• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,064 questions , 2,215 unanswered
5,347 answers , 22,734 comments
1,470 users with positive rep
818 active unimported users
More ...

  What are cosmological "firewalls"?

+ 10 like - 0 dislike

Reading the funny title of this talk, Black Holes and Firewalls, just made me LOL because I have no idea what it is about but a lively imagination :-P (Sorry Raphael Bousso but the title is just too excessively funny). Since this was a session talk targetted at the experts, I did not dare to watch the video, but still I am curious to know a little bit more about it.

So can somebody explain to me what firewalls are used for in cosmology? What do they have to do with black holes and how are they related to complementarity?

Edit to include followup questions

From just reading the abstracts of the papers the first two links in Lumo's answer point to, I'm wondering if such firewalls would contradict the holographic principle? In one paper they say holographic spacetime has no firewalls but what about the converse line of thought? And what happens to the microscopic degrees of freedom of the black hole, would the firewall change the description of them?

asked Sep 22, 2012 in Theoretical Physics by Dilaton (6,240 points) [ revision history ]
Thanks to all for the still incoming interesting answers and the new questions related and linked to my post, I like this a lot :-). In addition to the discussion of firewalls here on Physics SE, Lumo reports from time to time about what is going on in the arxiv too, see for example here and here‌​.

This post imported from StackExchange Physics at 2014-03-09 15:48 (UCT), posted by SE-user Dilaton

7 Answers

+ 7 like - 0 dislike

The firewall is a new term in an extremely provocative paper

Black Holes: Complementarity or Firewalls?

by Ahmed Almheiri, Donald Marolf, Joseph Polchinski, James Sully that claims that an observer who falls into a black hole gets burned at the horizon, after all. So the event horizon – the surface of a black hole – is a burning place you don't ever want to hit, therefore called "firewall".

Most other physicists – including Susskind, Banks, Fischler, and others – disagree with the conclusion. After all, we've been explaining for many decades why the "burnout" at the would-be "firewall" is exactly what doesn't happen. See the followups:


These disagreeing physicists usually point out that the black hole complementarity – the idea that the degrees of freedom (fields) inside a black hole aren't quite independent from those outside but they're highly scrambled versions of them – has everything it needs to preserve the conclusion of classical general relativity, namely that nothing special happens at the horizon (to a large enough observer), without violating any insights about the quantum theory (such as the existence of the Hawking radiation and its ability to preserve the information).

Whether the bizarre argument by Almheiri et al. would apply to other horizons, such as the cosmic horizons in de Sitter space, is an interesting question. That would be pretty bad because the firewall could be everywhere around us. ;-)

I have written a more detailed review of the followups as of today here:


A few days later, I decided that Raphael Bousso's resolution was right:


This post imported from StackExchange Physics at 2014-03-09 15:48 (UCT), posted by SE-user Luboš Motl
answered Sep 22, 2012 by Luboš Motl (10,278 points) [ no revision ]
Thanks Lumo, now I know a bit better what it is supposed to be :-). From just reading the abstracts (since these are HEP-TH papers :-P ...), I'm wondering if such firewalls would contradict the holographic principle? In one paper they say holographic spacetime has no firewalls but what about the converse line of thought ?

This post imported from StackExchange Physics at 2014-03-09 15:48 (UCT), posted by SE-user Dilaton
... and what happens to the microscopic degrees of freedom of the black hole, would the firewall change the description of them?

This post imported from StackExchange Physics at 2014-03-09 15:48 (UCT), posted by SE-user Dilaton
Dear @Dilaton, I have written an entry on that on TRF. Various people say that the firewalls contradict holography or are needed by holography (or something similar to holography). There seems to be a chaos in what people think. I think that the natural "feelings" of an infalling observer will try to extrapolate the semiclassical environment so the observer will feel nothing special but it's plausible that these perceptions will be insanely scrambled relatively to many seemingly nearby observers...

This post imported from StackExchange Physics at 2014-03-09 15:48 (UCT), posted by SE-user Luboš Motl
Wow thanks a lot for the extended answer on TRF Lumo, I really appreciate this :-)

This post imported from StackExchange Physics at 2014-03-09 15:48 (UCT), posted by SE-user Dilaton
+ 6 like - 0 dislike

Let's clarify some common misconceptions here.

Suppose we have a spherically symmetric black hole. Let's perform a mode analysis here. For simplicity, work with l=0 spherical harmonics for massless fields first. The same conclusion still applies to higher harmonics or massive fields, but the analysis is more complicated.

Work in Eddington-Finkelstein coordinates. The modes are complex oscillatory modes of the form $A(r,t)\exp[-i\theta(r,t)]$. Even if a quantum field is real, we still need a complex mode basis for a Bogoliubov analysis using creation and annihilation operators. Roughly speaking, we call a mode infalling if the wavefronts of constant phase $\theta$ are infalling. Ditto for outgoing modes. In the eikonal approximation, the wavefronts are null. Outgoing null wavefronts can either lie outside the event horizon, or inside it. If inside, even though they are locally outgoing, they will still hit the singularity.

If $\theta$ increases with time, we say it's a positive frequency mode. If it decreases, it's a negative frequency mode. The former is associated with creation operators, while the latter is associated with annihilation operators. The positive/negative frequency analysis is independent of whether a mode is infalling/outgoing.

Suppose $\theta$ goes exponentially in time $t/R$. Then, even though $\theta$ increases monotonically with time, the Fourier transform of $A(t)\exp[-i\theta(t)]$ will contain negative frequency components. This is the case for outgoing modes for black holes in the near horizon region where t is the Eddington-Finkelstein time, for locally "natural" looking modes according to freefalling observers.

In general, there are a couple of distinct cases of entangled Hawkng pairs:

  1. Both Hawking quanta are locally outgoing, but one lies outside the horizon while the other lies inside.
  2. Both Hawking quanta are locally outgoing, and both lie outside the horizon. So, the entanglement is between two outgoing external Hawking quanta.
  3. A Hawking quanta of the pair is locally outgoing and lies outsiide the horizon, while the other is locally infalling.
  4. Other miscellaneous cases, like both infalling, or both locally outgoing but inside the horizon.

These cases must not be confused with each other! Some authors have claimed entanglement between Hawking pairs have transplankian origins at the event horizon. This is the case for case 1. The origins for case 1 and case 2 entanglements come from frequency mixing due to the Fourier transform of $A(t)\exp[-i\theta(t)]$ with exponentially varying $\theta$ in time.

Of course, we can ask why a freefalling observer ought to detect no local excitations outside the horizon. This isn't the case for a nonblack hole metric which is Schwarzschild up till a short distance $d < h$ above where the horizon ought to be, but with a massive shell at that height so that the shell's interior contains no black hole. Then, it's the external observer static relative to the shell who detects no quanta while freefalling observers will detect quanta.

For a black hole which formed a finite time in the past, we can trace locally outgoing modes outside the event horizon to transplankian modes present during the formation of the black hole a long long time in the past in semiclassical analysis. If such transplankian analyses are valid, we can conclude a freefalling observer will detect no local quanta outside the horizon. However, we ought to make transplankian cutoffs. In that case, we still need to justify why a freefalling observer ought to detect no locally outgoing quanta at a height of h above the horizon. The equivalence principle by itself isn't enough.

Case 3 can only come about from mixings between outgoing and infalling modes, which is distinct from positive/negative frequency mixings. This doesn't happen in 1+1D for CGHS models with a massless scalar/fermion. Such fields in 1+1D only couple to the conformal metric modulo surface Lagrangian terms. So, there are no firewalls in the CGHS model.

However, let's look at spherically symmetric 3+1D black hole metrics. Then, a near horizon mode which is locally outgoing is backscattered by the curved metric when it is redshifted to around the black hole radius R so that it becomes a linear combination of far away outgoing modes and a backscattered infalling mode. The mode goes as $$e^{-i\omega t}\frac{1}{r} e^{i\omega r},\;r\gg R$$ and $$e^{-i\omega t}\left[ A(r)e^{i\theta_1(r)} + B(r)e^{-i\theta_2(r)} \right],\; 0<r-R\ll R$$ A gives the initial outgoing near horizon mode. B gives the backscattered mode. Both $\theta_1,\theta_2$ increase monotonically with r. No significant backscattering occurs during the period when the locally outgoing wavelength is still much less than R.

What happens is due to frequency mixing among the locally outgoing modes when we move from freefalling proper time to EF time while counting the number of wavefronts intersected, we initially get case 2 entanglement between two near horizon locally outgoing modes after a Bogoliubov transformation. Then, when these modes are redshifted to the R scale, one of the quanta escapes far away while the other backscatters. We end up with case 3 entanglement.

So, if we have an outgoing external Hawking quanta, it could be entangled with another locally outgoing Hawking quanta behind the horizon which is eventually dragged to the singularity (case 1), or another outgoing external Hawking quanta (case 2), or a locally infalling Hawking quanta (case 3).

Suppose a freefalling observer at a height of $\ell_P \ll h \ll R$ above the horizon observes no excited modes with wavelength around $h$. Then, a Bogoliubov transform from freefalling local time to EF external time means locally, there ought to be case 2 entanglements at a height of h according to an external observer. Then, after a time of order $R\ln(R/h)$ according to distant clocks, some of the outgoing quanta get backscattered.

Suppose the information contained in the totality of all outgoing external Hawking quanta which never get backscattered is a very scrambled version of all the infalling information which ever enter the black hole throughout its lifetime. Wait until more that 1/2 of all the outgoing Hawking quanta has been emitted. Then, the totality of all the external Hawking quanta radiated after then has to be maximally entangled with the totality of all the external Hawking quanta emitted before then. This is a debatable assumption. If so, by the monogamy of entanglement, any external Hawking quanta can't be entangled at all with any nonexternal Hawking quanta. There's no way an external observer can test for case 1 entanglement. However, he can test for case 3 backscattered entanglement before the backscattered infalling mode crosses the horizon. In fact, he can do so while the backscattered quanta is still above the horizon by much more than the Planck scale.

So, if we wish to avoid a violation of the monogamy of entanglement between modes which are all outside the horizon by far more than the Planck scale, and are hence accessible to an external observer, we have to conclude there can't be any case 3 entanglement according to an external observer, not even those dynamically generated from backscattering. This means, evolving backward in external time by $R\ln(R/h)$, if a freefalling probe beams the results of measuring the presence of excited modes with local wavelength of order h at a height of h above the horizon, the beamed results ought to report the detection of excited quanta. This is the firewall. To translate back from the external frame to the freefalling frame requires another inverse Bogoliubov transformation.

Evolving further backward in external time by $R\ln(h/\ell_P)$, this firewall is at a Planck height above the horizon. We expect transplanckian cutoffs in quantum gravity, so we should not evolve any further backward according to semiclassical gravity. The best we can say is that this firewall came from a stretched horizon hovering a Planck distance above where the horizon ought to be.

This post imported from StackExchange Physics at 2014-03-09 15:48 (UCT), posted by SE-user Gidom Mera
answered Nov 30, 2012 by Gidom Mera (240 points) [ no revision ]
+ 5 like - 0 dislike

Quoting a deleted post of Dilaton:

Joe Polchinski has just written a very nice guest post at Cosmic Variance titled "Black Holes, Complementarity and Firewalls".... Lumo has written a new article about what Raphael Bousso has to say about thi issue here:

In response to Polchinsky, I gave this comment in the link, which I think answers the question here:

I am puzzled by the separation of radiation into “early” and “late”— you are saying “wait for 99% the radiation to come out, then consider something dumped into the remaining black hole, and it is entangled with the early radiation, and this means that the late radiation is determined”, but this implicitly assumes that you can do detailed experiments on the precise entangled quantum state of the outgoing radiation, even after knowing it is early (this is a brutal measurement already— you have learned to a certain extent when the radiation has come out! Why should you still be able to extract anything now? This measurement already ruins the phase coherence of the late state), and you also implicitly assume you know the late black hole thermal ensemble ( you know about where the black hole is at late times, and about how big it is—- this is also an extremely brutal measurement).

Given that you assume you have these brutal bits of knowledge, namely which photons are early, and which are late, and how big the black hole is and where it is, I don’t see any reason to suppose you can entangle the remaining coherence in the early radiation distant from the black hole and learn anything at all about the emissions of the late-classical black hole from the radiation. I think all you have shown is that the separation “early” and “late” is just not compatible with a unitary S-matrix for the black hole.

Just by measuring a black hole’s approximate position and approximate horizon location, you are restricting it’s thermal ensemble in a way that prevents certain kinds of entanglement from surviving. While I don’t see any proof that what I am saying is right, I also don’t see any guarantee that the implicit entanglement involves in measuring that the radiation is early leaves the early radiation state pure enough to do the measurements you need. Obviously if it does, your argument goes through, but the very fact that your have a paradox must mean it is not so— in order to determine the late radiation, you need to mae measurements on the “early” radiation over such a very long time that you aren’t even sure when you are done if it is early or late. This means that you are working over an entire black hole S-matrix event, not separating it into a two-step scattering where you know something about the intermediate black hole state (that it is a certain size, with a certain amount and kind of early radiation).

So, as far as I see, there is an additional unjustified assumption here, which is common to all the referenced literature about the Page time, namely that it is possible to simultaneously produce semiclassical black hole states entangled with pure-enough early Hawking radiation to make measurements on the whole set of early Hawking radiated particles which determine something about the late radiation. This is a heuristic assumption, and I think all you are doing is showing that it is false.

The only case in which I can see this early/late separation is completely justified is if you throw something into a highly charge black hole, and wait for the hole to decay to extremality, and look at all the radiation emitted during this process (so all the radiation is “early” in this definition, since once the black hole is extremal again, it’s cold asymptotic S-matrix state). In this case, the arguement is surely completely coherent, and the end-state of the black hole is a known pure-state, it’s an extremal black hole with charge Q and velocity V (assuming a perfectly BPS model black hole, so there is no further decay). Then you can measure the outgoing radiation state, and determine the Q and V of the final state.

But in this case, the end result is no longer decaying at all, so there is no paradox, no thermal horizon and no Hawking radiation. The only time you get a paradox is when the late-state black hole is truly thermal and truly macroscopically entropic, so any intuition that associates the GR solution to a quantum state of some sort is not particularly clear (you have to associate the GR solution to a thermal ensemble).

So I can’t internalize the argument enough to see whether it is correct, it seems obviously wrong (but that’s only because the holographic complementarity seems obviously right to me), and the sticking point in understanding the argument for me is the heuristic regarding describing hugely entropic black holes using some sort of unknown quantum state for the black hole alone, rather than an entangled state of the black hole and all the radiation, early and late, with no way to make the distinction between early and late without completely ruining the ability to measure anything interesting at all about the late state.

So while I don’t find the argument persuasive, it’s only because I don’t buy the assumptions in the related literature on Page times (assumptions which don’t appear in Page’s original paper, I should add). I am questioning these obscure assumptions, not the detailed stuff in the latest paper.

In Susskind’s reply, since he at times made similar arguments about early and late radiation, he also ends up using a classical black hole picture, and pretends that you can talk about the state of infalling mater and outgoing early/late radiation separately and coherently. So Susskind already implicitly internalized this framework, and perhaps this is the reason the argument was persuasive for him. I would not give up complementarity for this, or honestly, just about anything barring someone taking an instrument and throwing it into a black hole and getting a contradiction with complementarity. It’s just too obviously correct to be false.

Regarding the “firewall” resolution, it is not satisfactory, because the firewall stress, in the same semiclassical approximation, is nonzero on the horizon, and falls inward along with anything else. This means that the singularity needs to constantly replenish the firewall with new stress by some crazy mechanism, something which is not really reasonable at early times. To see this, consider charged black holes, because the domain of communication with the singularity does not extend past the Cauchy horizon (which degenerates to r=0 in the neutral limit).

I really think that this is finding an inconsistency in the implicit assumptions in the Page time literature, not in black hole theory itself. This is very interesting and important, but please don’t discard complementarity, as I think it is almost surely fine as is.

This post imported from StackExchange Physics at 2014-03-09 15:48 (UCT), posted by SE-user Ron Maimon
answered Oct 3, 2012 by Ron Maimon (7,730 points) [ no revision ]
Ha ha Ron, thou shalt not cite my hidden answers LOL :-D. I deleted it since it was no longer needed as Lumo has updated his answer to include his post about Raphael Bousso's take on the firewall issue (which contains a link to Polchinski's CV guest blog too). Anyway, thanks for your interesting answer, your objections to the principle feasibility to separate the Hawking radiation into an early and a late part makes sense to me. Maybe you could throw a paper into the arxiv discussion too ;-) ?

This post imported from StackExchange Physics at 2014-03-09 15:48 (UCT), posted by SE-user Dilaton
And I am curious if Joe Polchinski writes an answer to you on CV, ... if certain "off topic discussions" have not driven him away :-/ (?)

This post imported from StackExchange Physics at 2014-03-09 15:48 (UCT), posted by SE-user Dilaton
@Dilaton: Whether he does or doesn't, I had this concern in the 90s when the page-time style Susskind stuff was current--- I wasn't sure how to do early and late, and it smelled inconsistent even then. I think Polchinski has just shown that this Page-time stuff definitely doesn't work, that you can't do crude early/late analysis using heuristics where the BH is classical, and we need more careful tools. I have a job now, and I spend most of my time staring at "Chromosome 1: the greatest hits", which is fantastic, but it doesn't leave much time to write a proper analysis which can go on arxiv.

This post imported from StackExchange Physics at 2014-03-09 15:48 (UCT), posted by SE-user Ron Maimon
glad you got yourself a job, you need to update your profile ;)

This post imported from StackExchange Physics at 2014-03-09 15:48 (UCT), posted by SE-user Larry Harson
@LarryHarson: I work at a fellow's lab, doing some bioinformatics stuff which is my main field, but I didn't want to say his name, for fear he would get tainted by association. I updated my profile.

This post imported from StackExchange Physics at 2014-03-09 15:48 (UCT), posted by SE-user Ron Maimon
+ 4 like - 0 dislike

There is a resolution using the two-state formalism of quantum mechanics. Instead of a single state, we have two: $|\psi_i\rangle$ and $\langle \psi_f |$. $\langle \psi_f |$ is the postselected state. If $\rho_f$ is the postselection operator, then $\langle \psi_f | = \langle \psi_i | \rho_f$.

Let the subscripts $e, f, i, o$ represent the early Hawking radiation and anything maximally entangled with the infalling information, the freefalling information dumped into the black hole, the ingoing Hawking radiation of entangled Hawking pairs, and the outgoing Hawking radiation respectively. Consider a very simplified toy model where each are represented by a single qubit. Assume all infalling information is maximally entangled with some system $e$ outside the black hole, and this quantum system also collects all of the outgoing Hawking radiation emitted during the time period when the hole emits half of all the Hawking radiation it will ever emit. Just outside the black hole horizon after it has emitted half of its Hawking radiation, entangled Hawking pairs are constantly generated due to the redshifting effect of the warped geometry for outgoing field modes, and the resulting frequency mixing once the redshifted wavelength becomes comparable to the Schwarzschild radius. In the simplified toy model, $|\psi_i\rangle_{efio} = \frac{1}{2} \left[ |00\rangle_{ef} + |11\rangle_{ef} \right] \otimes \left[ |00\rangle_{io} + |11\rangle_{io} \right]$. Here, the subscripts indicate which subsystem each qubit belongs to. The postselection occurs at the singularity, and it projects the freefalling information and the ingoing Hawking radiation hitting the singularity at the same location to the Bell state $\frac{1}{\sqrt 2} \left[ \langle 00|_{fi} + \langle 11|_{fi} \right]$ in our simplified toy model. The corresponding final state is then given by $$\langle \psi_f|_{efio} = \frac{1}{2} \left[ \langle 00|_{fi} + \langle 11|_{fi} \right] \otimes \left[ \langle 00 |_{eo} + \langle 11|_{eo} \right] = \frac{1}{2} \left[ \langle 0000|_{efio} + \langle 0110|_{efio} + \langle 1001 |_{efio} + \langle 1111|_{efio} \right].$$ $\langle \psi_f |$ is not the conjugate of $| \psi_i \rangle$. The monogamy of entanglement needs to be modified in the two-state formalism. Here, in $|\psi_i\rangle$, $e$ is maximally entangled with $f$. However, in $\langle \psi_f |$, it's maximally entangled with $o$. Things get more interesting after taking a partial trace over the interior. An external observer has no direct access to the interior, and so, he needs to trace over $f$ and $i$. The reduced state is then given by $|\psi_i \rangle_{eo}\langle \psi_f |_{eo} = Tr_{fi} \left[ |\psi_i \rangle_{efio}\langle \psi_f |_{efio} \right] \propto \left[ |00\rangle_{eo} + |11\rangle_{eo} \right] \left[ \langle 00 |_{eo} + \langle 11 |_{eo}\right]$. We might as well normalize the tensor product so that $Tr\left[ |\psi_i \rangle \langle \psi_f |\right] = \langle \psi_f | \psi_i \rangle =1$. Before taking the partial trace, for the initial state $|\psi_i\rangle$, $e$ is maximally entangled with $f$. After taking the partial trace, it's maximally entangled with $o$! In the two-state formalism, entanglement isn't invariant under the operation of taking partial traces.

Suppose we have a freefalling probe just above the horizon where the entangled Hawking pairs are produced, and it measures the total particle number. $\frac{1}{\sqrt 2} \left[ |00\rangle_{io} + |11\rangle_{io} \right]$ corresponds to no particles, $\frac{1}{\sqrt 2} \left[ |01\rangle_{io} \pm |10\rangle_{io} \right]$ to one particle, and $\frac{1}{\sqrt 2} \left[ |00\rangle_{io} -|11\rangle_{io} \right]$ to two particles. As $|\psi_i\rangle$ is an eigenstate of this measurement, this doesn't alter $|\psi_i \rangle$, and hence, the postselection process either, and the analysis proceeds as before. The probe will detect no particles at the horizon. No firewalls. As the probe measures just outside the horizon, it can broadcast this measurement outcome both to the interior and the exterior.

Consider the case without the system $e$. The freefalling qubit is in a superposition $c_0 |0\rangle_f + c_1 |1 \rangle_f$. By the same kind of analysis, we get $|\psi_i\rangle_{fio} = \left[ c_0 | 0\rangle_f + c_1|1\rangle_f \right] \otimes \frac{1}{\sqrt 2} \left[ |00\rangle_{io} + |11\rangle_{io} \right] $ and $\langle \psi_f |_{fio} = \frac{1}{\sqrt 2}\left[ \langle 00|_{fi} + \langle 11|_{fi}\right] \otimes \left[ c_0^* \langle 0 |_o + c_1^* \langle 1 |_o \right]$. This isn't exactly quantum cloning. In $|\psi_i\rangle$, the quantum information is stored in the $f$ qubit, but in $\langle \psi_f |$, it's stored in the $o$ qubit. However, $|\psi_i\rangle$ isn't $\langle \psi_f |$. It's just that the same quantum information is stored in different qubits for the initial and final states. Taking the partial trace over the interior, we get $|\psi_i \rangle_o \langle \psi_f|_o = Tr_{fi} \left[ |\psi_i \rangle_{fio} \langle \psi_f|_{fio} \right] \propto \left[ c_0|0\rangle_o + c_1|1\rangle_o \right] \left[ c_0^*\langle 0 |_o + c_1^*\langle 1|_o\right]$. After taking the partial trace over the inaccessible interior, the qubit has been unitarily teleported to the outgoing Hawking radiation thanks to postselection!

This is the two-state generalization of GHZ entanglement. Recall GHZ entanglement goes as $\frac{1}{\sqrt 2} \left[ |0000\rangle + | 1111\rangle\right]$. In the two-state modified version, which isn't really GHZ anymore, we have an even number of qubits 2N. The preselected state is a tensor product state of the Bell entangled pairs $\frac{1}{\sqrt 2}\left[ |00\rangle + |11\rangle \right]$ between the $2i^{th}$ and $2i+1^{th}$ qubits, and the postselection is for $\frac{1}{\sqrt 2}\left[ \langle 00| + \langle 11| \right]$ between the $2i+1^{th}$ and $2i+2^{nd}$ qubits. Which other qubit a given qubit is maximally entangled with depends upon whether we're looking at the initial or final state. It also changes after taking partial traces. What this corresponds to in the black hole scenario is a zigzag of causal and retrocausal influences along a chain of entangled Hawking pairs. The warped geometry of a black hole causes entangled Hawking pairs to be produced all the time in its interior, and just outside the horizon. For most of the entangled pairs, both particles hit the singularity, but at different locations. Only for a small fraction of cases does only one particle of the pair hit the singularity while the other escapes the black hole permanently. Such pairs have to be produced just outside the event horizon. The other pairs, where both particles hit the singularity can be produced at other locations. So, we have a zigzag chain. The freefalling information hits the singularity at the same location as one particle of an entangled pair of infalling Hawking particles. The chain is of the order of $\mathcal{O}(A/\ell_P^2)$ long where $A$ is the black hole area. The last particle of this long chain is an outgoing Hawking particle. Because of retrocausal effects, this might happen before or after the original freefalling information crosses the horizon.

This post imported from StackExchange Physics at 2014-03-09 15:48 (UCT), posted by SE-user Loony physics crank
answered Oct 3, 2012 by Loony physics crank (160 points) [ no revision ]
Added +ve rep - You should use your real name if this is an original idea, or you're just throwing away any vestige of credit for nothing.

This post imported from StackExchange Physics at 2014-03-09 15:48 (UCT), posted by SE-user John R Ramsden
+ 3 like - 0 dislike

Maimon and the crank's proposals have problems with them.

Take Maimon's proposal first. Let's say there is an external system which is initially fully entangled with some infalling information. Then, the black hole fully evaporates, and assuming purity, this external system is now fully entangled with the outgoing Hawking radiation. Maimon's assertion is that during intermediate periods before the black hole has fully evaporated, there is some tripartite entanglement between the external system, the Hawking radiation which has already been emitted, and the metric outside which determines the mass, location and velocity of the black hole. This tripartite entanglement means neglecting the external metric, there is no entanglement between the external system and the outgoing Hawking radiation. Well, just after the information has passed through the horizon, there could be a random snap decision by the external system whether to jump into the hole and catch up with the infalling information and check if they are still maximally entangled, or remain outside and collect all the remaining Hawking radiation. It takes a time period of about the Schwarzschild radius before it's no longer possible to catch up with the infalling information, which leaves that as a critical window of time. If he decides to jump in and compare notes, he will find maximal entanglement with the infalling information inside. This would seem to suggest even if he chooses to remain outside, he is still maximally entangled with the information inside. By the monogamy of entanglement, he can't also be correlated at all with anything outside the black hole, and this includes the metric outside. Of course, this is a counterfactual conclusion, and maybe such counterfactual reasoning is inadmissible. But if it fails, it can only be because there is no tensor product structure for the Hilbert space for internal and external black hole states.

The crank's proposal has the following difficulty; suppose there is no black hole initially. A control qubit is set up in the superposition $\frac{1}{\sqrt 2}[|0\rangle +|1\rangle ]$. If its value is zero, don't create a black hole. If it's one, create a black hole. The initial state is $|\psi_i\rangle\otimes \frac{1}{\sqrt 2}[|0\rangle +|1\rangle]$. If there's no black hole, presumably, $\rho_f=1$. If there's a black hole, presumably $\rho_f=\rho_s$ is some positive operator less than the identity operator. So, $\rho_f = |0\rangle\langle 0|+\rho_s\otimes|1\rangle\langle 1|$. So, $\langle \psi_f | = \frac{1}{\sqrt 2} \left[ \langle\psi_i|\otimes\langle 0| + \langle \psi_i|\rho_s \otimes\langle 1|\right]$. Now, measure the relative probabilities for the value of the control qubit. It no longer interacts with the system after determining whether or not to create a black hole. True, we normalize $|\psi_i\rangle\langle \psi_f|$, but that doesn't alter the relative probabilities for getting 1 as opposed to 0. The ratio is given by $\langle \psi_i|\rho_s|\psi_i\rangle/\langle\psi_i|\psi_i\rangle$. In general, this ratio will be less than one. This violates unitarity. One way out is to assume $\rho_s$ actually has some eigenvalues greater than 1. Another way out is to assume time evolution is nonunitary inside the black hole.

This post imported from StackExchange Physics at 2014-03-09 15:48 (UCT), posted by SE-user Ned Pederson
answered Oct 9, 2012 by Ned Pederson (110 points) [ no revision ]
Your complaint doesn't work (but thanks for the input), it is well known that there is no tensor product structure for the Hilbert space for internal and external black hole states--- this is holography. The internal states are just duplicates of the outside states, in a different basis, that's the whole point of holography. The "tripartite entanglement" is a fine way to say it, but it's not just of the metric exactly, but the BH degrees of freedom, with all the radiation, early and late, that doesn't allow separation of radiation to early and late keeping a classical black hole background.

This post imported from StackExchange Physics at 2014-03-09 15:48 (UCT), posted by SE-user Ron Maimon
+ 3 like - 0 dislike

as for entaglement issues, observer centred perspective and the equivalence principle



...Recently, it was argued that an observer falling towards a suciently old black hole encounters a rewall at the horizon. This would amount to a gross violation of the equivalence principle in low-curvature regions...

Borun D. Chowdhury, Andrea Puhm


...We would like to point out a confusing use of the term equivalence principle in [5] where the notion of observer complementarity is introduced as a response to AMPS's result in order to "save\ the equivalence principle. AMPS found that there is a stress tensor at the horizon and therefore concluded that an in-falling person does not fall freely. This is no more a violation of the equivalence principle than an astronaut not feeling weightless upon re-entry into the Earth's atmosphere is. We thank David Turton for this analogy.



...Is there an alternative to the firewall conclusion ? Perhaps.

...Now suppose Alice is capable of extracting the information in RB from the early Hawk- ing radiation. She can then throw it back into the black hole where it meets its alter ego, A: This has a distinct similarity with time travel. The late bit A has somehow been transported to the early Hawking radiation, and then brought back to the black hole, where it meets itself. there are are two ways that nature could provide chronology protection in this context. It may simply be physically impossible to distill RB out of the Hawking radiation in time to bring it back to meet A ...The other possibility is that firewalls play the role of Hawking's chronology protection enforcer.

This post imported from StackExchange Physics at 2014-03-09 15:48 (UCT), posted by SE-user user12103
answered Oct 9, 2012 by user12103 (150 points) [ no revision ]
+ 1 like - 0 dislike

Loony's idea is very interesting, but there are some mild variants of the grandfather paradox if part of the outgoing information is emitted before information passes through the horizon.

Consider a black hole which has almost entirely evaporated away. It's only a few qubits away from final evaporation. Throughout its lifetime, all of its outgoing Hawking radiation has been collected with quantum coherence preserved and fed into a quantum computer. All matter falling into the hole throughout its entire lifetime is strictly controlled, and maximally entangled with some internal states of this quantum computer as well. At this moment, the quantum computer decides to throw a qubit into this hole. If information about this qubit only appears in the radiation after it passes through the horizon, there is no "paradox". However, the last few remaining qubits presumably also contain part of the information of information falling in earlier, and so, there might not be enough "room" for that. Suppose part of the information is encoded in the radiation coming out earlier, which now resides in the internal states of this quantum computer. Essentially, if the qubit dumped in at the last moment has value 0 or 1 respectively, this induces a POVM $\{C_0, C_1\}$ in the internal states of the quantum computer corresponding to the respective outcomes. If no information is emitted retrocausally, then $C_0=C_1=I/2 $. Otherwise, diagonalize $C_0$, which is equivalent to diagonalizing $C_1$ as they sum up to the identity operator. Measure along this basis. If the corresponding measured eigenvalue is less than 1/2, set the dumped qubit's value to 0. Otherwise, set it to 1. This appears to violate unitarity, but not purity, if this is indeed possible. Maybe there's always not enough running time for the quantum computer to do this in time?

This post imported from StackExchange Physics at 2014-03-09 15:48 (UCT), posted by SE-user Meiyun
answered Oct 6, 2012 by Meiyun (40 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights