When I look for the van der Waals interactions, they are defined as they are non-local interactions but no explanation for what they mean by non-locality. What would be the best way to understand this confusing local and non-local terms.


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Local interactions are interactions that are limited to a certain volume/distance. Let's examine the Coulomb repulsion between two particles. The physical interaction is $V(r_1, r_2) = q^2/|r_1-r_2|$ which is non-local. Two particles at very far positions will interact with each other via this interaction.

Now, let's consider the case where we have many of these particles and they are embedded in a medium with opposite equal total charge homogeneously distributed. This is the case, for example, when considering electrons in a metal. They are free to move but in the background there are the positives charges of the atoms that make the metal, such that in total the charge is neutral. The effect is a screening effect, and two electrons far apart from one another will not really feel each other, as on average the positive charge between them will cancel the negative charge. However, two electrons close by will feel each other. So to approximate this behavior we can write $V(r_1, r_2) = U \delta(r_1-r_2)$. Now this is a local interaction. One can consider other local type of interactions: for example, on a lattice, where each electron can be at a position $(i,j)$ we can imagine that it can feel some of its neighbors, and let the interaction term run for $(i\pm n, j\pm n)$ for some finite $n$. This will still be considered local, as for distance larger than this $n$, the interaction is cut-off.

  • $\begingroup$ So, this positively charged background cancel out the exchange of the virtual photons which form the nonlocal process? $\endgroup$ Feb 10, 2020 at 13:51
  • $\begingroup$ yes. Imagine in classical electrostatics, how a negative charge inside a sphere with a positive charge that cancels it exactly has no net electrical field outside the sphere. This is just an application of Gauss law, that is true for quantum mechanics as well. In the abstraction level of photons your description is correct - the effect of the virtual photons emitted by the electron will be cancelled by the effect of the virtual photons emitted by the positive background $\endgroup$
    – user245141
    Feb 10, 2020 at 14:17

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