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.

Off Topic:

ReplyDeleteLumo, I would be very interested in a blog about the Ads/Cft correspondence. I get a lot of it, but I get lost on how the S(U) symmetry is related to supersymmetry. How are the number or fermions and bosons and N number of the SU(N) determined by the supersymmetry. Also, how is the SU(N) dual to some kind of monopole. If there is some kind of visual analogy that would help too. I think a lot of your readers would be interested.

Thanks

Physics Jumke

Yep me interested too :-))) !

ReplyDeleteDear Lubos,

ReplyDeleteI suspect that something is wrong here. If it would be so easy to approximate mixed states (large entropy) and pure states (zero entropy) with each other, than that would render the concept of entropy meaningless.

Dear Lubos,

ReplyDeleteI suspect that something is wrong here. If it would be so easy to approximate mixed states (large entropy) with pure states (zero entropy), that would render the concept of entropy (and the second law) totally meaningless.

How can rho_corr be a density matrix and have negative eigenvalues? If you form a convex combination of two 'true' mixed states you can never purify it. It only gets more mixed or at most remains equally noisy (rare and not true for qubits).

ReplyDeleteThis seems vaguely reminiscent of the Poincare recurrence. theorem. Can the result be extended to more ordinary thermalized systems?

ReplyDeleteThanks for the great article. I think the following argument about the order of \rho_{correction} may be a bit simpler. Let's write

ReplyDelete\rho_{pure} = \rho_{mixed} + \epsilon \rho_{correction}

and see of what order the small parameter \epsilon should be, assuming that \rho_{correction} is O(1). We know that the trace of \rho_{pure}^2 is 1, so for a generic \rho_{mixed},

\epsilon \Tr \{\rho_{mixed}, \rho_{correction}\}

must be O(1). What's in the trace is O(1), because the entries of \rho_{mixed} are O(exp(-S)) but the matrix product involves a summation over O(exp(S)) entries. Therefore, the trace is O(exp(S)) and \epsilon is O(exp(-S)).

Dear Guest, there is nothing wrong here - just verify the three steps of maths – and indeed, the value of von Neumann entropy depends on these tiny adjustments. The lesson isn't that entropy is meaningless in general; the lesson is that it is operationally meaningless to assign zero entropy to pure states that naturally belong to large ensembles. Those near-thermal pure states are "operationally" indistinguishable from thermal ones, anyway.

ReplyDeleteSome intuition "why" this approximation is possible; all the corrections to matrix entries are tiny, of order exp(-S), but there are exp(2S) entries in the whole square matrix that are corrected by exp(-S). You may imagine that you are doing some random walk, most of the exp(2S) matrix entries average out, but the result of the matrix walk is exp(S) times the step and the step is exp(-S) so it's of order one.

So in the correction above, a "squared, very large" number of matrix entries is modified by a small amount. The effect on each single measurement is nonperturbatively tiny, but the effect on properties of the density matrix that depend on "everything" - such as its purity (encoded in the eigenvalues) - is of order one and qualitative. Just go through the simple lines that I deliberately isolated from the paper (it's just 1/50 of the paper) to see that it is the case.

Dear Rezso, there is nothing wrong here - just verify the three steps of maths – and indeed, the value of von Neumann entropy depends on these tiny adjustments. The lesson isn't that entropy is meaningless in general; the lesson is that it is operationally meaningless to assign zero entropy to pure states that naturally belong to large ensembles. Those near-thermal pure states are "operationally" indistinguishable from thermal ones, anyway.

ReplyDeleteSome intuition "why" this approximation is possible; all the corrections to matrix entries are tiny, of order exp(-S), but there are exp(2S) entries in the whole square matrix that are corrected by exp(-S). You may imagine that you are doing some random walk, most of the exp(2S) matrix entries average out, but the result of the matrix walk is exp(S) times the step and the step is exp(-S) so it's of order one.

So in the correction above, a "squared, very large" number of matrix entries is modified by a small amount. The effect on each single measurement is nonperturbatively tiny, but the effect on properties of the density matrix that depend on "everything" - such as its purity (encoded in the eigenvalues) - is of order one and qualitative. Just go through the simple lines that I deliberately isolated from the paper (it's just 1/50 of the paper) to see that it is the case.

Dear lotus, rho_corr isn't a usable density matrix by itself, it's just a difference of two admissible density matrices and such a difference doesn't have to be positively definite.

ReplyDeleteDear Charles, Poincare recurrences also play a role in this story, a kind of "annoying and dangerous" role that forces them to be careful what's happening at very long times and very long frequencies omega.

ReplyDeleteThe formulae for the density matrix above could apparently be density matrices for "any" thermalized systems. However, other systems than black holes actually never have this property because they - like burning coal or paper that they mention - aren't so intensely entangled. So individual emitted quanta are much more independent from each other, and one may show purity by looking at smaller packages of the quanta that are both near-thermal yet pure as well as independent from each other.

So this form of the density matrix is special for the Hawking radiation from black holes - and anything that is equivalent to it.

Thanks for your interest and calculation - and that's pretty much their calculation from the paper! For some reason, I find mine simpler or more constructive or more straightforward or whatever makes me prefer it.

ReplyDeleteDear Lubos,

ReplyDeleteI understand the math, but I think that your handling of the basis transformation is too handwaving. In QFT, you never consider arbitrary unitary transformations on the full Fock space. Instead, you define the basis transformations on the one particle subspace, i.e. creations operators. The transformation of the creation operators then completely determines the transformation of the many particle states. This is my mathematical problem.

And my physical problem is that the coherent field of a laser is clearly not the same as the thermal field of a simple light bulb. For example, one can take a look at the Wigner functions to mathematically see this.

I understand the math, but I think that your handling of the basis transformation is too handwaving.

ReplyDeleteDear Rezso, I don't understand this sort of reasoning. It's like you're saying "I know that 2+2 isn't equal to 5 but anything based on 2+2=4 is handwaving for me, anyway". If you understand the maths, how can you oppose it? What does it mean?

In QFT, you never consider arbitrary unitary transformations on the full Fock space. Instead, you define the basis transformations on the one particle subspace, i.e. creations operators.

It's just not not true. One may consider unitary transformations acting in the most general way on the Hilbert space - not necessarily Fock space because the theory is not necessarily free - and indeed, it's these most general unitary transformations that don't commute with the operator N, the number of particles, that are relevant for the map between the CFT degrees of freedom and local AdS/black hole degrees of freedom.

In essence, the correct version of your statement would be "I, Rezso, have never seen general unitary transformations acting on the Hilbert space." And you use this fact to conclude "So I won't ever consider them and I will spit on anything that uses it."

But this is an utterly irrational behavior - I wouldn't call it reasoning because it's not reasonable at all. The general unitary transformations manifestly exist in the mathematical sense and they're important in this business, so if you want to understand quantum mechanics of black holes, you should better learn how to deal with them instead of writing idiotic excuses why you don't want to do that.

Right, LASERs emit coherent light while light bulbs emit thermal light and they're different. The Hawking radiation emitted by a black hole is strictly speaking pure if the initial state is pure but operationally, it's even more thermal than the radiation emitted from a light bulb. These are different adjectives. The state may be pure and approximately calculable as a mixed one. But no sensible person would call the Hawking radiation "coherent". So the Hawking radiation is surely not exactly like LASER's radiation. It is more like light bulb's radiation because it's approximately thermal.

And be sure that if the initial state of a light bulb is pure, the radiation it emits will be in a pure state, too! So there is absolutely no difference in this totally basic question between light bulbs and black holes. But there are more subtle questions in which differences exist. Your idea that everything must behave exactly like a LASER or exactly like a light bulb is simply way too naive.

All I can say about this response by you Lubos is: what a powerful psychologically disarming effect it ought to, or hopefully will, have on Rezso.

ReplyDelete(However, some people are very stuck and stubborn.....).

Thanks for this user friendly manual to the paper Lumo (density matrices etc are my friends), I'm looking forward to reading the next one :-)

ReplyDeleteTroublesome off topic :-(

ReplyDeleteIt is probably time to leave Physics SE for all reasonable people left. The SE overlords from outside our physics community have now definitively overdone it by showing us their power to shoot down whoever and whenever they want to:

http://meta.physics.stackexchange.com/q/2713/2751

They dont give a damn about any site inside the SE network nor any of its community.

They dont care about the fact that many reasonable people and good contributors strongly disagree with Shog9's decision. On the risk of destroying Physics SE, because many good people (including Anna v.) probably will leave now they, stubbornly stay with their horrible and unjust verdict to dont lose their face by admitting that they have made a severe mistake by there negative interference.

This is somewhat unfortunate and too bad, since I planed to flood the site with (slightly technical) questions about dual spinor models (I wanted to create such a tag ;-P) as needed, when rereading a certain demystified book sonn :-D

But with this horrible and (still ongoing!) brutal interference of the overlords with our site, all the fun of it has gone.

I apologize for the reality, Dilaton, but the Stack Exchange simply belongs to someone so the users over there aren't real owners. It's not trivial to launch and sustain a site that is as successful as Stack Exchange, and somewhat less so, Physics Stack Exchange.

ReplyDeleteThe detailed politics might be complicated but I don't understand what's Ron's problem with the Larian guy. More precisely, I think it is illegitimate for users to question the legitimacy of a candidate for a moderator. In fact, before I saw the page you linked to, I was voting for moderators - a few hours ago - and (I hope it's not secret) picked Qmechanic, Crazy Buddy, and Larian as bronze because he described himself as a skeptic and he seemed to be a soldier or something (picture), and these things look like a good general background for a moderator to me.

I have very mixed feelings about the ban - which is supposed to be month-long temporary, right? But there are some rules and if someone has the right to do so, he has the right. At the end, I think that there must be some rules for such servers to work, and any rules will ultimately lead to events not popular among some people. So even if I mostly disagreed with the suspension, I think that the right reaction for Ron is to suffer through the month, calm down, and return. The reaction of others - and your proposed exodus - seems like an overreaction to something that is probably an inevitable kind of an event that must be regulated by the real people.

If you approximate ρ_mixed by its single large entry which you approximate by exp(S)

ReplyDeleteDidn't you mean p_correction?

You can't approximate p_mixed by single large entry when it is "identity-like" (up to overall factor)...

It would be ok if the suspension came from our own mods and were the will of our community. So the bad thing is that our community likes and needs run and invadors from the outside come to shoot our best contributers down against the will of the physics se community. Our mods are obviously just puppets too, if needed they have no real power to defend people against such attacks from the outside. Shog9 and have no right to step in that brutally and destroy the site. If they hat physics and physicist.s like that they should not have allowed the site to go out of the proposal state

ReplyDeleteAbsolute, thanks for the catch.

ReplyDeleteDear Dilaton, I understand you have some emotions but I think that your approach just contradicts the laws of physics valid on that server.

ReplyDeleteYou can't declare "independence" of that server because it's just not independent. It belongs to someone and it's not the users whom you know. It's not me, either, despite my #1 reputation score there. You (or we) just can't steal the server, and even if we could, we shouldn't.

True, the difference is not guaranteed to be positive. So suppose that only the order of the object matters (tiny corrections). Then the correction is always of the order of the pure state, no? When you rotate rho_corr by U you are rotating a difference of two density matrices (that's how rho_corr is defined). So after the rotation the order of rho_pure will be exp(-S) - comparable to rho_corr and so not that tiny.

ReplyDeletethere is a similar thing in dS context in a recent paper 1209.4135 by Polyakov. Around eq.9 he discusses a pure state Greens function which looks like a thermal one.

ReplyDeleteDear lotus, your intuition about mathematics is simply invalid. It's not such a rare situation; that's why maths must be calculated and not "intuited". Try to follow the calculation quantitatively and you must be able to discover where your reasoning goes wrong.

ReplyDeleteThe individual large entries of the matrix are corrected by corrections that are much tinier than the original but the eigenvalues of the matrix - which determine its "character" - are nevertheless corrected by O(1) corrections. The difference itself clearly has eigenvalues that I indicated and they have both signs.

There's no contradiction between these two statements because the "order of the matrix" (i.e. understood as the "order of its eigenvalues") isn't the same thing as "the order of the matrix entries". Indeed, I've shown that the matrix entries are exp(-S) times smaller than the eigenvalues.

If it were a vector and not matrix, you could understand that the "order of the vector" should be calculated from the inner product V.V. The order is therefore given by sqrt(V.V).

Similarly, the order of the matrix should be calculated by sqrt(Tr(rho.rho)), assuming rho is Hermitian here. This is a sum over a large number of terms because the matrix is large: it's exp(2S) terms. So if all the matrix entries are comparable, the order of the matrix is exp(S) times larger than the order of the individual entries.

So even when each entry changes by a number much smaller than one, all the eigenvalues may change by O(1).

AMPS isn't about how a pure state with many degrees of freedom can look like a mixed state if one doesn't have access to all the qubits. No one is disputing that.

ReplyDeleteAMPS is about the monogamy of entanglement with near horizon modes.

Dear Quantum, this blog entry isn't about AMPS or any other invalid paper, for that matter. It's about the pure state predictions' being equal to the mixed state predictions up to all orders of perturbation theory, up to exponentially tiny nonperturbative corrections.

ReplyDeleteHowever, concerning your claims, monogamy is guaranteed by black hole complementarity which indeed says nothing else than that the low-energy modes in the interior are a subset of the modes outside (they are "the same wife"), in a strongly scrambled basis, and this fact in no way requires the equivalence principle to be violated near the event horizon.