The 4th desirable property of a norm-conserving pseudopotential given by Hamann et al is for the 'logarithmic derivatives of the real and pseudo wave function and their first energy derivatives agree for $r>r_c$'. However I do not understand how the energy derivative of the log derivative $$ \frac{d}{d\epsilon}r\frac{d}{dr}\ln\psi(\epsilon;r) $$ can be defined. In the all-electron case, my understanding is that the valence states are discretised (traditionally labelled with $(n,l,m)$). So what are the states $\psi(\epsilon;r)$ with continuous $\epsilon$ that are presumably required to allow us to define the energy derivative of the log derivative above?


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1 Answer 1


Related prior question about norm-conserving potentials. Hopefully someone more informed on pseudopotenials can give a more descriptive answer. However if I'm understanding the question correctly, I think the issue is what you are considering discretized.

While the states for an atom/molecule are discretized, the energy of any given state can still vary continuously as you perturb the atom or molecule. Taking the example of an atom, while it has discrete set of orbitals, the energies of these individual orbitals can continuously vary if an electric field was applied to the atom or you embedded it within some larger molecule or structure.

  • $\begingroup$ Thanks for your answer. Are the perturbed states uniquely defined by their energy? Which specific perturbation do we consider applying to the potential when defining the energy derivative of the log derivative? And how does this relate to the scattering properties of the all-electron- and pseudo-potentials? Can we have scattering states with energy near the valence eigenenergies? $\endgroup$ Commented Dec 3, 2021 at 4:15
  • $\begingroup$ @user2582713 that seems like a lot of new questions. Feel free to ask questions you might have, on the main site, because comments can be temporary and deleted without you being notified. The answerer might answer some of your follow-up questions in comments, but I wouldn't totally rely on that being a guarantee. $\endgroup$ Commented Dec 3, 2021 at 6:11
  • $\begingroup$ @NikeDattani Thanks for the suggestion. I've written a new follow-up question on the main site and added a link. Do you happen to have any insights into this? $\endgroup$ Commented Dec 4, 2021 at 10:22

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