Dependence of electron density functional on the number of electrons

Question

The Hohenberg-Kohn theorem, the foundation of DFT, is based on the energy functional of electron density $n(\mathbf{r})$ for a system of Coulomb-interacting electrons, placed in in the external electrostatic potential $V(\mathbf{r})$: $$ E[n]=-e\int d\mathbf{r}\:V(\mathbf{r})n(\mathbf{r})+\frac{e^2}2\int d\mathbf{r}d\mathbf{r}'\:\frac{n(\mathbf{r})n(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}+F[n] $$ The theorem states that:

1) the part $F[n]$ of the functional is universal [i.e. independent on $V(\mathbf{r})$];

2) the minimum of $E[n]$ is achieved on exact electron density $n(\mathbf{r})$ in the ground state of the system.

The proof of the theorem is based on existence of one-to-one mapping from the density $n(\mathbf{r})$ on the external potential $V(\mathbf{r})$ and then (via solution of the Schrodinger equation) on the ground state $N$-electron wave function $\Psi(\mathbf{r}_1\ldots\mathbf{r}_N)$.

However, it is clear that this mapping $n(\mathbf{r})\Rightarrow V(\mathbf{r})\Rightarrow\Psi(\mathbf{r}_1\ldots\mathbf{r}_N)$ is different for different numbers of electrons $N$, and thus we must use different universal functionals $F_N[n]$ for each $N$.

For example, two electrons ($N=2$), placed in the potential $V(\mathbf{r})$, will have the ground state wave function $\Psi(\mathbf{r}_1,\mathbf{r}_2)$, found from the Schrodinger equation: $$ \left\{-\frac{\hbar^2\nabla_1^2}{2m}-\frac{\hbar^2\nabla_2^2}{2m}-eV(\mathbf{r}_1)-eV(\mathbf{r}_2)+\frac{e^2}{|\mathbf{r}_1-\mathbf{r}_2|}\right\}\Psi(\mathbf{r}_1,\mathbf{r}_2)=E\Psi(\mathbf{r}_1,\mathbf{r}_2), $$ and their ground state electron density will be: $$ n_{N=2}(\mathbf{r})=2\int d\mathbf{r}_2\:|\Psi(\mathbf{r},\mathbf{r}_2)|^2 $$ (antisymmetry of the wave function is implied).

In the case of three electrons ($N=3$) in the same potential $V(\mathbf{r})$ we should solve \begin{eqnarray} \left\{-\frac{\hbar^2\nabla_1^2}{2m}-\frac{\hbar^2\nabla_2^2}{2m}-\frac{\hbar^2\nabla_3^2}{2m}-eV(\mathbf{r}_1)-eV(\mathbf{r}_2)-eV(\mathbf{r}_3)\right.\nonumber\\ \left.+\frac{e^2}{|\mathbf{r}_1-\mathbf{r}_2|}+\frac{e^2}{|\mathbf{r}_2-\mathbf{r}_3|}+\frac{e^2}{|\mathbf{r}_3-\mathbf{r}_1|}\right\}\Psi(\mathbf{r}_1,\mathbf{r}_2,\mathbf{r}_3)=E\Psi(\mathbf{r}_1,\mathbf{r}_2,\mathbf{r}_3)\nonumber \end{eqnarray} and find the density $$ n_{N=3}(\mathbf{r})=3\int d\mathbf{r}_2d\mathbf{r}_3\:|\Psi(\mathbf{r},\mathbf{r}_2,\mathbf{r}_3)|^2. $$

It is clear that the mappings $V(\mathbf{r})\Rightarrow n_{N=2}(\mathbf{r})$ and $V(\mathbf{r})\Rightarrow n_{N=3}(\mathbf{r})$ should be different, i.e. the same $V(\mathrm{r})$ will generate different density profiles in two- and three-electron systems.

Now the question: why the dependence of the universal density functional $F_N[n]$ on the number of electrons $N$ is usually ignored?

Of course, in DFT calculations for many-electron systems (e.g., heavy atoms, complex molecules or solids) the difference between $F_N[n]$ and $F_{N+1}[n]$ can be negligible, thus we can use the limiting functional $F_\infty[n]$. However, the same functionals are often used and tested even for helium atom ($N=2$), for neon atom ($N=10$) and for solids ($N\rightarrow\infty$), and nobody bothers about different $N$ [see, for example F.G. Cruz et al., J. Phys. Chem. A 102, 4911 (1998)].

Answer 1

It is true that $F_{N}[n]$ is different for all $N$, but nothing prevents defining

$$ F[n] = \left\{ \begin{array}{c} F_1[n], \int d{\bf r} n({\bf r})=1, \\ F_2[n], \int d{\bf r} n({\bf r})=2, \\ \ldots \end{array} \right. $$ and calling $F[n]$ the universal functional. In other words, the domain of the universal functional consists of densities integrating to arbitrary integer.

Hence, it is quite clear that the universal functional can be defined robustly. However, in your mapping $n \rightarrow V_{\rm ext} \rightarrow \Psi$, it is true that $V_{\rm ext}$ alone cannot be used to deduce the number of electrons. Perhaps a more robust way of defining this mapping would be $n \rightarrow (N, V_{\rm ext}) \rightarrow \Psi$.

It seems that everything in this question is related to electron number, so it it worthwhile to point out that the electron number is a very special 'direction' in the uncountable space of representable densities. This is already seen in the Euler-Lagrange equation, where a Lagrange multiplier (ending up to be the chemical potential) is used to fix the electron number. For example, perturbing $E[n]$, where n is the ground state density, ${\delta}E=0$ for density perturbation satisfying $\int d{\bf r} \delta n({\bf r})=0$, but $\delta E = \mu_{\pm}$, for $\int d{\bf r} \delta n({\bf r})=\pm \epsilon$.

Regarding your comment on the difference of $F_N$ and $F_{N+1}$ being negligible on solids, this is not true with semiconductors. In a semiconductor, removing an electron from valence band yields I (ionization potential) and adding an electron to conduction band yields A (electron affinity). Hence, $\frac{\delta F_N}{\delta N+} \neq \frac{\delta F_N}{\delta N-}$. In literature, this is called the derivative discontinuity on integer occupation, and this research is mostly focussed on the discontinuity of the exchange-correlation potential called $\Delta_{\rm xc}$. It holds that

$$ E_{g} = \frac{\delta E[n]}{\delta N+} - \frac{\delta E[n]}{\delta N-} = \frac{\delta E[n]}{\delta n({\bf r})+} - \frac{\delta E[n]}{\delta n({\bf r})-} = E_{\rm KS} + \Delta_{\rm xc}, $$ for any ${\bf r}$. Note that these definitions require extending the definition of the universal functional to ensemble densities. Some people like this, but others prefer not to define energies for fractional numbers. In solids, one is however fortunate in such way, that the integer additions/substractions of electrons become infinitesimals on density.