---
title: Van der Waals Forces
module: Molecules and Chemical Bonding
moduleNumber: 1
lessonNumber: 4
order: 104
summary: >
  The bond of last resort acts between all atoms, even closed-shell noble gases
  with no permanent moment. This lesson separates the three van der Waals
  contributions — Keesom orientation, Debye induction, and London dispersion —
  derives the London 1/r⁶ attraction from the coupled-oscillator and
  second-order perturbation pictures, and assembles the Lennard-Jones potential
  to compute the equilibrium spacing and cohesive energy of the noble-gas
  crystals, argon in particular.
topics: [Molecules and Chemical Bonding]
draft: false
sources:
  - book: Ashcroft & Mermin
    ref: "Ch. 20 — The Cohesive Energy; molecular crystals"
  - book: Kittel
    ref: "Ch. 3 — Crystal Binding; van der Waals interaction"
  - book: Hook & Hall
    ref: "Ch. 2 — The Bonding of Solids; inert-gas solids"
---

The [ionic, covalent, and metallic
bonds](/condensed-matter/molecules-and-bonding/bonding-mechanisms) all involve
electron transfer or sharing. A weaker attraction acts between every pair of
atoms and molecules, including the closed-shell noble gases that neither share nor
transfer electrons. This **van der Waals** attraction holds argon and methane
crystals together, condenses the noble gases at low temperature, and sets the
short-range shape of every intermolecular potential. It has three physically
distinct sources, all decaying as $1/r^6$.

## The three contributions

Write $p$ for a permanent electric dipole moment and $\alpha$ for the electric
polarizability, the constant relating an induced dipole to the field that induces
it, $p_{\text{ind}} = \alpha E$. The three long-range attractions between two
molecules a distance $r$ apart are the following.

- **Keesom (orientation) energy** — between two permanent dipoles. Each dipole
  torques the other into a favorable alignment; thermal motion partly randomizes
  the orientation, and the Boltzmann average of the dipole-dipole energy leaves a
  net attraction
  $$
  U_{\text{K}} = -\frac{1}{3 k_B T}\,\frac{p_1^2\,p_2^2}{(4\pi\epsilon_0)^2\,r^6}.
  $$
  It is temperature-dependent because it relies on thermal averaging.
- **Debye (induction) energy** — between a permanent dipole and the dipole it
  induces in a polarizable neighbor,
  $$
  U_{\text{D}} = -\frac{p_1^2\,\alpha_2 + p_2^2\,\alpha_1}{(4\pi\epsilon_0)^2\,r^6}.
  $$
  It survives at $T=0$ and needs only one of the molecules to be polar.
- **London (dispersion) energy** — between the instantaneous dipoles of two
  neutral, nonpolar atoms, present even when both permanent moments vanish. It is
  the only van der Waals term available to the noble gases and is usually the
  largest of the three for small molecules.

Only the London term requires quantum mechanics; the other two follow from
classical electrostatics plus thermal averaging. The remainder of the lesson
derives the London energy, because it is the term that bonds the noble-gas
solids.

## The London dispersion energy

A noble-gas atom has zero average dipole moment, but its electron cloud
fluctuates, so its instantaneous moment is nonzero. A momentary dipole on one atom
produces a field that polarizes the neighbor, and the induced dipole is correlated
with the original in just the way that lowers the total energy. The correlation of
zero-average fluctuations produces a nonzero average attraction.

$$
% caption: A fluctuating instantaneous dipole on the left atom polarizes the
% right atom; the induced dipole points so that the pair attracts, and the
% correlation persists on average even though each moment averages to zero.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % left atom
  \draw[black] (0,0) circle (0.8);
  \fill[acc, opacity=0.25] (0.25,0) circle (0.55);
  \draw[acc, very thick, ->] (-0.5,0) -- (0.7,0);
  \node[acc, anchor=south] at (0,0.95) {instantaneous};
  % right atom
  \draw[black] (3.6,0) circle (0.8);
  \fill[black, opacity=0.25] (3.35,0) circle (0.55);
  \draw[black, very thick, dashed, ->] (3.1,0) -- (4.3,0);
  \node[black, anchor=south] at (3.6,0.95) {induced};
  % separation
  \draw[black, <->] (0,-1.15) -- (3.6,-1.15) node[midway, below] {r};
  \draw[black, dashed] (0.8,0) -- (2.8,0);
\end{tikzpicture}
$$

The cleanest derivation models each atom as a one-dimensional charge $-e$ on a
spring of natural frequency $\omega_0$, the **Drude oscillator**. With the atoms
on the $x$ axis a distance $r$ apart and displacements $x_1,x_2$ along the line of
centers, the two oscillators couple through the dipole-dipole interaction, giving
the Hamiltonian

$$
\hat H = \sum_{i=1,2}\left(\frac{p_i^2}{2m} + \tfrac12 m\omega_0^2 x_i^2\right)
- \frac{2ke^2}{r^3}\,x_1 x_2 .
$$

The normal coordinates $x_\pm = (x_1 \pm x_2)/\sqrt 2$ diagonalize the coupled
system into two independent oscillators with frequencies

$$
\omega_\pm = \omega_0\sqrt{\,1 \mp \frac{2ke^2}{m\omega_0^2\,r^3}\,}.
$$

The zero-point energy of the coupled pair is $\tfrac12\hbar(\omega_+ + \omega_-)$.
Expanding the square roots to second order in the small coupling $\beta =
2ke^2/(m\omega_0^2 r^3)$, the linear terms cancel in the sum and the shift
relative to the two uncoupled atoms is

$$
\Delta E = \tfrac12\hbar(\omega_+ + \omega_- - 2\omega_0)
\approx -\frac{\hbar\omega_0}{8}\,\beta^2
= -\frac{\hbar\omega_0\,\alpha^2}{2\,r^6}\equiv -\frac{C_6}{r^6},
$$

using the Drude polarizability $\alpha = e^2/(m\omega_0^2)$ (in Gaussian units,
where $k=1$). The energy is negative — an attraction — and scales as $1/r^6$. The
coefficient $C_6 = \tfrac12\hbar\omega_0\alpha^2$ is set by the polarizability and
a characteristic transition energy $\hbar\omega_0$ of order the atom's ionization
energy.

> **Result (London dispersion law).** Two neutral, spherically symmetric atoms
> attract with an energy
> $$
> U_{\text{L}}(r) = -\frac{C_6}{r^6},\qquad C_6 \approx \tfrac12\hbar\omega_0\,\alpha^2 ,
> $$
> arising from the correlated zero-point motion of their charge clouds. The
> attraction requires no permanent moment and vanishes only as $\hbar\to 0$.

The same result follows from second-order perturbation theory with the
dipole-dipole interaction $V \propto 1/r^3$ as the perturbation. The first-order
shift $\langle 0|V|0\rangle$ vanishes because each atom has no permanent moment.
The second-order shift is a sum over virtual excitations of both atoms,

$$
\Delta E^{(2)} = \sum_{n\neq 0}\frac{|\langle n|V|0\rangle|^2}{E_0 - E_n},
$$

in which every denominator $E_0 - E_n$ is negative because the ground state is
lowest. The sum is therefore negative, $|V|^2 \propto 1/r^6$, and the dispersion
attraction is recovered with the same $1/r^6$ dependence.

$$
% caption: Second-order perturbation theory: the dipole-dipole coupling admits no
% first-order shift, and every virtual excitation to a state above the ground
% state contributes a negative term, so the net energy shift is an attraction.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % ground state
  \draw[acc, very thick] (0,0) -- (2.4,0);
  \node[acc, anchor=east] at (-0.05,0) {ground state};
  % excited manifold
  \draw[black, thick] (0,2.0) -- (2.4,2.0);
  \draw[black, thick] (0,2.5) -- (2.4,2.5);
  \draw[black, thick] (0,3.0) -- (2.4,3.0);
  \node[black, anchor=west] at (2.5,2.5) {excited states};
  % virtual transitions
  \draw[black, ->] (1.2,0) -- (1.2,2.0);
  \draw[black, ->] (0.7,0) -- (0.7,2.5);
  \draw[black, ->] (1.7,0) -- (1.7,3.0);
  % lowered level
  \draw[acc, very thick, dashed] (0,-0.7) -- (2.4,-0.7);
  \draw[acc, ->] (2.9,0) -- (2.9,-0.7);
  \node[acc, anchor=west] at (2.95,-0.35) {net lowering};
\end{tikzpicture}
$$

## The Lennard-Jones potential

At short range the electron clouds overlap and the exclusion principle produces a
steep repulsion. Combining it with the London attraction gives a total potential.
The repulsion has no simple closed form, but the twelfth power is a computationally
convenient choice that makes the whole potential a function of $r^{-6}$ and its
square.

> **Definition (Lennard-Jones 6-12 potential).**
> $$
> U(r) = 4\epsilon\left[\left(\frac{\sigma}{r}\right)^{12} - \left(\frac{\sigma}{r}\right)^{6}\right],
> $$
> where $\epsilon$ is the depth of the well and $\sigma$ is the separation at
> which $U=0$. The $r^{-6}$ term is the physical London attraction; the $r^{-12}$
> term is an empirical stand-in for the exclusion repulsion.

Setting $\d U/\d r = 0$ locates the minimum at $r_m = 2^{1/6}\sigma \approx
1.122\,\sigma$, where $U(r_m) = -\epsilon$. The two parameters $\epsilon$ and
$\sigma$ are fit to gas-phase data (second virial coefficient, viscosity) and then
used to predict the solid.

$$
% caption: The Lennard-Jones potential in reduced units: the depth epsilon sits
% at the minimum near 1.12 sigma, and sigma is the inner separation where the
% curve crosses zero between the repulsive wall and the attractive tail.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (1.6,0) -- (7.6,0) node[right] {$r$};
  \draw[->, black] (2.47,-2.0) -- (2.47,2.6) node[above] {$U(r)$};
  % LJ curve, reduced r in [0.96, 2.75], scaled x=2.6*r, y=1.55*U
  \draw[acc, very thick, domain=0.96:2.75, samples=160, variable=\x]
    plot ({2.6*\x},{1.55*4*(pow(1/\x,12) - pow(1/\x,6))});
  % zero crossing at r=sigma -> x=2.6
  \fill[black] (2.6,0) circle (1.6pt);
  \draw[black, dashed] (2.6,0) -- (2.6,-0.5);
  \node[black, anchor=west] at (2.72,-0.3) {sigma};
  % minimum at r=1.122 -> x=2.917, y=-1.55
  \fill[black] (2.917,-1.55) circle (2pt);
  \draw[black, dashed] (2.917,-1.55) -- (4.6,-1.55);
  \node[black, anchor=west] at (4.65,-1.55) {depth epsilon};
\end{tikzpicture}
$$

## Cohesive energy of the noble-gas crystals

The noble gases (except helium) crystallize in the face-centered cubic structure.
The cohesive energy is the sum of Lennard-Jones pair energies over the lattice.
With nearest-neighbor spacing $R$ and each pair separation written as $p_{ij}R$
for dimensionless $p_{ij}$, the total energy of $N$ atoms is

$$
U_{\text{tot}} = 2N\epsilon\left[A_{12}\left(\frac{\sigma}{R}\right)^{12}
- A_{6}\left(\frac{\sigma}{R}\right)^{6}\right],
\qquad
A_{n} = {\sum_{j}}'\, p_{ij}^{-n},
$$

where the primed sums run over all neighbors of a fixed atom. For the FCC lattice
the lattice sums converge to $A_{12} = 12.13$ and $A_6 = 14.45$. Minimizing over
$R$ gives the equilibrium spacing and cohesive energy per atom,

$$
\frac{R_0}{\sigma} = \left(\frac{2A_{12}}{A_6}\right)^{1/6} = 1.09,
\qquad
\frac{U_{\text{tot}}(R_0)}{N} = -\frac{A_6^2}{2A_{12}}\,\epsilon = -8.6\,\epsilon .
$$

The predicted nearest-neighbor spacing $R_0 = 1.09\,\sigma$ is independent of the
element, a nontrivial prediction that the measured spacings confirm.

> **Worked example.** For argon the Lennard-Jones parameters fit from gas data are
> $\epsilon = 0.0104\ \text{eV}$ and $\sigma = 0.340\ \text{nm}$. The lattice
> model predicts
> $$
> R_0 = 1.09\,\sigma = 0.371\ \text{nm},\qquad
> \frac{U_{\text{tot}}}{N} = -8.6\,\epsilon = -0.089\ \text{eV/atom}.
> $$
> The measured nearest-neighbor distance in solid argon is $0.376\ \text{nm}$ and
> the cohesive energy is $0.080\ \text{eV/atom}$. The small overbinding is the
> neglected zero-point energy of the atoms in their potential wells, a quantum
> correction that grows for the lighter, more weakly bound neon.

$$
% caption: Predicted (8.6 epsilon) versus measured cohesive energy per atom for
% the heavy noble-gas solids; agreement improves down the column as the atoms
% grow heavier and zero-point corrections shrink.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (0,4.0) node[above, black] {cohesive energy};
  \draw[black] (0,0) -- (8.4,0);
  % scale: 20 units per eV; values Ne .027/.020, Ar .089/.080, Kr .120/.116, Xe .172/.17
  % Ne
  \fill[acc!30, draw=acc] (0.6,0) rectangle (1.2,0.54);
  \fill[black, draw=black] (1.25,0) rectangle (1.85,0.40);
  \node[black, anchor=north] at (1.2,-0.05) {Ne};
  % Ar
  \fill[acc!30, draw=acc] (2.6,0) rectangle (3.2,1.78);
  \fill[black, draw=black] (3.25,0) rectangle (3.85,1.60);
  \node[black, anchor=north] at (3.2,-0.05) {Ar};
  % Kr
  \fill[acc!30, draw=acc] (4.6,0) rectangle (5.2,2.40);
  \fill[black, draw=black] (5.25,0) rectangle (5.85,2.32);
  \node[black, anchor=north] at (5.2,-0.05) {Kr};
  % Xe
  \fill[acc!30, draw=acc] (6.6,0) rectangle (7.2,3.44);
  \fill[black, draw=black] (7.25,0) rectangle (7.85,3.40);
  \node[black, anchor=north] at (7.2,-0.05) {Xe};
  % legend
  \fill[acc!30, draw=acc] (0.4,3.7) rectangle (0.75,3.9);
  \node[acc, anchor=west] at (0.8,3.8) {predicted};
  \fill[black, draw=black] (3.0,3.7) rectangle (3.35,3.9);
  \node[black, anchor=west] at (3.4,3.8) {measured};
\end{tikzpicture}
$$

The Lennard-Jones fit reproduces the noble-gas solids to a few percent with two
parameters. The residual overbinding is systematic and understood: the atoms are
not fixed at $R_0$ but oscillate with zero-point energy, which the classical
lattice sum ignores. That zero-point contribution is largest for the lightest
atom, and for helium it is so large — larger than the shallow Lennard-Jones well —
that helium does not solidify at atmospheric pressure at any temperature, and its
liquid is the setting for superfluidity.

The van der Waals bond completes the catalog of molecular bonding. The overlap
repulsion and the pair potentials assembled here also fix the crystal-structure
problem: the same lattice sums, applied to the Coulomb energy of ions rather than
the London energy of neutral atoms, give the Madelung constant and the cohesive
energy of ionic solids, the subject of the [structure of
solids](/condensed-matter/crystal-structure/structure-of-solids).

## Summary

- Van der Waals attraction has three $1/r^6$ contributions: the temperature-
  dependent Keesom orientation energy between permanent dipoles, the Debye
  induction energy between a permanent and an induced dipole, and the London
  dispersion energy between fluctuating instantaneous dipoles.
- The London energy follows from the coupled Drude-oscillator model as
  $U_{\text{L}} = -\tfrac12\hbar\omega_0\alpha^2/r^6$, and equivalently from
  second-order perturbation theory, where every virtual-excitation term lowers
  the energy.
- The Lennard-Jones potential $U = 4\epsilon[(\sigma/r)^{12} - (\sigma/r)^6]$
  combines the London tail with an empirical $r^{-12}$ repulsion; its minimum is
  at $r_m = 2^{1/6}\sigma$ with depth $\epsilon$.
- Summing over the FCC lattice gives $R_0 = 1.09\,\sigma$ and a cohesive energy
  $8.6\,\epsilon$ per atom, matching argon ($R_0 = 0.371\ \text{nm}$,
  $0.089\ \text{eV/atom}$) to within the neglected zero-point energy.