Degeneracy of hydrogen energy levels

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SUMMARY

The discussion focuses on the degeneracy of hydrogen energy levels when a hydrogen atom is subjected to an infinite potential barrier on one side of the nucleus. The requirement for odd wavefunctions leads to the conclusion that only odd values of the orbital quantum number \( l \) are permissible, resulting in ground and first excited states with quantum numbers \( n=2 \) and \( n=4 \), respectively. The degeneracy formula derived is \( g_n = \lfloor{n/2}\rfloor(\lfloor{n/2}\rfloor+1) \), confirming that the energy eigenvalues retain the same form as the standard hydrogen atom despite the boundary condition restrictions.

PREREQUISITES
  • Quantum mechanics principles, particularly wavefunctions and boundary conditions.
  • Understanding of quantum numbers \( n \) and \( l \) in atomic physics.
  • Familiarity with spherical harmonics and their parity properties.
  • Knowledge of differential equations in quantum mechanics.
NEXT STEPS
  • Study the implications of boundary conditions on wavefunctions in quantum systems.
  • Explore the derivation of energy levels in quantum mechanics, focusing on the hydrogen atom.
  • Investigate the role of spherical harmonics in determining quantum states.
  • Learn about potential barriers and their effects on quantum particle behavior.
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Physicists, quantum mechanics students, and researchers interested in atomic structure and the effects of boundary conditions on energy levels in quantum systems.

Silicon-Based
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Homework Statement
Find the energy degeneracy of a hydrogen atom with infinite potential barrier
Relevant Equations
##|m| \leq l < n##
##P = (-1)^l##
I'm considering a hydrogen atom placed in an infinite potential on one side of the nucleus, i.e. ##V(x) = +\infty## for ##x < 0##. I require the wavefunctions to be odd in order to satisfy the boundary condition at ##x=0##. By parity of the spherical harmonics only states with ##l## odd are allowed, so the ground state and first excited state should respectively have ##n=2## and ##n=4##. Since even ##l## are excluded, the number of possible values of ##l## is ##\lfloor{n/2}\rfloor##, so the degeneracy is:

$$
g_n = \sum_{l=1,\, l \,\text{odd}}^{n-1 \, (n\, \text{even}),\,n-2 \, (n \,\text{odd})} (2l+1) = \lfloor{n/2}\rfloor(\lfloor{n/2}\rfloor+1)
$$

Is my reasoning above correct? Is it reasonable to assume that the energy eigenvalues have the same form as for the usual hydrogen atom?
 
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Yes. The differential equation is same as original Hydrogen atom problem with the restriction. As you have correctly determined only odd values are the solutions so as to satisfy boundary conditions. The degeneracy is now over those restricted ##l##
 
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