advanced28 min studyLesson 470 in Physics

Van der Waals forces and long-range potentials

Derive the attractive nonretarded dispersion tail from virtual dipole coupling, distinguish energy from force scaling, and interpret empirical short-range models cautiously.

Before you begin

On this page

The question for today

What you’ll understand

Explain why neutral ground-state species can attract without permanent dipoles.
Derive the inverse-sixth energy tail and inverse-seventh force.
Identify overlap, retardation, anisotropy, and empirical-model limits.

Two neutral ground-state atoms can attract even when neither has a permanent electric dipole. Their quantum charge fluctuations are correlated by the electromagnetic interaction, lowering the combined ground energy. In a useful distance window this produces the familiar inverse-sixth dispersion potential. We will derive the sign and distance power with second-order perturbation theory, distinguish that potential from its force, and then examine why an empirical Lennard–Jones curve is a convenient fitting model rather than an exact universal molecular interaction.

A zero average dipole can still couple virtually

Take two well-separated neutral species in nondegenerate ground states with no permanent electric dipole. Their electronic clouds overlap negligibly, and initially use the nonretarded electrostatic interaction. A multipole expansion has a dipole-dipole operator proportional to inverse separation cubed. Its first-order expectation vanishes for the chosen nonpolar ground-state product. The operator still connects that product to virtual excited configurations, so the absence of a permanent dipole does not make every interaction matrix element zero.

In second order, each excited configuration contributes its squared coupling divided by the ground-minus-excited energy difference. The denominator is negative, so the ground-state correction is attractive. Squaring an inverse-cubed matrix element gives inverse-sixth energy scaling. Summing the electronic response defines a positive dispersion coefficient C six for this isotropic ground-state setting. This is not classical thermal motion that must be supplied externally; quantum ground-state fluctuations and their virtual couplings are sufficient.

V_{dd}=\frac{\mathbf d_A\cdot\mathbf d_B-3(\mathbf d_A\cdot\widehat R)(\mathbf d_B\cdot\widehat R)}{4\pi\epsilon_0R^3},\qquad \Delta E^{(2)}=\sum_{n\ne0}\frac{|\langle n|V_{dd}|0\rangle|^2}{E_0-E_n}\simeq-\frac{C_6}{R^6}

R is species separation, d A and d B are dipole operators, and the ground product is nondegenerate. C six has units of energy times length to the sixth power. The asymptotic expression assumes negligible electronic overlap and negligible retardation.↔ Scroll wide equations horizontally.

Different mechanisms and distance windows differ

The radial force is minus the derivative of potential with respect to increasing separation, so the dispersion force points inward and scales as inverse seventh power. Permanent dipoles can instead interact anisotropically already in first order with an inverse-cubed potential. Thermal orientation averages or induced-dipole effects can produce other effective inverse-sixth contributions with different coefficients and temperature dependence. Thus “van der Waals” is often an umbrella term, and the mechanism and averaging assumptions should accompany a stated power law.

The nonretarded dispersion window requires separation large compared with electronic size yet small compared with light-travel scales associated with important electronic transitions. At much larger separation, retardation changes the ground-state atom-atom asymptote toward an inverse-seventh potential in vacuum under the usual conditions. Resonantly excited species can exhibit inverse-cubed energy splitting instead. Molecular anisotropy, multiple coupled channels, and a surrounding medium can also alter the simple isotropic pair description. A power law without its state and distance regime is incomplete.

U(R)=-\frac{C_6}{R^6},\qquad F_R=-\frac{dU}{dR}=-\frac{6C_6}{R^7};\qquad U_{\mathrm{LJ}}(R)=4\epsilon\left[\left(\frac\sigma R\right)^{12}-\left(\frac\sigma R\right)^6\right]

F R is positive outward, so its negative sign denotes attraction. In the empirical Lennard–Jones model epsilon is a positive energy scale and sigma a length; its attractive tail corresponds to C six = 4 epsilon sigma to the sixth power.↔ Scroll wide equations horizontally.

A short-range fit is not the dispersion derivation

When electronic clouds overlap, antisymmetry and electronic rearrangement introduce exchange and other short-range effects. The dispersion tail alone would decrease without bound as separation shrinks and cannot describe that region. Lennard–Jones adds an inverse-twelfth repulsive term for convenience and a finite minimum. Actual exchange repulsion is not universally an exact inverse-twelfth law, and the two parameters must be inferred or computed for a stated species pair and purpose rather than assigned the status of fundamental constants.

For the Lennard–Jones model, the zero occurs at sigma while the minimum is at $2^{1/6}\sigma$ with depth epsilon. These are distinct distances. A potential minimum still does not guarantee a deeply bound nuclear level: reduced mass and zero-point motion matter, particularly for weak complexes. Pairwise additivity can also fail in clusters because correlated polarization creates many-body contributions. The following numbers are hypothetical scaling exercises, intended to test signs and model interpretation rather than report an empirical force between named molecules.

Work it through

Double the separation

In a valid nonretarded dispersion regime, a hypothetical pair has $U(R_0)=-64\,\mathrm{\mu eV}$ . Find $U(2R_0)$ and the ratio of attractive-force magnitudes at $2R_0$ and $R_0$ , keeping the same C six.

Energy magnitude scales as $R^{-6}$ , so doubling separation multiplies it by $2^{-6}=1/64$ . The new potential is therefore $-1\,\mathrm{\mu eV}$ .
Force magnitude scales as $R^{-7}$ after differentiation, giving the ratio $2^{-7}=1/128$ . Using the energy ratio for the force would miss one inverse-length factor.
Both radial forces are negative under an outward-positive convention, so both are attractive. The ratio is meaningful only if both distances remain in the negligible-overlap, nonretarded window used to define the tail.

The result

The doubled-distance energy is −1 microelectronvolt, while the force magnitude falls to 1/128 of its original value. Potential and force obey different inverse powers.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

Differentiate the chosen potential, not a remembered power

Use the parabola only as a differentiation analogy. Reset, set Coefficient a = −1 and Position x = 1, and predict slope −2. For a fresh comparison Reset, set a = −1 and x = 2, and predict slope −4 while the function value changes by a factor of four.

y = ax²TangentSigned integral

y at chosen x: 0.608
Tangent slope · 2ax: 1.35
Signed area · ax³/3: 0.182

Coefficient a0.75Position x0.9

t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forFunction and derivative need not share a scaling exponent. This instrument plots only a x² and does not contain inverse-power dispersion, a Lennard–Jones well, molecular separation, or an electromagnetic force calculation.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

Differentiating U gives positive 6 C six/R to the seventh, and force is its negative. The derivative introduces another inverse length and leaves the force directed inward.

Your turn

Try it yourself

A hypothetical Lennard–Jones pair has $\epsilon=10\,\mathrm{meV}$ and $\sigma=0.30\,\mathrm{nm}$ . Find the potential at sigma, the minimum separation and depth, and its attractive-tail C six in $\mathrm{meV\,nm^6}$ .

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

At the minimum, (sigma/R) to the sixth is one half. The zero-crossing length and minimum length differ.

Compare with the solution

At R = sigma the two powers cancel, so U = 0. The minimum lies at $2^{1/6}(0.30)\simeq0.337\,\mathrm{nm}$ and has U = −10 meV. The tail coefficient is $C_6=4(10)(0.30)^6=0.02916\,\mathrm{meV\,nm^6}$ . These results belong to the specified empirical model; they do not establish exact short-range electronic physics or a nuclear dissociation energy.

Keep the idea

Take it with you

Virtual dipole coupling can lower a nonpolar ground-state pair’s energy in second order.
The inverse-sixth potential and inverse-seventh force apply within a stated nonretarded regime.
Empirical short-range potentials, nuclear binding, and many-body corrections require additional assumptions beyond the dispersion tail.

Read further

Original sources and deeper explanations to follow at your own pace.

Wolfgang Ketterle, MIT 8.422 Lecture 10 transcript: ground-state and resonant perturbation, nonretarded and retarded potentials, pp. 1–9

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.