The hydrogen atom is a quantum mechanical system. However, it is also attracted by the gravitational pull of the earth. Therefore, do we need quantum gravity to model its behavior correctly? Conversely, can we study hydrogen atoms on earth to obtain new information about quantum gravity?

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    $\begingroup$ For an incomplete list of gravitational effects observed in manifestly quantum-mechanical systems, see this answer. $\endgroup$
    – rob
    Dec 16, 2022 at 13:48
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    $\begingroup$ Since the electron and the proton in the hydrogen atom form a bound state because of electromagnetism... "Do we need a quantum electromagnetism theory to model a hydrogen atom ?" (the answer is: it depends on how tiny the details you are interested in: "quantum electrodynamics" accounts for tiny details at the hydrogen atom level, "quantum gravity" even smaller ones... so practically speaking NO, it is not needed). Related: physics.stackexchange.com/q/215997/226902 $\endgroup$
    – Quillo
    Dec 16, 2022 at 14:07

3 Answers 3


We don't need a quantum theory of gravity to model a hydrogen atom in the gravitational field of the Earth. In general, it's understood how to model quantum fields on a fixed gravitational background (see, e.g., the book by Birrell and Davies) — at least at a mathematical level; experimentally it's very hard to probe those kinds of situations. The hydrogen atom on Earth would be a special case of that framework, where the quantum degrees of freedom (the hydrogen atom) and the gravitational field are also non-relativistic. In that case, the formalism simplifies and becomes the Schrödinger equation with a potential due to the Earth's gravitational field, like \begin{equation} i \hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m}\nabla^2 \Psi + mg z \Psi + V\Psi, \end{equation} where the second term on the right hand side, $mgz$, is the potential energy due to being on the Earth's surface. Meanwhile, $V$ describes any other sources of potential energy.

We would need a quantum theory of gravity to describe the gravitational field of the hydrogen atom itself. Actually even that isn't quite true, since there are actually approximate methods to deal with the interaction of matter and gravity quantum mechanically, as well, so long as the gravitational interaction is small. However, at large enough energies (near the Planck scale — much, much larger energies than we can probe experimentally), the gravitational interactions of the hydrogen atom become strong. This is really the regime where we need quantum gravity.


Physics is all about formulating what exact aspect of a given phenomenon you would like to describe/understand, and then introducing a minimal model, which is just complex enough to mimic the desired behavior.

For instance, if you encounter pianos falling through roofs, and you want to have a description of pianos specifically as roof-breakers, then a sphere in a vacuum with a mass equal to that of the piano is often a good enough model. The piano material might be the next level of detail sophistication in your model. However, you can imagine that the color of the piano would hardly be a necessary parameter for this particular aspect piano's interaction with the rest of the Universe. On the other hand, for other "interaction channels" such as visual appeal, the color will be a primary parameter.

Coming back to your original question, gravity might or might not matter, depending on the questions you ask. If you are interested in the color of the glow of an arc-discharge lamp filled with Hydrogen gas, then gravity was not observed to have any measurable effect. Most likely this is what you meant by studying hydrogen atom.

However, there are all kinds of other questions you can ask about a hydrogen atom which does involve gravity. For example, do hydrogen and anti-hydrogen atoms experience similar free fall (https://www.nature.com/articles/nphys2624)? Actually, outside physics labs, gravity-related properties of hydrogen atoms have much more importance for our everyday lives: abundance of hydrogen and its derivatives like water on Earth's surface and its atmosphere; density of Hydrogen plasma in stars, etc.


No, we do not need quantum gravity to model a hydrogen atom, at least not for any practical purpose. We are in fact able to model hydrogen atoms extremely well with existing theories, so well that quantum electrodynamics is among the best verified theories in all of science.

Gravity is an extremely weak force, and its contributions to the behavior of an individual atom are so incredibly tiny that they produce no practically measurable effects. The error introduced by just ignoring gravity completely is well below any experimental uncertainty. Moreover, as @Andrew said in his answer, if we did want to include gravity in our model of the atom we could use a classical or semi-classical approximation which would reduce the error even further.

  • $\begingroup$ Why would the contribution of gravity be weak? The gravitational acceleration on Earth would still be 10 meters per square second. $\endgroup$
    – Riemann
    Dec 16, 2022 at 13:13
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    $\begingroup$ I guess it depends on what you mean by "model the behavior of a hydrogen atom". Usually that means determining spectral lines, chemical properties, and so on. Those are pretty much independent of gravity. If you also mean determining how the atom moves in the neighborhood of the Earth, then yes, gravity needs to be taken into account, but the classical approximation is more than good enough for that. Gravity is still a pretty small effect: the thermal velocity of hydrogen at room temperature is about 1750 m/s, so 10/s^2 is a relatively small influence. $\endgroup$
    – Eric Smith
    Dec 16, 2022 at 14:18
  • $\begingroup$ @Riemann We can generally ignore gravity entirely when we're dealing with the interaction between the nucleus of an atom and its electrons, so the differences between Newtonian, GR, and QG are generally irrelevant to chemistry & electronics. However, in some extremely precise applications we do take gravity into account, eg the atomic fountain, which uses laser-cooled caesium atoms in freefall, and which is more accurate than older caesium clock designs. $\endgroup$
    – PM 2Ring
    Dec 16, 2022 at 21:36
  • $\begingroup$ (cont) Newtonian gravity is adequate for that application, since the exact speed that the atoms fall doesn't matter. OTOH, all atomic clocks are precise enough to be used in GR time dilation measurements, and perhaps some future atomic clock design will need to worry about the difference between Newton & GR: cutting edge clocks are already several orders of magnitude more precise than caesium clocks, and can detect GR time dilation differences on the order of a metre of height, or less. $\endgroup$
    – PM 2Ring
    Dec 16, 2022 at 21:38
  • $\begingroup$ @Riemann If you plug in the numbers, you will find that the (classical) acceleration of electron at the Bohr radius, due to the attraction of the proton is $9 \times 10^{22} m/s^2$. Gravity is extremely weak compared to that. $\endgroup$ Dec 16, 2022 at 21:56

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