 


yes how i can use it in this case?
 


generally, a perturbation lifts the degeneracy.

The additional factor 1/r^2 is related to the spin-orbit interaction, which will have a 1/r^2 dependence [tex]\frac{1}{r}\frac{dV_C}{dr}(vec{L}\cdot \vec{S})[/tex] The derivative of the couloumb potential is 1/r^2 and L is proportional to r
 


I suggest that you refer to textbook.
(For example, Griffiths chapter 6)

By using perturbation theory, you can calculate approximate energy and fine slit of energy level on same value of L^2.

<1/r^2> can be calculated and it's value contains n,l so you may take good quantum number n,l ...
 


thanks for all :)
 

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