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Reconstruction of the initial state from Hawking radiation?

+ 2 like - 0 dislike
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I hear that unitary evolution and information conservation must imply that information about information content that defines the initial state of matter used to create a black hole can be inferred from hawking radiation of the black hole.

Now if I measure the hawking radiation that comes out from a black hole, Is it at all possible at all to infer the Initial state of matter used to create the black?

What is the algorithm involved in the reconstruction of initial state if at all it is possible?

This post imported from StackExchange Physics at 2015-03-23 11:09 (UTC), posted by SE-user Prathyush
asked Mar 25, 2013 in Theoretical Physics by Prathyush (695 points) [ no revision ]
First step: figure out the exact physical laws governing quantum gravity.

This post imported from StackExchange Physics at 2015-03-23 11:09 (UTC), posted by SE-user Peter Shor
Related (but NOT a duplicate) physics.stackexchange.com/q/18369/2751

This post imported from StackExchange Physics at 2015-03-23 11:09 (UTC), posted by SE-user Dilaton
@PeterShor Yes, I agree, Though I did have some hope that something can be said about the initial state apart from charge mass and angular momentum without invoking the full theory of Quantum gravity.

This post imported from StackExchange Physics at 2015-03-23 11:09 (UTC), posted by SE-user Prathyush
@Dilaton Very interesting question and an extremely interesting reply from Ron. It would take a (perhaps long)while to understand the part about extremal blackholes.

This post imported from StackExchange Physics at 2015-03-23 11:09 (UTC), posted by SE-user Prathyush

1 Answer

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To be sure, this exercise can't be done in practice for a macroscopic black hole. But the unitarity of the evolution guarantees that for any observable $A(t_i)$ describing the initial state that you may want to learn (for example, the location of a nuclear bomb that sparked some explosion that was used to help a star to collapse into a black hole), there exists an observable $A(t_f)$ acting on all the degrees of freedom in the Hawking radiation – some measurement of a correlated property of all the Hawking particles – that has the same value. So by measuring the latter, you measure the former. These two observables are simply evolved from each other by the Heisenberg equations of motion of the quantum gravity theory.

The only known – and quite certainly, the only mathematically possible – consistent theory of quantum gravity is string theory, in one of its forms or vacua. But even in string theory whose dynamics is sort of known, it's clear that $A(t_f)$ corresponding to rather simple and natural observables $A(t_i)$ is an extremely complicated operator that measures some correlation/correlated property of pretty much all the Hawking particles that were emitted by the black hole. In fact, black holes are the fastest and most efficient "scramblers" of information. So the measurement can't be done in practice. However, all the information about properties of the initial state – as expressed by the eigenvalues of operators $A(t_i)$ in the text above – are included in the final Hawking radiation just like they would be included if the black hole were replaced by a simple furnace that just burns the initial matter.

The black holes differ from furnaces because at the level of the classical theory, it seems plausible that there's a one-to-one correspondence between the initial and final states in the case of the furnace; but classically, it seems inevitable that the map isn't one-to-one in the case of the black hole because the information about the initial state is destroyed in the black hole singularity and can't get out of the black hole again, for causal reasons. However, quantum mechanics modifies this conclusion and allows the information to "tunnel out" essentially by the quantum tunneling effect (one way to describe it), thus making the black hole qualitatively identical to a furnace.

This post imported from StackExchange Physics at 2015-03-23 11:09 (UTC), posted by SE-user Luboš Motl
answered Mar 26, 2013 by Luboš Motl (10,248 points) [ no revision ]
Nice explanation :-).You should not write too many nice answers to things at the same time or I can not like them all in the course of one day without the system thinking I am serially apvoting you ... :-D

This post imported from StackExchange Physics at 2015-03-23 11:09 (UTC), posted by SE-user Dilaton
@Lubos Thank you for your interesting answer, I always wondered how the black hole having an entropy proportional to area was consistent with an expected entropy of zero, say in a case where a black hole was prepared using a pure state of matter. From what I understand from your answer, the "scrambling" is so fast and efficient that for all practical purposes one cannot unscramble the information that goes into the black hole, therefore what comes out appears to be in a thermal mess. But in principle it must be possible.

This post imported from StackExchange Physics at 2015-03-23 11:09 (UTC), posted by SE-user Prathyush
Dear @Prathyush, there's an implicit confusion about the definition of entropy in your comment above - an issue not specific to black holes at all. For a pure state, the von Neumann entropy is zero but it is really meaningless to call this zero "the [physical] entropy of the state". The physical entropy should always be considered to be $k\ln N$ where $N$ is the number of macroscopically indistinguishable states from the given one, assuming it has a macroscopic appearance at all. This definition isn't zero for a large black hole.

This post imported from StackExchange Physics at 2015-03-23 11:09 (UTC), posted by SE-user Luboš Motl
And yes, the unscrambling is possible in principle. Let me stress again that if the initial state (of the star etc.) is a pure state, so is the final state of the Hawking radiation. So it just appears to be a thermal state. In the full exact precision, it is a pure state that is just close to a thermal one. Pure and (maximally) mixed states may have really really close density matrices, see e.g. motls.blogspot.com/2012/12/…

This post imported from StackExchange Physics at 2015-03-23 11:09 (UTC), posted by SE-user Luboš Motl
@Lubos Thank you, that was a very important clarification.

This post imported from StackExchange Physics at 2015-03-23 11:09 (UTC), posted by SE-user Prathyush
@Lubos: pure and maximally mixed states can't possibly have really, really close density matrices. At least, not if you're talking about standard quantum mechanics. They can, however, appear indistinguishable to an observer who doesn't have access to microstates.

This post imported from StackExchange Physics at 2015-03-23 11:09 (UTC), posted by SE-user Peter Shor

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