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Is a Pseudo-Wavefunction Really a Mixed State?

  1. Aug 28, 2009 #1
    Hello,

    I'm studying pseudopotentials right now, and I had an epiphany that the pseudo-wavefunction is really mixed-state of the original Hamiltonian. Has anyone ever thought about a pseudo-wavefunction that way? Just curious.

    modey3
     
  2. jcsd
  3. Aug 29, 2009 #2
    I'm don't think that's correct. Pseudopotentials remove nodes that are near the core, so a pseudopotential for Nb (say) will have 4d valence states, however the 4d states will not have a node near the core so they really will look like 3d states. I don't think you can accomplish elimination of a node by mixing states with n > 4.
     
  4. Aug 29, 2009 #3
    kanato,

    If you take the overlap integral between the pseduo-wavefunction and a particular all-electron valence and the core wavefunctions you get non-zero terms. This means that the pseudo wavefunction can be represented as a sum of core-states and the valence states each scaled by their overlaps with the pseduo-wavefunction. The OPW-method does this.
     
  5. Aug 29, 2009 #4
    Sure, there's non-zero overlap, but I am unconvinced as to whether the all-electron basis is "complete enough" to represent the pseudo-wavefunction.
     
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