Hydrogen degeneracy

  • Thread starter Allday
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  • #1
Allday
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The hydrogen energy levels (when only considering the coulomb field of an infinitely massive proton - or when using the reduced mass for the electron) only depend on the principle quantum number n. Can someone give me a qualitative argument why the orbital angular momentum quantum number l doesn't come into play here?

related questions:

do hydrogen like atoms (ionized to the point of having only one electron but containing more than one proton) break this degeneracy between the different orbital angular momentum states?

do atoms with the same number of protons and electrons, but with one electron outside of completely full shells break this degeneracy?

gonna read some and see if i can find the answer
 
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Answers and Replies

  • #2
Allday
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It turns out the 1/r nature of the coulomb potential is the reason why all the states with the same n and different l have the same energy. The degeneracy is called accidental to differentiate it from the essential degeneracy of the 2l+1 states with the same l but different m. This essential degeneracy is due to the rotational invariance of any central potential.

This leads me to believe that the atoms with one electron maintain the degeneracy, while those with full shells + 1 don't because the electron shielding will alter the 1/r potential.
 
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  • #3
dextercioby
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The "accidental" degeneracy is due to a conserved quantity corresponding to a symmetry of the system via quantum Noether theorem.

See https://faculty.washington.edu/seattle/physics324/Lenz.pdf [Broken] for details.
 
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  • #4
Allday
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thanks for the link to the paper dextercioby. I remember learning about that Lenz vector when I was doing classical mechanics orbits. Interesting to see its quantum equivalent in play. I think it would be nice if they included a discussion of this in more of the standard quantum books.
 
  • #5
dextercioby
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I've seen this issue treated in some lecture notes before, but i can't remember any textbook author discussing it.
 

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