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How would the quantum mechanical treatment of the moon as a gravitationally-bound object differ from the usual treatment of the hydrogen atom using Schrödinger's equation?

[The earth's gravitational potential V(r) = -GM/r where M is the mass of the earth: 6 x 1024 kg. The mass of the moon is 7.4 x 1022 kg and its orbital velocity is 1,023 metres/sec. The moon's mean distance from the earth is 385 million metres.]

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  • $\begingroup$ This question is intended to be relevant to Einstein's famous question to Abraham Pais: “Do you really believe the moon is not there when you are not looking at it?” by suggesting that QM does predict that the moon is there when humans don't look at it. $\endgroup$
    – Nigel Seel
    Commented May 29, 2011 at 20:21

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Dear Nigel, your question is about the analogy between the mathematical form of potentials Coulomb and Newton, but physical situations are very different.

Yes, you could do it, use wave functions on the Hydrogen atom for solve your problem, but where is the sense?

You can do it if you consider Earth and Moon like material points, but their magnitude is very bigger than their quantum wave lenght.

Of course, you can try to enjoy yourself applying quantum mechanics for problem like this, but in this situation you must consider magnitude of Earth and Moon...so your problem is very difficult. And the wave lenghts in this problem make you consider the particles (electrons, protons, neutrons,...) that constitute Earth and Moon too...But thi is like use Relativistical Theory for calculate the motion when you walk on the road...An unuseful and terrible complication ;-)

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    $\begingroup$ Yes. One has to think in terms of the size of hbar which is order of 10^-34joulesecond or so, to think of applying quantum mechanical equations. Macroscopic objects are way off the scale. The only way to have macroscopic quantum mechanical solutions is where coherence applies, as in superconductivity superfluidity , and the behavior of crystals. There can be no coherent solutions in this case. $\endgroup$
    – anna v
    Commented May 29, 2011 at 8:05
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    $\begingroup$ Thanks, Anna, for pointing out that the key is decoherence. The electron wavefunction in the hydrogen atom can occupy much of the spherical volume centred on the proton. When the moon is observed, its wavefunction is localised at its observed position. But since it's "observed" by endless interactions with its environment (the sun, the CMB, ...) its localised wave packet never gets a chance to spread. This is unlike the situation of an electron in a hydrogen atom, where even after observation the wavefunction would have ample opportunities to rapidly diffuse back to its stationary state. $\endgroup$
    – Nigel Seel
    Commented May 29, 2011 at 13:11

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