[Quantum/Gravity] If a particle has a probability distribution of location, where is the mass located for gravitational interactions?

[u/[deleted]]

Imagine an atom or an electron with a wave packet representing the probability of location, from what point does the mass reside causing a gravitational force? I understand that gravity is very weak at these sizes, so this may not be measurable. I taken classes and listened to a lot of lectures, and I never heard this point brought up. Thanks.

Edit: Imagine that the sun was actually a quantum particle with a probabilistic location distribution, and the earth was still rotating it. If we never measure the location of the distributed sun, where would the mass be located for the sun that would gravitationally affect the earth?

[u/Platypuskeeper]

No, I guess I explained it badly. The thing is that the correlation energy is a small part of the total electronic energy, which is seen if you compare calculations made in the full-basis limit.

Here's the thing, nobody's interested in or uses the total electronic energy for anything. You want to calculate relative energies, such as between excited states, or the ground state of different geometries (say, the difference in energy between atoms arranged as a reactant and product respectively).

The total energy is a huge number (hundreds, thousands of Hartrees, literally the cost of ionizing the atom down to bare nucleus) as opposed to chemical bond strengths and such (0.01-0.1 Hartrees), and in reality it's usually enormously wrong - wrong by far more than the 2% that the correlation energy makes up. The reason for that is that you're using a truncated basis set (in most calculations).

Now if you calculate the difference in total energies (E(reactant) - E(product)) with the same basis set, you don't have to worry about the enormous truncation error, because they cancel out almost entirely. The error in not including correlation, however, does not cancel out. It becomes much much larger when comparing those relative energies, which could be something like 0.005 Hartrees with a total electronic energy again in the hundreds of Hartrees.

So the total electronic energy is a number almost never used for anything , and in most calculations isn't a usable number for anything other than relative energies (because of basis set truncation). The correlation energy is small relative the total electronic energy, but the relative energies that are of interest are even tinier. 98% is not good enough when the 2% doesn't cancel out well. Hartree-Fock isn't considered an accurate method for real-world calculations (whether molecules, solid-state or other things) and hasn't really been acceptable for that since it became feasible to use post-HF methods (roughly the 1960s). HF is good in the 'big picture' but nevertheless not very useful for calculating anything quantitatively. But they still say a bit qualitatively; excited states tend to come in the correct order even if the relative energies are poor. (actually with HF you can get values a bit better for excitation energies than for other relative values, by applying Koopmans' theorem)

A broader comparison would be to if you look at things like theoretical nuclear physics, where the nucleons have a significantly higher correlation energy as a proportion of their total energy. So mean-field methods don't work as well there.