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We know that electron trapped by nuclear, like the hydrogen system, is described by quantum state,and never fall to the nuclear. So is there any similar situation in the case of electron near the black hole but not fall into it? And what is "falling " in gravitation mean when considering quantum mechanics? What does the equivalence principle mean in quantum cases?

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This is asking for coherent bound motion near a black hole, and it is very much current research. It is an important question +1, but it is hard to answer without stepping on toes and entering into realms where calculations have not been done properly and not everything is 100% clear. –  Ron Maimon Apr 13 '12 at 6:01

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There exists semi-classical computations of the orbit of an electron in Schwarzschild spacetime, such as "The Gravitational Analogue to the Hydrogen Atom" by Koch, Kober and Bleicher. The electron has a non-vanishing wavefunction at the singularity, but on the other hand, the Hamiltonian isn't hermitian. This corresponds to the case of the electron hitting the singularity, and as such, "disappearing", hence the non-conservation of probability.

I'm not aware of any non-semiclassical calculations for it, though I suppose it will depend a lot on the quantum gravity theory.

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Consider a region of energy near the event horizon of a black hole. Through particle pair production, it is possible to create $e^+,e^-$. We can idealize the situation so that one of these two particles enters the black hole, and the other may or may not do the same. In the classical case, the equivalence principle is the same as that formulated by Einstein:

A little reflection will show that the law of the equality of the inertial and gravitational mass is equivalent to the assertion that the acceleration imparted to a body by a gravitational field is independent of the nature of the body. For Newton's equation of motion in a gravitational field, written out in full, it is: (Inertial mass) $\cdot$ (Acceleration) $ =$ (Intensity of the gravitational field) $\cdot$ (Gravitational mass). It is only when there is numerical equality between the inertial and gravitational mass that the acceleration is independent of the nature of the body.

(From A. Einstein. “How I Constructed the Theory of Relativity,” Translated by Masahiro Morikawa from the text recorded in Japanese by Jun Ishiwara, Association of Asia Pacific Physical Societies (AAPPS) Bulletin, Vol. 15, No. 2, pp. 17-19 (April 2005).)

The quantum scenario is a bit different. If the view of LQG is adopted, then the idea is essentially the same as GR, since it is only the spacetime that is quantized. ``Falling'' in QG is quite complex. In GR, objects "fall" because of the Ricci tensor, $R_{\mu\nu}$ in the EFEs. Therefore, in GR, due to the presence of the source $T_{\mu\nu}$, gravity acts on objects.

In the quantum case, if one considers QFT in curved spacetime (which is essentially what your question is about), then the path integral approach says that the same particle ($e^-$) takes multiple different paths, so the source $T_{\mu\nu}$ is smeared out all over the region in which the source is observed. Thus the equivalence principle, in a way, tells us that the source $T_{\mu\nu}$ has a probability distribution over the whole region of the "quantum spacetime" and this in turn induces a new probability distribution for the motion of $e^-$.

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In hydrogen atom, an electron in ground state has non-zero probabilty of being found on top of the nucleus (all s-orbitals start with const. value at r=0, and then decay for higher distances). so first of all, in your language, it does kind of fall to the bottom of potential well. QM then prevents it from sitting still there, and it is in fact distributed with some PDF. Now I suppose that, for "hydrogen" being kept in some kind of bound state by gravitational force, something analogous would happen, it's just that gravity is not yet quantized, so you have to deal with Schwarzschild radius somehow. At that point you encounter hawking radiation and all that. Other things are already explained well enough by other comments, I think.

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The equivalence principle is a classical principle, it does not apply in quantum cases. Several specific violations of the principle are reported





In the general case we wait the equivalence principle to not hold when the general relativistic spacetime metric $g_{ab}$ is corrected by the higher order graviton corrections.

The black hole exists if you treat gravitation classically or semi-classically


When graviton corrections are considered both the horizon event and the spacetime singularity disappear.

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It would be REALLY helpful if downvoters say why, for the good of nonspecialists in the field. I don't understand much after "....corrected by the higher order graviton corrections." but it seems to me from general reading that there is a problem reconciling quantum mechanics and classical equivalence. I'm guessing many workers in the field believe that the inconsistencies must be reconciled to have a workable quantum theory of gravity: is this what the downvotes are for? If so, then say so - I wouldn't have called juanrga's assertion in the first sentence outright wrong: perhaps it could ... –  WetSavannaAnimal aka Rod Vance Jun 16 at 0:17
... be better worded as the principle of equivalence has only been worked into consistent classical theory, namely GTR, but there are as yet unresolved difficulties with reconciling it with quantum mechanics. Incidentally, see this answer, which references a very clear article by Paul Davies. –  WetSavannaAnimal aka Rod Vance Jun 16 at 0:20
@WetSavannaAnimalakaRodVance Thanks by kindly words. Davies paper just confirms that the equivalence principle applies when quantum effects are neglected, which was my point. –  juanrga Jun 23 at 14:11
That's what I thought. And, incidentally, that Davies paper, with its stunningly clear and simple example, was the first thing that made me I realise that equivalence principle is violated in QM. –  WetSavannaAnimal aka Rod Vance Jun 23 at 14:29

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