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The first Hohenberg-Kohn (HK) theorem: The external potential $v(\vec{r})$ is determined, within a trivial additive constant, by the ground-state electron density $\rho(\vec{r})$.

From basic quantum mechanics, we know that: $v(\vec{r})\rightarrow \hat{H} \rightarrow \psi_0\rightarrow \rho$. Accoridng to the first HK theorem, we can further know that $\rho \rightarrow v(\vec{r})\rightarrow \hat{H} \rightarrow \psi_0,\psi_1,\cdots$. In essence, the first HK theorem proves one-to-one mapping between the external potentials and the ground-state densities $\rho$ in many-electron systems.

The second HK theorem: There exists a universal functional of the density, $F_{HK}[\rho']$, such that for any $N$-representable density ($\textit{i.e.}$, any density that comes from some wavefunction for an $N$-electron system) $\rho(\vec{r})$, which yields a given number of electrons $N$, the energy functional is, $$E[\rho'] = F_{HK}[\rho']+\int \rho'(\vec{r})v(\vec{r}) d\vec{r} \geq E_g \tag{1} $$ in which $E_g$ is the ground-state energy and the equality holds when the density $\rho'(\vec{r})$ is the, possibly degenerate, ground-state density $\rho_0'(\vec{r})$ for the external potential $v(\vec{r})$.

From the two statements, I can't see any connection between the two theorems. So what's the relation between the two theorems? If $F_{HK}(\rho')$ is the functional of the ground-state density, I can build a connection between the two theorems. But the density in $F_{HK}[\rho]$ is not necessary ground-state density.

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2 Answers 2

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Using your notation, the definition for the universal functional is

$$ F_{HK}[\rho] = \left< \psi_0[\rho] \right| \hat{T} + \hat{W} \left| \psi_0[\rho] \right>, $$

where $\hat{T}$ and $\hat{W}$ are kinetic and electron-electron interaction operators, respectively. This definition is possible because of the one-to-one mapping between densities and their corresponding ground state wavefunctions (i.e., because $\psi_0$ is a functional of $\rho$), which I believe is the connection you are seeking.

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    $\begingroup$ According to the first HK theorem, $F_{HK}[\rho]$ should only be the functional of ground-state density? $\endgroup$
    – Jack
    Jan 21, 2021 at 0:48
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    $\begingroup$ Suppose $\rho'$ is any reasonable density (in a sense that can be made precise). It probably won't be the ground state density for the potential $v(r)$. However, it will be the ground state density for some other potential $v'(r)$, and it will correspond to a ground state wave function $\psi_0' = \psi_0[\rho']$. So, $F_{HK}[\rho']$ makes sense for all densities, not just the ground state density. You can find the ground state energy with $$ E_g = \mathrm{min}_{\rho'} \left[ F_{HK}[\rho'] + \int v(r) \rho'(r) dr \right]. $$ $\endgroup$
    – wcw
    Jan 21, 2021 at 1:29
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    $\begingroup$ So the $\rho'$ of equation (1) in my post can be considered a ground-state denisty corresponding to a external potential $v'(\vec{r})$ (guaranteed by the first HK theorem). If $v'(\vec{r})-v(\vec{r}) \neq C$, then $\rho'$ can be considered as density rather than ground-denisty of $v(\vec{r})$ (What the second HK wants to solve). Is that right? $\endgroup$
    – Jack
    Jan 21, 2021 at 1:45
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    $\begingroup$ Yes, that sounds right $\endgroup$
    – wcw
    Jan 21, 2021 at 1:46
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A formal connection is that the first theorem is used in the proof of the second one. Indeed, the second is a translation of the principle that $E[\Psi']$ has a minimum at the correct ground state wave function $\Psi$, using the one-to-one correspondence $\rho \leftrightarrow \Psi$ known from the first theorem.

The derivation can be found in the original paper by Kohn and Hohenberg (part I-2.). It's quite short and easy to read, so it's worth a look.

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    $\begingroup$ Can $F_{HK}[\rho]$ be a functional of non-ground-state electron density? $\endgroup$
    – Jack
    Jan 21, 2021 at 0:49

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