Are Autoionizing States Necessary for Accurate Wavepacket Representations?

  • Context: Graduate 
  • Thread starter Thread starter chafelix
  • Start date Start date
  • Tags Tags
    States
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
chafelix
Messages
26
Reaction score
0
In terms of stationary states we have bound states and free(plane wave or Coulomb) states
Autoionizing states are free(E>0), e.g. continuum, but localized states(resonances). Hence
wavepackets consisting of a band of plane or Coulomb waves. I understand they may be long-lived, longer than the experiment and that ionization may occur by a bound state acquiring some energy, staying localized for a while in an autoionizing state. But why do we need to include them and not use the simpler bound+free(plane or Coulomb ) basis? After all autoionizing states should be nothing more than a linear combination of plane or Coulomb waves
 
Physics news on Phys.org
I think because the plane wave states are simply not valid solutions in the neighborhood of, say, a hydrogen nucleus. How can they be when they are the solutions for free space?

To put it more clearly, the free state solutions can be re-expressed in terms of spherical Bessel functions rather than plane waves. Now add a proton to the center of the coordinate systems. Do you not think the Bessel functions will be distorted?
 
I don't think that's right. Coulomb waves have the correct asymptotic behavior, but the point is that this is merely a basis; you can use any basis you want to represent the states and clearly bound states +say Coulomb waves is a complete basis. What's the point of non-stationary autoionizing states?