# I How does the London dispersion force really work?

1. Feb 13, 2017

### sshai45

I wonder about this. The explanation that I keep finding is that "dipoles" occur "randomly" when "electrons move" to different sides of the atom. Yet I find this difficult to reconcile with what I understand about quantum mechanics -- so I must be missing something, on either side or both.

In particular, in quantum mechanics electrons are not "moving" like classic particles as seems to be suggested by this "explanation", not unless you subscribe to those theories like Bohm's or similar, but rather are described by wave functions and the Schrodinger equation, and as far as I can tell these do not "randomly" "concentrate" in some fashion. How is this effect explained in a proper quantum-mechanical treatment, in a more mainline view of quantum theory? I've tried searching around but have not found anything satisfying.

2. Feb 13, 2017

### Staff: Mentor

That explanation is indeed (semi-)classical, although still useful for visualizing the process.

The full QM approach would be to simply solve the Schrödinger equation for two atoms at a certain distance from each other, where you would see that the wave function does show a polarization. More simply, you can a hydrogen atoms in its ground state, take a second atom as a perturbation and see how the wave function of the first atom is modified by the presence of the other.

3. Feb 13, 2017

### Demystifier

Suppose, for simplicity, that dipole can have only two polarizations, namely $\uparrow$ and $\downarrow$. And suppose that the polarization is random. This means that the wave function is a superposition of two polarizations, i.e. something like
$$|\uparrow\rangle+|\downarrow\rangle$$

Now suppose that you have two dipoles, both of which are random and mutually independent. Then the wave function is
$$(|\uparrow\rangle+|\downarrow\rangle) (|\uparrow\rangle+|\downarrow\rangle)$$
In this case there is no London force.

Finally, suppose that you have two dipoles which are random but not independent. Instead of being independent, they are correlated so that they always point to the same direction. In this case the wave function is
$$|\uparrow\rangle |\uparrow\rangle+|\downarrow\rangle|\downarrow\rangle$$
In such a case there is a London force.

4. Feb 14, 2017

### DrDu

In QM you can calculate the mean value of the dipole moment, which vanishes for closed shell atoms, and also the variance of the dipole moment, i.e. the expectation value of $d^2$ or more generally the time dependent autocorrelation function $< d(0)d(t)>$ which do not vanish. From these expressions it is clear that the dipole moment is fluctuating also in QM. Furthermore, in the classical limit, the autocorrelation function will converge against the classical expression.

5. Feb 15, 2017

### sshai45

Thanks, these explanations definitely help a lot.