Comprehensive Study Notes

Hellmann and Feynman independently applied Eq. (14.123) to molecules, taking $\lambda$ as a nuclear Cartesian coordinate. We now consider their results.

As usual, we are using the Born-Oppenheimer approximation, solving the electronic Schrödinger equation for a fixed nuclear configuration [Eqs. (13.4) to (13.6)]:

\(
\left(\hat{T}{\mathrm{el}}+\hat{V}\right) \psi{\mathrm{el}}=\left(\hat{T}{\mathrm{el}}+\hat{V}{\mathrm{el}}+\hat{V}{N N}\right) \psi{\mathrm{el}}=U \psi_{\mathrm{el}}
\)

where $\hat{T}{\mathrm{el}}, \hat{V}{\mathrm{el}}$, and $\hat{V}{N N}$ are given by (14.78), (14.79), and (14.89). The Hamiltonian operator $\hat{T}{\mathrm{el}}+\hat{V}{\mathrm{el}}+\hat{V}{N N}$ depends on the nuclear coordinates as parameters. If $x_{\delta}$ is the $x$ coordinate of nucleus $\delta$, the generalized Hellmann-Feynman theorem (14.123) gives

\(
\begin{equation}
\frac{\partial U}{\partial x{\delta}}=\int \psi{\mathrm{el}}^{} \frac{\partial\left(\hat{T}{\mathrm{el}}+\hat{V}{\mathrm{el}}+\hat{V}{N N}\right)}{\partial x{\delta}} \psi{\mathrm{el}} d \tau{\mathrm{el}}=\int \psi{\mathrm{el}}^{*}\left(\frac{\partial V{\mathrm{el}}}{\partial x{\delta}}+\frac{\partial V{N N}}{\partial x{\delta}}\right) \psi{\mathrm{el}} d \tau_{\mathrm{el}} \tag{14.126}
\end{}
\)

since $\hat{T}{\text {el }}$ is independent of the nuclear Cartesian coordinates, as can be seen from (14.78). [If we had omitted $V{N N}$ from $V$, we would have obtained Eq. (14.82), which was used in deriving the molecular electronic virial theorem.] From (14.79) we get in atomic units

\(
\begin{equation}
\frac{\partial V{\mathrm{el}}}{\partial x{\delta}}=-\sum{i} \frac{Z{\delta}\left(x{i}-x{\delta}\right)}{r_{i \delta}^{3}} \tag{14.127}
\end{equation}
\)

where $r{i \delta}$ is the distance from electron $i$ to nucleus $\delta$. To find $\partial V{N N} / \partial x_{\delta}$, we need to consider only internuclear repulsion terms that involve nucleus $\delta$. Hence

\(
\frac{\partial V{N N}}{\partial x{\delta}}=\frac{\partial}{\partial x{\delta}} \sum{\alpha \neq \delta} \frac{Z{\alpha} Z{\delta}}{\left[\left(x{\alpha}-x{\delta}\right)^{2}+\left(y{\alpha}-y{\delta}\right)^{2}+\left(z{\alpha}-z{\delta}\right)^{2}\right]^{1 / 2}}=\sum{\alpha \neq \delta} Z{\alpha} Z{\delta} \frac{x{\alpha}-x{\delta}}{R{\alpha \delta}^{3}}
\)

where $R{\alpha \delta}$ is the distance between nuclei $\alpha$ and $\delta$. Since $\partial V{N N} / \partial x{\delta}$ does not involve the electronic coordinates and $\psi{\mathrm{el}}$ is normalized, (14.126) becomes

\(
\begin{equation}
\frac{\partial U}{\partial x{\delta}}=-Z{\delta} \int\left|\psi{\mathrm{el}}\right|^{2} \sum{i} \frac{x{i}-x{\delta}}{r{i \delta}^{3}} d \tau{\mathrm{el}}+\sum{\alpha \neq \delta} Z{\alpha} Z{\delta} \frac{x{\alpha}-x{\delta}}{R{\alpha \delta}^{3}} \tag{14.128}
\end{equation}
\)

Consider the integral in (14.128). Using Eq. (14.8) with $B\left(\mathbf{r}{i}\right)=\left(x{i}-x{\delta}\right) / r{i \delta}^{3}$, we get

\(
\begin{equation}
\frac{\partial U}{\partial x{\delta}}=-Z{\delta} \iiint \rho(x, y, z) \frac{x-x{\delta}}{r{\delta}^{3}} d x d y d z+\sum{\alpha \neq \delta} Z{\alpha} Z{\delta} \frac{x{\alpha}-x{\delta}}{R{\alpha \delta}^{3}} \tag{14.129}
\end{equation}
\)

The variable $r_{\delta}$ is the distance between nucleus $\delta$ and point $(x, y, z)$ in space:

\(
r{\delta}=\left[\left(x-x{\delta}\right)^{2}+\left(y-y{\delta}\right)^{2}+\left(z-z{\delta}\right)^{2}\right]^{1 / 2}
\)

What is the significance of (14.129)? In the Born-Oppenheimer approximation, $U\left(x{\alpha}, y{\alpha}, z{\alpha}, x{\beta}, \ldots\right)$ is the potential-energy function for nuclear motion, the nuclear Schrödinger equation being

\(
\begin{equation}
\left(-\frac{\hbar^{2}}{2} \sum{\alpha} \frac{1}{m{\alpha}} \nabla{\alpha}^{2}+U\right) \psi{N}=E \psi_{N} \tag{14.130}
\end{equation}
\)

The quantity $-\partial U / \partial x{\delta}$ can thus be viewed [see Eq. (5.31)] as the $x$ component of the effective force on nucleus $\delta$ due to the other nuclei and the electrons. In addition to (14.129),
we have two corresponding equations for $\partial U / \partial y{\delta}$ and $\partial U / \partial z{\delta}$. If $\mathbf{F}{\delta}$ is the effective force on nucleus $\delta$, then

\(
\begin{gather}
\mathbf{F}{\delta}=-\mathbf{i} \frac{\partial U}{\partial x{\delta}}-\mathbf{j} \frac{\partial U}{\partial y{\delta}}-\mathbf{k} \frac{\partial U}{\partial z{\delta}} \tag{14.131}\
\mathbf{F}{\delta}=-Z{\delta} \iiint \rho(x, y, z) \frac{\mathbf{r}{\delta}}{r{\delta}^{3}} d x d y d z+\sum{\alpha \neq \delta} Z{\alpha} Z{\delta} \frac{\mathbf{R}{\alpha \delta}}{R_{\alpha \delta}^{3}} \tag{14.132}
\end{gather}
\)

where $\mathbf{r}_{\delta}$ is the vector from point $(x, y, z)$ to nucleus $\delta$,

\(
\begin{equation}
\mathbf{r}{\delta}=\mathbf{i}\left(x{\delta}-x\right)+\mathbf{j}\left(y{\delta}-y\right)+\mathbf{k}\left(z{\delta}-z\right) \tag{14.133}
\end{equation}
\)

and where $\mathbf{R}_{\alpha \delta}$ is the vector from nucleus $\alpha$ to nucleus $\delta$ :

\(
\mathbf{R}{\alpha \delta}=\mathbf{i}\left(x{\delta}-x{\alpha}\right)+\mathbf{j}\left(y{\delta}-y{\alpha}\right)+\mathbf{k}\left(z{\delta}-z_{\alpha}\right)
\)

Equation (14.132) has a simple physical interpretation. Let us imagine the electrons smeared out into a charge distribution whose density in atomic units is $-\rho(x, y, z)$. The force on nucleus $\delta$ exerted by the infinitesimal element of electronic charge $-\rho d x d y d z$ is [Eq. (6.56)]

\(
\begin{equation}
-Z{\delta} \frac{\mathbf{r}{\delta}}{r_{\delta}^{3}} \rho d x d y d z \tag{14.134}
\end{equation}
\)

and integration of (14.134) shows that the total force exerted on $\delta$ by this hypothetical electron smear is given by the first term on the right of (14.132). The second term on the right of (14.132) is clearly the Coulomb's law force on nucleus $\delta$ due to the electrostatic repulsions of the other nuclei.

Thus the effective force acting on a nucleus in a molecule can be calculated by simple electrostatics as the sum of the Coulombic forces exerted by the other nuclei and by a hypothetical electron cloud whose charge density $-\rho(x, y, z)$ is found by solving the electronic Schrödinger equation. This is the Hellmann-Feynman electrostatic theorem. The electron probability density depends on the parameters defining the nuclear configuration: $\rho=\rho\left(x, y, z ; x{\alpha}, y{\alpha}, z{\alpha}, x{\beta}, \ldots\right)$.

It is quite reasonable that the electrostatic theorem follows from the BornOppenheimer approximation, since the rapid motion of the electrons allows the electronic wave function and probability density to adjust immediately to changes in nuclear configuration. The rapid motion of the electrons causes the sluggish nuclei to "see" the electrons as a charge cloud rather than as discrete particles. The fact that the effective forces on the nuclei are electrostatic affirms that there are no "mysterious quantum-mechanical forces" acting in molecules.

Let us consider the implications of the electrostatic theorem for chemical bonding in diatomic molecules. We take the internuclear axis as the $z$ axis (Fig. 14.4). By symmetry the $x$ and $y$ components of the effective forces on the two nuclei are zero. [Also, one can show that the $z$ force components on nuclei $a$ and $b$ are related by $F{z, a}=-F{z, b}$ (Prob. 14.40). The effective forces on nuclei $a$ and $b$ are equal in magnitude and opposite in direction.]

From (14.134) and (14.133), the $z$ component of the effective force on nucleus $a$ due to the element of electronic charge in the infinitesimal region about $(x, y, z)$ is

\(
\begin{equation}
-Z{a} \rho\left[\left(z{a}-z\right) / r{a}^{3}\right] d x d y d z=Z{a} \rho\left(\cos \theta{a} / r{a}^{2}\right) d x d y d z \tag{14.135}
\end{equation}
\)

FIGURE 14.4 Coordinate system for a diatomic molecule. The origin is at $O$.
since $\cos \theta{a}=\left(-z{a}+z\right) / r{a}$. ( $z{a}$ is negative.) Similarly, the $z$ component of force on nucleus $b$ due to this charge is

\(
\begin{equation}
-Z{b} \rho\left(\cos \theta{b} / r_{b}^{2}\right) d x d y d z \tag{14.136}
\end{equation}
\)

A positive value of $(14.135)$ or $(14.136)$ corresponds to a force in the $+z$ direction, that is, to the right in Fig. 14.4. When the force on nucleus $a$ is algebraically greater than the force on nucleus $b$, then the element of electronic charge tends to draw $a$ toward $b$. Hence electronic charge that is binding is located in the region where

\(
\begin{equation}
Z{a} \rho\left(\cos \theta{a} / r{a}^{2}\right) d x d y d z>-Z{b} \rho\left(\cos \theta{b} / r{b}^{2}\right) d x d y d z \tag{14.137}
\end{equation}
\)

Since the probability density $\rho$ is nonnegative, division by $\rho$ preserves the direction of the inequality sign, and the binding region of space is where

\(
\begin{equation}
Z{a} \frac{\cos \theta{a}}{r{a}^{2}}+Z{b} \frac{\cos \theta{b}}{r{b}^{2}}>0 \tag{14.138}
\end{equation}
\)

When the force on $b$ is algebraically greater than that on $a$, the electronic charge element tends to draw $b$ away from $a$. The antibinding region of space is thus characterized by a negative value for the left side of (14.138). The surfaces for which the left side of (14.138) equals zero divide space into the binding and antibinding regions. [T. Berlin, J. Chem. Phys., 19, 208 (1951); J. Hinze, J. Chem. Phys., 101, 6369 (1994); Berlin's ideas are extended to polyatomic molecules in T. Koga et al., J. Am. Chem. Soc., 100, 7522 (1978); X. Wang and Z. Peng, Int. J. Quantum. Chem., 47, 393 (1993).]

Figures 14.5 and 14.6 show the binding and antibinding regions for a homonuclear and a heteronuclear diatomic molecule. As might be expected, the binding region for a homonuclear diatomic molecule lies between the nuclei. Charge in this region tends to draw the nuclei together. Bonding leads to a transfer of charge probability density into the region between the nuclei because of the overlap between the bonding AOs. Electronic

FIGURE 14.5 Cross section of binding and antibinding regions in a homonuclear diatomic molecule. To obtain the three-dimensional regions, rotate the figure about the internuclear axis.

FIGURE 14.6 Binding and antibinding regions for a heteronuclear diatomic molecule with $Z{b}>Z{a}$.

charge that is "behind" the nuclei (to the left of nucleus $a$ or to the right of nucleus $b$ in Fig. 14.5) exerts a greater attraction on the nucleus that is nearer to it than on the other nucleus and thus tends to pull the nuclei apart.

From the Hellmann-Feynman viewpoint, we seem to be considering chemical bonding solely in terms of potential energy, whereas the virial-theorem discussion involved both potential and kinetic energy. For the purposes of the Hellmann-Feynman discussion, we are imagining the electrons to be smeared out into a continuous charge distribution. Hence we make no reference to electronic kinetic energy. The use of the electrostatic theorem to explain chemical bonding has been criticized by some quantum chemists on the grounds that it hides the role of kinetic energy in bonding. [See the references cited after Eq. (13.66).]

In 1939, Feynman conjectured that the dispersion attraction between two molecules A and B at relatively large intermolecular distances is explainable as follows: The interactions between the two molecules cause the electron probability density of each molecule to be distorted and shifted somewhat toward the other molecule. The attractions of the nuclei in molecule A toward the distorted (polarized) electron density of molecule A and the attractions of the B nuclei toward the polarized B electron density then draw the two molecules together. In 1990, Hunt proved that the dispersion interaction between any two molecules in their ground electronic states results from the attractions of the nuclei in each molecule to the polarized electron density of the same molecule [K. L. C. Hunt, J. Chem. Phys., 92, 1180 (1990)].

For further applications of the Hellmann-Feynman electrostatic theorem, see B. M. Deb, ed., The Force Concept in Chemistry, Van Nostrand Reinhold, 1981.