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I wanted to re-open the question of quantum measurements across event horizons. If I set up two slits or more generally a diffraction grating which crosses a black hole event horizon, and I shoot a stream of appropriate particles at the slits/grating, will I observe an interference pattern on a detector placed on the far side of the slit/grating and also crossing the event horizon?

It seems to me that the previous discussion was focused on the perspective of the observer or observers within or without the horizon, not the particles themselves. A double slit experiment does not involve EPR-like entanglements of multiple particles, if I understand correctly it works through the adding up of detections of "individual photons" (whatever that means; otherwise we can posit electrons or buckyballs which certainly seem to have some kind of particulate behavior at least sometimes). Very naively, if a photon or other particle is fired at a screen with a double slit, in which one slit is just outside and the other just inside the event horizon, it seems to me that there is still a potential problem regarding whether interference will occur. The wave function itself is not a mass-energy construct, hence does not obviously respond to the structure of spacetime. Given that the nonlocal aspects of QM leading to interference fringes are a problem even in flat spacetime, I don't see how they become any less difficult in curved spacetime. Is there a trivial objection to proposing that interference fringes would build up with multiple particles, just as in any double slit experiment, indicating that QM waves, whatever those are as well, are immune to event horizons? Clearly only that part of the detector screen lying outside the horizon would be accessible to the observer. Note that spin is not an issue in this experimental design, so some of the papers I have read which deal with the interaction between particles with spin with black holes (especially rotating ones) would not seem to be relevant in this design.

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Can you please state the question in the form "If I do this and that, what will I observe in this or that detector?" –  Ron Maimon Oct 22 '11 at 5:58
    
Have revised as requested, hope it is now ok. –  MES Oct 23 '11 at 0:27
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"Reopen"? It wasn't ever "closed"... anyway, it seems that your question is asking pretty much the same thing as physics.stackexchange.com/questions/2750/…. Could you take a look at that and clarify how the answers to that question are insufficient for what you want to know? –  David Z Oct 23 '11 at 0:46
    
I agree this is basically the same question. It seems to me that none of the answers in the previous thread were apropos, they were focused on the observers and not the actual particles, moreover they did not seem to address the issue of quantum non-locality in a two-slit design, being sidetracked by EPR designs and particle spins, which add unnecessary complexities and are not directed to the original question. Should I create an "answer" to the previous question, even though I don't have one? –  MES Oct 23 '11 at 3:35
    
The question is stated better now, but the issue with the observer status of the particles is dubious. Do you mean to make a conscious system and diffract it through two slits? This has problems of principle, due to the constant internal changes of state in a conscious system required to stay conscious--- these make it so that it cannot interfere, because only the same state can interfere--- no changes. If you mean to shoot an atomic scale structure through a diffraction grating and observe the result, the answers overlap those of the previous question. –  Ron Maimon Oct 23 '11 at 7:41

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When a diffraction grating is crossing an event horizon, it can't stay put. It has to keep falling to keep from flying apart. If you hold the part that is outside the horizon so that it isn't falling inward, the part that is past the horizon gets torn off and falls inside. If you shoot an electron past the diffraction grating (and apply an electric field to keep the electron from falling into the black hole) you will see diffraction only from the slits outside the black hole. The part of the electronic wavefunction which passes the horizon will not interfere with the outside part for any reasonable macroscopic size black hole, but will fall inward.

This is a near horizon phenomenon, and near the horizon, space is flat. So the effect doesn't require a black hole--- the exact same thing happens when you accelerate fast in Minkowski space. If you hold a large diffraction grating past your acceleration Rindler horizon, the apparent horizon you see when you accelerate, it will get torn off just the same.

In order for interference to happen, the two paths must be indistinguishable in a quantum sense, so that you can't know which one happened. If you have a particle with some miniscule internal consciousness of any sort, and you want it to diffract, you need to cool it down to absolute zero, shutting off any internal computation for a while. Then you do the diffraction, then you heat it back up again so that it turns back on. The diffraction is always necessarily happening on something unaware. If you isolate a person in a chamber, and you try to get the chamber to diffract, you will need a perfect isolation system to prevent any leaking of information about the internal quantum state to the exterior of the chamber. Then, in theory, you can get two paths of the chamber to diffract, but the person inside can't have any idea which path was taken, and wouldn't even know anything about the exterior world at all.

The issues for the diffraction grating which falls through the black hole are fully adressed by the viewpoints of the observer that falls along with it, and the one that stays outside, and are covered by this earlier question: Double slit experiment near event horizon

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Thanks for the interpretation Ron. I understand that the issue does not require a black hole, any accelerating frame of reference will suffice. But the event horizon does seem to be a more extreme case so better for the thought experiment. There is no issue of consciousness here. I did not follow your argument about "perfect isolation", but in the first section you said clearly that the part of the wave function inside the horizon is off-limits to the part outside. That is what I am still wondering about. Are we really sure that GR "trumps" QM non-locality? I thought that the real problem. –  MES Oct 24 '11 at 0:26
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@MES: quantum nonlocality does not allow you to access the wavefunction of a far away thing--- only the reduced density matrix of the things you can experiment on. Quantum nonlocality doesn't care about whether the other thing is inside a black hole or on the other side of town. The local thing is always described by its own density matrix, without regard to the other thing, unless you go to your friend on the other side of town and compare notes. –  Ron Maimon Oct 24 '11 at 3:59
    
I think I follow, Ron. But why doesn't the detector respond to the total wave function, of which part is spatially across the event horizon? –  MES Oct 24 '11 at 12:26
    
@MES: If it does, the part across the event horizon vanishes--- that's collapse, not interference. –  Ron Maimon Oct 24 '11 at 13:36
    
Aha, now that makes sense. Crossing the horizon itself collapses the wave function so that interference is no longer possible. I wish I could follow the math behind this interpretation, maybe someday. Thanks so much for staying with me. –  MES Oct 25 '11 at 0:16

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