The Hamiltonian of the Earth in the gravity field of the Sun is the same as that of the electron in the hydrogen atom (besides some constants), so why are the energy levels of the Earth not quantized?
(of course the question is valid for every mass in a gravity field).

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    $\begingroup$ Why do you say that the energy levels aren't quantized? $\endgroup$
    – user93237
    Commented Feb 20, 2018 at 21:23
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    $\begingroup$ Another problem is that the Hamiltonian you are referring to is just an approximation to the underlying general relativity framework, and there's no consensus on the procedure we should take to quantize that... $\endgroup$
    – valerio
    Commented Feb 21, 2018 at 0:03
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    $\begingroup$ To quantize the Earth's energy level you need an unambiguous definition of "Earth", one which accounts for every single quantizable particle which belongs to that "Earth". I'm sure you can see how a not-completely-quantized approximation is far more useful and achievable than any attempt to identify every last particle belonging to some particular notion of "Earth" at some particular instant in time. $\endgroup$
    – Beanluc
    Commented Feb 21, 2018 at 20:23
  • $\begingroup$ The earth is not in a stationary state $\endgroup$ Commented Feb 22, 2018 at 19:28

4 Answers 4


The orbital energy of the Earth around the Sun is quantized. Measuring this quantization directly is infeasible, as I'll show below, but other experiments with bouncing neutrons (Nature paper) show that motion in a classical gravity field is subject to energy quantization.

We can estimate the quantized energy levels of the Earth's orbit by analogy with the hydrogen atom since both are inverse square forces--just with different constants. For hydrogen: $$E_n = -\frac{m_e}{2}\left(\frac{e^2}{4\pi\epsilon_0}\right)^2\frac{1}{n^2\hbar^2}$$ Replacing $m_e$ with the mass of Earth ($m$) and the parenthesized expression with the corresponding expression from the gravitational force ($GMm$, where $M$ is the mass of the sun and $G$ is the gravitational constant) to get $$E_n = -\frac{m}{2}\left(GMm\right)^2\frac{1}{n^2\hbar^2}$$ Setting this equal to the total orbital energy $$E_n = -\frac{m}{2}\left(GMm\right)^2\frac{1}{n^2\hbar^2} = -\frac{GMm}{2r}$$ Solving for $n$ and plugging in values gives: $$n = \frac{m}{\hbar}\sqrt{GMr} = 2.5\cdot 10^{74}$$ The fact that Earth's energy level is at such a large quantum number means that any energy transition (which are proportional to $1/n^3$) will be undetectably small.

In fact, to transition to the next energy level, Earth would have to absorb: $$\Delta E_{n \to n+1} = m\left(GMm\right)^2\frac{1}{n^3\hbar^2} = 2\cdot 10^{-41}\ \textrm{J} = 1\cdot 10^{-22}\ \textrm{eV}$$ For a sense of how little this energy is, a photon of this energy has a wavelength of $10^{16}$ meters--or, one light-year.

Solving for $r$: $$r = n^2\left(\frac{\hbar}{m}\right)^2\frac{1}{GM}$$ An increase in the principal quantum number ($n$) by one results in a change in orbital distance of \begin{align} \Delta r &= \left[(n+1)^2 - n^2\right]\left(\frac{\hbar}{m}\right)^2\frac{1}{GM} \\ &= \left[2n + 1\right]\left(\frac{\hbar}{m}\right)^2\frac{1}{GM} \\ &= 1.2\cdot 10^{-63}\ \textrm{meters} \end{align} Again, way too small to measure.

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    $\begingroup$ That's so preposterously small I laughed out loud! $\endgroup$ Commented Feb 21, 2018 at 5:43
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    $\begingroup$ Something that might go missing along all the maths: this answer does not claim that the earth resides in quantized energy states, it merely explains why we can currently not verify (or refute) such a claim. $\endgroup$ Commented Feb 21, 2018 at 7:51
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    $\begingroup$ Neutrons have been shown to have quantized energy levels in a gravitational field, yes, but this is not enough to answer by the affirmative: Earth is a classical object and there is no experimental proof that it can behave in any way as a quantum one. This is only a leap of faith extending the validity of QM well beyond its tested domain. $\endgroup$ Commented Feb 21, 2018 at 9:44
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    $\begingroup$ Note that, to actually claim that the orbital energy is quantised in any observable sense, one also needs to show that the linewidth associated with transitions between different orbital states is smaller than the energy difference between them. Since it appears that the stimulated emission of even one photon into space is enough to cause a transition, the linewidth must be far larger than the spacing and therefore the "spectrum" would be continuous at such high $n$ even if it could be plausibly measured (which of course it can't, as this nice answer shows). $\endgroup$ Commented Feb 21, 2018 at 13:54
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    $\begingroup$ @zibadawatimmy That article links to nature.com/news/2002/020117/full/news020114-8.html which seems more directly on point, and cites physi.uni-heidelberg.de/~abele/nature.pdf for more detail. $\endgroup$
    – zwol
    Commented Feb 21, 2018 at 21:24

They are. It is just that they are so closely spaced between each other that we can't observe it. Note that we do not yet have a good theory of quantum gravity.

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    $\begingroup$ It's not clear that they must be—perhaps the planet is too massive and energetic to maintain the required coherence—but this consideration explains why it wouldn't matter if they were. $\endgroup$ Commented Feb 20, 2018 at 21:25
  • $\begingroup$ @Iván Mauricio Burbano Why they are so closely spaced between each other? The quantization needs to be: C/n^2 like the Hydrogen. $\endgroup$
    – Jacob
    Commented Feb 20, 2018 at 21:26
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    $\begingroup$ I'm not sure about what the constant is for gravitational purposes. However, any object with enough mass surely has energy corresponding to very high $n$ under normal conditions. That's why I presume they are so closely spaced. $\endgroup$ Commented Feb 20, 2018 at 21:30
  • $\begingroup$ @Jacob When you have a bunch of interacting atoms, they no longer have the same energy levels as the separate atoms themselves. When atoms interact, their electron wavefunctions distort, which drastically changes the spacing of the energy levels. So no, there's no reason to expect it should be a 1/n^2 quantization. $\endgroup$ Commented Feb 20, 2018 at 21:35
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    $\begingroup$ @Jacob, at high $n$ and $l$ (which must be true of states representing the Earth) the quantum and classical solution are identical and we haven't learned anything new. But you must estimate those quantum numbers to get an appreciation for how true that is. Don't guess and don't try to intuit it: calculate and see. $\endgroup$ Commented Feb 21, 2018 at 0:06

tl;dr- In principle, quantization might still apply. Scientifically speaking, we have no idea yet.

We don't know how far down our current quantum theories might hold.

To draw an analogy, Newton's laws of motion predict that things can move faster than the speed of light, $c$. But, turns out that that wasn't right; Newton's laws kinda fell apart at the relativistic limit, and today we know that that prediction wasn't meaningful.

So, as described in @MarkH's answer, the energy levels are separated by

$$\begin{align} \Delta r &= \left[(n+1)^2 - n^2\right]\left(\frac{\hbar}{m}\right)^2\frac{1}{GM} \\ &= \left[2n + 1\right]\left(\frac{\hbar}{m}\right)^2\frac{1}{GM} \\ &= 1.2\cdot 10^{-63}\ \textrm{meters} \end{align}$$

In terms of the Planck length,$$ \ell_{\mathrm{P}} ~ {\approx} ~ 1.616229{\times}{10}^{−35}\textrm{meters}, $$that'd be about $7.4{\cdot}{10}^{-29}\ell_{\mathrm{P}}$.

As a rule of thumb, any prediction that's astronomically smaller than the Planck length falls into the realm of speculation as opposed to verified scientific models.

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    $\begingroup$ Your rule of thumb is a little bit... conservative... $\endgroup$
    – user541686
    Commented Feb 22, 2018 at 0:58
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    $\begingroup$ @Mehrdad Hah, just a little, right? In all seriousness, I think that SE.Physics tends to lean toward Platonic realism in some cases, so answers of the form, "Our scientific models are really just correlations that hold over the limited domain in which they've been verified" tend to be poorly received. $\endgroup$
    – Nat
    Commented Feb 22, 2018 at 1:02
  • $\begingroup$ Maybe we need more businesspeople on this site who can sell the theories better... =P $\endgroup$
    – user541686
    Commented Feb 22, 2018 at 1:12
  • $\begingroup$ To me, quantum mechanics is a verified scientific model, which allows for extrapolation to unusual situations. It does show that quantum mechanics actually describes the everyday world, since the quantum weirdness becomes undetectable for large systems. The accusation of Platonism is rather libelous, though. :) $\endgroup$
    – Mark H
    Commented Feb 22, 2018 at 9:40

Summary of this answer: the earth is nothing like an electron in a hydrogen atom, and using the hydrogen atom solution to calculate energy levels of the earth doesn't make sense.

Quantization of energy levels isn't a postulate of quantum mechanics. It's a consequence of solving the Schrödinger equation and finding that the solutions are discrete harmonics. This doesn't always happen—for example, the kinetic energy of a free particle isn't quantized. When it does happen, it's because there is some sort of periodicity in the system. Examples from QM 101 are the infinite square well (where a classical particle would bounce back and forth off the edges), the $kx^2$ harmonic oscillator potential (where a classical particle would oscillate sinusoidally), the hydrogen atom (where a classical particle would loop around the nucleus in some orbit), and so on. While these problems are usually analyzed in a wave picture, you can also think of them from the perspective of a particle sum-over-histories picture, in which the quantization happens because of constructive/destructive interference between histories in which the particle oscillates/loops different numbers of times.

Although it's not normally mentioned in QM 101, adding a detector to any of these systems has the same effect as adding a detector to the double-slit experiment: it prevents interference. It's also not always mentioned that a measurement (and wavefunction collapse) isn't necessary for interference to disappear. All that's necessary is that which-path information be preserved in any form anywhere in the universe. Even if the effect of passing one of the slits is as small as the emission of a single additional photon, and even if that photon is never seen by anyone or never absorbed at all, the analysis is the same as if a measurement had occurred.

The earth emits, if I calculated correctly, around $10^{37}$ infrared photons per second, or $10^{44}$ per year. The position of these photons in spacetime—or the unique patterns of heat caused by their absorption—constitute a permanent record of the earth's position as a function of time. Imagine a quantum harmonic oscillator packed so densely with detectors that the particle passes $10^{44}$ of them on each oscillation. It's clear that this system would be essentially classical in its behavior, even if the particle had a very low mass (and large de Broglie wavelength) and would otherwise show obviously quantum behavior. This is just one of many sources of information leakage from the earth; even the records of the passage of the year within the earth itself, from calendars to tree rings, preclude interference between histories in which different numbers of years have passed.

One of the others answers plugs the earth and sun masses and the gravitational constant into the QM 101 formula for the hydrogen atom and concludes that the earth's orbital energy should be theoretically quantized with the levels separated by $10^{-22}\text{ eV}$. That is completely wrong. It would perhaps be correct for a stable quantum particle of the mass of the earth, in a Newtonian universe where there is no gravitational radiation. But if there's any record of the earth's motion around the sun, then it is in effect measured and the hydrogen atom solution doesn't apply.

Of course, we have no hope of detecting the quantization or lack of it experimentally. But there isn't even any theoretical reason to expect it to exist. The calculation is based on a misunderstanding of QM.

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    $\begingroup$ In my defense, my answer originated from a homework problem in Griffith's "Introduction to Quantum Mechanics" (problem 4.17, pg. 159). It is somewhat fanciful, even more so than many other idealized physics problems. $\endgroup$
    – Mark H
    Commented Aug 18, 2020 at 21:33

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