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Is decoherence even possible in anti de Sitter space? The spatial conformal boundary acts as a repulsive wall, thus turning anti de Sitter space into an eternally closed quantum system. Superpositions remain superpositions and can never decohere. A closed system eventually thermalizes and undergoes Poincare recurrences. Whatever the dual is to an interacting conformal field theory, is it really a theory of quantum gravity, or does it only look like one superficially?

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In large enough de Sitter space, like our Universe, the local physics is surely de facto indistinguishable from that of a flat space or a large AdS space, isn't it? Why do you find the cosmic horizon relevant for decoherence? Macroscopic things decohere within $10^{-50}$ seconds, long before things reach the cosmic horizon. And yes, dS space eventually thermalizes which in its case means that it gets empty, except for thermal quanta of wavelength comparable to the dS radius. What's the problem? Recurrence is just a very low-probability effect of a dropping entropy. Is this problematic? –  Luboš Motl Dec 21 '11 at 19:31
    
@Lubos: dS and AdS are separate issues entirely, the difference between an open and closed system. –  Ron Maimon Dec 22 '11 at 0:50
    
theorist says "Superpositions remain superpositions and can never decohere." They certainly can, at least in a large enough AdS space, as Luboš points out. Decoherence means that wavepackets in the combined system+environment configuration space have spread out and aren't interfering. Once that has happened, decoherence has occurred, regardless of whether unusual dynamics or Poincare recurrence later cause "recoherence" by bringing them back together. –  Mitchell Porter Dec 22 '11 at 3:19
    
@Mitchell: Lubos is using dS space to illustrate, which is a bad illustration, because dS thermalizes and is weird, while AdS is well defined. deSitter won't have poincare recurrences, which is why Susskind says it must always be unstable to decay to something else. I don't know the answer, because QM in dS space is not worked out. This is not about the question, but about the comments. I agree that the idea of decoherence is that the observer will decohere the system with information gained, later Poincare recurrence will happen long after tha observer's information is lost again. –  Ron Maimon Dec 22 '11 at 5:16
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@Lubos: I see now, you wanted a finite number of degrees of freedom. I imagine that a suitable box will do, or a toy model, you don't need a full cosmology. But it has to be fully unitary and AdS-like in this sense. I am aware that AdS has infinitely many degrees of freedom, but they are unitarily evolving, and there are suitable truncations of the transverse coordinates, just take N=4 SU(N) gauge theory in a periodic box. –  Ron Maimon Dec 22 '11 at 8:43

3 Answers 3

Your question is not about AdS at all, it is about a closed quantum system with a finite number of degrees of freedom.

If you examine the space-time near a finite area quantum black hole, you will see an approximate AdS space. For this reason, my original answer included AdS spaces in the list of finite systems, although this is completely false for what people would call a normal classical AdS space. Thse spaces are unbounded and make open systems. The near horizon extremal black hole is presumably sitting in an infinite spacetime, so it is also an open quantum system. So it is best to reformulate the question for the domain you intended:

How can you have irreversible decoherence in a closed quantum system with a finite number of degrees of freedom?

The same question can be formulated to a certain extent in a closed classical system, where Poincare recurrences still occur. How can you reconcile poincare recurrence with irreversible information gain by an observer?

I think the many-worlds point of view is most useful for understanding this. If you have a closed quantum system that contains an observer, it is the observer's gain in information that determines which relative branch of the wavefunction becomes the real one for this observer, a branch selection from the decohered branches which occurs outside the theory. The branches are still effectively dechered so long as the arrow of time is still forward, because the wavefunction is spreading out into new regions of the enormous phase space. It is reasonable to think that the system will need to thermalize and destroy all the observers long before the recurrence makes the entropy decrease to the initial value.

This observer can only exist while there is an arrow of time, so there is no real paradox. You can have effective decoherence and measurements for the thermodynamic arrow-of-time stage. The same is true in classical closed system, the observer will have to dissociate and reform in order for the system to have a poincare recurrence.

But perhaps you meant this as a counterexample to the claim that irreversible decoherence can occur in pure closed-system quantum mechanics. In this case, I agree that it is obvious that the closed system cannot really decohere irreversibly, since it will eventually recohere to close to its initial state.

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There are no Poincare recurrences in the AdS space or any other non-integrable system with infinitely many degrees of freedom (dS space is the only exception with a "finite maximum entropy" where it may occur), so your whole answer is totally wrong. Moreover, the very idea that decoherence depends on some big-scale cosmological properties of the spacetime is entirely misguided. One may say that you copied both of these misconceptions from the OP but that doesn't change the fact that your answer lacks any positive value. –  Luboš Motl Dec 22 '11 at 8:29
    
@Lubos: I agree with you. I was assuming that the OP was interested in the case where the system is not open, so I said "your question is not specific to AdS space" right at the beginning, but I did not make it clear that AdS is open, and I should have. I immediately thought of finite degree of freedom truncations. But this answer explains what the OP is confused about. I edited it to incorporate the valid points you made. I will point out that I always think of AdS as shorthand for "near horizon", and to get a finite system, just think of a finite area extremal black hole. –  Ron Maimon Dec 22 '11 at 8:50
    
Thanks, @Ron. I still think that there are many confusions, however. AdS is a near-horizon geometry of a black hole geometry but AdS itself doesn't have any horizon. Also, the reversibility may require the system to be closed and finite, and AdS isn't one. More generally, I don't understand the logic of the original question because it mixes cosmology with local quantum prcesses, but that's not really your fault. –  Luboš Motl Dec 23 '11 at 6:46
    
@Lubos: That there are confusions is true, but these confusions are not solely mine--- there isn't a good picture of the classical AdS/CFT world, or else one would be able to follow a massive object falling into a black hole and reemitted. Gubser analyzes this, but I am sure wrongly, by considering a stack of branes with one brane moving slowly to a direct collision with the others. I thought about this also. The solution should be oscillatory and reversible, corresponding to a harmonic oscillation in the AdS space (at least when the brane has a small velocity and starts out tightly bound) –  Ron Maimon Dec 23 '11 at 8:14
    
But Gubser suggests that the noncommutative description makes this irreversible for arbitrarily small temperature on the brane-stack. his argument is that there are many more noncommutative degrees of freedom, so that when the single brane hits the stack, it will mix up into a noncommutative mess in an irreversible way, corresponding to irreversible absorption. This idea is not at all plausible to me, although I can't rule it out definitively. The picture should have a classical geodesic interpretation with classical in-out motion. That explains the mismatch between the horizon structures. –  Ron Maimon Dec 23 '11 at 8:20

This depends on whether your space includes the observer. In presence of the observer there will be decoherence. Without the observer there will be unitary evolution.

I am ready to answer any additional questions that can arise from my answer.

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The question was about decoherence of small observer-less systems. It was about whether there is a sensible decohered limit in a closed system (and whether AdS counts as a closed system). The OP is aware of the observer issues in QM, as were the previous answers, they are just not the topic of the question. –  Ron Maimon May 21 '12 at 2:58

Decoherence is more than anything a matter of what you define to be the "environment". The environment is supposed to be external to the system of interest and entangling interaction with it produces decoherence. If the environment in question is a part of the adS space then the subsystem can certainly decohere. If what you are asking is whether the space as a whole can decohere when it's isolated from its environment then the answer must by definition be no.

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