0
$\begingroup$
If we consider it is all at one point, then there is no superposition, it is just exact.

If we consider it to be at one point, just that it develops only when we observe, then how does the gravitational force and electric force work in this scenario?

If we consider it to be same distribution as the wave function, it might work, but it does not seem right to me.

(edit: Rephrased)

If we consider mass/charge is all at one point, then there is no superposition, it is just exact, which should not be the case.

If we consider mass/charge to be at one point, just that it develops only when we observe, then how does the gravitational force and electric force work in the unmanifested stage?

If we consider mass/charge to be following same distribution as the wave function, it might work, but it does not seem right to me.

Improve this question
$\endgroup$
6
  • 2
    $\begingroup$ there is no superposition of what? A superposition or not is just a choice of basis states. $\endgroup$
    – JEB
    Commented Jul 24 at 13:45
  • $\begingroup$ Superposition is not a physical property of one electron. It's a mathematical property of the wave function that describes the ensemble of the electron, i.e. an infinite number of copies of the system. As JEB said, superposition is very much like a choice of coordinates for a vector in three dimensional space (except that the Hilbert space can have an infinite number of dimensions). All by itself it contains no useful information. $\endgroup$
    – FlatterMann
    Commented Jul 24 at 16:54
  • $\begingroup$ @FlatterMann My question is regarding the mass distribution and charge distribution in accordance with this same theory. $\endgroup$
    – Someone
    Commented Jul 24 at 17:24
  • $\begingroup$ @JEB just to confirm, when we say that this electron is in a superposition state of being on the left and on the right, we mean that the spatial distribution of $\Psi$ has 2 clouds, one on the left and one on the right, but that crucially there is only one $\Psi$ for this electron, and "superposition states" are really our analytical terminology for chopping up this single $\Psi$ for analysis, would this be correct? $\endgroup$
    – James
    Commented Jul 24 at 17:38
  • $\begingroup$ @FlatterMann If you adjust an interferometer so that it has probability 1 of the photon going out of a particular port, what prevents it from going out of the other port? $\endgroup$
    – alanf
    Commented Jul 24 at 20:43

1 Answer 1

Reset to default
0
$\begingroup$

Saying how the gravitational force would work is a bit tricky because there is currently no accepted theory of quantum gravity for reasons that are somewhat complicated. So I will just consider electromagnetism.

You wrote:

If we consider it to be at one point, just that it develops only when we observe, then how does the gravitational force and electric force work in this scenario?

In practice there is no such thing as an electron at one point in quantum theory. Rather, the electron's state is spread out over space. If you measure whether the electron is in a particular region the state will give you the probability of finding it in that region. If you perform an interference experiment with the electron all of the possible paths the electron could go down will contribute to the result. There is no account of what is happening in reality in which the electron is just at one point.

Nor does the electron develop only when you observe. There are well known quantum equations of motion and they don't say an electron develops only when observed. An observation is just an interaction that produces a record and it can be treated by the same equations of motion. It doesn't magically make electrons change. For some more details on how measurement can be understood see

https://arxiv.org/abs/quant-ph/0306072

There is a quantum theory of electromagnetism. I'm going to oversimplify a lot here. Suppose the electron is in a superposition of being in region 1 and region 2, its state might be a bit like this: $$|\psi\rangle_e=|R_1\rangle_e+|R_2\rangle_e$$

If the electron interacts with another charged particle in the state $|B\rangle_{other}$ by its electromagnetic field, then if it was in region 1 the interaction would go like this: $$|R_1\rangle_e|B\rangle_{other}\to|R_1\rangle_e|B_1\rangle_{other}$$ and likewise for region 2.

The relevant interaction is linear so you would get $$|\psi\rangle_e|B\rangle_{other}\to|R_1\rangle_e|B_1\rangle_{other}+|R_2\rangle_e|B_2\rangle_{other}$$

The result of this would be that if you measured the electron and found it in state $|R_1\rangle_e$ you would find the other particle in state $|B_1\rangle_{other}$ and likewise if you found the electron in region 2.

Improve this answer
$\endgroup$
1
  • $\begingroup$ An electron is a quantum of energy, momentum, angular momentum and charge. While the physical vacuum has these physical properties in general, the "actual" electron is the outcome of a measurement. We can't say that "there was an electron" in the vacuum unless we measure one. The standard ontology of Copenhagen eliminates all the problems with electrons as points, states or particles. One electron is never in superposition. It's simply the final outcome of the actual measurement process. What is in superposition is the abstract wave function of the ensemble, but nature knows nothing about it. $\endgroup$
    – FlatterMann
    Commented Jul 24 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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