Can two particles remain entangled even if one is past the event horizon of a black hole? If both particles are in the black hole?

What changes occur when the particle(s) crosses(cross) the event horizon?

I have basic Physics knowledge, so I request that answers not assume an in-depth understanding of the field. Thank you all in advance!

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    $\begingroup$ in both cases measurements are impossible, hence, even if we come up with some sensible argument about what should happen, its slightly outside of the realm of science. And i say 'slightly' because this conclusion is based on our current understanding of black holes. $\endgroup$
    – lurscher
    Commented May 14, 2012 at 20:56
  • $\begingroup$ @lurscher: This is not completely true, since we can study model black holes in AdS/CFT. $\endgroup$
    – Ron Maimon
    Commented May 15, 2012 at 6:39

2 Answers 2


This question is the black hole information paradox. If you take two entangled particles, make a black hole by colliding two highly energetic photons, throw in one of the two entangled particles, and wait for the black hole to decay, is the remaining untouched particle entangled with anything anymore?

In Hawking's original view, the infalling particle would no longer be in communication with our universe, and the entanglement would be converted to a pure density matrix from our point of view. The particle outside would no longer be entangled with anything we can see in our causal part of the universe. Then when the black hole decays, the outgoing Hawking particles of the decay would not be entangled with the untouched particle.

This point of view is incompatible with quantum mechanics, since it takes a pure state to a density matrix. It is known today to be incorrect, since in models of quantum gravity when AdS/CFT works, the theory is still completely unitary.

This means that the particle stays entangled with something as its partner crosses the horizon. This "thing" is whatever degrees of freedom the black hole has, those degrees of freedom that make up its entropy. When the black hole decays completely, the outgoing particles are determined by these microscopic variables, and at no point was there ever a loss of coherence in the entanglement.

This point of view requires that the information about the particle that fell through the horizon is also contained in the measurable outside state of the black hole. This is t'Hoofts holographic principle as extended into Susskind's black hole complementarity, the principle that the degrees of freedom of a black hole encode the infalling matter completely in externally measurable variables. This point of view is nearly universal today, because we have model quantum gravity situations where this is clearly what happens.

The details of the degrees of freedom of a four dimensional neutral black hole in our universe are not completely understood, so it is not possible to say exactly what the external particle is entangled with as the infalling particle gets to the horizon. But the basic picture is that the infalling particle doesn't fall through from the external point of view, but smears itself nonlocally on the horizon (like a string theory string getting boosted and long). The external particle is still entangled with this second representation of the infalling particle.

This means that the same thing is described in two different ways, the interior and the exterior. But since no observer measures both at the same time, it is consistent with quantum mechanics to just have a unitary transformation that reconstructs the interior states from those states of the exterior that can be measured at infinity by scattering experiments.

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    $\begingroup$ But by making a measurement on the external particle, we can immediately deduce the state of the "internal" particle. That implies we've been able to extract information from a black hole, which isn't supposed to be possible, right? $\endgroup$ Commented May 14, 2016 at 18:16
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    $\begingroup$ I've never taken college level physics and I suck at math, but... If I understand correctly, from outside of a black hole, we never actually witness anything cross the event horizon of a black hole, right? If you drop something in, it take forever for it to actually fall in, right? So from our perspective, the electron never really leaves our universe, does it? I may well be misunderstanding this concept, though. $\endgroup$
    – Pete
    Commented Sep 11, 2017 at 14:40

That isn't actually an issue. You have no control over what answer you will get when you make a measurement on the external particle, so you get one of several possible answers; then you are deducing the only logically possible answer for the internal particle. That isn't information transfer from the inside to the outside - for that someone inside the black hole needs to intentionally send you something specific that they direct.

As an example, if your friend could be in Dallas or London with 50% probability, if you discovered that person in London, the probability of them being in Dallas drops to zero instantaneously. That doesn't violate relativity - information wasn't actually transferred anywhere.

  • $\begingroup$ But isn't there a difference in your analogy in QM? In QM your friend IS only in London immediately when you measure Dallas. In your analogy your friend was in London the whole time $\endgroup$
    – K2xL
    Commented May 11 at 12:11

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