Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose a particle has 50% probability of being at location $A$, and 50% probability being at location $B$ (see double slit experiment). According to QM the particle is at both $A$ and $B$ at the same time, so is there a force of gravity between the two particle superpositions? Is there self-gravity when a wave-function reaches over a finite distance?

I cannot seem to wrap my head around this. Is the gravity a proportional fraction of the entire mass based on the probabilities. How do you combine a wavefunction with Gauss' law of gravity? I have being trying to think about self-gravity for a long time now.

share|cite|improve this question
Recent related article on self obscuring gravity measurements… – ja72 Oct 12 '15 at 21:59
up vote 7 down vote accepted

There is some work by Roger Penrose on the subject. The papers title is, "On Gravity's Role in Quantum State Reduction", and it discuses how the interaction of two states that have different mass distributions with spacetime can cause the wavefunction to collapse in the one state or the other. There is also a following paper that discuses the same thing in Newtonian gravity, "Spherically-symmetric solutions of the Schrödinger-Newton equations" (and there is also this that you could have a look).

There is one thing that I should point out that is also pointed out by David. In a situation as the one described in the question (double slit experiment), the particle is not at two different places at the same time and interacts with it self. It is the two states (wavefunctions) that interact to give you the interference.

share|cite|improve this answer
+1 for the reference, although I would doubt (as most other physicists) that gravity have anything to do with the wavefunction collapse (Zurek's decoherence is way more plausible). – Piotr Migdal Nov 21 '10 at 10:53
Thanks for the links. I am a great admirer of Sir R. Penrose and his insights. – ja72 Nov 22 '10 at 3:14

I'm fairly sure it's not correct to say that the particle is at both A and B at the same time. If it interacts with something at A, then it's at A, not at B. And vice-versa. I believe the production of a gravitational field would be one such interaction (although perhaps we might need a quantum theory of gravity to be truly sure), so when you detect the gravitational field produced by the particle, it will appear to be "emanating" from either A or B, but not both.

This would mean that the particle can't interact with itself, since if it exists at point A to be "emitting" the gravitational field, it can't also exist at point B to be reacting to the gravitational field.

I believe the same question could apply to electromagnetic self-interaction of a charged particle. But for that case we do have a theory that should explain what happens, namely quantum electrodynamics. Perhaps someone else can explain that case in detail, or if I can figure out something, I'll edit it in here.

share|cite|improve this answer
The particle does interact with itself in the double slit experiment (not gravitationally) when in produces an interferance pattern on the back wall, even when the emision rate is slowed down to 1 particle at a time. If you detect the combined effect of gravity of the particle being at both location then you have not collapsed the wave function (still don't know where the particle is). – ja72 Nov 20 '10 at 5:49
If I remember correctly, the results drawn from the double split experiment are not conducted using a double split as depicted in textbooks. So I think that gravity is out of question in real life. – Robert Smith Nov 20 '10 at 5:54
@jalexious: How can you slow down the emission rate to 1 particle at a time? – Robert Smith Nov 20 '10 at 5:56
@jalexiou: that's not the particle interacting with itself, that's wavefunction interference. – David Z Nov 20 '10 at 6:05
@jalexious: No, measurement have nothing to do with virtual particles. I do not know what do you mean by 'particle represented by 2 wavefunctions'. However, there is a huge difference between: a) having particle at A and another at B b) having a one-paticle wavefunction with with non-zero values at A and B. – Piotr Migdal Nov 20 '10 at 9:44

First a general comment - everything in the world is described by either classical or quantum fields. Point particles are a fiction, sometimes useful, sometimes not. Starting with classical field theories like Maxwell equations or general relativity, you find that you are forced to forget about point sources and repalce them by continuous charge or mass distribution, otherwise you get all kinds of nonsense (non-locality, acausality, etc. etc.). One of the reasons for that is the infinite self-force or self-energy problem that crops up already at the classical level.

We can approximate a continuous distribution of matter by a "particle" if it obeys certain conditions, roughly speaking it has to be localized and weakly interacting. By "localized" I mean that all relevant observable quantities (expectation values of operators) are localized. This is not the situation you describe - the wavefunction is approximately localized (with two centres) but it is not observable. Relevant observable quantities like the expectation value of currents will not necessarily be localized.

So, what you are asking in effect is the self-force for a particular distribution of mass (or charge). There is an answer for that, but since you are asking a question that had to do with short distance physics, the quantum mechanics of the gravitational (or electromagnetic) field comes into it. Probably not enough space-time to elaborate on this here.

share|cite|improve this answer

This is similar to what my old boss called the "Lafyatis Problem," because it was first posed to him by Greg Lafyatis at Ohio State, which involved light emission from a particle in a "Schroedinger cat" state. The question is, if you have an atom in a superposition of two position states, and it emits light, should you expect to see an interference pattern in the emitted light due to the light being emitted by the two different possible locations of the atom? This got debated periodically, and as I recall the consensus answer that was settled on was that there wouldn't be any interference, but it was never fully settled, and kept being brought up from time to time.

I know that Dave Wineland's group did experiments in which they put trapped ions into a "cat" state, a superposition of two positions in the trap (Science 272, 1131 (1996)). They didn't deal with this issue experimentally, but I think it's what prompted the discussion, and might be a place to start looking for information.

share|cite|improve this answer

We don't know, really.

Thinking in Quantum Mechanical terms, you should proceed like this:

  • The setup of the experiment is represented in quantised form in Schrödinger's equation (it gets more complex with QED and QFT, but the principle is pretty much the same). This includes the physical setup (screens, etc.) and all the fields you wish to account for. This is where you should include gravity.

  • The solution of the equation gives you a function which ultimately describes your experiment. You apply specific operators which correspond to measurements and obtain a series of possible outcomes and the probabilities of each.

The thing of note here is that when you add gravity initially, its effects are taken into account globally (in Schrödinger's equation). So this would take into account all the effects you are talking about. What doesn't work in your line of thinking is that you account for gravity locally (i.e. with particle at point A or B or both), whereas, in a QM world you can't do that.

Now, the problem is that obviously we don't have a good and valid representation of gravity in QM terms.

share|cite|improve this answer

Let's first recognize that the gravitational force is 10^{-42} times weaker than the electrostatic force. Now pause to ingest that number. The earth is only about 10^{23} times more massive than we are.

So the question is not a practical concern, but one of fundamental physical theory. We don't have a good theory in which we can analyze dynamical gravitational and quantum-mechanical processes. Hawking radiation is perhaps the best blend of the two, but it takes place in an extremely large gravitational field, where the discrepancy between the forces is much less. (Within string theory, which is a consistent quantum theory which includes gravity, the best quantum-gravity calculations regard the number of microstates of a black hole, i.e. its entropy.)

To guess that an incoming state's gravitational self-interaction will affect its evolution -- and that this effect will be different for superpositions of localized states -- but I have no idea what would actually happen. I'm sure that people more connected with experiments will have a better feel, but a fundamental understanding is currently beyond our ken.

share|cite|improve this answer
Obviously this is a theoretical question to be asked. There is either an effect to be considered, or not. – ja72 Nov 21 '10 at 5:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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