Non central potential and energy dependence

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SUMMARY

The discussion centers on the implications of non-central Hamiltonians in quantum mechanics, specifically how energy levels depend on both the principal quantum number (n) and the angular momentum quantum number (l). It is established that for Hamiltonians lacking spherical symmetry, l is not a good quantum number, as these Hamiltonians do not commute with rotational operators. The unique degeneracy observed in hydrogen-like atoms, which is a result of the 1/r potential, is attributed to a hidden symmetry known as the Runge-Lenz vector, distinguishing it from other potential forms.

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly Hamiltonians.
  • Familiarity with quantum numbers, especially principal quantum number (n) and angular momentum quantum number (l).
  • Knowledge of spherical symmetry in quantum systems.
  • Concept of the Runge-Lenz vector and its significance in classical and quantum mechanics.
NEXT STEPS
  • Research the properties of non-central Hamiltonians in quantum mechanics.
  • Study the implications of angular momentum in non-spherically symmetric potentials.
  • Explore the concept of the Runge-Lenz vector and its role in quantum systems.
  • Investigate various potential forms beyond the 1/r potential and their energy level structures.
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Quantum physicists, students of quantum mechanics, and researchers exploring advanced topics in Hamiltonian systems and energy level dependencies.

naruto365
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Could anyone tell me where can i find more about the fact that if your Hamiltonian is non-central then the total energy is not dependent only on principle quantum number n but also on l.
 
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If your Hamiltonian is not spherically symmetric, then l is not a good quantum number in the first place: If H does not commute with rotations, you cannot find a common eigensystem of H and the L^2 operator[1].

I guess what you mean is "why is the energy dependent on n and l for for spherical potentials which are not 1/r" (i.e., not hydrogen). If this is the case: Because there generally is no reason for all energy to be independent of l. Rather, this degeneracy in the hydrogen atom is a special case for 1/r potentials, ultimately related to a hidden additional conserved symmetry (the so called Runge-Lenz-Vector already known from classical mechanics). So it is the 1/r potential which is the special case, not the other ones.[1] of course you could still prepare a wave function in a specific l state with respect to some point, but such states would not be eigenstates.
 

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