I Classical Hayden-Preskill Randomizer

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The discussion centers on the concept of a classical black hole acting as a randomizer, as described in the referenced paper. It explains that after thermalization, Bob can receive additional bits from the black hole, allowing him to potentially access Alice's original message. The key point is that Bob does not need to decode the message to retrieve the information; instead, he can leverage the inherent randomness of the black hole. This randomness ensures that the bits are uncorrelated, enabling Bob to access the message's content without traditional decoding methods. The conclusion emphasizes that the black hole's properties facilitate information retrieval despite the challenges posed by classical error-correcting codes.
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Why is it possible for Bob to decode Alices ##k##-bit message after only reading ##k+c##
I am reading this paper: https://arxiv.org/abs/0708.4025. In section 2, they describe a classical black hole as a classical randomizer. The black hole initially is a ##n-k## bit string, and Alice sends a ##k##-bit string into the black hole. A permutation of all ##2^n## possible strings is the state after the black hole thermalizes. Bob then receives ##k+c## bits via radiation, and it is assumed that Bob knows exactly the dynamics of the black hole. They claim that Bob can decode Alices message with high probability with only ##k+c## bits. But there are classical error correcting codes that make it very difficult to decode a message even will all bits, so how is this possible?

The only solution I can think of is that Bob can't necessarily decode the message but only that with ##k+c## bits the information contained in the message is still there. What am I missing?
 
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What you are missing is that Bob does not need to decode the message in order to access the information contained in it. Bob can use the randomness of the black hole to extract the information without decoding it. In other words, Bob can use the randomness of the black hole to gain access to the message's content without having to decode it. This is possible because the black hole's randomness makes it so that no two positions in the bit string are correlated with each other. This allows Bob to access the information contained in the bits without having to decode them.
 
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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