A Problem about Kohn Hohenberg theorem

[georg gill]

This proof is taken from this site:

http://cmt.dur.ac.uk/sjc/thesis_ppr/node12.html

I get that $v_{ext}(r)$ and $N$ determines $H$ from proof 1. But why is $\Psi$ determined by $H$? Can someone derive a path to prove it mathematically?

 

[hilbert2]

Unless the ground state is degenerate, which is very rare, there's no ambiguity as to the ground state wave function obtained for a given Hamiltonian (except for an arbitrary choice of the constant complex phase). You can't have a spectrum that's not bounded from below (in that pathological case there wouldn't be a ground state).

 

[georg gill]

As far as I am concerned the wave function is definded in the derivation of the schrødinger equation as $\Psi=\Psi_oe^{kx-\omega t}$. Of course there are others but thoose are derived explicitly for a given problem. For example the wave functions for the hydrogen molecule. I don't get why they can derive the wave functions by using $H$

 

[hilbert2]

As far as I am concerned the wave function is definded in the derivation of the schrødinger equation as $\Psi=\Psi_oe^{kx-\omega t}$. Of course there are others but thoose are derived explicitly for a given problem. For example the wave functions for the hydrogen molecule. I don't get why they can derive the wave functions by using $H$

What you're describing are momentum eigenstates, not bound states that DFT is used for finding.