Question on the Proof of Hohenberg -Kohn Theorem Between Equation (5) and Equation (6) of the the original paper titled as "Inhomogeneous Electron Gas" by P. Hohenberg and W. Kohn,  there is a sentence stating that: "

Now clearly (unless $v'(\mathbf{r}) - v(\mathbf{r})=\text{constant}$) $\Psi'$
  cannot be equal to $\Psi$ since they satisfy different Schrodinger
  equations."

This statement basically says: two different Hamiltonians have different ground-state wavefunctions. 
This is a necessary condition of Hohenberg-Kohn theorem 1.
But I'm not convinced by this statement, nor can I find a proof of it.
Is there a proof of this statement?
Thank you in advance for providing any references or comments.

Correction: After thinking about this question, I find that this statement has a strict condition: these two Hamiltonians are differ by an external potential $v(\mathbf{r})-v'(\mathbf{r})$, which is not a constant. With this condition, the proof is not hard.
 A: (The notations follow the original paper by Kohn and Hohenberg.)
Suppose there are two Hamiltonian $H_1 = T+U+V_1$ and $H_2=T+U+V_2$,
where
$T = \frac{1}{2}\int\nabla \psi^\dagger\nabla \psi d^3 r  $
$ U = \frac{1}{2}\int\frac{1}{|\mathbf{r}-\mathbf{r}'|}\psi^\dagger(\mathbf{r})\psi^\dagger(\mathbf{r}')\psi(\mathbf{r}')\psi(\mathbf{r})d^3 r d^3 r'$
and
$ V_i = \int v_i(\mathbf{r}) \psi^\dagger(\mathbf{r})\psi(\mathbf{r}) d^3 r $, $(i=1,2)$.
Note the precondition is: $v_1(\mathbf{r})$ and $v_2(\mathbf{r})$ differ by more than a constant.
We can prove $\hat{H}_1$ and $\hat{H}_2$ don't have the same ground-state wavefunction by reductio ad absurdum.

Let's assume they have the same ground-state $\Psi$, i.e., $\hat{H}_1 \Psi = E_1 \Psi$ and $\hat{H}_2 \Psi = E_2 \Psi$.
$\Rightarrow (\hat{H}_1 - \hat{H}_2) \Psi = (E_1 - E_2) \Psi$
$\Rightarrow (V_1-V_2) \Psi = \epsilon \Psi$, ($\epsilon = E_1 - E_2$ is a constant.)
Now plug in expression of $V_i$ and $\psi^\dagger(\mathbf{r})\psi(\mathbf{r})=\sum^{N}_{i=1}\delta(\mathbf{r}-\mathbf{r}_i)$:
$\Rightarrow \int \bigg(v_1(\mathbf{r}_i) - v_2(\mathbf{r}_i)\bigg) \sum_{i=1}^N \delta(\mathbf{r}-\mathbf{r}_i)\Psi(\mathbf{r}_1,\ldots,\mathbf{r}_N)d^3 r = \epsilon \Psi (\mathbf{r}_1,\ldots,\mathbf{r}_N)$
$\Rightarrow \Psi(\mathbf{r}_1,\ldots,\mathbf{r}_N) \bigg(\sum_{i=1}^N \big(v_1(\mathbf{r}_i) - v_2(\mathbf{r}_i) \big) - \epsilon\bigg) = 0 $
Since $\Psi(\mathbf{r}_1,\ldots,\mathbf{r}_N)$ is not zero, so:
$\sum_{i=1}^N \big(v_1(\mathbf{r}_i) - v_2(\mathbf{r}_i) \big) = \epsilon \Rightarrow v_1(\mathbf{r}) - v_2(\mathbf{r})=\text{constant}$, which contradict with our condition.
