Van der Waals interaction

Today - the Van der Waals interaction between two atoms. Two physical pictures illuminate the richness of the problem. Consider two atoms $a$ and $b$ separated by distance $R$. The Coulomb interaction between these is

$H_{I}=\int {\frac {\rho _{a}\rho _{b}}{r_{ab}}}d^{3}r_{a}d^{3}r_{b}\,.$

Integrating leaves us with two interacting dipoles

$H'={\frac {1}{R^{3}}}({\vec {d}}_{a}\cdot {\vec {d}}_{b}-3({\vec {d}}_{a}\cdot {\hat {R}})({\vec {d_{b}}}\cdot {\hat {R}}))\,.$

In the quantum picture, these dipoles become operators. First order perturbation theory tells us that if one atom is excited and the other is in the ground state, this gives an interaction term

$-{\frac {C_{3}}{R^{3}}}\,.$

But if both atoms are in the ground state, then the leading term is given by second order perturbation theory, as

$-{\frac {C_{6}}{R^{6}}}\,.$

Let us interpret this result now (see Kleppner's class notes) - the physics of the van der Waals interaction are all due to quantum fluctuations. Consider two resonant circuits that are capactively coupled. The coupling goes as $1/R^{3}$. Each resonator has frequency $\omega _{0}$, and classically coupled oscillators split their frequencies, giving

$\omega _{\pm }=\omega _{0}\pm (term){\frac {1}{R^{3}}}-(term){\frac {1}{R^{6}}}\,.$

Quantum-mechanically, the $1/R^{3}$ term cancels out leaving just the $1/R^{6}$ term, when the oscillators are each in the ground state. An equivalent description focusing on the atomic dipole moment is as follows. A neutral atom has no expectation for the dipole operator, $\langle d_{a}{\rangle }=0$, but the variance for this is nonzero, $\langle d_{a}^{2}{\rangle }$. This is just as the simple harmonic oscillator in its ground state has zero $\langle p{\rangle }$, but nonzero $\langle p^{2}{\rangle }$ in the ground state. The fluctuating dipole moment of atom $a$ creates a fluctuating field at position $b$,

$E_{b}\propto {\frac {d_{a}}{R^{3}}}\,.$

This induces a dipole in atom $b$. The interaction energy between two dipoles is

${\frac {d_{a}d_{b}}{R^{3}}}={\frac {\langle d_{a}^{2}{\rangle }}{R^{6}}}\,.$

So far, we've neglected fluctuations of the electromagnetic field. What could the vacuum fluctuations of the field do to neutral atoms? Fluctuations at position $a$ and $b$ can induce dipole moments, since the atoms are polarizable:

${\begin{array}{rcl}d_{a}&=&\alpha _{a}E(r_{a})\\d_{b}&=&\alpha _{b}E(r_{b})\,.\end{array}}$

The corresponding interaction is

$H_{I}\propto {\frac {\alpha _{a}\alpha _{b}\langle E(r_{a})E(r_{b}){\rangle }}{R^{3}}}\,.$

See Larry Spuch, Physics Today article, posted on web site for more about this. Conceptually, one sums over all modes of the electromagnetic field - this time in free space, without the boundary condition as we did in the Casimir force case. The long wavelengths do not average out to zero, when they are of wavelength comparable to the distance between the two atoms. Their contribution, when density of states are included, and cutoff appropriately, gives

$H_{I}\propto {\frac {1}{R^{7}}}\,.$

So how does the $1/R^{6}$ term disappear? If you focus on the fluctuations of the instantaneous dipole moments, then you get $1/R^{6}$; this is an appropriate result for short range interactions. It can be derived, as we showed above, by conceptually using the uncertainty principle for the atomimc dipoles. In contrast, when the uncertainty principle for the electromagnetic field is used, we get a $1/R^{7}$ force; this is applicable at long ranges, and is known as the Casimir-Polder effect, or the retarded van der Waals interaction. Note that a fully quanutm theory of both atoms and field are needed to get both of these interactions properly. The crossover distance between the two interactions is set by $R\sim \lambda _{res}$, the resonant wavelength of that atoms.

The entirety of the van der Waals interaction is a quantum phenomenon (thus, the reason your table is solid is because of quantum mechanics!). Classically, there is no average value of the dipole moment, and also no fluctuations - the classical van der Waals force is zero. But quantum-mechanically, vacuum fluctuations are always present, giving rise to the van der Waals force.

Absorption and emission

Returning to our theme of perturbation theory for this chapter, let us consider the van der Waals interaction from a diagramatic approach. (This follows API pp 121-126). The fundamental idea is an exchange of virtual photons. The Hamiltonian in the dipole approximation is

$H'_{I}=-dE_{\perp }(R)-d'E_{\perp }(R')\,.$

In deriving the QED Hamiltonian, we discarded the longidutional fields because they were determined by the charges and not a degree of freedom, and we kept the transverse fields, as the radiation field. In the qualitative discussion above, we use ${\vec {d}}$ as a dipole moment, which involves static Coulomb interactions; that is fine, because it is a near field picture, involving interactions inside an atom. Outside the neutral atoms there only exists radiation field, however. The only way two neutral atoms can attract each other is by exchange of virtual photons. See Cohen-Tanoudji for hundreds of pages of illuminative discussion about this.

File:Van der Waals interaction-c3-dipole-dipole

The language of Feynman diagrams can be used to understand the van der Waals interaction. Consider two identical atoms spaced apart by distance $D$. The atoms initially start in state $a$, the ground state. One emits a virtual photon, going into a virtual excited state (with an energy below the ground state). This photon is absorbed by the second atom. After a short time, the first atom emits another photon, which is absorbed by the second one, returning both atoms to the ground state:

File:Van der Waals interaction-c3-vdw-diagrams

This picture indicates that to leading order, we need diagrams with two photons, and four vertices. The dominant contribution comes from photons with $\lambda \sim D$, that is $\omega \sim 2\pi c/D$. Another important piece of guiance is that the virtual process is admixing with states of higher energy; energy shifts in second order processes go as

$\Delta E\sim {\frac {V_{kj}V_{0i}}{E_{i}-E_{j}}}\,,$

characterizing the higher price involved in mixing with higher energy states. With extra intermediate states, for example, this becomes

$\Delta E\sim {\frac {V_{lk}V_{kj}V_{0i}}{(E_{l}-E_{k})(E_{i}-E_{j})}}\,.$

When we emit virtual photons, there are two energies involved; one is $\omega _{0}$, one is the exchanged photon energy. The energy defects in the above picture are thus as shown in the figure above: initially $\omega _{0}+\omega$, then $2\omega _{0}$, then $\omega _{0}+\omega$. A second diagram involved is as if one atom emitted two photons simultaneously:

File:Van der Waals interaction-c3-vdw-diagrams

Which of these two diagrams is preferable depends on how large $\omega$ is compared to $\omega _{0}$. Ultimately, that distinguishes short and long range interactions, and $1/R^{6}$ versus $1/R^{7}$. The energy defects in the three regions are $\omega +\omega _{0}$, then $2\omega$ (versus $2\omega _{0}$ for the first diagram), and then $\omega +\omega _{0}$. Note we do not consider the cross diagrams, because it turns out they do not contribute to leading order.

Focusing on these two diagrams, let us consider what happens at short and long ranges. Let $\lambda _{0}$ be the wavelength of the radiation. Short range is defined by $D\ll \lambda _{0}$ (meaning $\omega \gg \omega _{0}$), and long range is $D\gg \lambda _{0}$ (meaning $\omega \ll \omega _{0}$). The contribution of each diagram is proportional to the inverse product of the energy defects:

$H_{I1}\sim {\frac {1}{\omega _{0}+\omega }}{\frac {1}{2\omega _{0}}}{\frac {1}{\omega _{0}+\omega }}\sim {\frac {1}{\omega _{0}\omega ^{2}}}\,.$

In contrast, for the second diagram, we get

$H_{I2}\sim \sim {\frac {1}{\omega _{0}^{2}\omega }}\,.$

Thus, going from long range to short range involves an additional factor of $\omega$ in the interaction. Since $\omega \sim 1/D$, this gives the change from $1/D^{6}$ to $1/D^{7}$ in the potential.

Casimir forces revisted

Consider three scenarios:

• atom-atom. At long distances $V\propto {\frac {\alpha _{1}\alpha _{2}}{R^{7}}}$.
• atom-wall.

Recall that the polarizibility of an atom has units of volume, $R^{3}$. A large metal sphere also has polarizability $R^{3}$. Thus, at long distances, $V\propto {\frac {\alpha R^{3}}{R^{7}}}\sim {\frac {\alpha }{R^{4}}}$.

• two walls. ${\frac {V}{A}}\propto {\frac {1}{RR^{2}}}\sim {\frac {1}{R^{3}}}$.

At short distances:

• atom-atom. $V\sim {\frac {1}{R^{6}}}$
• atom-wall. $V\sim {\frac {1}{R^{3}}}$
• two walls. Always in the far field.