(Solved) Is Kohn-Sham approach in DFT compatiable with Hohenberg-Kohn theorem?

Question

From DFT theory, the Euler equation of an interacting system is:

$$\begin{align*} \frac{\delta E_{v}[\rho]}{\delta \rho(\vec{r})} &= \frac{\delta}{\delta \rho} \left[T_{s}[\rho]+\int \rho(\vec{r})v(\vec{r}) d \vec{r} +\frac{1}{2}\int\int \frac{\rho(\vec{r}) \rho(\vec{r}')}{|\vec{r}-\vec{r}'|}d \vec{r} d \vec{r}'+E_{xc}[\rho] \right] \\ &=\frac{\delta T_{s}[\rho]}{\delta \rho}+v(\vec{r})+\int \frac{\rho(\vec{r}')}{|\vec{r}-\vec{r}'|} d \vec{r}'+\frac{E_{xc}[\rho]}{\delta \rho}=\mu \end{align*}$$

In Kohn-Sham approach, we introduce the auxiliary non-interacting system such that

$$ \frac{\delta E_{v_{s}}}{\delta \rho(\vec{r})}=\frac{\delta T_{s}[\rho]}{\delta \rho} + v_{s}(\vec{r}) = \mu $$

Then, we assume that 1st and 2nd equations are the same. In other words, two variation problems have the same solution, ground-state electron density $ \rho_{0}=\rho_{s0} $.

That's why we can get the ground-density from the auxiliary system i.e. $$\rho_{0} = \rho_{s0} = \sum_{i=1}^{occ} |\varphi_{i}(\vec{r})|^{2}$$

According to Hohenberg-Kohn theorem, an electron density $\rho$ has the one-to-one correspondence with a unique external potential $v[\rho]$.

However, $\rho_{0} = \rho_{s0}$ imply two different external potentials, or $v_{0} \neq v_{s}$.

I think that this would violates HK theorem. What mistakes did I made?

Answer 1

The central idea of density functional theory can be outlined in just one line

$$\text{external potential} \rightarrow \text{tiem-dependent Schrodinger equation} \rightarrow \text{wavefuntion} \rightarrow \text{density}.$$

We know the total energy of the system can be expressed as the functional of the wavefunction and hence the functional of the density.

$$ E[\rho(r)]=E_k[\rho(r)]+E_{ee}[\rho(r)]+\int\rho(r) V_{ext}(r) dr .$$

where the first part on the right hand is kinetic energy functional of the system, the second part is the interaction part functional and the third part is related to the external potential.

The Kohn-Sham approach just maps your original interacting many-body system (many-body problem) into a non-interacting many-body system (single particle problem) by introducing the exchange-correlation functional

$$ E[\rho(r)] = E'_{k}[\rho(r)] + E_{ee}[\rho(r)]+\int\rho(r)V_{ext}dr + \{E_k[\rho(r)]-E_{k}'(\rho(r))\} = E'_{k}[\rho(r)] + \int\rho(r)V_{ext}dr + E_{xc}[\rho(r)]$$

if you take the variation for the modified total energy functional and then you will get the Kohn-Sham equation, which is just a single-particle Schrodinger equation. In K-S DFT, all the difficult parts are how to find the $E_{xc}[\rho(r)]$.

So it doesn't spoil the H-K theorems.

Here is a good tutorial for the Kohn-Sham approach.

Hope it helps.

Answer 2

The non-interacting system has the same density as the real system by construction. The variational problem is to find the external potential that has mimics the effective potential of the real system. The H-K theorem is not violated because the ground-state electron density is a constraint.

Answer 3

Given the same density, $v_0 \neq v_s $ and it doesn't mean that it violates the HK theorem. You can think of the strength of electron-electron interaction can be scaled by a parameter $\lambda$, $0 \le \lambda \le 1$. For every $\lambda$, the potential is different from the others, but the $\lambda$ connects these two potentials in a continuous way(better assume so).

Answer 4

I had a silly mistake. I should've considered the proof more seriously.

Write two equations from the $\hat{H}$ and $\hat{H}_{s}$variation principle.

$$ E < \langle{\Psi_{s}|\hat{H}|\Psi_{s}}\rangle = E_{s} + \langle \Psi_{s}|\hat{H}-\hat{H_{s}}|\Psi_{s} \rangle $$

$$ E_{s} < \langle{\Psi|\hat{H}_{s}|\Psi}\rangle = E + \langle \Psi|\hat{H}_{s}-\hat{H}|\Psi \rangle $$

Assume that $ \rho_{s0}= \rho_{0} $. In other words, let $\rho_{s0}$ be the ground state electron density from KS approach. Then,

$$ E < E_{s} + \langle \Psi_{s}|\hat{V}_{ee}|\Psi_{s} \rangle + \int (v-v_{s})\rho_{0} d \vec{r} $$

$$ E_{s} < E -\langle \Psi|\hat{V}_{ee}|\Psi \rangle + \int (v_{s}-v)\rho_{0} d \vec{r} $$

Adding two expression makes no contradiction as HK proof do, because $\Psi \neq \Psi_{s}$ in general.

HK theorem guarantees the one-to-one correspondence between the external potential and the electronic density in the set of many-body systems with the same particle-particle interaction.