Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What happens if you perform a double slit experiment near an event horizon, if one of the slits is outside, one is inside the event horizon?

share|cite|improve this question
I suspect the formulation of the question may need refinement, but I haven't made progress in the last few days so I posted it anyway. – mtrencseni Jan 12 '11 at 19:49
don't forget that you can always edit it later if you come up with useful revisions. While a little more detail never hurts, I do think this is definitely a very interesting question :) – David Z Jan 12 '11 at 20:07
not sure how a photon could escape from either path. any photon passing within $3GM/c^2$ of a Schwarzschild BH (horizon is at $2GM/c^2$) will get captured, unless it is on an outward-directed trajectory. In other words, the light source would have to be coming from inside the black hole to reach an observer at infinity. – Jeremy Jan 13 '11 at 16:58
up vote 9 down vote accepted

This is just a little bit subtle, because at first it looks like nobody has causal access to both light rays, which is required for the existence of interference pattern. On the other hand the principle of equivalence seems to assure us that local experiments done by a freely falling observer would not be affected by the presence of the horizon. There is a sense in which this is true, as follows.

For an observer freely falling into the horizon of a large black hole while conducting the double slit experiment, there will not be any change in the results of the experiment. Gravity is weak and nothing special happens when they cross the horizon, they will only discover they fell into a black hole sometime in the future when tidal forces will make life very uncomfortable. Any experiment localized within space and time will be the same up to tiny corrections due to weak gravity. This is true also if one of the slits (and therefore necessarily the light source) happens to be outside the horizon at the time of the experiment. Of course, for the freely falling observer that statement does not mean much, only the outside observer will be able to make meaningful distinction between inside and outside the horizon.

On the other hand, for an observer staying outside the horizon, one of the slits is invisible, they don't have access to all the light rays, and they will not see any interference. In fact they will not even be aware that there is a double slit experiment going on. There is no problem there because they are not freely falling observers, and there is nothing that states that the result of their experiment is identical to the same experiment made in flat space.

The fact that infalling and outside observers describe the same phenomena so differently lead to many black hole paradoxes, where quantum mechanics and GR (or the principle of equivalence) just barely co-exist without contradiction. Lenny Susskind's popular book on black holes has a good discussion, I think.

Edit: If you jump into the black hole you can have access to the interference pattern after the fact. So, it looks like you can generate another paradox, in addition to the original one. Look at for the resolution of that one. Roughly speaking, after you jump into the black hole you have only finite time to make measurements before you encounter the singularity. This time is insufficient to get more information than you are entitled to by the rules of quantum mechanics. This statement requires a detailed calculation which is in that paper.

share|cite|improve this answer
...and you can't maintain a situation where one slit is inside the horizon and one is outside, once inside things only go down and they do it fast. Plus there are no stable orbits inside 3/2 of the event horizon radius. – dmckee Jan 12 '11 at 20:15
Sure, it cannot be a completely static situation, but it can move really slowly initially, for sufficient amount of time so you can conclude the experiment. Remember, these are all thought experiments... – user566 Jan 12 '11 at 20:22
I"m not sure this is right--it's true that nothing (locally) special happens as you cross the horizon, but it's also true that no null ray inside the horizon can exit the horizon. I don't think you would see an interference pattern on the half of the screen outside the horizon, at least classically--the wavefronts inside the horizon would be unable to leave the horizon, and thus would be unable to interact with the wavefronts outside the horizon. – Jerry Schirmer Jan 12 '11 at 23:03
By the principle of equivalence every local experiment, viewed by a freely falling observer, will not be affected much by it crossing the horizon. What I have in mind is moving lab, which momentarily (at the time of the experiment) has one slit inside and one slit outside the horizon. The interference pattern is seen later by the inside observer, and it would be pretty much the same as we'd see it. An outside observer is not freely falling, their perspective is completely different, they would not see any double slit, interference pattern or any of that. – user566 Jan 12 '11 at 23:40
@Moshe: Look at the Penrose-Carter diagram for the Schwarzschild spacetime--the entire outside of the hole is outside of the domain of dependence of the entire inside of the hole. Nothing that happens at $2M+\delta$ can depend on anything that happens at $2M-\epsilon$ (I wasn't the downvote, though) – Jerry Schirmer Jan 13 '11 at 18:48

You get two, single slit experiments.

One on either side of the event horizon, by definition.

Light in, Light out.

share|cite|improve this answer

In order to perform a double slit experiment, you need to recombine the beams passing through both slits, and that is where the interference occurs. Since one slit is inside the event horizon, this interference location has to be inside the event horizon, in order to be reachable by photons/electrons/whatever passing through the "inside slit".

An observer outside the black hole will therefore not be able to see whether the interference happened or not. And an observer inside the horizon can see the interference pattern, but it is not problematic because he can see both slit. However, he will not be able to describe us what he saw ;-)

Edit: Clarify sentence 1 of §2, to answer @kakemonsteret's comment.

share|cite|improve this answer
This seems like pure speculation to me – TROLLKILLER Jan 21 '11 at 19:03
I don't see why it's more speculative that the initial question. What do you fine speculative ? The fact the the recombining area is inside the event horizon ? – Frédéric Grosshans Jan 21 '11 at 19:10
This thing "recombine the beams passing through both slits". And also, if there is interference or not can be recorded in the detector, so if you are outside you say you cannot observe interference, so if that is stored in the detector, once you pass inside the black hole and you compare the recordings of the detector with the inside observer then you got a paradox – TROLLKILLER Jan 21 '11 at 19:17
For me, that's exactly right, for what it's worth. – user566 Jan 21 '11 at 23:52
@kakemonsteret At the risk of giving an unnecessary advice: the confusion in such cases often comes from assuming there is a n objective, observer-independent meaning to statement like "there is interference". Try to think about the process of observing the interference - it is a different process for both observers, and they get different results. You need access to both light rays to see interference, so only the inside observer can see them. – user566 Jan 22 '11 at 18:40

I strongly suspect there would be no interference, as we would be able to determine which slit the particle went through.

share|cite|improve this answer

The question hinges upon how EPR pairs behave in the presence of an event horizon. The operators become Bogoliubov transformed $$ b~=~a~cosh(g)~+~a^\dagger sinh(g),~b^\dagger~=~ a^\dagger~cosh(g)~+~a sinh(g). $$ These transformed operators mean that a Minkowski vacuum composed of regions inside and outside the horizon will involve terms ~ $sech(g)$. A superposed or entangled state across the horizon will contain this factor, which for a large $g$ or acceleration near the horizon. An EPR pair near the event horizon becomes entangled with the black hole, where this is similar to a measurement. Another way of looking at this is the stationary observer near the horizon experiences a set of thermal states with a black body spectrum $$ cosh(g)~=~\prod_{\omega}\frac{1}{\sqrt{1~-~e^{2\pi\omega}}} $$ for $\omega$ the frequency of a Rindler particle. The black hole acts as a thermal decoherence bath. This loss of coherence of an EPR pair can occur outside the BH, for the path integral of the system contains paths with “probe” the black hole interior.

share|cite|improve this answer
I am a little confused about the setup here and how it relates to the OP's question. I think that even classically, the observer firing his rockets to stay outside of the event horizon will fail to register any measurement that a detector screen that has passed the horizon can send.. So it strikes me that at least for this non inertial observer, there is nothing that requires a semiclassical calculation? – Columbia Jan 22 '11 at 6:29
If one part of an EPR pair actually passes through the event horizon the entanglement is completely lost. For the experiment done above the event horizon the quantum states travel on a path integral which fills space with probability amplitudes. Some of these amplitudes cross the even horizon. This results in quantum noise which is a source of decoherence. – Lawrence B. Crowell Jan 22 '11 at 22:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.