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.

  • 1
    $\begingroup$ This question seems to be tightly related and might be even a duplicate. Looks like, in general, different Hamiltonians can share even all eigenfunctions, so that the Hohenberg-Kohn theorem is just a special case. $\endgroup$
    – Wildcat
    Nov 29, 2016 at 16:47
  • $\begingroup$ @Wildcat Thank you. This discussion (physics.stackexchange.com/q/293013) is very helpful. And I suddenly realized that "the statement" has an additional requirement, which make the statement valid. $\endgroup$
    – user24399
    Nov 30, 2016 at 0:25

1 Answer 1


(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'$


$ 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.

  • $\begingroup$ Actually you don't need reductio ad absurdum: your argument shows that same ground state $\Rightarrow$ same Hamiltionian, to within a constant, which is the contrapositive of (and therefore logically equivalent too) the OP's assertion. Not a big criticism of course, but it is tidier style IMO to avoid RAA if one can. $\endgroup$ Nov 30, 2016 at 0:53

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