# Incommensurability between different observers describing the same universe?

According to black hole complementarity, if there is a black hole and Alice falls into it carrying a qubit, but Bob stays out, then Alice can measure the qubit inside the black hole, and confirm it exists inside, but Bob can observe the Hawking radiation coming out, collect all the data, feed them into a giant quantum computer, and extract the qubit from the outgoing Hawking radiation. Either the no xerox theorem is violated, or Alice' and Bob's observations are incommensurable.

In plain old quantum mechanics, it is possible to set up an experiment where we can choose between measuring the position, or the momentum of a particle, but once that choice is made, all observers will agree upon whether it's the position or the momentum which is being measured. But not for black holes? Is the qubit inside, or outside?

Bob has a problem. The qubit is scrambled and deeply encoded within the Hawking radiation. After collecting all the relevant Hawking radiation, would Alice have hit the singularity by then? If so, Alice measured first.

Things are actually much worse. According to complementarity, encoded within the Hawking radiation is also an extremely scrambled version of Alice as she would look like after drinking too many "Drink me"s, eating too many "Eat me"s, and nibbling way too many mushrooms. To make matters even worse, the scrambling scrambles the scrambled Alice with the scrambled qubit, and to extract the qubit, his quantum computer would also have to unscramble Alice from the qubit.

PS: If Alice is encoded holographically within the Hawking radiation, does that mean Bob can reconstruct what Alice did in the hole by observing all the Hawking radiation and feeding them into his quantum computer? And if he can reconstruct what Alice did, in what sense can't she communicate with Bob?

Is it possible to simultaneously measure observables which don't commute with each other?

Does Alice commute with Bob? They can't compare notes, that's for sure.

According to Niels Bohr, something is objective only if an observer can measure it and unambiguously communicate the results in public using common language. If Alice can't communicate anything with the outside world, is anything she observes objective? Is she even qualified to be considered an observer?

Let's say Humpty-Dumpty had a great fall to the black hole. According to Bob, as he approaches the hot stretched horizon, he becomes heated and scrambled, until we get scrambled eggs. The scrambled eggs evaporate as Hawking radiation. What Bob wishes to do is put Humpty-Dumpty back together again. To do that, he has at his disposal the entire royal team consisting of all the king's horses, and all the king's men. They engineer a device which can detect and measure every single quanta of Hawking radiation from the beginning right until the very end. Not only that, the measuring devices can preserve the quantum coherence of all the Hawking radiation quanta. All this data is transmitted along a coherence preserving perfect quantum channel to a quantum computer far away.

Is it possible to reconstruct Humpty-Dumpty inside the quantum computer? Would this violate the second law of thermodynamics? According to Szilard, Landauer and Bennett, no, because we can convert entropy into information. Bob's quantum computer has enough memory to store all the information of the black hole. Not only that, it's much larger than the black hole in size so it doesn't collapse into a black hole of its own. It preserves quantum coherence. In fact, it's so large, and we are so patient to wait for an astronomically long time that all the quantum computer processes are low energy, and never remotely approach Planck energies at all. All the redshifted Hawking radiation are also low energy. Surely the universe is Turing complete enough for Bob to construct and program such a quantum computer?

Can Bob reconstruct Humpty-Dumpty in principle? What about in practice? If something can be done in principle, but not in practice, is it still possible?

According to Alice inside the hole, Humpty-Dumpty said a word inside the hole. This word is holographically encrypted within the Hawking radiation, and is all there outside in encrypted form for Bob to read. The information is all out there, and if Bob takes the detected encrypted pattern as the word itself, he gets the word. But to find out what that word means, he needs a "common language", and he can use the quantum computer as the translator. If translated, did Humpty-Dumpty send a word faster-than-light? If untranslated, did he still send a word faster-than-light?

Inside the hole, Alice measures the qubit (Tweedle) along a basis consisting of the states "DEE" and "DUM". Outside, Bob detects all the Hawking radiation and feeds them into his quantum computer. He runs a program to extract Tweedle. But Tweedle is scrambled with Alice and her measuring apparatus, and in fact, just about almost everything else inside the hole. That's the nature of holographic encryption. According to Bohr, we have to include the measuring apparatus in the description. To unscramble Tweedle requires the quantum computer to extract a simulated Alice and simulated apparatus as well, and store them in the memory needed according to Landauer and Bennett. We can't erase the information! But the holographic Alice has already measured the holographic Tweedle using the holographic apparatus, and now, they are all entangled with each other. From the perspective of these holographic entities, this entanglement has decohered. To extract the pure qubit, Bob has to run another problem implementing the unitary transformation reversing the measurement and run them on the simulated Alice-apparatus-qubit system. This recombines all the decohered many worlds, recohering them. After extracting Tweedle, he measures it along an incompatible basis. But the simulated Alice now no longer has any memory of the measurement after the time-reversed transformation. The simulated Alice can remember the future, but not the past.

Think about it this way; the Hawking radiation is what we get after evolving time forward for the system consisting of the exterior and the stretched horizon. To reconstruct Alice requires the quantum computer to reverse time in the simulation right back to the moment when Alice and the qubit fall into the hole, then run the simulation forward in time again from the perspective of Alice. To simulate this time evolution, the quantum computer also has to include a significant part of the state outside the hole to take into account infalling matter, but this can be done in principle.

There was a time when a simulated Alice knew whether the simulated Tweedle was DEE or DUM, but Bob had no clue whatsoever about Tweedle, and can't without decohering the simulation along the DEE/DUM basis. After reversing time in the simulation, the simulated Alice had her memory of the value of Tweedle wiped clean, but now, Bob can measure the simulated qubit along the incompatible basis. There was never any time when both the simulated Alice, and Bob knew what the value of Tweedle was along incompatible bases simultaneously.

Was Alice teleported from the interior into the simulation faster than light? The simulated Alice would have no clue she is inside a simulation, and would think she exists in an objective black hole interior. The faster-than-light interpretation only happens if we mix up incompatible frames. From Bob's frame consistently, it's the holographic Alice which was teleported, and this never happens faster than light.

Do Alice and Tweedledee/Tweedledum really exist, or are they all a simulated dream inside the quantum computer? Remember that for Bob, the interior doesn't exist; only the stretched horizon does. Are the holographic Alice and Tweedledee/Tweedledum distinct as objects from the simulated Alice and Tweedledee/Tweedledum? Which of them is the real Alice? If you say neither of them are the real Alice, by the same reasoning, Bob doesn't have the real qubit either.

Actually, as Alice and Tweedle are quantum systems, they obey the no Xerox principle. So, the holographic Alice and the simulated Alice can' both exist at the same time, although they can exist at different times. What no Xerox is telling us is from a quantum information theoretic point of view, both the holographic and simulated Alice are one and the same.

According to Bob, the disembodied information encrypted within the stretched horizon and the Hawking radiation hallucinated itself as Alice, and this illusion hallucinated a nonexistent black hole interior. This disembodied delocalized illusion is transfered and reconstructed as a simulation inside a quantum computer, all for the sake of getting the qubit, but only after the original disembodied illusion had dematerialized to satisfy the no Xerox principle. The simulated Alice had her memory erased, all for the sake of getting the qubit.

Things get even more exciting. After measuring the value of the simulated Tweedle along an incompatible basis, Bob has to decide what to do with the simulation running on his quantum computer. He could decide his task is done, and end the simulation, effectively killing the simulated Alice. But what if he decided to keep on running the simulation, this time with the simulated time running forward? The simulated Alice, who still thinks she's inside an objective black hole, would notice a violation of the laws of physics leading to an unexpected change of the qubit, projecting it into one of Bob's basis states.

There is an entirely incommensurable point of view to this story. According to Alice, she is objectively real, and is inside an objectively real black hole. The part of spacetime outside the black hole which exists before the black hole finally evaporates away is also real. More precisely, those events in spacetime for which there exist future directed trajectories which can end inside the black hole. So, Bob and his instruments measuring the Hawking radiation are real before the final evaporation. Their reality can be confirmed by messengers (maybe photons, or something more elaborate) from Bob falling into the black hole to be picked up by Alice. But time would end at the singularity, and with it, the exterior of the black hole too, because messengers can't reach beyond the singularity. Actually, there is no beyond the singularity. But Alice can imagine an extrapolation of Bob and his quantum computer into the future beyond that, but alas, not simulate them because the computational resources inside the hole are too limited. Maybe she does this because she's a cosmologist, or a philosopher ;) She knows from the messengers that the real Bob before the final evaporation had built a quantum computer and detectors collecting all the Hawking radiation with the express purpose of reconstructing her and Tweedle. She also knows that if time didn't end at the singularity, this Bob would actually have the resources to reconstruct her, and will, but she also knows time will end, so that won't really happen in reality. But she's free to imagine what would happen if time didn't end for Bob. The imaginary future Bob will be under the delusion he's the one who is real, and so is his quantum computer, but Alice, who is very real, is but a simulation running on his quantum computer. Well, Alice has already measured Tweedle, and observes it's in a definite state, either DEE or DUM. So, in her imagination, if she were in a simulation, which she's not, this would be the time before the imaginary future Bob made the measurement, when the simulation is being run backward... Ah, what about earlier, before she measured Tweedle?

If the imaginary Bob decides never to run the simulation forward in time all the way to the singularity, then she won't experience the singularity if her imaginary scenario were actually true, and that would be a prediction distinguishing whether she was real, or simulated, but that's in the future and she hasn't experienced the future yet...

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Is there some reason why you create a new account for every new question (or, one could say, a new copy of your question)? – Luboš Motl May 16 '11 at 8:27
I read the first paragraph and it seemed abit scientifically eccentric... sorry! because i only have layman's understanding. I wouldnt expect a science writer to phrase the conjecture as you did. Also, i couldn't find that xerox theorem is commonly used concept... No-Cloning Theorem ??? and also, as a layman, i have read that no quantum information can escape via Hawking Radiation. so... tripply confused and 27 more paragraphs to read? – comprehensible Jan 25 '15 at 20:36

But not for black holes?

Not for black holes because for black holes, the infalling and exterior observers are spatially separated, so they can't communicate about what they should be measuring. Moreover, whatever A and B measure are observables that don't strictly commute with each other.

Is the qubit inside, or outside?

Both. This answer "both" is the very content of the black hole complementarity principle.

Bob has a problem. The qubit is scrambled and deeply encoded within the Hawking radiation. After collecting all the relevant Hawking radiation, would Alice have hit the singularity by then? If so, Alice measured first.

As a simple look at the causal diagram of an evaporating neutral black hole shows,

the black hole singularity is spacelike-separated from the exterior Minkowski spacetime - even the portion of the Minkowski spacetime "after" the black hole evaporated - so one cannot say that Alice measured the qubit first. To say that Alice did something near the singularity before Bob detected the last quanta of the Hawking radiation, the detection of the last quanta by Bob would have to belong to the future light cone of the singularity and it obviously doesn't.

Because the two measurements are spacelike-separated, there can't be a classical causal relationship between them, in either direction, and whether $t_A < t_B$ or $t_A > t_B$ depends on the coordinate system.

Things are actually much worse.

Things cannot be "worse" before they are "bad", and you haven't offered a single glimpse of any inconsistency of the black hole complementarity principle. There is no inconsistency exactly because of the basic point that you are fruitlessly attempting to deny - namely that the observations of Alice and Bob are complementary. They will never be compared with each other in the future because Alice is going to be destroyed at the singularity, without any hope of informing the external post-black-hole world about her perceptions. That's why there's no contradiction in assuming that Alice and Bob did measure observables that didn't commute with one another.

P.S. No, Bob can't reconstruct what Alice has actually measured inside the black hole because the corresponding observables don't commute with those that Bob inevitably measures when he wants to have any local device that is located outside the black hole event horizon. Alice's degrees of freedom - those from the black hole interior - are holographically encoded in the degrees of freedom outside the black hole. But that doesn't mean that they're actually being measured by Bob. At least if Bob is doing some doable low-energy local experiments outside the black hole, he can't measure the natural low-energy degrees of freedom relevant for Alice.

In principle, if Bob could make a sophisticated correlated measurement of everything, such a measurement could perhaps be made equivalent to what Alice can measure. In that case, he would get the same results as Alice has. However, this could only be done after all the Hawking radiation has been emitted.

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We'd expect physical black holes to have timelike singularities, though--the spacelike nature of the Schwarzschild singularity is something inherited from the inner Cauchy horizon, and even a small spin will make the singularity timelike. (although this means that observers won't be forced to intersect the singularity, either). – Jerry Schirmer May 16 '11 at 15:22
Tx for your refinement, Jerry. Still, one can't get out of the black hole. – Luboš Motl May 16 '11 at 19:11
If the horizon is shrinking, yes you can--a shrinking horizon is a timelike surface in the spacetime, so it is, in principle, transversible. If it evaporates completely, then every generator that enters the horizon will eventually leave. All of this is dependent on the degree to which the semi-classical approximation is still valid while the black hole is emitting Hawking radiation, though. – Jerry Schirmer May 16 '11 at 19:39
Dear @Jerry, the term "horizon" above always refers to the event horizon which is a null hypersurface. That's the relevant type of horizon that is helpful to answer questions such as "can Alice inform Bob", your horizon isn't, and the answer is No. – Luboš Motl May 17 '11 at 9:18
Lubos: if the hole evaporates completely, then there is no event horizon in the spacetime, only apparent horizons. The union of the past development and future development of the singularity is the whole manifold. – Jerry Schirmer May 17 '11 at 14:25

This question strikes right at the heart of the black hole information paradox. As I like to think of it, consider a Mach-Zehnder interferometer experiment with the first beam splitter well outside the black hole but the second beam splitter inside. Of course, due to the black hole geometry, this setup will break up fairly soon, or the first beam splitter will fall inside, but the only thing which matters is this apparatus holds up for long enough for a photon to pass through and interfere on the other side. Most people will swear the photon will interfere with itself inside the black hole. Quantum mechanics tells us this interference pattern means the which-way information of the photon trajectory is not entangled with anything at all outside the black hole, or inside for that matter. No decoherence by anything.

Then, the photon falls into the singularity and is annihilated. Supposedly, the information about the state of the photon is encoded within the outgoing Hawking radiation. If it is actually possible to absorb every single outgoing Hawking radiation and feed them into a quantum computer, it is possible to reconstruct the state of the photon just before it passes through the second beam splitter, and measure the path it took in the quantum computer simulation. This is nothing other than the qubit thought experiment rephrased in terms of quantum optics.

Unfortunately, the photon path is not entangled with anything outside the black hole and supposedly, any information inside the black hole can only get out superluminally. Surely if we modify the experiment a bit to get rid of the first beam splitter and replace it with a photon entering the black hole from one of the two directions with the direction not recorded anywhere outside the black hole, the same arguments indicating there is no decoherence will tell us no trace of which direction the photon came in is recorded anywhere in any form outside the black hole. Quantum information has really passed completely from outside the black hole to inside it with no trace outside. So why does the outgoing Hawking radiation know about which direction the photon came in? Clearly, only superluminal nonlocality can explain this state of affairs.

Entanglement is ruled out. The possibility that the interference pattern is washed out by a quantum computer simulation much much later is implausible. Cloning is also ruled out by the no-cloning theorem.

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The question posed here is not entirely vacuous. The hole in the BH complementarity is with black holes that are composed of a few Planck units of mass, say less than $10^3M_p$. For black holes of this scale the event horizon is probably not a well defined null surface, but is quantum mechanically blurred. The radius of the event horizon classically determines the mass of the BH $R~=~2M$, but quantum mechanically there is an uncertainty relationship which will adulterate this. The uncertainty in the $\Delta E~\simeq~\hbar/\Delta t$ does have a funny situation, where if the energy is $\Delta E$ is larger than the Planck mass there is the occurrence of an event horizon. The horizon has a radius $R~\simeq~2G\Delta E/c^4$, which is the uncertainty in the radial position $R~=~\Delta r$ associated with the energy fluctuation. Putting this together with the Planckian uncertainty in the Einstein box we then have $$\Delta r\Delta t~\simeq~\frac{2G\hbar}{c^4}~=~{\ell}^2_{Planck}/c.$$ This connects with noncommutative coordinates in quantum gravity. This means exterior and interior states are not sharply distinguished from each other. An exterior observer who measures states of the black hole has a non-zero probability of measuring a state that would be associated with the classical black hole interior.

The black hole is a sort of processor, and it processes exterior states into interior states and then into exterior states as Hawking radiation. Of course this statement betrays one problem with understanding this, for it is tensed. The English language tends to be very time ordered, and yet this is a quantum process that nonlocally connects the interior of the BH that is causally disconnected from the exterior. The exterior states have an S-matrix description which extends from $\infty$ to $2M$, or more properly for a stationary observer to $\lim_{r\rightarrow 2M}~r^*~\rightarrow~-\infty$ for $$r^*~=~r~+~2Mlog\Big(\frac{r~-~2M}{2M}\Big)$$ This choice of coordinates extends the S-matrix to a domain $-\infty,~\infty$. The interior observer observes quantum states on a different domain $-\infty,~0$. The two sets are inequivalent, and so the black hole entropy which connects interior microstates with exterior QM is not complete. The classical black hole radius = mass relationship is used to get the area-entropy relationship, which is an imposed rule and in some ways ad hoc.

To return to the question at hand, even with a quantum mechanical black hole if an observer in the exterior measures quantum states associated with the BH the observer might have a nonzero probability of measuring an interior state, but that state is not equivalent to a measurement of an exterior state. Further, this observer has no way of determining whether that state pertains to the exterior or the interior. What is not well understood is how the BH complementarity operates in this case.

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