I What is the origin of Van der Waals force?

Tags:
1. May 26, 2017

blue_leaf77

Let's an easy example, a hydrogen atom in the ground state and the proton fixed at the origin. Calculating the expectation value of the electric field at a point $\mathbf R$ in space is equivalent to calculating the electric field due to a continuous charge distribution $-e|\psi_{100}(r)|^2$ and a point charge $+e$ at the origin. We can use Gauss law to calculate the field since the density is spherically symmetric. It can easily be seen that the electric field does not vanish unless $|\mathbf R| \to \infty$. Anyway, a molecule or atom being neutral is not always a good approximation, when you go closer and closer and eventually arrive in a location between the orbitals, you can no longer say that the atom is neutral. The same case as in a molecule interacting with another one to possibly form a bonding, as one of them gets closer to the other it will feel a net force due to the other molecule's electrons and nuclei and it is this "residual" force that is responsible for the bonding formation. Classically speaking, depending on whether this net force is repulsive or attractive, a bond may form.

2. May 27, 2017

Demystifier

Well, my $x(t)$ was only time-dependent in the Heisenberg picture, so it was not really an explicit time-dependence.

The quantum fluctuations of dipole moment are essential for van der Waals force, but their time dependence is not essential. After all, the average force does not depend on time.

Maybe it looks natural because it looks classical in spirit, but, as you certainly know, atoms are not well described by classical physics. Attempts to describe intermolecular forces with such classical models are not in good agreement with observations.

3. May 27, 2017

TeethWhitener

I'm pretty sure you can't use Gauss' Law here. (Though I'm happy to be proven wrong) For the expectation value, you have to integrate over all space. Once you perform the integration, assuming properly normalized orbitals, you get a contribution of $-e/\mathbf{r}$ from the electron and $e/\mathbf{r}$ from the proton. They cancel to give zero. Using Gauss' Law means that you're unphysically constraining the electron to be within a finite sphere.

4. May 27, 2017

blue_leaf77

No, that's not true. Remember when you calculated the electric field inside a dielectric with a spherically symmetric charge density. What matters is only the charges inside the spherical surface containing the observation point. Of course, however Gauss law is not the only tool to calculate the E field. One can use the basic equation
$$E(\mathbf R) \propto \int_0^\infty \int_0^{2\pi}\int_0^{\pi} \frac{\rho(r)}{|\mathbf r - \mathbf R|^3} (\mathbf r - \mathbf R) d^3\mathbf r$$
Now the expectation value of electric field operator in the ground state hydrogen atom should be
$$\langle E(\mathbf R)\rangle \propto \int_0^\infty \int_0^{2\pi}\int_0^{\pi} \frac{|\psi(r)|^2}{|\mathbf r - \mathbf R|^3} (\mathbf r - \mathbf R) d^3\mathbf r$$
You see that the two equations are different only by interchanging $|\psi(r)|^2$ and $\rho(r)$. So, I don't see why the second equation cannot be alternatively calculated using Gauss law.

Last edited: May 27, 2017
5. May 28, 2017

TeethWhitener

I thought about it a little more and you're probably right.