Correct me if I'm wrong, but if you have a proton that is in superposition, you don't know where exactly it is; it is everywhere but with different probability. Couldn't you measure the gravity field it creates? And if you could, you would know where it is and it wouldn't be in superposition. I'm new to superposition but this keeps confusing me.

  • $\begingroup$ Related: physics.stackexchange.com/q/275711 $\endgroup$ – Dvij Mankad May 24 at 13:17
  • $\begingroup$ The question I linked is essentially the same question but since it has been phrased with unnecessary and sloppy remarks about the superposition of macroscopic bodies, the answers are not distinctively addressing the question of the gravitational field of a particle in a superposition of position eigenstates. Thus, I feel hesitant to mark your question as a dupe of the linked question. $\endgroup$ – Dvij Mankad May 24 at 13:19
  • $\begingroup$ @probably_someone And I think it does, I mean how else do we measure the position of anything? But I don't think it makes the argument wrong, it just suggests that a low-energy theory of gravity with quantized matter is not a big deal. I am not at all well-read here but I think it is the case. $\endgroup$ – Dvij Mankad May 24 at 13:21
  • $\begingroup$ Can't gravity also act as if there was a probability distribution? I don't know how even classical gravity fits into quantum, so correct me if I'm wrong. $\endgroup$ – acarturk May 24 at 14:44

I think the title and the question are a bit conflicted. In the example you’re considering, it wasn’t gravity that collapsed a superposition— it was your measurement of the particle’s state via a measurement of the gravitational field it was generating that would collapse the superposition. Such a measurement, if sufficiently precise, could cause a collapse. But this isn’t special to gravity, you could measure the coulomb field of a charged particle in a superposition of position states and obtain the same result.

The distinction between “the field collapsing the state” and “the measurement collapsing the state” is that if you don’t bother measuring the gravitating states in question, then a collapse won’t transpire. There’s nothing wrong with simply plugging a gravitational potential term $V(X)$ into the Hamiltonian, H(X,P), and evolving the state forward in time unitarily.

But this doesn’t mean that gravitational (or even electromagnetic) interactions between particles cannot themselves produce over time a loss of coherence between multiple interacting states. For a really brilliant demonstration of this happening in a gedankenexperiment I refer to this paper by Alessio Belenchia, Robert M. Wald, Flaminia Giacomini, Esteban Castro-Ruiz, Časlav Brukner, and Markus Aspelmeyer.

  • $\begingroup$ "if you don’t bother measuring the gravitating states in question" Where "measuring" is quite broader than the lay use of the term. You don't need to actually engage in activity directed towards discovering the states for the wavefunction to be collapsed; all you need is for the state to become entangled with macroscopic properties. $\endgroup$ – Acccumulation May 24 at 15:59

Not the answer you're looking for? Browse other questions tagged or ask your own question.