# Effect of gravity on quantum probability

As a thought experiment: The nucleus of a hydrogen atom is outside the event horizon of a black hole such that 40% of the 1s electron shell is within the event horizon. The single electron that is "entangled" ( I avoid the use of the word "orbit" to avoid the circular orbit visualization of the problem) with this atom has its position governed by the quantum mechanical wave function/ probability of its position. This would mean that a 40% probability of the entangled electron being present inside the event horizon exists. According to relativistic theory, it should not be able to leave, yet its position as an entangled electron is governed by the wave function. This would appear only be possible if gravity can change the wave equation, or if gravity can ionize( effectively "un-entangling" that electron from the nucleus) the atom, or if a virtual particle pair can replace the entangled electron with another and a positron inside the event horizon( which is still probably changing the wave equation. Or does the electron , by virtue of its probability function appear back across the event horizon( there are other "forbidden" events allowed by wave functions that this might represent. OR something else I have not thought of.

• Note that you are trying to use a standard nonrelativistic quantum model in an environment that is explicitly relativistic. It should not be that surprising that it does not make much sense. The full treatment would have to be quantum gravity, which we don't have. One can can also try a semiclassical approximation: given a black hole spacetime, solve the Schrödinger equation in that spacetime and see what happens. I do not know if anybody has done that, but for uniform acceleration the result is sciencedirect.com/science/article/pii/S0375960116307083 – Anders Sandberg Apr 11 at 1:31