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 ?
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6$\begingroup$ I don't really understand your question. The electron density $\varrho (\vec{r})$ is defined to be a function of the many-particle wave function, i.e. $\varrho (\vec{r}) = \langle \Psi | \hat{\varrho} (\vec{r}) | \Psi \rangle$, where $\hat{\varrho} (\vec{r})$ is the density operator, $\vec{r}$ is the position vector, and $\Psi$ is the many-particle wave-function. But that has nothing to do with the definition of a density functional. A density functional in its most general meaning is just a functional of the aforementioned electron density. Maybe you should provide a bit more context. $\endgroup$– PhilippCommented May 23, 2016 at 8:20
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1$\begingroup$ @Mr.Why The energy is a functional of the wavefunction, i.e. $E=E[\psi]$. $\endgroup$– user23061Commented Jun 7, 2016 at 9:24
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$\begingroup$ You are right . $\endgroup$– M.ghorabCommented Jun 7, 2016 at 23:45
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$\begingroup$ Eignvalue is not as same as function $\endgroup$– M.ghorabCommented Jun 10, 2016 at 12:01
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2 Answers
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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.
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$\begingroup$ (i.e. density functional is a functional of the electron density). So cannot it be said that density functional means that density is it self a functional of some thing else (orbitals=wavefunctions) ? $\endgroup$– M.ghorabCommented Aug 2, 2017 at 21:02
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$\begingroup$ The density functional is distinct from the density itself. The density maps from a position vector (3 real numbers) to a single real number $\rho:\;\mathbb{R}^3 \rightarrow \mathbb{R}$. The density functional maps from an entire function to a single real number $F:\;\mathcal{X} \rightarrow \mathbb{R}$. Mapping from a function to something else is what makes something a functional. $\endgroup$ Commented Aug 2, 2017 at 22:36
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$\begingroup$ Also, the orbitals themselves are fictitious; there are infinitely many choices of orbitals for the same wavefunction but some are more sensible than others. Also, although the density can be expressed in terms of the wavefunction: $\rho(r) = |\psi(r)|^2$ both side just equal a number and require us to input a position vector so the density is still a function mapping from vectors to numbers, as is the wavefunction: $\rho : \mathbb{R}^3\rightarrow \mathbb{R}$, $\psi : \mathbb{R}^3\rightarrow \mathbb{C}$. $\endgroup$ Commented Aug 2, 2017 at 22:42
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Electron density, is defined as the probability density of finding any of the $n$ electrons with arbitrary spin at some point $\vec{r}_{1}$ in space, $$ \newcommand{\el}{_\mathrm{el}} \newcommand{\dif}{\mathrm{d}} \rho(\vec{r}_{1}) := n \sum_{m_{s1}} \sum_{m_{s2}} \dotsi \sum_{m_{sn}} \iint \dotsi \int | ψ\el(\vec{q}_{1}, \vec{q}_{2}, \dotsc, \vec{q}_{n}) |^{2} \dif \vec{r}_{2} \dif \vec{r}_{3} \dotsb \dif \vec{r}_{n} \, . $$
From the first Hohenberg–Kohn theorem, it is known that electronic energy is a functional of electron density, $$ E\el = E\el[\rho(\vec{r}_{1})] \, . $$ i.e. electronic energy $E\el$ is a function that takes another function, namely, electron density $\rho(\vec{r}_{1})$, as its input argument and returns a scalar value (real number). So, density functional is a functional of electron density that returns (possibly approximate) electronic energy or a part of it, if $E\el$ is subdivided into parts.
