# I Help with Hohenberg-Kohn Theorem

1. Aug 29, 2017

### TeethWhitener

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?

2. Aug 29, 2017

### TeethWhitener

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.

3. Sep 9, 2017

### hilbert2

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.