Orbitals while transitioning from free electron to ground state

  • Thread starter rconde01
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  • #1
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I was thinking of putting together a visualization of electron orbitals as it transitions from unbounded or weakly bounded state to the ground state. However, it occurred to me that orbitals are symmetric about the proton. At some point the probability distribution must become asymmetric eventually approaching a point. What am I missing?
 

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  • #2
DrDu
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The Coulomb potential always has an infinite number of bound states however weak you chose the central charge. So there won't be a transition from unbound to bound.
 
  • #4
Khashishi
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Dr. Du, of course you can transition from unbound to bound. This is called recombination emission.

The "s", "p", "d", "f"... orbitals are just one way of representing the state, which can have almost arbitrary shape of the electron distribution. The arbitrary shape will be some superposition of orbitals. It is analogous to how an arbitrary wave is a superposition of frequency components. Just as a localized wave-packet doesn't have a single frequency, a localized electron doesn't have a single orbital state.

A weakly bounded atom is called a Rydberg atom. http://en.wikipedia.org/wiki/Rydberg_atom
This is a case where a classical representation (planetary model) of an atom can be useful. The corresponding quantum state is some kind of coherent state of the Rydberg atom which mixes angular momentum states.

You might be able to find something by searching for coherent states of Rydberg atom. Good luck. It's not an easy problem. I'm interested in how the results look.
 
  • #5
DrDu
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Dr. Du, of course you can transition from unbound to bound. This is called recombination emission.
Of course. Apparently I was thinking in something completely different to what the OP was asking.
 
  • #6
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It occurs to me if you stick to the simple model of Couloumb potential you are never going to do so. Since central potential has parity symmetry, which means the wave functions chosen are always symmetric or antisymmetric... I think for the more realistic situation, models in condensed matter might be more applicable.
 

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