Why aren't the energy levels of the Earth quantized?

Question

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).

Answer 1

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.

Answer 2

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.

Answer 3

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

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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.

Answer 4

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.

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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.