quantum chemistry - Density Functional Definition

Is it wrong to say that Density Functional means that Electron Density is a function of the orbitals (wave function) of all electrons in 3 dimensions, if so, why ?

Answer

In general, a functional $F$ is a mapping from an arbitrary set $\mathcal{X}$ of functions to the set of complex numbers $\mathbb{C}$ or the set of real numbers $\mathbb{R}$: $$F : \mathcal{X} \mapsto \mathbb{R}.$$ or $$F : \mathcal{X} \mapsto \mathbb{C}.$$

For example, if you consider $\mathcal{X}$ as the set of polynomials with real coefficients, you can define a functional $F$ as $$F[f] = \int_0^1f(x)\,dx$$ i.e. your functional $F$ takes a polynomial function $f\in\mathcal{X}$ (for example $f(x)=3x+1/2$) as an argument and returns a scalar (2 for $f(x)=3x+1/2$, as you can easily verify).

A density functional is simply a functional $F[f]$ where the argument $f$ is the electron density $\rho(\vec{r})$ (i.e. a density functional is a functional of the electron density). For example Hohenberg and Kohn showed that the energy $\epsilon$ of a quantum system is a functional of the density $$\epsilon=E[\rho]$$ This means that when you plug the electron density of your system $\rho(\vec{r})$ into the energy functional $E[\rho]$ you get a number $\epsilon$, which is the energy of your system. The whole energy functional is not known explicitly, but some of its components are known. For example for the external potential energy we have $$V[\rho] = \int v(\vec{r})\rho(\vec{r})d\vec{r}$$ and for the Coulomb interaction between electrons we have $$J[\rho] = \frac{1}{2}\iint \frac{\rho(\vec{r})\rho(\vec{r}')}{|\vec{r}-\vec{r}'|}\,d\vec{r}d\vec{r}'$$ which are clearly functionals of the electron density.