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A problem about Kohn Hohenberg theorem

  1. Jul 4, 2017 #1

    georg gill

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    upload_2017-7-4_21-50-10.png

    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?
     
  2. jcsd
  3. Jul 4, 2017 #2

    hilbert2

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    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).
     
  4. Jul 4, 2017 #3

    georg gill

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    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 dont get why they can derive the wave functions by using ##H##
     
  5. Jul 4, 2017 #4

    hilbert2

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