What does the equivalence principle mean in quantum cases?

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

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

http://prd.aps.org/abstract/PRD/v85/i4/e044052

http://iopscience.iop.org/0264-9381/29/2/025010

http://prd.aps.org/abstract/PRD/v78/i6/e064002

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

http://arxiv.org/abs/0902.0346

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

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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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