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I Help with Hohenberg-Kohn Theorem

  1. Aug 29, 2017 #1

    TeethWhitener

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    Hi everyone, I was going through the derivation of the first Hohenberg-Kohn theorem (see here under eqn 1.31 for reference) when I noticed a once-obvious statement that didn't seem so obvious anymore. Namely, the proof requires that if you have two Hamiltonians ##H_1 \neq H_2##, then their ground states are not equal ##\Psi_1 \neq \Psi_2##. Maybe this statement is just completely obvious and I'm missing some PDE or linear algebra uniqueness theorem that I should have remembered from college, but for the life of me, I can't prove it. I know that 2nd order ODE's have unique solutions (given the proper assumptions), but why can't we have:
    $$H_1\Psi = E_1\Psi$$
    $$H_2\Psi = E_2\Psi$$
    where ##H_i## is a partial differential operator?
     
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  3. Aug 29, 2017 #2

    TeethWhitener

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    Nevermind, I think I figured it out. Let:
    $$H_1 = -\frac{\hbar^2}{2m}\nabla^2+V_1(\mathbf{r})$$
    $$H_2 = -\frac{\hbar^2}{2m}\nabla^2+V_2(\mathbf{r})$$
    Then subtracting ##(H_2-E_2)\Psi=0## from ##(H_1-E_1)\Psi=0## gives
    $$V_1(\mathbf{r})-V_2(\mathbf{r}) = E_1-E_2$$
    So the potentials are only allowed to differ by a constant.
     
  4. Sep 9, 2017 #3

    hilbert2

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    Note that here it's assumed that ##\Psi \neq 0##. If you have a potential like ##V(x) = \frac{1}{2}kx^2## for some dimensionless position coordinate ##x## (with the unit of ##x## being same order of magnitude as the std. dev. of the ground state probability distribution) and then change it to something like

    ##V(x) = \left\{\begin{smallmatrix}\frac{1}{2}kx^2 ,\hspace{20pt}x<1000\\ 10kx^2 ,\hspace{20pt}x\geq 1000\end{smallmatrix}\right.##

    the change doesn't really affect the ground state, as the wave function will be practically zero for values of ##x## larger than 1000 anyway.
     
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