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  What exactly does the holographic principle say?

+ 8 like - 0 dislike

Does the holographic principle say given a spatially enclosing boundary satisfying the Bousso condition on expansion parameters, the log of the number of microstates in its interior is bounded by $\exp\{A/4\}$ where $A$ is its area? Or does it say something stronger, namely the state restricted to the boundary uniquely determines the state restricted the interior? Can we have the former without the latter?

This post imported from StackExchange Physics at 2014-04-04 16:24 (UCT), posted by SE-user joobjoob
asked Aug 16, 2012 in Theoretical Physics by joobjoob (40 points) [ no revision ]

4 Answers

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The holographic principle is a stronger proposition than the entropy bounds but what is the strongest possible formulation that is still valid is disputable. Nevertheless, some facts are well-known.

The entropy bounds say that the maximum number of distinguishable microstates that may describe the interior of a region with a fixed boundary is $C\cdot \exp(A/4G)$ where $C$ is some subleading, non-exponential correction (or exponential in lower powers of $A$). The entropy bounds hold for light sheets etc. that satisfy the appropriate criteria – which of them are the "best ones" is a bit disputable, especially for time-dependent spacetime geometries (which were the focus of Bousso's work: Bousso wanted to generalize previous, much older results on the entropy bounds from stationary to time-dependent geometries and to some extent, he succeeded, although it's not clear whether his particular prescription is unique or canonical in any sense).

However, such entropy bounds are largely classical in character – they pretty much say that the objects with connected event horizons maximize the entropy and nothing else.

The holographic principle, as originally formulated, was and is meant to be a stronger proposition than the entropy bounds. It says that not only the number of "bits" needed to describe the interior is bounded by a linear function of the surface area; in fact, there exists a "natural" theory living on the surface area that is fully equivalent to quantum gravity in the interior even though the surface area has a smaller dimensionality (by one)!

I used the adjective "natural" because the theory on the surface has to satisfy some properties for the statement to go beyond the entropy bounds. Why? If we allow the theory on the boundary qubits to be "arbitrary", we may write down any Hamiltonian as some complicated operator on the space of all the qubits, assigning any basis vector of the Hilbert a general linear superposition of the basis vectors.

It's always possible to write down a Hamiltonian on the Hilbert space of a given dimension as "a theory". However, in general, such a theory will be totally nonlocal. The ordering of the qubits along the surface will generically be just a bookkeeping device to remember the number of the degrees of freedom but the Hamiltonian won't respect the arrangement in any way.

To make the holographic principle stronger than the entropy bound, the theory on the surface has to be required to be "natural" which ideally means "local". Indeed, this is exactly the case of the AdS/CFT correspondence where the boundary sits at infinity of the AdS space and the dual conformal field theory, CFT, is indeed exactly local.

However, there are no known examples in which the boundary would be at a finite distance. It's rather obvious that if there were a boundary theory for a finite region, it would be at least somewhat nonlocal because the geometry of the surface is fuzzy, with the fuzziness length scale equal to the Planck scale (or much longer than that). The local character of the CFT in the AdS/CFT probably depends on the infinite scaling of the metric: infinitely short distances in the CFT theory may still correspond to finite or large proper distances in the AdS space (because they're distances between points near the boundary at infinite where everything is long) and that's why we get locality.

All the known versions of the AdS/CFT correspondence are therefore the only known examples in which the holographic principle in the form saying something about the actual "theory on the boundary" – i.e. saying more than the entropy bounds – seems to be correct.

This post imported from StackExchange Physics at 2014-04-04 16:24 (UCT), posted by SE-user Luboš Motl
answered Aug 17, 2012 by Luboš Motl (10,278 points) [ no revision ]
+ 0 like - 0 dislike

There's no a priori reason why we can't have the former without the latter. The stronger version was only introduced to resolve the black hole information loss "paradox".

This post imported from StackExchange Physics at 2014-04-04 16:24 (UCT), posted by SE-user Tweeter
answered Aug 16, 2012 by Tweeter (0 points) [ no revision ]
+ 0 like - 0 dislike

The stronger version violates the no-cloning theorem of quantum mechanics. The only way out is complementarity a la Susskind, which is extremely radical.

This post imported from StackExchange Physics at 2014-04-04 16:24 (UCT), posted by SE-user Hunter
answered Aug 18, 2012 by UnknownToSE (505 points) [ no revision ]
It is not radical, it is well accepted. The interior and exterior of black holes are not separate places.

This post imported from StackExchange Physics at 2014-04-04 16:24 (UCT), posted by SE-user Ron Maimon
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I have to agree with Hunter's answer. But let me elaborate on it.

A proposed way to get around the no-cloning objection is to assume there is an orthonormal preferred basis (different from the preferred basis of the measurement problem) for boundary states, given by $\{|\chi_\lambda\rangle\}_{\lambda\in C}$, and for interior states, given by $\{|\psi_\lambda\rangle\}_{\lambda\in C}$ for some set $C$ such that a generic state is given by $\sum_{\lambda\in C}\,c_\lambda|\chi_\lambda\rangle\otimes|\psi_\lambda\rangle$. However, this breaks rotational invariance. Consider a spherically symmetric boundary with a massive spin-s particle right at the center of its interior. s is taken to be nonzero. Any choice of basis will have to explicitly break this rotational symmetry.

What complementarity says is there are two "pictures". In one, there is just an interior description with no extra states at the spatial boundary. In the other, the interior is removed and replaced with a stretched horizon boundary with additional horizon degrees of freedom. Both pictures are each internally consistent by themselves, but can't be mixed (or so claims Susskind). It's not valid to consider the interior and the stretched horizon simultaneously. That would only violate no-cloning. In the same manner, it's not valid to remove the interior while not including the stretched horizon. That would only lead to a loss of information, as well as Hawking radiation coming out unpredictably from the boundary.

If there is a stretched horizon, the degrees of freedom on it are not like that of a local quantum field theory. This has been argued in Susskind's articles on fast scrambling. Information falling to the stretched horizon has to be "reflected" within a time period of $R\ln(R/\ell_P)$ or so where $R$ is the black hole radius in a "maximally scrambled" manner. A brane with local QFT degrees of freedom and no superluminal interactions needs more time to scramble. Contrary to what Susskind thinks, this actually strongly argues against black hole complementarity.

Also, complementarity, if it holds, might not apply to any spatial boundary satisfying covariant entropy bounds. In Susskind's version, it only applies to stretched horizons hovering a Planck distance above the event horizon. In the AdS/CFT correspondence, it only applies to spatial boundaries at the conformal boundary. It doesn't apply to enclosing boundaries in between.

This post imported from StackExchange Physics at 2014-04-04 16:24 (UCT), posted by SE-user Charging Bull
answered Aug 22, 2012 by Charging Bull (0 points) [ no revision ]

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