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Let's say I have a hydrogen atom in the lab. "Normally", we would say its electron is delocalized across the ground state orbital.

Because electrons have mass, we expect it to exert a gravitational force. However, the location of the electron is uncertain (since it is delocalized). Does this mean the gravitational field of the hydrogen atom fluctuates?

The obvious possibilities I see are:

  1. No it does not fluctuate, because electrons are not point particles until they are directly measured, so they have no well-defined location. Instead they are 'somewhere' in the s-orbital, which does have a well-defined shape, and the resulting gravitational field is gravitational field averaged over the s-orbital.
  2. Yes it does fluctuate. When you measure the gravitational field you collapse the wavefunction and the electron ceases to be delocalized. This gives it a well-defined location and therefore a well-defined gravitational field.
  3. We don't know until we have a theory of quantum gravity.

Problem with #1 is that it seems by measuring the gravitational field, we are measuring, so it sounds like it will collapse the wavefunction. But if it actually does collapse the wavefunction (as in #2) then quantum systems would always decohere, because there's no way to shield gravity, so the system would always interact with the environment. This sounds observationally excluded. #3 is also not ideal since hydrogen atoms don't sound like a system where we'd expect quantum gravity effects either (since those are expected where the gravitational field is very strong & the size of the system is very small).

Option #4 would be that I am missing something, in which case: what?

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    $\begingroup$ I've removed some comments that should have been posted as brief answers, and replies to them. Please use comments to improve or clarify the question. To post a brief answer, please post a brief answer. $\endgroup$
    – rob
    Commented Jun 6, 2023 at 10:50
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    $\begingroup$ Well, I would like to repeat my comment that one may as well answer the easier question "Does the electromagnetic field of a hydrogen atom fluctuate depending on where the electron is?", because I don't see that things would be any different for gravitation. But I would like @Allure to say if there's a reason why we need to focus on gravitation. $\endgroup$ Commented Jun 7, 2023 at 2:09
  • $\begingroup$ @MitchellPorter I don't see any reason to focus on gravitation, but does using the electromagnetic field change anything? If it doesn't, then how does it make the question easier? $\endgroup$
    – Allure
    Commented Jun 7, 2023 at 2:34
  • $\begingroup$ With electromagnetism, there's a well-understood theory that we can agree to use, leaving us to focus on core conceptual issues. With gravitation, there are a lot of extra distracting directions that the discussion can take. $\endgroup$ Commented Jun 7, 2023 at 3:36
  • $\begingroup$ @MitchellPorter I suppose with electromagnetism, an extra issue is that electrons already have the smallest electric charge possible (discounting quarks), so a "test charge" that one might use to measure the electromagnetic field doesn't exist. But that seems like sidestepping the question. If you think the question is better phrased with the electromagnetic field, feel free to edit it (or write an answer for EM fields - I'd be interested). $\endgroup$
    – Allure
    Commented Jun 7, 2023 at 3:40

3 Answers 3

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A similar question could have been asked about electromagnetism before we knew about QED. But it was well understood how electrons, described by the Schrodinger equation, moved in response to an electric field: you insert the classical electromagnetic potential as the potential $V$. Nobody was saying "We can't know how the electron responds to the electric field of the proton... because we don't have a theory of quantum electrodynamics."

We can just use classical Newtonian gravity

In other words, you only need quantum gravity if you're considering a quantum mechanical situation where the effect of gravity is not approximated as a potential which is a function of the electron's position. For example "What is the potential between the proton and the electron when they are within the electron Schwarzschild radius". The specific question you're asking about is okay with that approximation.

For example, we can be sure that the effect of gravity on an electron in a hydrogen atom is given by perturbation theory, with the perturbing Hamiltonian being the gravitational potential between the proton and the electron.

It's experimentally verified even

Consider also the experimental observation of quantum states emerging within a gravitational potential (link to abstract page). Ultracold neutrons bounce against some kind of solid and their distribution in observed heights is consistent with the quantum mechanical calculation, and not the classical one. This proves experimentally that the Schrödinger equation is valid even if the potential in question comes from gravity.

It's just quantum mechanics with a gravitational potential

So to answer your specific question let's just see what the Schrodinger equation plus the postulates of quantum mechanics say. We have a test mass $t$ in a wavepacket a distance $R$ away from a hydrogen atom. It is neutral and only responds to gravity, so it moves in a potential given by $$ V(x_t,x_e,x_p)=-G\left(\frac{m_em_t}{|x_t-x_e|}+\frac{m_pm_t}{|x_t-x_p|}\right) $$ The whole system evolves under the Schrodinger equation $$ i\hbar\frac{d\Psi(x_t,x_e,x_p)}{dt}=(V+K)\Psi $$ Where $V$ is the potential (including gravity!) and $K$ is the kinetic energy.

And if you do a measurement on $\Psi$, you project onto the subspace of the Hilbert space which is consistent with the result of your observation (please don't debate interpretations of quantum mechanics in this comment section - yes I know in your favorite QM interpretation this event would be phrased differently).

So after all that text, here's your answer

After you measure the electron's position, you project onto the subspace consistent with that being the electron's position. Your system after the measurement will be consistent with the electron being in that position at the time of the measurement. In some sense, you retroactively change the gravitational effect of the electron by removing from your multi-particle wavefunction the parts where the electron pulled on the test mass and was measured in the end as being in a different position.

An easier toy problem

This is tough to give a really clear answer to because even if the electron is observed in one position, its wavefunction evolves with time (very quickly too!). So I can't just say something like "It's as if it was there the whole time"... no, because one nanosecond ago it could still have been in any position, and one nanosecond after the observation it will again be in a wide-ranging position-space superposition.

But consider as an alternative question an atom interferometer, where an atom is in a superposition where it travels down two radically different paths with equal probability separated by, in some cases, a whole meter. Let's say when the atoms are separated by a meter, we observe whether the atom is on the upper or lower path, and we find it is on the lower one. Then the moon's wavefunction (for example) is projected onto the subspace where it was gravitationally attracted to the atom following the lower path the whole time. It's as if the upper-path component of the wavefunction never existed (and never pulled gravitationally on anything).

However, as Andrew Steane points out, the moon's wave function will change by a very small amount upon this projection. If you subsequently did a measurement on the moon, you wouldn't be able to prove that this one atom had any gravitational effect until you repeat the measurement some insane number of times.

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    $\begingroup$ I like your answer; take my upvote; our answers are not that far from each other than you might be thinking. As for "you retroactively change the gravitational effect of the electron", that part is interpretation dependent. $\endgroup$ Commented Jun 6, 2023 at 7:52
  • $\begingroup$ @naturallyInconsistent yeah... I see that now. I've deleted that first sentence. Also, it was 80% directed toward a now-deleted answer that said we couldn't know without quantum gravity. While I do say in the answer "I don't want to debate interpretations of QM in the comments," it is worth noting that that step is pretty different in different interpretations. Could be "projecting onto the subspace," could be "you enter the universe where that was always the electron's position," could be "the observer entangles themselves with blah blah blah." $\endgroup$
    – AXensen
    Commented Jun 6, 2023 at 14:28
  • $\begingroup$ oh, I didn't see the other answer, so I thought you were commenting against mine. And blah and blahhhh $\endgroup$ Commented Jun 6, 2023 at 15:09
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A further consideration (in addition to those in existing answer) is the precision of any gravitational measurement, and consequent position measurement.

Consider the light beams in the two arms of an interferometer (e.g. the widely-discussed simple Mach Zehnder configuration). One might argue that the mirror in either arm undergoes a recoil when the light is reflected, and by measuring this recoil one could in principle detect which way any given photon went, thus collapsing the state. If so then it follows that the interference effects of such an interferometer would never be observed since it does not matter whether the measurement of the mirror recoil is actually carried out; it only matters that in principle the information is there.

What the above argument misses is that the reflection of the photon off the mirror does not produce a big enough recoil to change the quantum state of the mirror by a significant amount. The initial and final quantum states of the mirror overlap almost 100%. The same will be true of the kind of gravitational measurement proposed in the question, in ordinary circumstances.

This does not rule out that one could imagine experiments where gravitational effects are more significant and of course many people have come up with ideas.

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  • $\begingroup$ Good point. I've added two more sentences to the end of my answer to acknowledge this caveat. $\endgroup$
    – AXensen
    Commented Jun 6, 2023 at 15:36
  • $\begingroup$ Mine would not be suitable to incorporate your correct point. Take my upvote too. $\endgroup$ Commented Jun 6, 2023 at 15:48
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This question is interpretation dependent and incredibly difficult to answer, or even know what possibilities are out there.

For example, what if the wavefunction itself is the electron, and thus there is no localisation of the electron to speak of? Then the electron in the ground state is already this spherically symmetric cloud that is stationary, and thus there is no gravitational wave to speak of, even if it is to fluctuate.


EDIT: The next paragraph corresponded to an earlier version of this question that said a simultaneous determination of position and velocity. The OP later remembered that HUP forbids that, and edited his question accordingly. This part can thus be skipped.

If there is any localisation fluctuation, the measurement of this gravitational fluctuation is not going to localise both position and momentum. Only one linear combination of the two may be given a measurement result, not both observables at once. This is because we already know that, when pitting quantum theory and relativity to an experimental test, quantum theory is the one that wins.


That is, even if you take the view that gravity stays classical, the resulting hybrid theory must agree with basic quantum principles.

Whether or not the gravity is classical, maybe there are fluctuations, but those must yield a spherically symmetric, stationary solution in time of the electron. The wavefunction must stay spherically symmetric in it, even though the individual solution parts thereof are not. This is to meet up with known physics like the standard scattering theory.

In the end, we have to have the final theory, in order to know what actually is the answer.

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    $\begingroup$ You are giving the OP a lot of really poor ways to think about his problem. None of these are correct and I am pretty sure that you know it. There are no fluctuations in quantum mechanics, not even without gravity. It's not clear why they should be there with gravity. If there is a reason to believe that gravity is quantized, then it is exactly this: a coupling to a classical field should cause a steady loss of energy to a quantum system (which, of course, has not been observed). Quantum gravity would get around that in the most straight forward way. $\endgroup$ Commented Jun 6, 2023 at 1:41
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    $\begingroup$ I am not going to dignify your comment with a proper reply because you have consistently shown to not even attempt to understand theoretical issues. You do not even know that in Schrödinger's equation, the electrostatic Coulomb potential is purely classical, undermining your own argument for needing to quantise gravity. Unlike you, I actually talk to people on their own terms, entertaining ideas I might consider wrong, if I think the situation calls for it. $\endgroup$ Commented Jun 6, 2023 at 7:57
  • $\begingroup$ The non-relativistic Schroedinger equation is not a self-consistent theory. It's an ad-hoc quantization procedure that works halfway for half a dozen systems. One can't even to chemistry with it properly without additional assumptions. $\endgroup$ Commented Jun 6, 2023 at 9:09

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