# Doesn't gravity collapse superposition? [duplicate]

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.

• – Dvij Mankad May 24 at 13:17
• 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. – Dvij Mankad May 24 at 13:19
• @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. – Dvij Mankad May 24 at 13:21
• 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. – acarturk May 24 at 14:44

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.