Briefly, for microscopic systems be they nano-particles or electrons, that are massive (i.e. rest mass $m_r \neq 0$), we know that their behaviour is described using Quantum Mechanics (QM). Such small systems can be positionally delocalized in space, meaning that the corresponding position operator of the system is not in an eigen-state yet, but instead it is in a coherent superposition of different positions. With this in mind, given that our system is not massless, what happens to the gravity field it is generating?

In other words, what is our current understanding for the gravitational interaction of particles that are delocalized in space? This leads to a number of basic considerations: (which are only my guesses)

  • The resulting gravitational field is also in a superposition state, which I imagine would be a very imaginative statement as the question of quantum gravity is itself an open problem.
  • As soon as the gravitational field interacts with another object (particle), the source particle (of the field) is collapsed onto an eigen-position, so then there are no gravity related issues.
  • In the coherent state of our system, like any other state, we can compute an average position, namely, the expectation value of the position, then for the gravitational interaction of our system with any other, this average position determines the separation ($|\vec{r_2}-\vec{r_1}|$) between the systems, and then knowing the separation we can estimate the gravitational force each object is subjected to.
  • The graviational interaction is itself responsible for the decoherence of microscopic systems, therefore, due to the ubiquity of graviational fields, all such systems have very very short coherence times meaning they are almost always found to be in definite (eigen) position state. But I imagine this cannot be true, as e.g., in the sun itself, huge gravitational forces are at play near the core, and tunneling allows hydrogen atoms to get close enough to each other, such that the strong force dominates the electromagnetic interaction, and fusion can take place. This means that these hydrogen atoms, despite the present gravitational field, are not classical objects, instead they're very much delocalized (i.e., their position is described by a continuous wavefunction), as otherwise tunneling wouldn't take place.

  1. Are any of the above thoughts meaningful or has my physical intuition completely failed me?
  2. Ultimately, the question is, what is our current understanding of these kinds of interplays between gravity (as understood by GR) and quantum mechanics?
  3. Lastly, can we have a basic idea of what is going on without having to adhere to any specific interpretations of QM?
  • $\begingroup$ Comments: 1. We don't have a quantum theory of gravity in the sense of general relativity. 2. If we're talking about "gravitational field" in a Newtonian sense (apart from the tag your question doesn't really sound like it's talking about the relativistic notion of gravity), why do you think the treatment of such a gravitational interaction would be any different from the treatment of e.g. the electrostatic Coulomb interaction? $\endgroup$ – ACuriousMind Jul 4 '17 at 10:30
  • $\begingroup$ @ACuriousMind you're right, I admit I may be confusing GR and Newtonian gravity at some points. But then again, you're right that the same question can be asked for the Coulomb interaction, because there as well the interaction is dependent on the separation of the two particles/systems, so we can re-ask the question for the delocalized position of the particles. How do we remedy the situation for the Coulombian interaction of two quantum objects that have delocalized position states? $\endgroup$ – user929304 Jul 4 '17 at 11:35
  • $\begingroup$ What is there to remedy? You just use the Hamiltonian that has the Coulomb potential in it, why would that depend on what the states of the system are? $\endgroup$ – ACuriousMind Jul 4 '17 at 11:39
  • $\begingroup$ @ACuriousMind but do we not need to have exact position values for the pair in order to meaningfully calculate the Coulomb force between them? $\endgroup$ – user929304 Jul 4 '17 at 11:42
  • $\begingroup$ What do you mean by "force" in quantum mechanics to begin with? There is no force, there's just the time evolution as governed by the Hamiltonian. $\endgroup$ – ACuriousMind Jul 4 '17 at 11:55

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