# Concerning psi and black holes

shaan_aragorn
According to quantum theory the total probability of finding the particle somewhere in space is always zero. So if a particle vanishes in a black hole will that mean (psi)^2 = 0, which is illegal?

Staff Emeritus
shaan_aragorn said:
According to quantum theory the total probability of finding the particle somewhere in space is always zero.

Say what?

What version of "quantum theory" is this? Can you give an exact citation where you read this?

Zz.

Staff Emeritus
Gold Member
I think s/he meant one, not zero.

quetzalcoatl9
shaan_aragorn said:
According to quantum theory the total probability of finding the particle somewhere in space is always zero. So if a particle vanishes in a black hole will that mean (psi)^2 = 0, which is illegal?

i'm going to assume that the poster meant 1, and not 0.

I don't know if you'll find a sufficient answer to your question...if a particle falls into a black hole, is it still in this universe? if yes, then the probability density over all space holds. if not, then is it even relevant, since the equation says nothing about this?

so a) i don't think that your scenario violates the equation and b) i also don't think that we can even speculate since, to my knowledge, black holes are still largely not understood entities.

Antiphon
At the risk of re-railing this thread, I think over time the integrated probability
of the particle being found outside the black hole would decrease, so yes
the hole would be "sucking up" $$\Psi$$. If this is true, the inside of the hole should
be considered part of the "Universe" but an inaccessible part.

Let me add a serious twist to the question. What happens if one member
of an entangled pair of particles falls in? If even light can't escape, can
a measuement "from beneath" the event horizon bring about a new quantum state which
is still correlated with it's partner outside the black hole?

Conservation laws say it must.

General relativity says no information can come out of the hole. So
how does the change in the wavefunction take place outside based
on something that happens to the particle's correlated pair inside?

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Staff Emeritus
Gold Member
Gokul43201 said:
I think s/he meant one, not zero.

Yes, and even in that case, this is only the case in NON-relativistic QM - of a single particle - so we are far from a treatment where we have black holes :-)

I think that the answer is hence two-fold: the "psi defined everywhere and normalized to unity" is a concept that comes from non-relativistic quantum mechanics and hence not applicable to the situation at hand as such. Nevertheless, the crux of the problem remains, formulated differently: how can the motion of matter be described by a unitary operator in a black hole geometry (which is supposed to be reversible). That's the famous "quantum information riddle at a black hole".
As far as I know, we don't have a fully working theory of that. But some bricolage seems to show that the problem goes away (I think Hawking recently lost an encyclopedia over it, in a bet with another one) when you consider superpositions of "more or less flat space" and "hole forms and evaporates". These are exotic superpositions indeed, but I think Hawking had some indications that the overall operation can be unitary.

marlon
vanesch said:
(I think Hawking recently lost an encyclopedia over it, in a bet with another one)

Yes, he lost it to John Preskill. He was so glad he won, he dedicated part of his personal website to it :)

Preskill's VICTORY

regards
marlon

shaan_aragorn
I don't know what I was thinking when i wrote zero! I meant one.