However, I also think that the initial sections with the precise formulae for the bulk fields written in terms of the CFT variables – while being a professional piece of work that shows that the authors have been heavily trained in the AdS/CFT technology – are too long and tedious and may discourage many readers from getting to the most conceptually important part of the paper that may appear in Sections 4 and especially 5 and 6.
Because their picture seems to be the final word on many general puzzles concerning the infalling observer and black hole information – and I think that the people who won't be familiar with the basic results in a month shouldn't be counted as world's top quantum gravity experts – it may be meaningful to write short and separate summaries of some clearest results that everyone may read in minutes.
The following discussion covers some results of Subsection 6.2.2. See the paper for their original discussion and/or references.
Stephen Hawking proved that in the semiclassical approximation, the radiation emitted by a black hole is exactly thermal – i.e. described by a fully mixed state, a thermal density matrix. The information about the initial state – which may be pure – is totally lost. In fact, it's pretty obvious that even though general relativity has problems with renormalizability at multiloop level, Hawking's conclusion must be true to all orders in perturbation theory. Even if you consider some loop processes on the black hole background, it's still true that the information from the black hole interior – to be destroyed by the singularity – can't get out of the black hole. The reason is called causality, stupid.
When we expand around the black hole spacetime, the information just can't get out, and the Hawking radiation is exactly thermal. But we know that at the very end, the whole black hole evaporates and the full Hawking radiation must be nothing else than the evolved initial state which was pure – so by unitarity, it must be pure, too. Is it possible that the exact answer is pure but Hawking's thermal, mixed answer is valid up to all orders in perturbation theory?
The answer is Yes.
Mixed plus tiny corrections is pure
Assume that the black hole emits Hawking radiation whose total entropy is \(S\). So we want to describe it in the Hilbert space whose dimension is \(\exp(S)\). This entropy \(S\) of the Hawking radiation is proportional to the black hole entropy which is \(S_{BH}=A/4G\hbar\) in units with \(c=1\) where I restored \(G,\hbar\), however. But the evaporation typically increases the entropy so they're not equal.
At any rate, the detailed microstate of the Hawking radiation is described by density matrices whose size is\[
\rho:\quad \exp(S)\times \exp(S).
\] If you have trouble to calculate with fractional dimensions, think about the nearest integer to \(\exp(S)\) or, which is even easier to imagine, the nearest number of the sort \(\exp(S)\sim 2^N\) i.e. the nearest power of two which means that you imagine that the black hole has emitted \(N\) qubits (roughly one qubit per particle). Now, the simple and far-reaching point is that the exact pure state may be approximated by a mixed state \[
\rho_{\rm pure} = \rho_{\rm mixed} + \exp(-S) \rho_{\rm correction}
\] where the correction matrix has all matrix entries of order one at most. You might think that the identity above is impossible because it suggests that the distance between a "totally pure" density matrix and a "totally mixed" thermal density matrix is exponentially small, due to the \(\exp(-S)\) suppression of their difference. It would seem surprising because mixed and pure matrices seem to be very far – one of them has comparable and small eigenvalues; the other one has one large eigenvalue and multiple vanishing eigenvalues which seem "qualitatively different".
However, the formula above is totally possible. Let's diagonalize the mixed density matrix calculated by Hawking and subtract it from the pure one. The difference will be:\[
\rho_{\rm correction} = \pmatrix{
\exp(S)-1& 0 & 0&\cdots & 0\\
0&-1&0&\cdots &0\\
0&0&-1&\cdots&0\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
0&0&0&\cdots&-1
}
\] One obtains it by taking the difference between the thermal density matrix, whose eigenvalues are pretty much equal and of order \(\exp(-S)\) so that the trace equals one (we took all the eigenvalues being equal, a kind of a "microcanonical ensemble", but if you took a different ensemble, the main conclusions would be the same), and the pure matrix whose eigenvalues are zero except for one eigenvalue (in the left upper corner) that is equal to one.
You see that the matrix above has one entry, in the upper left corner, that is much greater than one. It equals \(\exp(S)-1\). Well, it may be a bit different, but still close to \(\exp(S)\), if \(\rho_{\rm mixed}\) and \(\rho_{\rm pure}\) are diagonalized in different bases, but the difference wouldn't matter for our conclusions. However, we must realize that the matrix has this form in some basis that is unnatural from the viewpoint of any low-energy observables you want to measure. What do the matrix entries look like in some more natural basis? Well, a more natural basis is related to the basis where the matrix is diagonal by the following conjugation:\[
\rho_\text{mixed, user-friendly basis} = U\cdot \rho_{\rm mixed} \cdot U^{\dagger}.
\] And similarly for the pure and correction matrices. The unitary matrix \(U\) has size \(\exp(S)\times \exp(S)\), too. We will assume it is rather "generic". If you approximate \(\rho_{\rm correction}\) (thanks, Karle) by its single large entry which you approximate by \(\exp(S)\), you may see that the matrix on the right hand side has matrix entries given simply by \(U_{i1}U^*_{j1}\exp(S)\): only one entry of \(U\) and one entry of \(U^\dagger\) "clicks" when you evaluate this product.
However, the matrix entries of \(U\) or \(U^{\dagger}\) are of order \[
U_{ij}\sim {\mathcal O}(\frac{1}{\sqrt{\exp(S)}})\sim{\mathcal O}(\exp(-S/2))
\] because the norm of each column (or row) of \(U\) has to be equal to one by unitarity and the (squared) norm is just the sum of \(\exp(S)\) terms (each of which is a square, from the complex Pythagorean theorem). That's why you may see that in \(U_{i1}U^*_{j1}\exp(S)\), the two factors of \(\exp(-S/2)\) cancel against \(\exp(S)\) and you get a number of order one. This correction to the density matrix is multiplied by \(\exp(-S)\) so each entry of this matrix-valued term is exponentially small!
Suvrat and Kyriakos offer you a different argument based the consistency checks for the trace of the "correction density matrix" and its square.
This general point – that a tiny modification of the large mixed density matrix is enough to "purify" it – has been known to me as a part of the lore but I have actually never seen this simple and unassailable calculation. Much of the rest of the paper is dedicated to calculating rather explicit forms of this correction matrix that "purifies" the maximally mixed state computed by Hawking. It really works.
Let me write down once again the relationship between the mixed, pure, and correction density matrices:\[
\rho_{\rm pure} = \rho_{\rm mixed} + \exp(-K\cdot A/4 G \hbar) \rho_{\rm correction}
\] Now I rewrote the entropy as a multiple of the black hole entropy. Note that this entropy scales like \(1/\hbar\). So if you use your quantum gravity theory to calculate anything for a black hole whose shape is fixed in some macroscopic SI units (imagine some occupation number for the Hawking radiation in a mode given by its wavelength), you may do so by a Taylor expansion in \(\hbar\) which adds increasingly quantum corrections to a classical result.
However, if you do this perturbative expansion, the result will be still mixed – all these terms will be a part of \(\rho_{\rm mixed}\). The correction term will be totally invisible – totally non-perturbative – because it goes like \(\exp(-K'/\hbar)\). Calculate the Taylor expansion of this exponential around \(\hbar=0\) and you will get zero plus zero plus zero, and so on. At the level of the perturbative accuracy, the correction needed to change the mixed density matrix to a pure one is so small that it is invisible. But the full theory of quantum gravity adds and has to add this contribution so that the exact state of the Hawking radiation is pure for a pure initial state.
Once again, note that the accuracy with which you would have to (statistically) measure the entries of the density matrix to determine that it's pure (and not mixed, so Hawking's qualitative conclusion has to be exponentially tiny) is exponentially high. It's so high that in the whole perturbative expansion, the whole unitary, information-preserving process of Hawking evaporation looks exactly thermal, mixed, and information-destroying. But the idea that the difference between the pure state of radiation and the mixed state of radiation is so great and "qualitative" that it must be seen through order-one corrections in observable quantities is an illusion.
I want and plan to write down a similar short text about the doubling of the degrees of freedom for pure states that resemble a mixed one.
Note that John Preskill wrote a blog entry on the black hole firewall saga on his/their blog. He clearly but somewhat uncritically summarizes the AMPS argument but also says that he probably doesn't believe the conclusions, without making the reasons for the disbelief clearly justified.
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