5
$\begingroup$

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.

$\endgroup$
2
  • 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

3
$\begingroup$

(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.

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.