The E = E[N, v] functional and the linear response function: a conceptual DFT viewpoint

Abstract

The energy (E) versus number of electrons (N) and external potential (v) functional E = E[N, v], which has proved to be of fundamental importance in conceptual density functional theory through the response functions it generates, has been examined concentrating on the concavity of the E = E[v] functional as opposed to the convexity of the E = E(N) function. The concavity of the E = E[v] functional is reflected in negative values of the diagonal elements of the linear response function $\chi ({\mathbf{r}},{\mathbf{r}}')$ comprising the second functional derivative of E with respect to $v({\mathbf{r}})$, whereas no sign can be retrieved for the off-diagonal elements. These findings are in agreement with recent computational studies in extracting the chemical content of the linear response function. The results for the diagonal elements can easily be interpreted in terms of electron depletion from regions where the potential is increased and are easily retrieved via the independent particle model expression and in agreement with the diagonal elements of the most general expression for the linear response function. These concavity-related issues are put in contrast with the positive $(\partial ^2E/\partial N^2)_v$ derivative, resulting from the convexity of the E(N) function, in line with the positivity of the chemical hardness as resulting from the I versus A (ionization vs. electron affinity) ratio. The first-order derivative $(\delta E/\delta v({\mathbf{r}}))_N$ is discussed within an iso-electronic atom series for which it is shown in detail that the potentials can be ordered univocally, through their Z-dependence. The sign of the slope of the E = E[v] curve is in agreement with the positivity of $\rho ({\mathbf{r}})$. The result for the E = E[v] functional is put in contrast with the negative $(\partial E/\partial N)_v$ derivative (identified as minus the electronegativity) for the E = E(N) function retrieved on the basis of experimental and theoretical data on ionization energies and electron affinities.

Appendix: The energy of iso-electronic atoms

Consider a neutral many electron atom, with nuclear charge Z and number of electrons N ($Z=N$). By virtue of the Hellman–Feynman theorem [56, 57], the derivative of the energy as a function of Z can be written as

$$\frac{\partial E}{\partial Z} = \left\langle \varPsi \left| \frac{\partial H(Z)}{\partial Z}\right| \varPsi \right\rangle$$

(23)

which reduces in the case of an atom to

$$\frac{\partial E}{\partial Z} = -\int \rho ({\mathbf{r}};Z)\frac{1}{r}{\text {d}}{\mathbf{r}}.$$

(24)

As compared to a reference state $E(0)=0$ an atom with nuclear charge zero, implying that all electrons are at infinity [63], the energy $E=E(Z)$ can be written as

$$\begin{aligned} E&= \int _0^Z \left( \frac{\partial E}{\partial Z}\right) {\text {d}}Z \\&= E(Z)-E(0) \\&=-\int _0^Z\int \rho ({\mathbf{r}},Z)\frac{1}{r}{\text {d}}{\mathbf{r}}{\text {d}}Z. \end{aligned}$$

(25)

retrieving Foldy’s result for the exact non-relativistic energy of a ground-state atom with atomic number Z and N electrons [58]. (In fact, this zero value as Z-reference could be replaced by the $Z_{{\text {min}}}$ in Fig. 3. This choice, however, does not alter the remaining part and the conclusion of the discussion.)

Consider now two iso-electronic systems with $Z=Z_1$ and $Z=Z_2$ where we choose $Z_2>Z_1$. One then has

$$E(Z_2)-E(Z_1) = -\int _{Z_1}^{Z_2}\int \rho ({\mathbf{r}},Z)\frac{1}{r}{\text {d}}{\mathbf{r}}{\text {d}}Z.$$

(26)

Due to the spherical nature of the electron distribution assumed throughout, one gets

$$\begin{aligned} E(Z_2)-E(Z_1)&= -\int _{Z_1}^{Z_2}\int \rho (r,Z)\frac{1}{r} r^2\sin \theta {\text {d}} \theta {\text {d}}\phi {\text {d}}r{\text {d}}Z \\&= -4\pi \int _{Z_1}^{Z_2}\int _0^\infty \rho (r,Z)r{\text {d}}r{\text {d}}Z \\&= -4\pi \int _0^\infty r{\text {d}}r\int _{Z_1}^{Z_2}\rho (r,Z){\text {d}}Z. \end{aligned}$$

(27)

Denoting the integral over ${\text {d}}Z$ as $f(r;Z_1,Z_2)$, we obtain

$$E(Z_2) - E(Z_1) = -4\pi \int _0^\infty f(r;Z_1,Z_2)r{\text {d}}r.$$

(28)

As it cannot be that for all r values the density $\rho (r,Z)=0$, $f(r;Z_1,Z_2)$ will be positive for a number of r values so that the integrand of the integral is positive. Denoting this integral as $F(Z_1,Z_2)$, we obtain

$$E(Z_2)-E(Z_1) = -4\pi F(Z_1,Z_2)<0$$

(29)

or $E(Z_2)<E(Z_1)$, yielding the result that upon increasing Z, the energy E decreases (i.e. becomes more negative).