- (Information) Paradox Lost

Tim Maudlin

arXiv:1705.03541 [physics.hist-ph]

Here is the problem. The dynamics of quantum field theories is always reversible. It also preserves probabilities which, taken together (assuming linearity), means the time-evolution is unitary. That quantum field theories are unitary depends on certain assumptions about space-time, notably that space-like hypersurfaces – a generalized version of moments of ‘equal time’ – are complete. Space-like hypersurfaces after the entire evaporation of black holes violate this assumption. They are, as the terminology has it, not complete Cauchy surfaces. Hence, there is no reason for time-evolution to be unitary in a space-time that contains a black hole. What’s the paradox then, Maudlin asks.

First, let me point out that this is hardly news. As Maudlin himself notes, this is an old story, though I admit it’s often not spelled out very clearly in the literature. In particular the Susskind-Thorlacius paper that Maudlin picks on is wrong in more ways than I can possibly get into here. Everyone in the field who has their marbles together knows that time-evolution is unitary on “nice slices”– which are complete Cauchy-hypersurfaces –

*at all finite times*. The non-unitarity comes from eventually cutting these slices. The slices that Maudlin uses aren’t quite as nice because they’re discontinuous, but they essentially tell the same story.

What Maudlin does not spell out however is that knowing where the non-unitarity comes from doesn’t help much to explain why we observe it to be respected. Physicists are using quantum field theory here on planet Earth to describe, for example, what happens in LHC collisions. For all these Earthlings know, there are lots of black holes throughout the universe and their current hypersurface hence isn’t complete. Worse still, in principle black holes can be created and subsequently annihilated in any particle collision as virtual particles. This would mean then, according to Maudlin’s argument, we’d have no reason to even expect a unitary evolution because the mathematical requirements for the necessary proof aren’t fulfilled. But we do.

So that’s what irks physicists: If black holes would violate unitarity all over the place how come we don’t notice? This issue is usually phrased in terms of the scattering-matrix which asks a concrete question: If I could create a black hole in a scattering process how come that we never see any violation of unitarity.

Maybe we do, you might say, or maybe it’s just too small an effect. Yes, people have tried that argument, which is the whole discussion about whether unitarity maybe just is violated etc. That’s the place where Hawking came from all these years ago. Does Maudlin want us to go back to the 1980s?

In his paper, he also points out correctly that – from a strictly logical point of view – there’s nothing to worry about because the information that fell into a black hole can be kept in the black hole forever without any contradictions. I am not sure why he doesn’t mention this isn’t a new insight either – it’s what goes in the literature as a remnant solution. Now, physicists normally assume that inside of remnants there is no singularity because nobody really believes the singularity is physical, whereas Maudlin keeps the singularity, but from the outside perspective that’s entirely irrelevant.

It is also correct, as Maudlin writes, that remnant solutions have been discarded on spurious grounds with the result that research on the black hole information loss problem has grown into a huge bubble of nonsense. The most commonly named objection to remnants – the pair production problem – has no justification because – as Maudlin writes – it presumes that the volume inside the remnant is small for which there is no reason. This too is hardly news. Lee and I pointed this out, for example, in our 2009 paper. You can find more details in a recent review by Chen

*et al*.

The other objection against remnants is that this solution would imply that the Bekenstein-Hawking entropy doesn’t count microstates of the black hole. This idea is very unpopular with string theorists who believe that they have shown the Bekenstein-Hawking entropy counts microstates. (Fyi, I think it’s a circular argument because it assumes a bulk-boundary correspondence ab initio.)

Either way, none of this is really new. Maudlin’s paper is just reiterating all the options that physicists have been chewing on forever: Accept unitarity violation, store information in remnants, or finally get it out.

The real problem with black hole information is that nobody knows what happens with it. As time passes, you inevitably come into a regime where quantum effects of gravity are strong and nobody can calculate what happens then. The main argument we are seeing in the literature is whether quantum gravitational effects become noticeable before the black hole has shrunk to a tiny size.

So what’s new about Maudlin’s paper? The condescending tone by which he attempts public ridicule strikes me as bad news for the – already conflict-laden – relation between physicists and philosophers.

Tim,

ReplyDeleteNow that we have settled the debate about Gauss' law versus Gauss' theorem, let us continue.

I brought up this EM example precisely to highlight the issue regarding the meaning of Gauss' law: div(E)-\rho=0. We now have to think very carefully about the distinction between formalism and physics. When we try to "quantize" the electromagnetic field we encounter the quandary that a photon has two physical states, distinguished by their helicity, yet we wish to describe the photon in terms of the 4-component vector potential A_\mu. The quandary is that 4>2. Our eventual goal is to obtain a theory of photons interacting with charged particles that respects Lorentz invariance, unitarity, and cluster decomposition (roughly, locality). There are two paths you can go by, but in the long run, they end up at the same place. Let me explain.

Option 1: This is the most familiar textbook approach in which we write down a Lorentz invariant Lagrangian ( -1/4 F^{\m\nu}F_{\mu\nu}+ ...) and then turn the quantization crank. This approach maintains manifest Lorentz invariance, but not unitarity. The theory is defined on a Hilbert space with 4 field theoretic degrees of freedom corresponding to the 4 components of A_\mu, and so it is not at all manifest that we will end up with a unitary S-matrix governing the 2 physical photon states. At the classical level we can say that we have an enlarged phase space, and we need to project it down to the hypersurface corresponding to the physical phase space. We do this by imposing gauge conditions and constraints, such as Gauss' law. The proper approach for dealing with a constrained phase space is via Dirac brackets and so on, but let's not get into that. At the quantum level we have an enlarged Hilbert space, and then we restrict to the physical Hilbert space by imposing Gauss' law on physical states: (div(E) - \rho)|\psi_\phys> =0. So in this formalism div(E)-\rho is certainly not the zero operator when defined on the enlarged Hilbert space, but it is zero (by definition) when acting on physical states. At the end of the day, one indeed gets an S-matrix respecting the fundamental principles noted above.

cont

cont

ReplyDeleteOption 2: Now let us maintain manifest unitarity but not Lorentz invariance. This is the approach masterfully laid out in Weinberg's QFT treatise. Throughout we work with only the physical photon and charged particle states, so every state in the Hilbert space is physical and there are no constraints. Lorentz invariance is not manifest because there is no way to package the 2 photon states (and nothing else) into a Lorentz covariant field. The photon field operator in this formalism transforms in a non-standard way under Lorentz transformations, and so for Lorentz invariance to be maintained we need to restrict the type of interactions that can appear in this theory. Without getting into details, we ultimately obtain S-matrices that respect the principles. My main point is the following. In this approach if you compute div(E)-\rho you find that this vanishes identically. This makes perfect sense, as in this approach we have directly constructed a theory on the physical Hilbert space, and so there is no need to impose constraints. Therefore in this approach div(E)-\rho really is the zero operator, and the charge operator really is a pure boundary term.

We see that the question, "is div(E)-\rho the zero operator?" is not well posed, as it depends on the formalism used to describe the theory. However, everyone agrees that div(E)-\rho acts as zero on physical states.

Now let's turn to AdS/CFT and think about how we describe the bulk EM field. In bulk language we typically describe the theory via Option 1, whereas the dual CFT describes the same theory via a version of Option 2. The CFT describes the physical Hilbert space directly: there is a 1-to-1 correspondence between CFT operators and physical states. The enlarged Hilbert space encountered if we use Option 1 in the bulk has no counterpart in the CFT.

The bulk theory and the CFT both have a charge operator Q, and we would like to claim that Q_bulk = Q_CFT. But what we really mean is that these two operators share the same matrix elements between any two physical states. It is meaningless to ask whether Q_bulk = Q_CFT on the enlarged Hilbert space, since the CFT doesn't have an enlarged Hilbert space to begin with.

Now, I think we have agreed that when acting on physical states Q_bulk is a pure boundary operator, i.e. a surface integral. This is what I mean when I say "the charge operator is a pure surface term". The retort: "no it isn't because it also has a volume piece that acts nontrivially on the enlarged Hilbert space", carries no force, as it is a statement that depends entirely on the formalism we use to describe the theory. If we restrict questions to those that have an invariant meaning regardless of formalism, then we will agree that Q is a pure surface term.

Of course, everything here has an analog in the case of the Hamiltonian, but before considering that I would like to reach consensus on the above points.

BHG,

ReplyDeleteWe agree about option 1, so that's fine. But your claim in option 2 that "Throughout we work with only the physical photon and charged particle states, so every state in the Hilbert space is physical and there are no constraints." does not seem to be accurate. Even if all the states have only physical photon and charged particle states, it obviously does not follow that every state is a physical state in the relevant sense, i.e. represents a physical possibility. In general, the kinematical space only involves physical particle states: in this sense the issue of photon helicity is rather unusual. The usual issue is to get only those states that obey the dynamics: the unphysical states in the kinematic space represent physical entities behaving in unphysical ways. And that is evidently the case just as well in Option 2. That's exactly why you need to "restrict the interactions" to get to the solution: states with only physical particles but unrestricted interactions are in the kinematic space but not the solution space. It is therefore not the case that "every state in the Hilbert space is physical" in the relevant sense. This seems evident from what you yourself have written, so is there any disagreement about that?

And of course any approach using the A_mu will have gauge degrees of freedom to kill off by gauge fixing or whatever. The states available before that killing off is done are in the kinematical space but not the solution space. So the attempt to somehow eliminate the kinematical space altogether, and not see that the whole point of various operators is to provide a means to reduce the kinematical space to the solution space (which—reaching far back in all this—may not even *be* a Hilbert space) is not illustrated in Option 2 at all.

Tim,

ReplyDeleteYour interpretation of Option 2 is not accurate. When I say "physical states" I mean those that are physical in every sense of the word: these are states that can be produced, observed, and measured by an experimentalist living in this world. These states do not need to be further restricted or constrained in any way. To make life even simpler, let's consider the theory of a spinless particle (a scalar). In this approach we don't quantize a classical theory, or start from a scalar field, rather we construct the quantum particle theory directly. The starting point is a quantum theory of free spinless particles. States in the Hilbert space are free particles with energy E = \sqrt{p^2+m^2}. We then want to turn on interactions for these states so that we end up with an S-matrix respecting the principles I noted. To do this, it turns out to be efficient to adopt the following strategty. First introduce creation/annihilation operators for the single particle states. Then collect these together into a scalar field. This scalar field obeys the Klein-Gordon equation *identically* by virtue of the particle states obeying E = \sqrt(p^2+m^2}. We then construct an interaction Hamiltonian as the integral of a polynomial built out of the scalar field. The states then evolve according to this interacting Hamiltonian, and so scatter, etc. Then one proves that the S-matrix of this theory is satisfactory. At no time do I ever make reference to any unphysical states or degrees of freedom. There is no larger "kinematical space" or anything of that sort in this approach.

We would like to do the same for the theory of photons interacting with charged particles. The complication that arises is that (for technical reasons having to do with representations of the Poincare group) the creation/annihilation operators for the physical photon states cannot be packaged into a Lorentz covariant field. Instead, one is forced to combine them into a field operator A_\mu, which is not actually a Lorentz 4-vector. Indeed, for each momentum, although A_\mu has 4 components these all depend on just the operators corresponding to the 2 helicity states of the photon so the 4-components are not independent. For this oddity not to spoil Lorentz invariance it turns out that one needs to require that A_\mu only appear in the interaction Hamiltonian when contracted against a conserved current: A_\mu J^\mu. THen all is well. Again, the point is that there is no reference made to any state that is not fully physical, and there is no reference to any "kinematical space". The price you pay is that the formalism is a bit messy, but conceptually it is extremely clear, much more so than in Option 1 and its reference to unphysical degrees of freedom.

BHG

ReplyDeleteMaybe we can get closer to understanding this dispute by being very careful about what one is after. In your last answer you say this:"The starting point is a quantum theory of free spinless particles. States in the Hilbert space are free particles with energy E = \sqrt{p^2+m^2}." Is it your intention here that this Hilbert space contains *only* free particles? Then of course this Hilbert space does not contain the solution space. But now it seems that you are not even trying to construct a solution space, i.e. a set of physically possible states: you are just trying to get an S matrix. Then you are not even trying to get a physical description of the interaction and of what happens between the incoming and outgoing states. The S matrix simply does not offer any account at all of what happens during the interaction. If we are interested in how to understand black holes, then the S matrix approach is a non-starter, since it only gives you statistics for the transitions between the asymptotic In and Out states.And if my analysis is correct, no regular S matrix will even give this correctly: just as Hawking originally argued, you will need a superscattering matrix, not a regular S matrix.

So here is a simple question: is all you are after an S matrix? And if so, how do you know that the right thing to use is an S matrix and not a superscattering matrix? From the point of view I am arguing, that assumption is already completely question-begging.

TIm,

ReplyDeleteThe point was to start from a free theory (set the fine structure constant to zero) with a specified Hamiltonian and Hilbert space, and then ask how to smoothly turn on interactions. Once interactions are turned on the particles are of course no longer free -- they evolve according to the interacting Hamiltonian.

I'm not sure what you mean by "solution space". Is this supposed to have a precise mathematical meaning? For the QM of a particle in a potential I know what the Hilbert space is -- the space of square integrable wavefunctions -- but what is the "solution space".

No, I wasn't restricting myself to just an S-matrix description, I was just using that as a particularly clean physical observable but you can just replace that by "time evolution" if you like. For sure, in the context of black hole evaporation it would be begging the question to assume a unitary S-matrix.

The point of this QED example was only to explain the sense in which the charge generator is a boundary operator. This is so despite the correct statement that in one formulation of QED one introduces an enlarged unphysical Hilbert space at an intermediate stage, and the charge operator has a volume piece when acting on this enlarged space. The point is that this enlarged Hilbert space is pure formalism: ultimately one restricts to the physical Hilbert space, and on this space the volume piece vanishes, and furthermore there is an alternate (but physically equivalent) formulation of the theory in which the enlarged Hilbert space is absent from the get-go. So the claim that the charge operator is not a boundary operator is not a claim about the physical content of the theory, but rather about some choice of formalism in how one constructs the theory. If you still object to the statement "the charge operator in QED is a boundary operator" please state your objections now since this is a key point.

BHG-

ReplyDeleteBy a solution space I mean the set of solutions of the fundamental equations, or, I take it, what you mean by the physical state space. I'm really puzzled by why this is causing any difficulties.

If I have a classical Special Relativistic theory of N point particles, then just from that information I have a kinematical space, which is just the set of all the ways to put N inextensible timeiike curves on Minkowski space-time. In the end, depending on the Hamiltonian, some of these sets of trajectories will belong to the solution space—the set of solutions of the dynamical equations of motion—and other won't. Similarly, the Kinematical space for classical E & M is the set of all ways to put a pair of vector fields E and B on the space-time, and the solution space is the set of all such pairs that satisfy Maxwell's equations. The whole point of writing down dynamical laws is to put constraints on the solutions, so that not every kinematically possible evolution is physically or dynamically possible. As I said, any dynamical law, or indeed any law, can be written in the form Something(spacetime + physical ontology) = 0, so the condition that distinguishes the solutions from the other elements of the kinematical space is just the condition that Something acts on the solutions to yield 0. Of course the zero operator also acts on the solutions to yield zero, so the action of the Something operator and the Zero operator on the set of solutions is the same. But they still are not the same operator, since the Something operator distinguishes solutions from non-solutions and the Zero operator does not.

I have no problem with showing that one can determine the net charge at the boundary in a nice space-time. Not if all of this so far has been aimed at showing that the total change is something that can be determined by operators in the asymptotic limit, that is not at all controversial. Even the no-hair theorems allow for that much hair: charge, total angular momentum and Energy content. These features of a black hole are determinable by operations carried out arbitrarily far away. But the amount of detailed information that reveals about the interior is very, very thin. Many different black holes can be formed with the same charge, angular momentum and energy content. That is uncontroversial. The controversial claim is of course that it is possible to determine the *entire bulk state* at the boundary. That just seems frankly incredible. Knowing those three numbers, that you can get, provides very minimal information about what is inside the region bounded by the surface integral.

Finally, although it is completely uncontroversial that one can determine the net charge, angular momentum and energy of a system by operations carried out at the asymptotic boundary, I don't see how it follows that the boundary operator is a "charge generator". As we know from Gauss's Theorem, the surface integral of the flux is equal to the volume integral of the flux. So I don't see any grounds for thinking of the surface operator "generating" the charge any more than the Div operator as "generating the charge" in the bulk. What force is "charge generator" supposed to carry with it?

BHG-

ReplyDeleteSorry for the typos. "Now" not ""Not", and of course I meant the volume integral of the divergence, not the flux.

Tim,

ReplyDelete"Solution space": if this is the same as what I call the physical Hilbert space that's fine. Just wanted to clarify, since this is nonstandard terminology in the quantum context.

The proper statement about charge as a generator is that "the Q operator is the generator of gauge transformations", with the precise meaning of this spelled out in what I derived. What I want you take take away from this QED example is this fact, together with the fact that Q is a boundary operator in the precise sense that all of its matrix elements between physical states can be measured at the boundary. So there is no physical information regarding Q that cannot be extracted at the boundary. In gravity, the analogous statement is that the Hamiltonian H is the generator of time translations, and that H is a boundary operator.

As to whether knowing the energy determines the state of the system, let me pose the following. Consider a collection of N distinguishable particles in nonrelativistic QM confined to an irregularly shaped box (so there are no symmetries) and interacting via some generic two-body (say) potentials. Do you agree that the energy spectrum of this theory will be discrete and nondegenerate? Do you further agree that if we put a QFT without symmetries in such a box then its spectrum will be discrete and nondegenerate?

BHG-

ReplyDeleteLet's just clarify what is not controversial, what seems sort of unnecessarily off-topic, and what is controversial.

What is not controversial already in classical E & M is that the flux of the electric field through a closed surface equals the enclosed electric charge, and hence if one can measure the flux at the surface one can determine the enclosed charge. By the no-hair theorems, this quantify of change is not masked from the surface even if it is inside the event horizon of a black hole. As I said, the total mass of the black hole is also determinable at spatial infinity, at least as ADM define it. And the angular momentum of a black hole can be determined. Regarding those three quantities: sure, you can determine them at infinity.

It is already a bit odd, as far as I can tell, to want to bring this generator of gauge transformations into the story. A gauge transformation will not change the electric field, and hence will not change the flux, and your whole argument turns on the flux equaling the enclosed charge. And you only get the value of the Q operator to be proportional to the charge q by demanding that lambda co to a constant, which it need not do. I'm not saying that there is anything technically wrong here, but it seems more confusing than need be.

I don't have any objection to what you say about the spectrum of the Hamiltonian in the cases you mention (I would have to think it through more carefully to be sure, but it's not worth fighting about in any case). What I entirely do not understand is this: Due to linearity, the superposition of any two solutions of the dynamical equations of motion is another solution. So even if the energy spectrum is discrete and non-degenerate, and even if the Hamiltonian is measurable at the boundary, it is not clear in what sense the bulk state is determinable by measurements on the boundary. There are all the superpositions of energy eigenstates, and indeed there is good reason to believe that we can't be in an eigenstate of energy. And there are many, many, many distinct superpositions of energy eigenstates that have the same expectation value for the energy. So I don't see in what sense the interior state can be determined by operations carried out on the boundary. If you knew the interior state were an energy eigenstate, then in this particular instance a measurement could have revealed the interior state. But not knowing whether the bulk state is in an energy eigenstate or not leaves you powerless to know the interior state from the boundary operations. Do we disagree about that?

Tim,

ReplyDeleteTo address your last point: I am certainly not claiming that in a system with a discrete non-degenerate spectrum it is possible to determine a quantum state by measuring the energy. Clearly this would only work if one were also told that the quantum state happens to be an energy eigenstate. No disagreement about that.

In the following when I say "state" I will always mean "physical state", and when I say "non-degenerate" I mean non-degenerate up to the obvious symmetries (rotations, etc.)

Let's now get back to the case of evaporating black holes in AdS. Let me break down the logic

1) I think we have agreed that the bulk Hamiltonian generates asymptotic time translations -- i.e. if we label a family of bulk slices by their asymptotic time coordinate, then H_bulk advances the slices with respect to this time. Also, H_bulk is a surface integral. Note that bulk states are all labelled by a boundary time t, but this leaves open what the form of the corresponding slice is when one moves into the bulk.

2) The CFT Hamiltonian H_CFT advances states in CFT time, and under the AdS/CFT dictionary we have an isomorphism under which H_bulk = H_CFT

3) H_CFT has a discrete non-degenerate spectrum, and hence so too does H_bulk. This fact is manifest in the CFT; it is not manifest in the bulk, but the rules of the game are that we are here assuming that AdS/CFT holds, so it must be the case

4) Now we form a large black hole which subsequently decays to Hawking radiation. We then consider some time t in the far future, long after the black hole has evaporated. At this time, the CFT is in a pure state, and we can in principle determine this state by measuring correlation functions of CFT operators.

The main question is : what is the structure of the Hilbert space and quantum state at time t, expressed in bulk language?

5) First of all it is a pure state by AdS/CFT.

6) Your claim is that this state cannot be determined solely by measurements made on a spacelike slice connected to the AdS boundary at time t (effective information loss). In other words, the state restricted to such a slice is described by a mixed state (density matrix).

Now, I do not see how to describe a Hilbert space and state in which the above statements are all true. The spacetime diagram suggests that the Hilbert space is a tensor product, with one factor describing the region behind the black hole horizon, and the other factor describing the spacelike slice attached to the boundary at time t. The full state is then entangled. This was Hawking's original scenario. This satisfies (5) and (6) but violates (3). This Hilbert space contains many states of the form |\psi_in> |E_out>, and so by changing |\psi_in> at fixed E_out we violate the assumption of a nondegenerate spectrum. So a tensor product seems like a non-starter. On the other hand, if the Hilbert space is not a tensor product, then apparently the operators that change the state behind the horizon no longer commute with the external operators, which would mean that an observer who fell through the horizon could send signals to the late time external observer.

If you have a proposal to get around these problems I am happy to address it. But note that it has to be a sharp proposal explaining what is the structure of the Hilbert space and quantum state at late times. As is clear, the basic tension is that information loss implies that we can make changes to the state that are hidden behind the horizon, but since the Hamiltonian lives at the asymptotic boundary this implies a degeneracy in the energy spectrum.

BHG

ReplyDeleteIt appears to me that you are not observing points you have been insisting on, so I am extremely puzzled by this post.

In particular, as you know I am not committed to what you call the physical space, or what I have been calling the solution space, even being a Hilbert space. I take the kinematical space to be a Hilbert space, but not all of the elements of the kinematical space are solutions of the dynamical equations and hence represent physical possibilities. Offhand, I can't even see that there is any argument that if the solution space contains many product states of the form you indicate after 6), that there must be other solutions where the "in" part is changed for fixed "out" part. I can see an argument that there must be such a state in the kinematical space, but not in the solution space. Indeed, since I think that the "in" and "out" parts of the solution will be entangled, I see no reason to accept that there are any solutions at all of the sort you write down. Why should there be solutions that are product states?

As for the commutation relations, I recently exactly appealed to the ETCRs in an argument I was giving and you said that they do not hold in quantum gravity. I did not pursue that point there, but am I to take it that you are withdrawing that claim?

I don't even quite understand why you take the space-time diagram a provide to imply that the space of solutions is a product space any more than it might suggest the same thing for a Cauchy slice that is connected but has a part inside the event horizon. Does that similarly suggest to you a product space with a part inside the event horizon and a part that runs from spatial infinity to the horizon? If it does, then we seem to have the same problem about degenerate spectra at early times, irrespective of my solution to the "paradox". If it doesn't, the why does it in the later slice? And if in the early case the solution is to deny the product space structure of the solution space, does that imply that a person behind the event horizon can signal out? Of course not. So why would it in the late case?

I can't follow this reasoning at all. Where in the argument does the disconnection of the Cauchy surface play a role?

Tim,

ReplyDeleteThe logic of my post was to establish a proof by contradiction, so indeed some of the assumptions I was making were self-contradictory -- that was the whole point. The question is which of these assumptions to give up?

Let's first discuss the Hilbert space. You say that you are not committed to there being a physical Hilbert space. As you know, one of the axioms of QM is that physical states correspond to rays in a Hilbert space. If you give this up you are giving up quantum mechanics. You are also giving up AdS/CFT, since the CFT does have a physical Hilbert space, and its rays are supposed to be in 1-1 correspondence with those of the bulk theory. So what mathematical framework are you proposing for the physical states if they are not described by a Hilbert space? We can't proceed until this question gets resolved, since otherwise I have no idea what your proposal is supposed to mean.

BHG-

ReplyDeleteWe have been through this before, but just to be clear. Of course I take the kinematical space to be a Hilbert space. Since every physical state, every solution to the dynamical and constraint equations, is in the kinematical space, every solution corresponds to a ray in the kinematical Hilbert space. I often will have a clear understanding of the detailed structure of the kinematical space. But the set of solutions is a subset of the kinematical space. Does that subset itself form a Hilbert space? My main reason for questioning this, as I said, is that Thiemann explicitly states that in some cases it does not. (It may be, as well, that I picked up the term "solution space" from Thiemann, since he uses it.) Since the solution space inherits an inner product from the kinematical space, the issue is obviously completeness. I have no intuition at all about whether the solution space is likely to be complete. Indeed, I have no intuitions about the structure of the solution space in general. In the case of quantum gravity, I have no idea where any such intuitions could come from. I know from Theimann that the structure of the solution space of, e.g., the Wheeler-deWitt equation is very complicated, and indeed no one knows exactly what it is.

This is especially so because in the case of quantum gravity so much work is being done by constraint equations. It is not at all obvious (to me, anyway) what kind of formal principles the set of solutions to those equations will have. I am curious about why it is obvious to you.

As far as I can tell, much of your assurance about the structure of the solution space in the bulk theory is parasitic on having accepted AdS/CFT. Since the CFT is much simpler and more familiar, you may be confident of properties of its solution space, its energy spectrum etc. This doesn't help me since I find AdS/CFT quite implausible. Or at least I find some understandings of what it claims to be quite implausible, but since I have not even been able to get a clean statement of what the thesis is I am not sure what to think. I know I find the Holographic Hypothesis, as it is usually presented, to be extremely implausible and the grounds put forward for believing it to be exceedingly weak. What I have in mind are the calculations of Black Hole entropy, which are what is commonly raised as evidence for Holography. I can go into my specific complaints about that argument if you like.

But this is my situation. I think that there is excellent reason to believe that the solution space for a theory of quantum gravity sophisticated enough to yield black hole evaporation is going to be very, very complicated even if the kinematical Hilbert space is not. So when you ask me whether the solution space is, e.g., a product space I have no real opinion on the matter. Could go either way as far as I know. If you have a compelling argument that it can't be a product space, or can't fail to be a product space, my reaction will be to accept the argument. If you have an argument that it both can't be and can't fail to be a product space, that is obviously trouble. Do you think you have such an argument? Does it essentially appeal to AdS/CFT? If the latter, then I might well take the whole thing to be an argument against AdS/CFT.

Tim,

ReplyDeleteI am sorry, but it is simply a fact that an axiom of quantum mechanics is that physical states are rays in Hilbert space. You are misunderstanding Thiemann, because nothing he says contradicts this fact. His point is that one attempt at quantization runs into trouble because it doesn't naturally lead to a Hilbert space structure on the space of physical states. That's hardly a surprise, given that trying to directly quantize the metric degrees of freedom runs into all sorts of technical problems, this being just one more example. The lesson you are meant to draw is that this approach runs into trouble, not that you should give up an axiom of quantum mechanics.

Why you care if the kinematical space is a Hilbert space is beyond me, since this space contains all sorts of unphysical garbage, like longitudinally polarized gravitons, that play no role in the final theory. My EM discussion was meant to make this clear, since I explained that one can formulate the theory such that the larger kinematical space is absent and only physical states appear.

If you insist (without support from any source that I am aware of) that physical states in quantum gravity are not rays in Hilbert space, then you need to explain precisely what mathematical structure they do obey. What properties of a Hlbert space are you giving up, and what replaces them? This is not a minor issue: the black hole information paradox is about the compatibility of quantum mechanics and gravity, so one clearly needs to be very precise about what quantum mechanics means in this context.

BHG

ReplyDeleteI really don't know how to make this any clearer, so I'll just make it shorter. Of course every physical state is a ray in a Hilbert space: namely in the kinematical space. It does not follow that the set of physical states *forms* a Hilbert space. In particular, it may not be complete. No idea why you think it is obvious that it does.

Why do I insist that we start with a kinematical Hilbert space and then use various constraints—which we call laws—to pick our the solution space? Because that's what we do. What, for example, do you think the constraints in the gravitational Hamiltonian constrain? The solutions, obviously.

This is only important because you keep asking me whether the *solution* space has such-and-such a structure, and the answer usually is that I don't know, having never thought about it. Is it a product space? No idea. So if you have some objection based on the solution space "obviously" having some structure, I can only say that it will not be obvious to me, so you will need an argument. Why is that a big deal?

Tim,

ReplyDeleteYou are still not getting it. The axiom of quantum mechanics that you breezily dismiss is that physical states are *one-to-one* correspondence with rays in a Hilbert space. Your claim that you need to enlarge the space of states beyond the physical states to get a Hilbert space is in flagrant violation of quantum mechanics. Do you really want to say that physical states are not complete in the Hilbert space? Then you have to give up some basic features such as the ability to insert a complete set of physical states in a transition amplitude, which is a key step in analyzing the implications of unitarity. If you think about this for a bit you'll realize that this is much more radical than you suppose.

Again, your focus on the kinematical space is entirely misguided, as it is unphysical and only appears in one particular approach to quantization but not in others, as I have clearly explained. You have the situation precisely backwards: there is no fundamental requirement that the kinematical space carry a Hilbert space structure (how could there be, since it by definition is full of unphysical garbage).

The bottom line is that if you are going to violate the axioms of quantum mechanics then you need to explain what you propose to their place, otherwise it's impossible to know what you are talking about.

This is just getting ridiculous. What in the world do you think the constraint and dynamical parts of the Hamiltonian are for? Why even have a Hamiltonian? The operator is used on the *kinematical* space to pick out *solutions* that are *physical*. If every state in the only Hilbert space you ever work with already satisfies the constraints, why even have the constraints? What in the world do you think they are constraining?

ReplyDeleteLook, how about this. Cite a reference that has as an axiom that every state in the Hilbert space represents a solution to the dynamical equations. Then we can check the reference.

Tim,

ReplyDeleteObviously if you start with a kinematical space then the constraints tell you how to distinguish the physical states from the unphysical ones. Who is disputing that? Alternatively, as in Weinberg's book, you can build up the physical space directly. Either way, you end up with the same theory with a physical Hilbert space, and how you got there is immaterial. Why you are having trouble understanding this simple point I do not know.

Let's be clear here: do you or do you not accept the QM axiom that physical states are in one-to-one correspondence with rays in Hilbert space? Please don't duck the question. This has nothing to do with quantum gravity: it is an axiom of QM that applies to any QM system.

What do you think people are talking about when they refer to a "physical Hilbert space"? Isn't it obvious that it refers to a Hilbert space in which the states are physical? What else could it possibly mean?

Again, none of my business, but wouldn't the set of physical solutions not being complete imply that physical solutions could combine and/or evolve into non-physical solutions?

ReplyDeleteBHG:

ReplyDeleteI have been trying to make a point about how much one should be expected to easily *know* about a certain mathematical structure. In particular, you have several times asked me whether the solution space for the theory of gravity in the bulk is a product space or not, to which my answer has been: I don't know. That is, you have been acting as if certain mathematical questions ought somehow to be obvious, when I can't see a reason why they should be.

In order to understand why this might even be expected, one has to consider *how the structure is arrived at*. In particular, the works on quantum gravity that I have been looking through proceed in the way I have outlined: they begin by defining a kinematical Hilbert space, then they impose some constraints on it, as codified in the Hamiltonian. This yields a set of solutions—the solution space—and that apparently does not always immediately have a Hilbert space structure. For example, it may not be complete. Now one can at that point formally complete the space into a Hilbert space. That's fine. Whether these additional states that are added in order to complete the space should really be considered physical states, that is; states that represent real physical possibilities I have no idea. One would have thought that any physical state would have already shown up in the kinematical space, and that imposing the constraints would just filter out the physical ones. But in that case, the states so filtered should be a Hilbert space already, and apparently that does not always happen. At least according to Thiemann. So mathematically one can then complete the space formally and get a physical Hilbert space. Fine. But there are two points I have been trying to make.

One, which we have been over and over, is that the constraint operators used in this construction are not the zero operators. That is just a mathematical fact. And it is essential that they not be zero operators if they are to do their job of filtering out the physical from the non-physical states in the kinematic space. The kernel of these operators is essential and the kernel of the zero operator is the whole space. I take it we have no dispute about that?

The other, which is more relevant at the moment, is that even though I may know something about the mathematical structure of the kinematical space that I start with, the structure of the solution space or physical space may be quite obscure. Why not? Is there any reason, for example, that just because I start with a tensor product kinematical Hilbert space I will end up with a tensor product physical space? If there is, I don't see why. So when you ask questions of this sort about the physical Hilbert space, my response is that I just have no idea. And, as I said, if you can prove, for example, that the physical Hilbert space is not a tensor product space then I will take that as any interesting discovery but not at all a refutation of anything. As far as I can tell, whenever you start to try to show that there is something unacceptable or unphysical or contrary to AdS/CFT about my solution to the information issue, it always begins with a question about the structure of the physical Hilbert space. And I have been trying to make clear why I haven't a clue about that structure.

If all of that is now clear, we can proceed to the argument. Because I still don't see, even granting AdS/CFT (which I am not committed to and find independently implausible) what the supposed contradiction is that you keep insisting on.

Jim V: if it were not complete, then certain Cauchy sequences would not have limits in the space. I can't see how that would imply that a physical evolution could take you out of the space. And in fact, in quantum gravity there really isn't any dynamical evolution at all: that is the "problem of time" that arises from demanding diffeomorphism invariance. In this case, one should really think of individual rays in the solution space as complete 4-dimensional specifications of the entire "block universe", rather than thinking a dynamical solution as a time-indexed sequence of spatial states.

ReplyDeleteJimV,

ReplyDeleteYes, you are basically correct. TIm thinks completeness is just some technical detail having to do with convergence, and doesn't realize that as soon as you give it up many other properties go out the window, such as the statement that eigenvectors of Hermitian operators form a complete set, and so one will typically find less than unit probability for a measurement of some observable to yield some answer, assuming the usual Born rule. Also, Tim is of course wrong that there is no time evolution in quantum gravity in a space with an asymptotic region (which is what is being discussed here): there is evolution with respect to the asymptotic time, as generated by the Hamiltonian.

Tim,

ReplyDeleteBased on past experience, I find it impossible to proceed unless you give a clear answer to my question. Namely, do you accept that the bulk theory obeys the axioms of quantum mechanics: there is a physical Hilbert space, where the latter means a Hilbert space whose elements are physical states?

Here by "accept" I of course don't mean this as a confirmed empirical fact about the world, but just that in this discussion we will assume that the rules of quantum theory hold and ask where it leads.

Every reference on quantum gravity that I know of, certainly including Thiemann and the others that have been mentioned, regard constructing a physical Hilbert space as part of what it means to have a quantum theory of gravity. It's pretty clear what has been getting you confused: Thiemann describes one attempt to construct such a physical Hilbert space and note that it runs into technical difficulties. You seem to infer from this that he is therefore proposing a theory in which there is no physical Hilbert space, whereas he of course means that it is the approach that is lacking.

So again: does quantum gravity have a physical Hilbert space? If yes, we can proceed. If no, then you are willing to entertain the breakdown of QM, which is fine, but it is hard to know where to go from there.

BHG

ReplyDeleteAs I said, if you want to go through a construction that completes to a Hilbert space, fine. Go right ahead. If you want me to affirm that every state in the so-completed space is really a physical possibility, I am not prepared to do that without looking at a particular theory. I can't imagine that anything at all hangs on this. Physicists use unphysical states for mathematical and analytical purposes all the time. Like fictive "states at infinity". Why don't you try to actually present some sort of argument, and we will see if these details (which I don't imagine will matter) make any difference.

I have yet to see the beginnings of a coherent argument. So far, you have always begun by asking about the structure of the physical Hilbert space, which I have repeatedly said I have no clue about. But the issue there is not whether it is a Hilbert space but whether, e.g., it is a tensor product space. Just use a Hilbert space for the argument: if the issue of whether all the states are really physical comes up, it does and if it doesn't it doesn't. I don't anticipate that it will. My insistence on distinguishing the kinematical space from the solution space was part of trying to get across that the Hamiltonian is not the zero operator. I still can't tell if you have accepted that trivial mathematical point.

But all of this is just tedious. If you actually have an argument, make it. If you don't, then stop wasting our time.

Tim,

ReplyDeleteI ask a simple, precisely stated, question and I get back a convoluted word salad. I will try again:

Do you accept that the bulk theory obeys the axioms of QM? Namely, that there is a physical Hilbert space: a Hilbert space in which all the states are physical (i.e. annihilated by the constraints)? Every reference on quantum gravity takes this as part of what the words "quantum theory of gravity" mean, but for some reason you continue to waffle. Very odd.

I have learned that to make progress with you I need to proceed very slowly and clear up your misconceptions one by one, or they start to multiply. So why not just answer the question and then we can move on? Actually, it sort of sounded from your message that you are willing to violate the quantum axioms, but the wording is so vague ("physical possibilities"?) that it's impossible to tell.

BHG

ReplyDeleteI believe I answered this weeks ago, but it seems not to have been posted. So here the answer: Just go ahead and try to state an argument already!

Do I accept that the bulk theory obeys the axioms of QM? If you are including in the axioms that the set of physically possible states forms a Hilbert space, in the sense of being complete, then I have no idea, since there is no actual bulk gravitational theory to examine! Or are you hiding an actual quantum theory of gravity up your sleeve? Then let's just talk about it directly, and leave the whole CFT thing out of it.

If you insist on going into this in more detail, which I see no point of, take plain vanilla QM. Does it obey the axioms you have in mind? In particular, are so-called "position eigenstates" and "momentum eigenstates" parts of the solution space of the theory? Even though neither in is the Hilbert space of square-integrable complex functions? Do you want to eliminate them as physical possibilities, or expand the Hilbert space to include distributions? We could get into all sorts of technical and mathematical detail here, to no obvious end. I can't tell at this point whether any issue like this is likely to come up, since I have no idea what sort of an argument you think you actually have.

All of this is now noted, OK? Just get on with it!

Tim,

ReplyDeleteLet's recall what this entire discussion is about. You are claiming that there is no obstacle to accepting the standard rules of QM along with the usual Penrose diagram for an evaporating black hole. I am claiming that if you add AdS/CFT to the mix then there is a clash, and something has to give.

To get anywhere I need you to commit to some definite statements about the rules of QM, since this is after all the entire point. The issue of completeness of the physical Hilbert space is not a minor technicality. Let me rephrase it this way. Suppose the bulk is in some physical state of my choosing, and suppose I choose to measure some physical observable associated with some physically accessible Hermitian operator. Question: is the state after the measurement a physical state, and (using the Born rule) do the probabilities add up to 1? If the physical states are not complete in the Hilbert space then the answer will be "no" for a suitable initial state, simply because the physical eigenvectors of the operator being measured are then not a complete set. If this is the path you want to go down then you have conceded that QM does break down after all, and the breakdown is rather spectacular.

Now, I have of course explained many times why your "scenario" (scare quotes here because it is not at all clear to me that you have anything more definite in mind than a vague picture) plus AdS/CFT leads to contradictions, but everytime I state it I end up having to argue about standard facts such as that the Hamiltonian in quantum gravity is a boundary operator (yes, of course what is implicit here is that I am only referring to the action of H on physical states, since unphysical states will never occur anywhere in my argument, for the very good reason that they are ... unphysical). That said, I hope it is obvious to you that there is at the very least some serious tension between the following claims

1) Trusting the basic accuracy of the usual Penrose diagram, the full quantum state at "very late times" in some sense lives on the union of the late time slice connected to the AdS boundary and a slice inside the horizon

2) Information is lost to the late time external observer in the sense that there are many ways to change the state inside the horizon that cannot be detected by the external observer. That is, the external observer has no way of measuring the part of the state behind the horizon, and since there are many possibilities for the state behind the horizon, information is effectively lost.

3) The Hamiltonian is defined in terms of quantities in the external region, and hence energy measurements are carried out in the external region. It follows from (2) that there are many way to change the state inside the horizon that have no effect on any measurement of the Hamiltonian

4) The spectrum of the bulk Hamiltonian coincides with the spectrum of the dual CFT Hamiltonian

5) The spectrum of the CFT Hamiltonian is discrete and nondegenerate (modulo the obvious symmetries).

You are the one claiming that there is no incompatibility, so pray tell, whatever your scenario is (and again, I'm not sure you have one) how does it avoid contradicting one of the above? In fact, I have left a loophole in the above, which I will close depending on what your proposal is.

Black Hole Gut,

ReplyDeleteExcellent! Progress! I assert 1 and 2 and 3. Note that 2 and 3 really don't have much of anything to do with 1, by the way. Even for a future eternal black hole, you can divide any Cauchy surface into a part inside the horizon and a part from the horizon on out, and the No Hair "Theorems" at least strongly suggest that information about the exact interior state is lost to the external region, out to spatial infinity. So the issue is either 4 or 5.

I don't know enough about CFT for it to be obvious to me that the spectrum is discrete and non-degenerate up to some symmetries. But I assume that's the sort of thing easy to check and prove. So obviously I reject 4). That was the whole point of my example with the six particles that you did not understand. So: why should I believe 4? If it is relevant, I also do not believe the Holographic Hypothesis and have never seen any cogent reason to believe it. 4) is strictly speaking logically weaker than full holography, but let's just focus down on it.

First, why should I believe it at all, even in the case of AdS?

Second, even if it is true for AdS, AdS is non-physical. Why believe anything like it for physical space-time?

Now if your only answer is "that's what AdS/CFT hypothesizes", then obviously that's no help: all the more reason to reject AdS/CFT. So how does the argument go that the spectrum of the CFT Hamiltonian and the spectrum of the AdS Hamiltonian must be the same, even up to degeneracies?

BHG

ReplyDeleteOn further reflection, While it is true that I don't see any reason to accept 4, the sequence does not constitute an incompatible set of claims. So I can actually accept all of them as far as logic goes. Is this the "loophole" you mentioned?

First, for 3, I take it you would be happy with the less contentious: "The total energy (whatever corresponds to the ADM mass in the case of AdS) can be "measured" by operations in the regions outside the event horizon, and indeed in regions arbitrarily far from the origin used to draw the Penrose diagram". That gives the proposition you want, yes? That is, if we regard H as an observable, it can be measured by boundary operations. Just like the net electric charge in the bulk can be measured by operations (measurement of the electric field) at the boundary. Right? We agree about that.

Now at best, all you will get of the whole set 1 through 5 is that *If the bulk state is an eigenstate of H* and if one knows antecedently that it is an eigentstate but not which one, then one can determine what the state of the bulk is (up to symmetry) by measurements at the boundary. But you of course do not get, and never can get *the state of the bulk can in general be determined by measurements at the boundary*, not from these premises. Or any premises.

To make this point sharp, do you accept this proposition:

6) The discreteness and non-degeneracy of the spectrum of the Hamiltonian (up to symmetries) in no way implies that the state of a system can be determined (up to symmetries) by a "measurement"of the Hamiltonian, no matter how or where it is carried out.

6) is obvious, but I want to make sure we are not disagreeing about it.

Tim,

ReplyDeleteYour point (6) is correct. To clarify, when I talk about "measuring a system" I have in mind the ability to prepare arbitrarily many copies of the same system in the same quantum state, and then to make measurements on each of them. The question is whether by measuring the energy of each system can I retrodict the state. The answer is indeed "no", even if the energy spectrum is discrete and non-degenerate. Counterexamples are easy to construct. So no disagreement about that.

Regarding your "six particle" example, I'm not sure what your point is here. I will say this, if you take 6 particles in AdS, interacting weakly via gravitational attraction, Coulomb forces, etc, then the energy spectrum will generically be discrete and the only degeneracies will be associated with the obvious symmetries of the system. This can be shown by direct computation.

A significant fraction of the literature on AdS/CFT is devoted to computations checking the claim that the energy spectra of the two sides match. In every case that computations can be done -- and there many -- the results match, often in highly nontrivial ways. So while there is certainly no proof of this in general you should appreciate that there is a mountain of evidence pointing in this direction.

I also don't get your criticism along the lines of AdS not being a physical spacetime. Recall that we have no idea what the actual spacetime we live in corresponds to: all we can observe -- and can ever observe -- is a little patch of the universe, and what spacetime looks like outside our cosmological horizon we have no way of knowing. Certainly, asymptotically flat spacetime is just as unphysical as AdS in this regard, as neither correspond to our world.

My point all along has been that the combination of certain reasonable sounding assumptions about quantum gravity *plus AdS/CFT* leads to contradiction. Of course, you can counter by saying that you therefore reject AdS/CFT, and believe that it will fail in some yet to be discovered fashion. It's hard to argue against this except to say that people who have studied AdS/CFT tend to believe that it "works". This includes people from the GR community who are naturally inclined towards accepting information loss for the external observer.

Finally, I do think this discussion reflects a split between the way a physicist vs. a philosopher views this problem. Part of the appeal of AdS/CFT is the mathematically tight structure it embodies. To digress slightly, I think an analogy can be made regarding interpretations of non-relativistic quantum mechanics. For example, while I think a case can be made for Bohm's interpretation being conceptually appealing in certain respects, to most physicists it is unappealing on a mathematical level. It looks like a human invention rather than a fundamental truth. AdS/CFT on the other hand shares the kind of mathematical elegance of successful theories.

BHG

ReplyDeleteI am still completely unable to see what the disconnectedness of the Cauchy surfaces has to do with any of this. My main claim has been that the relevant information that Hawking was puzzled about at late times is on the disconnected part of the Cauchy surface, as it were, relative to observers who have the evaporation even in their past. Now as far as I can tell, your response is this:

A) the spectrum of the Hamiltonian of the gravitational theory in AdS is discrete and non-degenrate, as is "shown" by AdS/CFT and clear facts about the CFT.

B) Hence if one could make repeated measurements of the total energy of the state on identically prepared systems, one could figure out at least a density matrix for the state. One would get a probability distribution over the energy eigenstates as an observational fact. This would typically not nail down the exact initial state since phase factors would be missing.But we are completely ignoring that.

C) such total-energy measurements can, in principle be made at spacelike infinity.

Hence (?), in a somewhat weird sense of "in principle" (since the "principle requires multiple measurements made on identically prepared entire worlds that are known to be identically prepared) information about the initial state is never lost to observers at "spatial infinity".

Now there are a bunch of lacunae in this line of argument.

Here you can help out with the technical questions. I have been assuming that the way possible initial states for the AdS theory are created is by first settling on a vacuum state and then acting on that vacuum state with creation operators. If that correct or incorrect? Maybe I should just stop at this point and wait for answer. If the initial states are not constructed in this way, how are they constructed? Can I operate at any point in the vacuum with a creation operator? Or is there some other method of constructing the initial state envisioned?

Tim,

ReplyDeleteGood, now I think I will be able to make it to the end of the argument. Regarding your question -- and this will be relevant for the argument -- bulk states that have an interpretation in terms of particles living on a background AdS space can indeed be created by acting with creation operators on the vacuum state. However, for our purposes it will be better to think about such states as being injected into the bulk from the boundary; we'll get to that in a moment. Again, I will try to break down the argument into distinct statements

1) The CFT has a unique zero energy vacuum state. This state maps to empty AdS in the bulk, which is therefore also the unique bulk zero energy state. (Throughout, I will assume AdS/CFT and then derive a contradiction).

2) Starting in the vacuum state, we create an excited state by injecting particles in from the AdS boundary. In the bulk, we do this by setting up time dependent boundary conditions for bulk fields, and in the CFT we do this by turning on the corresponding "sources" for CFT operators. This correspondence was described in the very first papers on AdS/CFT (e.g. first Witten's paper). The bulk state is chosen to correspond to particles that will collide to form a black hole.

3) We then evolve the system forward in time by Hamiltonian evolution to the distant future, long after the black hole will have evaporated.'

4) Now we suppose that things are as the usual spacetime Penrose diagram would have us believe. Namely, in the external region there is a diffuse cloud of quanta, which are the Hawking radiation particles that the black hole decayed into. Further, I will assume that the bulk obeys the rules of QM. I will also assume that the Hilbert space takes the form of a tensor product, with two factors corresponding to the slices inside and outside the horizon. This is certainly true in Hawking's computation and is certainly what the Penrose diagram depicts. If you want to depart from this you need to suggest a concrete alternative, otherwise saying that you respect the spacetime diagram has no content. The assumption is then that the quantum state is an entangled state in the tensor product. Again, this is what Hawking originally proposed.

5) Here is the new part which closes a prior loophole. We now apply a unitary operator, that acts purely in the external region, and that removes the energy from the external region. First consider the CFT. The CFT is in some excited state. The best way to talk about removing energy from the CFT is to think of coupling the CFT to another auxiliary quantum system, call it AUX. You can think of the original CFT as living in a box, and then we introduce another CFT (in its vacuum state) that lives in a much bigger box. I put the two boxes next to each other and then open up a little hole that allow energy from the original CFT to pass into AUX. If I wait long enough all the energy will be transferred, and then I close the hole. This is easily modeled in the bulk: I am again changing the boundary conditions, except now the parameters appearing in the boundary conditions are the quantum operators of AUX. All of this is just to say that there is no problem in talking about transferring energy from the system to another system by a unitary operation, and the bulk operation is done purely in the external region, since it just involves a change of boundary conditions.

6) Having done this, the CFT state is now given as |0>_CFT x |\psi>_AUX. Hence the bulk state is given as |0>_AdS x |\psi>_AUX.

cont

possible duplicate message here, but the second half of my last message has not appeared.

ReplyDeletecont

7) But this contradicts the assumption that the bulk was in an entangled state. For given such an entangled state there is clearly no unitary operator that acts as the identity on the internal factor and that takes the state to |0>_AdS. I.e. to "unentangle" a state the corresponding unitary operator obviously must act nontrivially on both factors.

The upshot is that the existence of a unitary operator that sucks all the energy out of the external region implies that there must be a huge degeneracy of zero energy states, corresponding to all the possible states behind the horizon. But this degeneracy contradicts the CFT. You might try to argue against the existence of this operator, but this will require some breakdown of low energy physics, since this operator is just acting on a state consisting of a diffuse cloud of quanta. There is no easy way out of this paradox. In the absence of AdS/CFT you could try to simply accept this huge vacuum degeneracy; it sounds problematic but it's hard to definitively exclude on general grounds. But it is impossible given AdS/CFT, since we are absolutely certain that the CFTs appearing in this duality have unique vacuum states.

BHG,

ReplyDeleteI am very glad to see that you have given some thought to trying to construct an argument here. That is, you recognize that the claims made until now were not being backed by any rigorous arguments. And I do appreciate that you are trying to fill one in. But this deus ex machina arrival of AUX, which is all new, obviously does not actually work. Append as big an AUX, initially in the vacuum state, as you like: the Hamiltonian will simply never yield the vacuum for the state on Sigma 2 out. Never. You may dilute the energy density on Sigma 2, but the dilution never get you to |0>. I'm sure you know this. So there just is no contradiction of the sort you are arguing for.

Tim,

ReplyDeleteOf course, the argument I give here is the same one I have been attempting to give all along, only slowed down by the need to explain the Hamiltonian formulation of GR, the causal structure of AdS, the term "Gauss' law", etc. etc.

Anyway, your statement that this "obviously does not work" is not exactly a counterargument. What is obvious is the following. Given an initial state |\psi>_CFT |0>_AUX, there exists a unitary operator U that takes this state to |0>_CFT |\chi>_AUX. I was trying to give an intuitive picture of this in terms of placing two boxes next to each other. Perhaps you are worried about the fact that if the AUX box is of large but finite size then there is always some probability for the energy to return to the CFT? That's true, but this is not going to change the conclusion. By taking AUX sufficiently large I can get the CFT to be in the vacuum state to arbitrary accuracy. Clearly this is good enough. The state based on the Penrose diagram has a huge entanglement entropy, and so all I need to do is to reduce the entanglement sufficiently to rule this out. And I can certainly do this by the construction I have described.

So your scenario is still ruled out.

BHG-

ReplyDeleteI do appreciate that you are trying to construct an argument here, as I said, but let's just be candid. The argument involving AUX simply is no version of any argument you have mentioned heretofore, because you never mentioned AUX. You are trying to fill a gap in your thinking, which is fine. And this just does not work.

All you have to work with is that the spectrum of the Hamiltonian is discrete and non-degenerate up to symmetries. That's it. It is consistent with that that the spectrum be arbitrarily dense in any interval of energy. Getting a state to be the vacuum "to arbitrary accuracy" simply is not getting it to be the vacuum. And the closer to the vacuum you want to get it, the bigger AUX has to be, and the bigger AUX has to be the denser the spectrum. You simply cannot achieve the result you announce.

Your comment about entanglement entropy is cryptic in every way, so maybe you want to try to explain that. If the state on Sigma 2in is entangled with the state on Sigma 2out, then couple Sigma 2out to whatever AUX you like, the state on Sigma 2in will remain just as entangled with the state on Sigma 2out U Aux. No entanglement is going away, although it may be transferred from Sigma 2out to AUX.

The issue of AUX is not that "there is some probability for the energy to return to the CFT", whatever that means. The evolution of the wave function is always deterministic and unitary, so there is no room for any probabilities here. The Hamiltonian describing the coupling to AUX that you postulate simply will never, ever lead to the vacuum state on Sigma 2out.

The simple fact is that this appeal to AUX is completely new. If I am wrong, just point to where you ever mentioned AUX. And as you have described it, it won't do what you want. And even if you can describe a unitary on CFT X AUX that swaps the state on the CFT for the state on AUX, so what? What does that have to do with anything?

Why can't I play the same trick without the Cauchy surface being disconnected? Just cut the Cauchy surface into 2 parts, one outside the event horizon and one inside (put the event horizon together with whichever). Why doesn't all of your argument go through unscathed: the Hilbert space of the bulk is a product space, I can couple the boundary to AUX and switch with a unitary, etc, etc, and reach exactly the same conclusion? Where does your argument get blocked in this case but not the other?

Tim,

ReplyDeleteOK, let's put the AUX construction aside for the moment. It is useful for addressing a higher order subtlety, but I can see that we haven't gotten there yet.

Given on your remarks, I now see that it is in fact simple to rule out what you are saying, as follows. Forget about black holes and quantum gravity for a moment. Consider a quantum system with the following properties

1) The Hilbert space is a tensor product: A x B

2) The Hamiltonian is of the form H = 1 x H_B; i.e. the Hamiltonian is built purely out of degrees of freedom in B

3) The system has a state with eigenvalue zero for H, and this state is unique.

4) The system is in an entangled state in the tensor product

I hope it is obvious to you that these four assumptions are mutually contradictory (if not I will supply a simple proof). No quantum system can have all these properties simultaneously. But now substitute: A-> state behind horizon, B -> state in region connected to boundary. AdS/CFT and general facts about gravity tell us that (1), (2), and (3) hold, hence we conclude that the state is not entangled, and hence there is no information loss.

To answer your question: this argument clearly doesn't hold if you simply chop a connected Cauchy surface in two pieces, since the Hilbert space will then not be a tensor product. This has nothing specifically to do with quantum gravity -- just consider electromagnetism coupled to charged particles. I have explained this before. If A represents the interior of a sphere and B the exterior, then the Hilbert space is not a tensor product, since if you create a charged particle in A the field lines necessarily extend into B. On the other hand, if the Cauchy surface is disconnected this doesn't happen, since the field lines then can't go from A to B. In the black hole case with a disconnected Cauchy surface the field lines of a particle inside the horizon can go into the singularity, and need not extend to the exterior region (the fact that stuff can flow into the singularity like this is indeed why there appears to be information loss in the first place). I should also say that the notion of a disconnected Cauchy surface is a classical (or semi-classical) one when it comes to gravity, and the precise statement at the quantum level is that the Hilbert space is a tensor product. Finally, in case it isn't clear, my point is not that *I* think there is actually a tensor product Hilbert space in the black hole case, but rather that that is what your scenario based on the Penrose diagram corresponds to, and is what proponents of information loss have in mind.

BHG-

ReplyDeleteWe are back to this again! 2) is false. Of course the Hamiltonian does not have the form you give: if it did, then the state in A for a product state would have no effect at all on the eigenvalues! So there could not be a unique vacuum. It is the same error about H that I have pointed out over and over and over.

Your argument, if it proved anything, would prove that the Hilbert space of the bulk cannot be a tensor product. Is that what you are asserting? Yes or no?

And your argument that if A is the interior of a sphere and B the exterior then it is not a tensor product applies just as well to the case with the disconnected Cauchy surface, If you create a charged particle in the interior of a black hole, then in the solution you change the field values outside. Charge is a hair that violates the no hair theorem. So if you create a charged particle inside the interior, the solution changes at the boundary. True for charge, true for mass, true for angular momentum. So your argument is broken-backed everywhere.

If it is by the argument you give that you conclude it is a tensor product, then you have no reason to conclude it.

As I said many times, I don't even have an opinion on the subject. I don't really care. But there is just no argument you can appeal to for the case of disconnected Cauchy surfaces that is not available for connected ones. You seem to be confusing the claim that the Cauchy surface is disconnected for the claim that the space-time is disconnected. But the space-time is obviously connected. Change the interior of the event horizon, and the constraints in the Hamiltonian will force changes outside. Obviously.

Is there any argument that does not rest on these mistakes, that we have been over and over?

Tim,

ReplyDeleteI must say that I have never encountered anyone who goes on with such self-confidence even after having been shown to be committing blunders time and time again. Usually some self-doubt starts to creep in.

One basic problem here is that you claim to have a concrete scenario, but when pushed for details you always respond with some version of "I have no opinion on that" . In fact, all you have is Hawking's orginal Penrose diagram. But physics is more than pictures. If you claim to have a concrete scenario involving the harmonious coexistence of gravity and quantum mechanics then you need to say more about what the quantum state looks like than just drawing a classical spacetime diagram and then waffling whenever you are pressed for details. It now seems clear that you are doing so because you realize your position makes no sense: either you agree that the state is a tensor product in which case it will be ruled out by the argument I gave, or it is not a tensor product in which case you have to explain how it is that information could be lost to the outside. It is on you to propose something here.

What every expert who has though about this realizes is that if you take the Penrose diagram at face value then the Hilbert space is a tensor product (as is manifest in Hawking's orginal computation), and this implies a huge degeneracy in the spectrum including the vacuum. By itself this is not necessarily a problem, but it clashes with AdS/CFT. Why can't you grasp this?

One thing you need to do is state what it means quantum mechanically for a Cauchy surface to be a union of disconnected components. The standard statement is that the Hilbert is a tensor product in this case. Do you dispute that? If so please provide an example where that is not the case. And please don't say "I have no opinion on that", since this is a key point.

Finally, your statement " If you create a charged particle in the interior of a black hole, then in the solution you change the field values outside" completely misses the point as usual. Consider a disconnected Cauchy surface, with one component inside the horizon and terminating at the singularity and the other component outside. Obviously, given any configuration on this Cauchy surface I can change the data inside by adding a charged particle without changing anything on the outer component. The only restriction on initial data is that they obey the constraints, and if the Cauchy surface is disconnected then the constraints decouple into two independent sets, one for each component. So it is a mathematical triviality that you can change the data on one component without changing it on the other. Your statement is just flat out wrong. It is only for initial data on a connected Cauchy surface that the constraints link what data I choose on one part of the surface to another. Again, this is a mathematical triviality, and it is also intuitively obvious by thinking about field lines. Are you sure you know what a Cauchy surface is?

BHG

ReplyDeleteI will continue to write in the same tone as, note, you do as well. It is only your own judgment that I have been shown to commit any blunders. So let's just stick to the issues, and who is blundering will become evident, if you stop deflecting.

Shall we do this in backwards order? Do in know what a Cauchy surface is? It is a set of space-time events that every inextensible timelike curve intersects exactly once. That is a standard definition, and the one I am using. Would you care to dispute it?

That definition says exactly zero about whether a Cauchy surface is connected or disconnected. In the space-time at issue, some Cauchy surfaces are connected and some are disconnected. That is manifest. Do you want to deny that?

Now you say that "every expert"sees that the Penrose diagram somehow makes obvious that "the Hilbert space" is a tensor product space. You give no clue at all about how the supposed experts conclude this. But let me note, for again the nth time, that usually when people write about "the Hilbert space" they are referring to the *kinematical* space, not the "solution" space. I certainly think that the *kinematical* space is a tensor product space. Whether the *solution* space is or is not I have no opinion. Never thought about it. What do you want me to do? I mean it's not even as if there is any concrete, exact theory of quantum gravity on the table here! Without a concrete set of dynamical equations, who knows what the solution space looks like? What are the equations you have in mind? And even with them, to determine the structure of the *solution* space you would have to have mathematical understanding of the complete set of solutions. Nobody knows any of this. And I am supposed to?

You want me to make an assertion about the *kinematical* space? Sure: it's a tensor product space. You want an assertion about the *solution* space? You have no right to demand such a thing in this case. If all the "experts" are sure it must be a tensor product space just by looking at the Penrose diagram, then please explain the reasoning. You have not come anywhere in the vicinity of that. Or let's make it easier: cite a single "expert" who clearly states this and also clearly has the *solution* space and not the *kinematical* space in mind. The I can at least read their reasoning.

As for your own argument, it makes no sense at all. You persist in writing as if there is some objective unique fact about whether an event lies on a connected or a disconnected Cauchy surface. But it is at this point that it is my turn to ask: are you sure you know what a Cauchy surface is? Take any event inside the event horizon. It, like every other event, lies on an infinitude of Cauchy surfaces. in this case, some are connected and some are disconnected. That is just plain to see. Do you deny it?

So if you claim that for any objective feature phi (e.g the feature of the solution space being a tensor product space or not), we have both "if E lies on a connected Cauchy surface then phi" and "if E lies on a disconnected Cauchy surface then not-phi", then your entire system is just incoherent. Your principles imply that E cannot both lie on a connected Cauchy surface and lie on a disconnected Cauchy surface, which plainly E does.

I have pointed this out before, but one more time: the only events that "objectively" lie on disconnected Cauchy surfaces, i.e. that *only* lie on disconnected Cauchy surfaces, are events *in the future light cone of the Evaporation Event*. And creating a charged particle *there* is obviously unproblematic: the constraints *on the connected part that E lies on* hook up to the boundary, even by your lights.

Con't

Con't

ReplyDeleteYour persistent confusion of the *kinematical* Hilbert space with the *solution* space is evident again, as is your persistent identification of the Hamiltonian operator with a very different operator that it happens to agree with it only on the space of solutions. I have pointed out these two errors over and over and over and here they are conspiring into a perfect storm of confusion.

In particular, you write: " Obviously, given any configuration on this Cauchy surface I can change the data inside by adding a charged particle without changing anything on the outer component". This is obvious in the *kinematical" space, not in the "solution* space. In the *solution* space, more charge inside means more field outside. period. And more field outside means a different physical state on the disconnected piece outside. Period. So as a statement about the *solution* space, your claim is nonsense. As a claim about the *kinematical* space it is true but irrelevant.

First acknowledge agreement or disagreement where noted, or "I don't know" if you don't know.

Tim,

ReplyDeleteYou have stated a correct definition of a Cauchy surface. I admit that this query was in response to your condescending tone, which I have had to put up with, including during the infamous Gauss' law episode.

Now I will show that your "I don't know" stance about tensor product Hilbert spaces fails to withstand any scrutiny.

Take a disconnected Cauchy surface and let's ask whether the Hilbert space of states defined on this surface is a tensor product. By your own statement, the kinematical space is a tensor product. Now we have to impose the constraints to get the physical Hilbert space. But the constraints, in every theory existing theory including gravity and electromagnetism, are local functions of data on the surface. Therefore, the constraints give absolutely no link between data on one component of the surface and that on another. That's why in every known theory the Hilbert space on a disconnected Cauchy surface will be a tensor product. And again, besides the mathematical proof I just gave the result is physically obvious. The converse would be astonishing: my freedom to change initial data here would depend on what the data looks like on some part of the slice disconnected from my own!

Note that I am being careful, as I have always done, to distinguish a disconnected Cauchy surface from a disconnected spacetime. Only the former is under consideration here.

Do you really still want to keep up your "I have no opinion" stance on whether Hilbert spaces on disconnected Cauchy surfaces are tensor products? Do you agree that this is true for every known theory, including perturbative quantum gravity, for the elementary reason I provided? If not, please explain. I am dying to know as this would revolutionize physics.

BHG-

ReplyDeleteAs long as we are getting conceptual issues out in the open, let's at least start to talk about the elephant in the room: diffeomorphism invariance. Diffeomorphisms are regarded as symmetries of a gravitational theory so states related by a diffeomorphism lie on the same gauge orbit and represent the same solution. And as a consequence, foliations by different sets of Cauchy surfaces are also regarded as yielding the same solution, up to a gauge symmetry. But what that means is that the picture presented in your last message is conceptually confused. You asked what happens if we operate on a solution with a creation operator to put more charge inside the event horizon. And you argued that if the Cauchy surface on which the new particle lies is disconnected, the physical state on the disconnected piece will not change. But, as I said, that event where the new particle is created lies on an infinitude of Cauchy surfaces, some connected and some disconnected. The reason this is important is that you seemed to think that constraints in the Hamiltonian only effect the state on the Cauchy surface, and if the Cauchy surface is disconnected, only on the connected piece where the particle is created, But that is just incoherent.

We can assign the event to any Cauchy surface it lies in, and impose the constraints on that. But the union of all of those Cauchy surfaces covers the whole of spacetime that is space like separated from the event. You are thinking of the constraints as somehow only affecting the connected piece of the Cauchy surface, but they obviously have a much wider range than that.

We key to understanding how this is possible is that in quantum gravity, *the dynamics itself is expressed as a constraint". So properly speaking, what is determined by the constraints is not just the physical state on a connected piece of a Cauchy surface, or even on the whole of the Cauchy surface, but rather on *the whole of the spacetime*. You have to think of a solution not as a series of physical states on a series of Cauchy surfaces, but rather as just the state on the whole 4-dimensional space-time.

This is important because you seem to have a picture like this: As time goes forward in some foliation of the CFT, that determines the state on the boundaries of a succession of Cauchy surfaces, and that boundary state in turn determines the interior state. But that whole framing is wrong. If your picture works at all, then the state on the boundary of the CFT does not merely determine the states on the successive Cauchy surfaces, and hence the state in the interior of the AdS, but the state of the entire interior of the AdS from any one boundary state. The state on one Cauchy slice at the boundary cannot possibly determine just the state of *that* one Cauchy surface that intersects the boundary there.For many different Cauchy surfaces will intersect the boundary there. No measurements made at the boundary can distinguish which Cauchy surface you are *really* talking about from any other surface. For that reason, the mapping from any *local* operator in the bulk to some local operator on the boundary has to be very complex and delicate. I get the sense that you a treating the mapping between local states in the bulk and the corresponding state on the boundary as much more straightforward than it can be.

Anyway, once you see that *applying the constraints* does not just generate the state on some one Cauchy surface but actually the state in the entire spacetime, your worry about the exterior electric field lines (and you argument for a product space structure of the solution space) just vanish.

Tim,

ReplyDeleteYour mental picture of how the Hamiltonian formulation of gravity is very confused. First of all, the correct statement is that small diffeomorphisms relate equivalent configurations. Large diffeomorphisms relate inequivalent configurations. Small vs. large refers to whether the diff is nontrivial at infinity or not. The generator of a time translation diff is the Hamilton and this is why the non-constraint part of the Hamiltonian is a boundary operator. So it is totally wrong to say, as you do, that a state of the CFT at a given time describes the entire bulk spacetime. The correct statement is that it describes the state of the bulk at the same time, where time refers the time on the boundary, i.e. where a Cauchy surface hits the boundary. THere are infinitely many such Cauchy surfaces, but they are all related by small diffs, and so they are identified with each other.

One thing you seem to be missing is the crucial distinction between spacetimes with and without asymptotic boundaries. In a closed universe there is no invariant meaning to a "t" coordinate, which is why people sometimes speak of a "problem of time". But there is no such issue in a space with boundary in the sense that the asymptotic conditions fix a physical notion of time on the boundary, and there is no subtle issue regarding evolution with respect to this time.

Coming back to the main point, a physical state in the bulk at a time t lives on a Cauchy surface that hits the boundary at that time t, and the state satisfies the constraint equation by virtue of being physical. At a sufficiently late time t all Cauchy surfaces in the BH evaporation context are disconnected, according to the Penrose diagram. The constraint equations are equations that act on data on a Cauchy surface. On a disconnected surface they yield no link between one component and another, so there is simply no way that the constraints are going to take you from a product space to a non-product space.

Actually, this has really nothing at all to do with the quantum theory. We can consider the classical theory, and ask about the description of the phase space. In general, the phase space is the space of allowed q's and p's that can exist on a Cauchy surface. On a disconnected Cauchy surface the phase space will be a product space. I can prove this with as much rigor as you like. And there is absolutely no conflict with the possibility that the spacetime has other connected Cauchy surfaces on which the phase space is not a product. There is no conflict because you can't evolve from one to the other without passing through a singularity.

Besides the mathematics, I again note that your claim is physically outlandish. If I sit on some component of a Cauchy surface, my freedom to manipulate the state can't possibly depend on what someone does on a disconnected component, but this is precisely what you are saying if you claim that the physical Hilbert space is not a tensor product in this case.

BHG,

ReplyDeleteOK, we can just match rhetoric here and get that part out of the way: you are completely confused as is evident in your post. It looks like we will have to review some basic material to make any progress here. But in order for you to even try to pay attention, you first have to realize that what you have posted makes no sense.

Taking "configuration" to mean "configuration of the state *on a Cauchy surface*" (as opposed to the entire 4-D configuration that is the complete solution), and taking "small diffeomorphism" a per your own definition, the claim you make (" small diffeomorphisms relate equivalent configurations") is either empty or false, depending on what you mean by "equivalent". For example, by your definition a "small diffeomorphism" can relate a state with no black hole and no event horizon to a state with a black hole and an event horizon. Are these "equivalent"? Or a small diffeomorphism can relate a state that only contains a uniform plasma to a state that contains many separate stars and galaxies. Are these "equivalent"? Or, in our case, a "small diffeommorphism" can relate a state that is connected to a state that is disconnected. Are those states "equivalent"?

If the answer is "yes", then "equivalent" means nothing more than that the the states are states taken from the same space-time on Cauchy surfaces that agree at infinity. So your "principle" is just a definition of "equivalent". But if "equivalent" has some intuitive force at all, and is not defined this way, then your principle is false. So let's go step by step: which is it? What, exactly does "equivalent" mean?

Tim,

ReplyDeleteYou are correct that you do need to review some basic material, which I will now provide. Let's talk first about the classical theory, and to make things even simpler start with electromagnetism. What is the physical phase space? We start by picking a Cauchy surface and consider all vector potentials A_i(x) and their canonical conjugates E_i(x) (i.e. the spatial components of the vector potential and electric field). First of all these must obey the constraint equation, which is Gauss' law (div E = 0, if no charged particles). Second, we need to form equivalence classes under the action of small gauge transformations. So any two A_i(x) related by \partial_i \lambda(x), where \lambda(x) vanishes at infinity, are taken to lie in the same equivalence classes. The phase space of the theory is thus the space of (A_i, E_i) which obey the constraints, modulo the equivalence relation. This is textbook stuff.

The story in gravity is analogous. We pick a spacelike hypersurface and consider the space of metrics on the surface and their canonical conjugates. There are again constraints that these must obey. Then we impose equivalence relations. Two points in phase space lie in the same equivalence class if they are related by a diffeomorphism that is trivial at infinity. To be even more precise, the motion of phase space points under such diffeomorphisms is given by taking the Poisson bracket with the constraints, just like the Gauss' law constraint generates gauge transformation via Poisson brackets: \delta_\lambda A_i = [ A_i, \int \lambda div(E)]_{PB}.

Those are the precise mathematical statements. Intuitively, you can say that Cauchy surfaces with two sets of data are equivalent if you can get from one to the other by a combination of a spatial coordinate transformation and time evolution that keeps fixed the location of the surface at infinity. I suspect you will object to this, but these are just standard facts about the canonical formalism for gauge theories and gravity.

BHG

ReplyDeleteYou have been taken in by a false analogy, which explains your missing the point. All the relevant facts are in your own post, but you have not put them together.

In classical EM, the physically real ontology is taken to be the E and B fields, or the electromagnetic tensor if you like. The vector and scalar potentials are regarded as mere mathematical conveniences. Different A and phi fields represent the very same situation if they yield the same E and B, i.e. if they are related by a gauge transformation. There is, of course, no requirement that the gauge transformation go to the identity at infinity, Standard gauge transformations do not do this. One can add it as an extra constraint, but as Itzykson and Zuber write, "there is actually no very convincing argument to justify this restriction". But whatever...it is not relevant to the point here.

The point is that *of course* in EM states on a Cauchy surface related by a gauge transformation are the same physical state! Because they are identical with respect to E and B.

But this is *not at all* what is going on in GR! As you know, pairs of states there can be generated from one another by application of the lapse and shift operators. The shift is a gauge transformation in the same sense as in EM: it takes a mathematical representation of a state on a given Cauchy surface and yields a different mathematical representation of that same state. Thought of passively, all you are doing is recoordinatizing the same state on the same surface. But the lapse operator is an entirely different beast. The lapse operator takes you from one Cauchy surface to a *different* Cauchy surface. The lapse operator gives the dynamics of the theory. The lapse operator takes you from uniform plasma to stars, from collapsing matter to a black hole, and even from a black hole to a post-evaporaration state. And these states, on *different* Cauchy surfaces are clearly physically different! A star is not a uniform plasma and collapsling matter is not a black hole.

So here the analogy to EM gauge transformations breaks down entirely. It is because people *call* the lapse operator a "gauge transformation" that they say that in GR the dynamics is "pure gauge". But this is an abuse of terminology.

If you want to call two such states "equivalent", you can make that stipulation. It means that the state of a Cauchy surface that passes below (in time) even the formation of the Earth can be "equivalent" to one that cuts through the earth today. But despite this "equivalence", the states on these surfaces are very, very physically different.

Of course, one could also decide to call the states on any two Cauchy surfaces taken from the same model "equivalent". That is how I would speak is I went this route. But then "equivalence" just means "Cauchy surface taken from the same model.

What does this prove? Well certainly that no measurement at infinity can tell you anything at all about the state on the Cauchy surface that the measurements at infinity lie on. Because the experiments will be identical and give identical outcomes no matter which Cauchy surface in the interior is chosen. But those various interior surfaces can be very, very different from each other.

Any disagreement so far?

Tim,

ReplyDeleteYes there is disagreement. First, in GR it is not possible to formulate the theory in terms of analogs of E and B like in E&M, so the entire point of my E&M analogy is missed unless you discuss how things work in terms of the vector potential.

Your description of GR is simply not how the mathematics is set up, and you are missing the key interpretational points that allow this mathematical description to make physical sense. First a point of terminology: you speak of a "lapse operator". In quantum GR the "lapse" is not an operator but rather a c-number function. I am going to assume that what you really mean here is an operator of the form \int N(x) H_t(x) d^3x, where N(x) is the lapse and H_t(x) is the corresponding constraint operator that annihilates physical states. This operator generates a time translation, and I will consider N(x) to go to zero at infinity so that the time translation is zero at infinity. This is our operator that moves the surface in time while holding it fixed at infinity.

Now let's consider your earth formation example, which in a physical sense describes the formation of the earth out of a ball of gas. We consider a family of hypersurfaces, some of which pass through the earth region before its formation, and others which pass through it after, but with all the surfaces going to the same fixed time at infinity. The wavefunction is a function of the positions of all the particles that are involved, along with the spatial metric on the surface. The key point: since the wavefunction is annihilated by the operator defined above, the wavefunction necessarily takes exactly the same value on a diffuse cloud of particles as on the configuration where they have coalesced to form the earth. Mathematically this is required since these configurations lie in the same gauge equivalence class.

So how do we describe in this context the fact that there is a physical planet formation process occurring? Answer: you have to ask physical questions! For example, let's suppose there is a clock in the vicinity of the collapsing cloud, and say that planet formation occurs when the clock reads 1 billion years. The way you see this in the wavefunction is to look for correlations in the clock and the particles. Given \psi(x_clock, x_particles) you will see that if you choose x_clock to correspond to a reading of much less than 1 billion years then the wavefunction is strongly peaked around particle configurations which are dispersed. But if you choose x_clock to be much later than 1 billion years then the wavefunction is peaked around particle configurations which are tightly bound into an earth shape. So even though these configurations are equivalent in the mathematical sense, there is no conflict with their obvious physical inequivalence, and classical time evolution emerges as it should in the hbar ->0 limit, where "time" means the reading on a clock.

Thit is how things work in AdS/CFT. Given a time on the boundary there will be a CFT state. This single CFT state at a given time can describe all the bulk Cauchy surfaces that attach to the corresponding time on the boundary. The physical evolution that goes on in the bulk is obtained by identifying a CFT operator that acts as a local clock in the bulk, and then asking about correlations of other observables relative to the clock variable.

The quickest way to see that your story is incompatible with the mathematical formulation is note that physical state wavefunctions are (by definition) invariant under the operator that generates time translations that are trivial at the boundary. Any interpretation has to be compatible with this mathematical fact, and the interpretation I have reviewed above (which is the standard one) is.

BHG-

ReplyDeleteSo we are making progress. Of course you realize that many of the things you say are not *correcting* my post but *repeating* it. That is, I said exactly that by your criterion different Cauchy surfaces that have the same behavior at spatial infinity are all "gauge equivalent" as the terminology is used here, but that this *does not* correspond to the usage in classical EM, where gauge equivalent states of A and phi on a slice correspond to physically identical situations on that slice, because only E and B are physically real. My whole point is that "gauge equivalent* in GR just does not mean that at all. Two slices with physical quantities on them can be "gauge equivalent" in GR even though the physical situation on the two slices is completely different, e.g. one with plasma and the other with stars. In your language, two slices being "gauge equivalent" just means that they are both slice from the same model and they overlap at spatial infinity. Now there is really, as I noted, no justification for imposing the restriction about overlap at spatial infinity. One might as well just take "gauge equivalent" to mean that the two Cauchy slices are Cauchy slices from the same complete 4-dimensional space-time, and lift the restriction about spatial infinity. That is cleaner and conceptually clearer. Of course, if one of these slices is annihilated by H the other will be too. And so too will other slices from different models. The gauge equivalence really arises because one can be generated from the other by applications of shift and lapse.

Now, everything goes OK until this sentence: "The key point: since the wavefunction is annihilated by the operator defined above, the wavefunction necessarily takes exactly the same value on a diffuse cloud of particles as on the configuration where they have coalesced to form the earth. Mathematically this is required since these configurations lie in the same gauge equivalence class." This is just a mistake, as is obvious. The fact that two wavefunctions are annihilated by the same operator in no way implies, mathematically or other wise, that the wavefunctions are the same! Nor is this required because they lie in the same "gauge equivalence class" given how you have decided to define "gauge equivalence class". Here you are being misled by the use of the word "gauge". In classical EM, two configurations of A and phi that are gauge equivalent are in every way physically equivalent because they agree exactly on E and B, and only E and B are real. But, as we have seen, the state of a slice that passes through dust is just not at all physically the same as one that passes through the solid earth. The *word* "gauge" does not have the magic ability to just make this case like the EM case, and it isn't.

Con't...

Then comes all this stuff about the "clock variable". This is, as you say, standard stuff. I am quite familiar with it. It evidently makes no sense at all. You write as if we can distinguish the slice that goes through the dust from the one that goes through the earth—even those these two states are "equivalent in the mathematical sense"!—by looking for correlations with a "clock variable". Now please do not just repeat this, or say that this is what all your books say. Stop for a moment and try to figure out what in the world this could mean.

ReplyDeleteFirst: if the two states are equivalent in the sense you defined above—" the wavefunction necessarily takes exactly the same value "—then obviously the values of any "clock variables" or any sort of variable at all will be the same on the two slices! Similar for any "correlations between variables": mathematically identical wavefunctions cannot take different values for anything.

Second: In fact, there were no clocks around in the dust from which the earth formed. Not in any literal sense of "clock".

Third: even if there were such a clock, why are you bringing it into the discussion? If you can somehow, by looking at the wavefunction, distinguish whether a clock is in one state or another, then why not just look directly for whether the matter is in the form of dust or in the form of a big solid ball? Presumably that would be a lot easier to tell then trying to pick the time off a clock if there were one.

So all of this "clock variable" stuff gets you nowhere. And if you insist that the wavefunctions on the two slices are the same, there is nowhere to go anyway. That would mean that nothing physically changes between the two slices, so there is no physical change in the inhabited universe. You have the problem of time in spades.

Now: in the Penrose diagram I have been using, every Cauchy surface will be part of a “gauge equivalence class” that saturates the bulk: every event in the bulk lies on many, many members of the class. I am not entirely sure if that is true in AdS or not. If it is, then I don’t see how moving from one boundary state to another corresponds to any change at all in the bulk: no matter what “time” it is on the boundary, If not, then at least there must be a lot of overlap: many events passed through by Cauchy surfaces in the equivalence class associated with one time are just as well passed through by Cauchy surfaces associated with a later time. So how can “passage at the boundary” in any way generate or correspond to passage in the bulk?

You final mathematical fact seems to require that nothing in the bulk change. Good luck with that physics.

Tim,

ReplyDeleteI wish we were making progress, but apparently we are not since you write

" This is just a mistake, as is obvious. The fact that two wavefunctions are annihilated by the same operator in no way implies, mathematically or other wise, that the wavefunctions are the same!"

You are arguing with yourself here, since I certainly claimed nothing of the sort. What I did say was the following:

If the generator of a symmetry annihilates a physical state wavefunction, then the wavefunction takes the same value on all points of the symmetry orbit. Obviously. In the case at hand, the generator is \int N(x) H_t(x) d^3x with N(x) going to zero at infinity. This operator annihilates physical state wavefunctions since H_t is a constraint. Points on the same orbit are configurations related by a local time evolution that keeps the surface at infinity fixed. These configurations are by definition gauge equivalent, since the definition of gauge transformations in the canonical formalism are transformations generated by the first class constraints, of which H_t(x) is one. How is this in any way controversial?

Contrast this with the case where N(x) does not vanish at infinity. Then the generator has an extra surface term, as Regge and Teitelboim showed, and this surface term does not annihilate physical state wavefunctions, and so the wavefunction does not take the same values on configurations related by these large gauge transformations.

I won't respond to the rest of your post right now, as there is no point if you don't understand these basics. It would be great if you respond to what I actually say rather than some distorted version thereof. Also, I earlier went through the trouble of giving a pedagogical explanation of gauge transformations and their generators in E&M to help you understand these issues, but that appeared to have little impact other than generating a polemic based on confusing Gauss' theorem with Gauss' law. Do you disagree with what I wrote there? I suggest you go back and read my explanation there if you are still confused and then we can continue the discussion.

BHG,

ReplyDeleteI am at present having a discussion with another physicist about the bearing (or otherwise) of AdS/CFT on my paper. He is making almost verbatim the same arguments are you, but because we have been able to exchange more quickly the progress is faster. So far, every version of an argument that he has tried to formulate has been easy to refute, but I'll see if he can come up with something new. If not, I will just sum up what he tried and you can see if you have anything else.

There are several key observations about what you have written. Start here:

If the generator of a symmetry annihilates a physical state wavefunction, then the wavefunction takes the same value on all points of the symmetry orbit. Obviously. In the case at hand, the generator is \int N(x) H_t(x) d^3x with N(x) going to zero at infinity. This operator annihilates physical state wavefunctions since H_t is a constraint. Points on the same orbit are configurations related by a local time evolution that keeps the surface at infinity fixed. These configurations are by definition gauge equivalent, since the definition of gauge transformations in the canonical formalism are transformations generated by the first class constraints, of which H_t(x) is one. How is this in any way controversial?"

First, you admit that it in no way follows that any pair of eigenstates of the same operator are the same state. Then you say that the "wave function takes the same value on all points of the symmetry orbit". Now the plain meaning of that claim seems to be that the all of the points in the symmetry orbit have the same wave function, which means that there is no orbit at all. Why in the world think that applying the Hamiltonian to a state yields *the very same state*? It would, in any case, only have a chance of being true for stationary states, which are not our concern.

Then you say that the configurations generated from each other by the Hamiltonian are "by definition" gauge equivalent. What sort of a "definition" is this? Just a convention for using words? Then nothing at all follows. And if all of the states are the same, then of course they are gauge equivalent: they are the same state!

You seem to be trying for the following sequence of inferences: psi is generated from phi by the action of the Hamiltonian, THEREFORE psi and phi are gauge equivalent, THEREFORE psi and phi are the same wavefunction or at lest are "physically equivalent". That is obviously silly. Especially if the first "therefore" is justified by a convention. Simply *calling* two states gauge equivalent means nothing. And the state of a Cauchy surface that cuts through the solar system before the planets have formed is obviously not "gauge equivalent" to the state of a Cauchy surface that cuts through after the planets have formed. Those are physically quite different states. Is that at all controversial?

Tim,

ReplyDeleteClearly my point is still not getting across, because there is zero correspondence between what I write and your interpretation thereof. I will try again, this time even more slowly . I assure you that everything I say below is totally standard knowledge, but until you understand it you don't stand a chance of appreciating my argument that rules out your scenario.

1) Let's start with single particle non-relativistic quantum mechanics and discuss the symmetry of translation invariance. There is a symmetry generator p, the momentum. THat p is the generator of translations is the statement that [\epsilon p, f(x)] = f(x+\epsilon) - f(x). ( Here x and p are operators of course). \epsilon is infinitesimal, and a finite translation is obtained by exponentiation. Two particle locations, x and x+\epsilon, are thus on the same orbit of the symmetry (in this case there is only one orbit). Now, suppose the system is in a state described by wavefunction \psi(x), and suppose that the symmetry generator p annihilates the wavefunction, p \psi(x) = 0. It follows that \psi(x+\epsilon ) = \psi(x). That is, it follows that the wavefunction takes the same value on all points of the symmetry orbit

2) Now consider gauge transformations in E&M. As above, we have a symmetry generator, which in this case is the operator X = \int d^3x \grad(\Lambda) . E = - \int d^3x \Lambda(x) div(E) (plus a matter term if charged particles are present). Here I am assuming that \Lambda goes to zero at infinity so that I can integrate by parts without a boundary term (we'll come back to that in a moment). That X generates a gauge transformation is the statement that [X, f(A_i)] = f(A_i + d\Lambda/dx^i) - f(A_i). Two gauge fields A_i and A_i + \delta A_i are on the same orbit if \delta A_i = d\Lambda/dx^i for some \Lambda(x) that goes to zero at infinity. The canonical formalism implies that the symmetry generator X is a constraint, so a physical state wavefunction must obey X \psi=0, which is mathematically equivalent to \psi(A_i + d\Lambda/dx^i) = \psi(A_i) for any \Lambda(x) that goes to zero at infinity. That is, the wavefunction takes the same value on all points of the symmetry orbit. Finally, if I allow \Lambda(x) to be nonzero at infinity then X picks up an extra surface term and this term need not annihilate physical state wavefunctions. So wavefunctions need not take the same value on gauge field configuration related by such a large gauge transformations.

cont

cont

ReplyDelete3) Now we come to GR. Mathematically the situation is no different than in E&M. Let's leave interpretational issues aside for the moment, so that we can make sure the mathematics is understood. The symmetry generator is X = \int d^3x N(x) H_t(x) with N(x) =0 at infinity. Instead of the gauge field we have the spatial metric g_{ij}(x) and the locations x_a of any matter particles (a labels the particle). What are the symmetry orbits? You just work this out by computing the commutator X with a function of the metric and particle operators. You of course find that two metric/particle configurations are on the same orbit if they are related by a classical local time evolution that goes to zero at infinity. Just work out the commutator and see this for yourself. Now, suppose that a specified asymptotic time the system is in a state corresponding to the wavefunction \psi( g_{ij}, x_a). For this to be a physical state wavefunction it must be annhilated by X: X \psi = 0. Just as in E&M, this equation is mathematically equivalent to the statement that \psi takes the same value on pairs (g_{ij}, x_a) and (g_{ij} +\delta g_{ij}, x_a + \delta x_a) if they are on the same orbit of the symmetry, i.e if they are related by a local time evolution that goes to zero at infinity.

I don't know how to make it any clearer than that, so please give this your best shot. I understand now that you were not familiar with terms like "symmetry generator", "gauge orbit" etc, but hopefully you are now.

BHG-

ReplyDeleteYes, we do have to get down to basics here, but I'm afraid that it is you who have the learning to do. All of these examples are quite different, and the fact that you would just line them up shows that you have not really thought deeply or clearly about any of this.

I will start with your first example, just to make the point. Maybe once you have absorbed that, we can make some progress. You write:

"1) Let's start with single particle non-relativistic quantum mechanics and discuss the symmetry of translation invariance. There is a symmetry generator p, the momentum. THat p is the generator of translations is the statement that [\epsilon p, f(x)] = f(x+\epsilon) - f(x). ( Here x and p are operators of course). \epsilon is infinitesimal, and a finite translation is obtained by exponentiation. Two particle locations, x and x+\epsilon, are thus on the same orbit of the symmetry (in this case there is only one orbit). Now, suppose the system is in a state described by wavefunction \psi(x), and suppose that the symmetry generator p annihilates the wavefunction, p \psi(x) = 0. It follows that \psi(x+\epsilon ) = \psi(x). That is, it follows that the wavefunction takes the same value on all points of the symmetry orbit."

This whole paragraph is really confused. You say that in what you write x and p are operators. Well, no. Every x in that paragraph is not an operator, it is a variable. p is an operator, x isn't. So you have to stop right now and reverse your attitude. Maybe we can sort you out here and maybe not: I have no idea. But let's see if we can clear up this paragraph, and then go on from there.

The relevant symmetry here is the spatial translation symmetry that exists because the spatial structure in the non-relativistic theory has that symmetry. Three-dimensional Euclidean space is symmetric under translations of any amount in any direction, rotations of any amount around any axis, as well as some discrete global symmetries like reflection. On the case of the continuous symmetries, like translation and rotation, that can be indexed by a continuous parameter, the symmetry operations form a Lie group, and there is a generator of that group. What the symmetry of the space means is that if you take any solution and operate on it with a symmetry operator, you get another solution. If the solution itself happens to also have the same symmetry, then you get the same solution. A "gauge orbit" is a set of solutions that can be generated from one another by the action of the symmetry operators. In the case of a Lie group, this set of finite symmetry operators can be derived from the generator by exponentiation, which yields a set of operators indexed by a parameter. Thus: the generator of the set of operators that implement translating the state by some distance d in the x direction are generated by a generator of infinitesimal translations in the x-direction which is exponentiated. That generator is the x-momentum operator, p_x.

The sentence "Two particle locations, x and x+\epsilon, are thus on the same orbit of the symmetry (in this case there is only one orbit)' Is just a complete mess. Two locations in space, for example (2, 3,4) and (3,3,4) (in the relevant Cartesian co-ordinates) are not "on the gauge orbit". That makes no sense at all: the orbits are collections of wavefunctions, not collections of locations. Did you mean Dirac delta functions concentrated on those points? Or what? In any case, the action of a x-transaltion operator for some finite distance d should take the function f(x,y,z) to the function f(x + d,y,z). Once again, the 'x' here is *not* an operator: it is a variable. If there is more than one particle then the function is a function over configuration space and all the x-values get shifted by d.

Con't

Con't

ReplyDeleteOK, so we have the p_x operator which is the generator of a group of symmetry transformations. Most states are not, of course, invariant under the transformations. That is, take a wave function and compare it to the same function shifted by a distance d in the x-direction. In general, the new and old functions are different. But in some cases they are the same, up to a multiplicative factor. In such a case the wave function is an eigenfunction of the operator and the multiplicative factor is the eigenvalue.N

Now we get this: " Now, suppose the system is in a state described by wavefunction \psi(x), and suppose that the symmetry generator p annihilates the wavefunction, p \psi(x) = 0. It follows that \psi(x+\epsilon ) = \psi(x). That is, it follows that the wavefunction takes the same value on all points of the symmetry orbit."

Now this is a strange thing to write. If the "symmetry operator" p "annihilates the wave function" psi(x), all that means is that psi is an eigenstate with eigenvalue 0. That is, 0-x--momentum eiqenstate. And sure enough, the 0-x-momentum eigenstates, alone of all states, literally take the same value at (x,y,d,) as they do at (x + d, y z). None of the other momentum eigenstates do that, and of course none of the physical solutions that interest us have that feature. So you are confusing the space-time having a symmetry, which it does no matter what, and the set of solutions having an orbit where the members of the orbit are related by the symmetry, with the issue of whether a particular state has the symmetry. If a state has the symmetry, then there is no "gauge orbit". The state maps to itself under the symmetry.

In the cases we are looking at, the possibility of the last kind is uninteresting. In particular, if a state were time-symmetric in this sense then it would never change.

Do you see all of this now, or can we move on to your second example?

Tim,

ReplyDeleteSo basically you are now saying that you don't even understand the quantum mechanics of a single non-relativistic particle? I started with the simplest conceivable example in the hope that you could follow, but apparently it was not simple enough, even though anyone (and I mean anyone) who has ever cracked open a book on quantum mechanics and understood anything could follow this. You write

"This whole paragraph is really confused. You say that in what you write x and p are operators. Well, no. Every x in that paragraph is not an operator, it is a variable. p is an operator, x isn't. So you have to stop right now and reverse your attitude. Maybe we can sort you out here and maybe not: I have no idea. But let's see if we can clear up this paragraph, and then go on from there. "

What in god's name are you talking about? I am dying to see you defend this statement. I write (obviously suppressing factors of i and hbar) [\epsilon p, f(x)] = f(x+epsilon) - f(x). This is OF COURSE an operator statement, with x and p being operators. Do you not even know the most basic of all QM formulas [p,x]=-ihbar, where x and p are operators? Commutators in QM are statements about operator multiplication. When I put in the comment about x and p being operators in reference to this equation, I worried that I was being condescending by assuming that you might get confused by the fact that in this equation x and p are operators whereas in the wavefunction \psi(x) x is not an operator but an eigenvalue of x. Now I see that I was assuming too much knowledge on your part. Are you really not able to follow that in the above x and p are operators, whereas in the wavefunction \psi(x), x is an eigenvalue of the x operator?

Anyway, for comic relief, please do elaborate on your confident assertion that in the equation, [\epsilon p, f(x)] = f(x+epsilon) - f(x) , x and p are not operators. This should be good.

And obviously I am not going to waste my time reading the rest of your post until we can get to the point where you understand the distinction between an operator and an eigenvalue.

BHG-

ReplyDeleteGood lord, you are joking, aren't you? I didn't hit you on suppressing the i and the h-bar. You are systematically mixing up the operator X with the variable x in that paragraph, and it has completely confused you. Since hats aren't available, use capital X for the operator and x for the variable and see if you can sort out what you have written.

Look: why are you even writing down f(X)? Which function of the X operator do you have in mind? Functions of the X operator are not very common in non-relativistic QM. On the other hand, the wavefunction, which is what I assumed you had in mind, is not f(X) but f(x). And just to be clear, the x in psi(x) is not an "eigenvalue of X"! an eigenvalue of X is a real number. The x in psi(x) is a variable. That's why you can differentiate the thing. So maybe you should crack a basic calculus book some time.

So your equation [\epsilon p, f(x)] = f(x+epsilon) - f(x) just looks pointless as an operator equation. I charitably assumed that what you meant to get to was epsilonP f(x) = f(x + epsilon) - f(x) (suppressing i and h-bar). That is at least a relevant equation, since f(x) can stand for the wavefunction. That gives us an infinitesimal translation and we can then expotentiate to get finite translations. And you understand, in this formulation, what f is doing there: it is standing in for psi. If I try to do that with your equation, then I somehow get not a psi function but a psi operator. You claimed you were talking about "single particle quantum mechanics", and in that theory psi is not an operator of any sort: it is the thing the operators operate on.

So no, I was not assuming the x and p were neither one operators, but that p was the operator P, what you really had in mind by f(x) was psi(x), and the whole commutator part was just a confusion that I set aside to make progress, fixing up the left side of the equation in my mind to have it make sense. The meat of the comment was what comes after that, of course, where the variable/operator confusion hit a crescendo.

So apparently that is not what you had in mind. OK: then do enlighten me. What in the world is the *operator* f(X) doing there,and what is the relevant f and what does this commutator have to do with anything? Commutators of generators of Lie groups have interesting commutation relations. But f(X), for some unspecified f, is obviously not a generator of anything. So what f do you have in mind here? A potential function? That is about the only operator of the form f(x) that I can think of arising at all. If you are not trying to represent psi there, what are you trying to represent?

As you say: this should be good.

Dear BHG,

ReplyDeleteI admire your patience with Tim. I would have given up long ago. Here again we have a low point in the "discussion"; the fact it isn't clear to Tim from the statement [ep,f(x)]=f(x+e)-f(x) that all x's here are operators (otherwise the commutator would make no sense) is amazing. Now he is right, though, that later in the same paragraph you keep using the same symbol, x, even when you mean it as an eigenvalue. All I can suggest is to use different symbols (say, denote the eigenvalue of the operator x by X) so as not to confuse poor Tim.

More importantly, it seems to me Tim often does his best to find the least charitable interpretation of what you write. That makes the whole discussion cringeworthy. Again, I would simply give up.

Chai tea,

ReplyDeleteIf you are paying attention at all, then learn something. Eigenvalues are not variables, they are numbers. If you want to take up the challenge I gave BHG, feel free. What in the world is the unspecified function f in that equation, and why it is relevant to the fact that P is the generator of spatial translations? BHG has made a hash of this whole thing. try to make sense of his entire paragraph, and report back.

Tim,

ReplyDeleteIt's true for any function f.

Sabine

ReplyDeleteI am not questioning its truth, I am questioning its relevance. I have provided what I take to be a quite standard account of P as the generator of the translations without ever mentioning some operator f(X). I still can't make out what f(X) has to do with anything here. I know what psi(x) has to do with the question, but not, of course, psi(X). And why the commentator with F(X)?

If there is some alternative way that this bears on generating a set of symmetry operators, I'm all ears. But it looks like just the first step in a series of confusions fueled by running together the operator X, the variable x, and the eigenvalues of X, ie. the real numbers.

Tim,

ReplyDeleteFor all I can see it's just a different way of saying that p is the generator of translations.

Sabine

ReplyDeleteTranslations of what? States or operators? At the end of the day, BHG talks about the translation of the state psi(x), not of any operator f(X). So what's the connection with f(X)?

Tim,

ReplyDeleteLook, I haven't followed the last 400 or so comments in this thread, but I can't see anything mysterious going on here. Last time looked at a qm book, the translation operator acted on states and I assume it still does. The function f is totally irrelevant, just ignore it if it bothers you. Put in the identity if you want.

How is this confusing? Take f(X) = X^2. Then just calculate: [- i epsilon * P, f(X)] = 2 eps X = f(X+epsilon * I) - f(X) up to O(epsilon^2). The finite translation generator is exp(-i epsilon * P), and -i epsilon * P is just its expansion for small epsilon. One can easily show that [exp(-i epsilon * P), f(X)] = f(X + epsilon I) - f(X) exactly, for any epsilon, for any function f.

ReplyDeleteIn this equation X is an operator, I is the identity and epsilon is an infinitesimal real number. This is undergrad-level quantum mechanics. The action of the generators the wavefunction follows from these properties -- the expressions are even given on Wikipedia (https://en.wikipedia.org/wiki/Symmetry_in_quantum_mechanics#Symmetry_transformations_on_the_wavefunction_in_non-relativistic_quantum_mechanics).

Part of me wants to believe that Tim is just trolling, which would be far less depressing than the alternative. The BHG is a brave soul.

My bad, and worth a correction since we are trying to minimize confusion: the generator is exp(i epsilon P), not exp(-i epsilon P), and it is [i epsilon P, X^2] that is equal to 2 epsilon X up to O(epsilon^2).The rest is the same.

ReplyDeleteAnd just to finish off what I (unfortunately) started: the translation operator T(epsilon) = exp(i epsilon P) also translates the position-space wave function psi(x) = . The wave function translated by epsilon in the x-direction is psi(x+epsilon) = . You can see that this is the right formula by calculating = delta(x' - (x+epsilon)): this implies T(epsilon)|x> = |x+epsilon>. Here the X-eigenstates |x> are defined as X|x> = x|x>, where x is used to label both the eigenvalue and the eigenstate. If you are unfamiliar with these facts you should do the calculation. This is really basic stuff.

ReplyDelete" THat p is the generator of translations is the statement that [\epsilon p, f(x)] = f(x+\epsilon) - f(x). ( Here x and p are operators of course). \epsilon is infinitesimal, and a finite translation is obtained by exponentiation."

ReplyDeleteIf x is an operator, and \epsilon is an infinitesimal what? in order to make sense of the sum ( x + \epsilon )? Further is \epsilon p a product of two operators?

I do understand the physicists' shorthand; but it is a problem whenever conceptual clarity is essential.

Dear Tim,

ReplyDeleteYou write

"If you are paying attention at all, then learn something. Eigenvalues are not variables, they are numbers."

Please pay attention, you might learn something:

psi(X) = , i.e., psi(X) is simply the position representation of the state |psi>.

X here is the eigenvalue of the operator x: x|X>=X|X>.

Now, I give up.

Dark Star

ReplyDeleteOne more time: I am not questioning the accuracy of the equation, but its relevance. You go through the derivation, which I am not interested in, and then end with "The action of the generators ['on", I assume] the wavefunction follows from these properties". If you are interested in the action on the wavefunction—which you are—and the wavefunction is not an operator—which it isn't—then starting with the action on an operator is at the least a peculiar thing to do. Exactly the same peculiar thing that you completely gloss over with the single word "follows". I am not trolling, and you haven't told me anything I don't already know. I am trying to make a point that you have to be careful here about keeping your operators, your functions and your numbers apart. If you start mixing them up then you are going to get hopelessly confused. BHG, Chai tea, and now you have demonstrated that you can't keep these things straight, and it you don't learn to you will never make any progress.

Just so you and the rest understand: the three examples that BHG have given all have a different character. The first one concerns a *physical symmetry*, namely the translational symmetry of *physical space*. Euclidean space and Minkowski space-time have lots of symmetries and these symmetries are connected to important conservation laws: conservation of linear momentum, angular momentum and energy. It is essential to understand these connections because you will eventually be working in space-times that no longer have these symmetries, and the principles that depend on them no loner hold.

The second example involves a *gauge symmetry*. In classical EM, that is not a physical symmetry at all. It is a purely mathematical artifact of using a representation of the electromagnetic situation that contains non-physical ("gauge") degrees of freedom. Because the word "gauge" is first learned when dealing with scalar and vector potentials in classical EM, there are a lot of features of things classed "gauge symmetries" that one can unconsciously expect, just because some degree of mathematical freedom in a representation is being called "gauge". This is an especially tricky case because, as the Aharonov-Bohm effect shows, the scalar and vector potentials have a completely different status in quantum theory than they do in classical EM. So already it is very easy to get confused between classical and quantum treatments of the potentials.

The third, and critical, example involves diffeomorphism invariance, which many people refer to as a "gauge symmetry", in virtue of which it is often said that in quantum gravity (e.e. Wheeler-deWitt) the dynamics becomes "pure gauge". Now unless you have been very precise and meticulous about the difference between a *physical symmetry" and a "gauge symmetry* and the difference between the status of the potentials in classical theory and in quantum theory, as well as having thought about what diffeomorphism invariance means, you are certain to get confused at this point. So I am trying to get BHG (and Chai tea, and now you) to attend to some distinctions that you have been running together in your minds.

Con't

Con't

ReplyDeleteIf you want to understand these exchanges, you have to start with the assumption that there are things I understand better than you do. I have spent my career thinking about foundational issues. You haven't. You haven't even ever been taught about what they are, much less how to think clearly about them. To be even more direct, a physics education not merely does not teach foundations, it positively discourages students for asking foundational questions. Physicists are sloppy about their math and sloppy about their concepts, and end up being very, very sloppy about their physics. So instead of sniping, try to learn something. Begin by sorting out what is an operator, what is a variable, and what is a number. Then do the exercise of trying to clean up and make sense of BHGs first paragraph.If you put some effort into this you will learn something.

Tim,

ReplyDeleteSorry to add to your pain, but in the hope of contributing to the progress of this miserably endless discussion, it really is undergrad stuff. You always start with defining the momentum operator by its commutation relation with the position *operator*. The two are conjugated to each other. You define the position space representation by the action of the position operator on the wave-function and this defines both the eigenstates and the eigenvalues. Then you can calculate how the momentum operator acts on the state, by which you learn that it's the translation operator.

You can do the same thing the other way round if you want to. Either way, there is nothing incorrect or fishy or even inaccurate here.

I'm not sure why bhg used a definition that's an epsilon more difficult than that and that threw you off track. Quite possibly he leans more towards sadism than masochism, as I originally assumed. In any case, I hope you can get over this and move on.

Sabine,

ReplyDeleteThat last comment to Dark Star applies to you as well. You have also never studied foundations and make rookie errors, rather embarrassing ones to be honest, whenever you drift into foundations. If you want to actually learn something, do the exercise. Go back to BHG's three examples and try to put them in al least vaguely correct and precise form. I doubt that there is one physicist in 1,000 who really understands what difffeomorphism invariance is and what physical significance it does (or does not) have. And this has helped make discussions of quantum gravity a complete disaster. This is serious stuff, Either take it seriously or don't act as if you understand it.

Tim,

ReplyDeleteI have now had several exchanges with you and I have observed the following strategy that you pursue when you make a mistake.

a) Deny the hell out of it. This sometimes works because your argumentative style is so unnecessarily verbose that few make the effort of digging through it.

b) If a) doesn't work, bring up plenty of other topics for distraction. Like you just did when you were referring to some three points someone has made well above, which neither I nor dark star nor the Chai Latte said anything about.

c) Retreat to personal insults.

I am summarizing this merely so you know it's obvious.

Sabine

ReplyDeleteSo what mistake it is, exactly, that I have made and am denying? I never denied a thing about the standard commutation relations, the stuff you call "undergrad stuff". What I have said from the beginning is the BHG is being sloppy about distinguishing operators, variables and numbers. And that showing that P is the generator of spatial translations for states can be done much more straightforwardly—and standardly is done much more straightforwardly, than BHG has done it, which is not a good sign. In fact, it is still not very obvious how to get from that commutation relation to the desired result.

But to be exact: what mistake precisely have I made? Please be explicit and careful here.

And, on the topic of personal style, do not respond that this reasonable request is "bullying" you. You've used that one twice on me already when you made an error, and it is extremely annoying.

Tim,

ReplyDeleteYou mistakenly think there is something wrong with the equation the bhg wrote down, as demonstrated for example in your sentence

"your equation [\epsilon p, f(x)] = f(x+epsilon) - f(x) just looks pointless as an operator equation"You have then tried to deny you ever said something like that and instead written that it's not the equation itself you doubt but its relevance. It's possibly the case that it isn't relevant for the discussion at hand but I wasn't commenting on that and have no intention to comment on it now.

I am sure I never said or wrote anything to the extent that you are bullying me (or anyone else for that matter) because it's not a word I commonly use. You seem to have simply made up this quote, or else please provide a source.

The crucial equation in my post got messed up: I wrote

ReplyDeletepsi(X)= bra(X) ket(psi)

i.e, psi(X) is just the position representation of |psi>.

Tim,

ReplyDeleteI did say "this should be good" and indeed you did not disappoint. Incredible. So you are apparently proudly standing by your statements " Every x in that paragraph is not an operator, it is a variable" and " your equation [\epsilon p, f(x)] = f(x+epsilon) - f(x) just looks pointless as an operator equation". I admire your persistence.

Thanks to Bee, Chai Tea, and dark star for confirming that I am not insane.

Since I can't resist twisting the knife, I will rehash what others have patiently explained to you. The only good that has come out of this is that I now realize I should use capital letters for operators, even though it is usually obvious to anyone by context when something is an operator or not. First of all, [\epsilon P, f(X)] = f(X+epsilon) - f(X) does of course make perfect sense as an operator equation,and is furthermore obviously correct, no matter how confidently you assert otherwise. Somehow, in your confused state you were trying to think of f(x) as a wavefunction, but that is really dumb since there is no sense in talking about the commutator of an operator with a wavefunction. That's like talking about the commutator of a matrix and a vector -- it's crazy talk. So, even with my explicit statement that X and P were operators, you chose instead to veer off into a bizarro interpretation and then went into a downward spiral of confusion from which you have not yet recovered.

I will now partially restate my QM example, although everything I said originally is completely correct, as well as perfectly clear in my opinion.

The reason I wrote down that equation is that it is the definition of p being the operator that generates translations. I can restate this in other ways, but you'll probably not be able to grasp this either unless you try to learn some basic QM. Repeating some of what has appeared in other comments (and not paying attention to signs, i or hbar, although I can add those in if requested) we define a translation operator T(a) = e^{iPa}. To say that this is a translation operator means, *by definition*, that T(a) f(X) T^{-1}(a) = f(X+a) for any function f(X). Now take a= \epsilon, expand to first order in \epsilon, and you will get the equation I wrote.

Now, if for some reason which we need not specify, a state |\psi> obeys P|\psi> = 0 then it follows from the above that the wavefunction for the state obeys \psi(x+a)=\psi(x). Any competent person can show this. This of course a trivial example, but the point here is to set it up in a way that will admit immediate generalization to more complicated cases.

What have we learned? The operator P generates a 1-parameter flow (an orbit) which translates the position. If P annihilates a wavefunction then the wavefunction takes the same value on all points of the orbit. That's the general story: if the generator of some symmetry annihilates a wavefunction, then the wavefunction takes the same value on all points of the symmetry orbit. This was just meant to establish some elementary points before turning to the case of gauge symmetries. One thing at a time.

Do you know see that [\epsilon P, f(X)] = f(X+epsilon) - f(X) is a perfectly correct equation with X and P being operators, or you are you still struggling to comprehend?

** By the way, in \psi(x) it is perfectly accurate to say that x is an eigenvalue, because the entire meaning of the wavefunction is that \psi(x) is the amplitude for a measurement of X to return the eigenvalue x. But let's not waste our time with this.

My matrix elements got cut off: my last post should have read

ReplyDelete"And just to finish off what I (unfortunately) started: the translation operator T(epsilon) = exp(i epsilon P) also translates the position-space wave function psi(x) = . The wave function translated by epsilon in the x-direction is psi(x+epsilon) = . You can see that this is the right formula by calculating = delta(x' - (x+epsilon)): this implies T(epsilon)|x> = |x+epsilon>. Here the X-eigenstates |x> are defined as X|x> = x|x>, where x is used to label both the eigenvalue and the eigenstate. If you are unfamiliar with these facts you should do the calculation. This is really basic stuff."

Hopefully you won't be confused by the difference in notation between my post and Chai Tea's, we both have made ours clear.

The relevance is that the BHG is trying to explain to you how gauge symmetries work in quantum mechanics, since the quantization of gauge theories leads to various standard properties (such as Gauss's law) that you do not understand at all and are relevant to your "solution" to the info paradox. Understanding these properties requires knowledge of how symmetries (both physical and gauge) in general work in quantum mechanics in the first place, which you apparently lack.

But as Sabine points out, you seem mostly interested in the rhetoric, not the science; it really does seem pointless to try. If you dropped your hubris for a moment you might learn something, but so far we have learned far more about your technical inability than you have learned about physics. Your combination of confidence and incompetence is truly astounding. I know you will try to turn this around or engage in some form of whataboutism, as usual -- I do not intend to respond to anything like that, you have wasted enough people's time.

"you have to start with the assumption that there are things I understand better than you do. I have spent my career thinking about foundational issues. You haven't. You haven't even ever been taught about what they are, much less how to think clearly about them. To be even more direct, a physics education not merely does not teach foundations, it positively discourages students for asking foundational questions. Physicists are sloppy about their math and sloppy about their concepts, and end up being very, very sloppy about their physics. So instead of sniping, try to learn something. Begin by sorting out what is an operator, what is a variable, and what is a number. Then do the exercise of trying to clean up and make sense of BHGs first paragraph.If you put some effort into this you will learn something."

This is hilarious. You think you work on foundations, yet do not even understand how symmetries work in quantum mechanics. As has been made abundantly clear throughout this exchange, physicists on the whole are far less sloppy than you (the psi(x) thing might have been BHG's single sloppiest statement in these many months, and it was an abuse of notation that is clear to anyone who knows basic quantum mechanics. since it was not clear to you Chai and I cleared it up). You do indeed seem to work on foundations of a sort: namely, anything that can be expressed with rhetoric and without understanding.

Argh, I see, it's thinking that my inner products are HTML. The same thing must have happened to Chai Tea. I'll use (x|y) for Hilbert space inner products.

ReplyDeleteShould read

"And just to finish off what I (unfortunately) started: the translation operator T(epsilon) = exp(i epsilon P) also translates the position-space wave function psi(x) = (psi | x). The wave function translated by epsilon in the x-direction is psi(x+epsilon) = (psi | T(epsilon) | x). You can see that this is the right formula by calculating = delta(x' - (x+epsilon)): this implies T(epsilon)|x> = |x+epsilon>. Here the X-eigenstates |x> are defined as X|x> = x|x>, where x is used to label both the eigenvalue and the eigenstate. If you are unfamiliar with these facts you should do the calculation. This is really basic stuff. "

Chai tea,

ReplyDeleteDo I really have to point out how what you have just written, supposedly to set me straight, is mathematical nonsense?

You write "

psi(X)= bra(X) ket(psi)

i.e, psi(X) is just the position representation of |psi>. "

And before that

"X here is the eigenvalue of the operator x: x|X>=X|X>."

There is no such thing as " the eigenvalue of the operator x". Strictly speaking, the operator x (warning: I am using your notation in which x is the operator, not mine in which X is the operator) has no eigenstates and hence no eigenvalues. What we often pretend is an eigenstate of x is a "delta function", which is, properly speaking, not a function at all and not an element of the Hilbert space. I am not being pedantic: this will be important in a minute. The thing you are calling X (which is a confusing name, for another reason we are about to see) is a distribution.

First important point: assuming that that is what you have in mind, there is no unique "eigenvalue of the operator x". x has a continuous spectrum. There are infinitely many such eigenvalues. Hence there is no "state" |X>". That makes no sense. If I follow your notation, then there is are states like |3>, which is really an awkward way to write the distribution delta(x-3), where my "x" here is a *variable*.

I can't make head or tail of "psi(X)= bra(X) ket(psi)" Since you don't seem to have any variables anywhere, I don't know even what you re aiming at deriving.

Now the serious point. If p is the momentum operator, and you are really interested in understanding how p can be the generator of a symmetry or of a gauge orbit, then using delta functions for your states is a really, really bad idea. We can usually figure out what a delta function is supposed to be doing, but you just can't take a derivative of a delta function—because it isn't a function. So this is not going to give any insight into the action of p on states.

I know that physics texts just gloss over these details, but you have written down equations where exactly those details are crucial. Read over what you wrote, again separating the x operator, the x variable and the real numbers. Try to be more explicit about what you have in mind. You will get something out of the exercise.

I have also been wondering if "Tim" is a troll, or perhaps a Russian bot set up to destabilize theoretical physics. It's hard to make sense of this otherwise. For example, take his statement: "Functions of the X operator are not very common in non-relativistic QM.". Could any real person who has looked at a book on QM write this? Doubtful. Recall that there exists an operator called the "Hamiltonian" which in NR QM takes the form H = P^2/2m + V(X), thus displaying the "not very common" feature that Tim mentions. And we also have the "obscure" Heisenberg equations of motion, obtained as dP/dt = -i[P,H] = -i[P,V(X)], which exhibits the supposedly strange feature of a commutator of P with a general function of X. Tim's knowledge of QM seems to be pre 1925.

ReplyDeleteIn all seriousness, I still think Tim and I could have a productive discussion if he could just get over himself. If he would agree to simply ask for clarification when needed rather than going off on one of his bizarre tangents, then I would be happy to return to neutral scientific discourse.

BHG,

ReplyDeleteWell I have to admire your chutzpah. I never denied that the equation was accurate as an operator equation. What I said was that it seems irrelevant as an operator equation, and that I could not see any point in worrying about the commutator of P with the operator f(X). I asked rather directly what f you had in mind and why you were mentioning it. And I explained why I just ignored the whole commutator and figured that what you meant to write down was what you have, at last, actually written down, which is well known and obvious. Notice that you do not even attempt to explain what f you had in mind or what relevance it had. And why you would be interested with the commutator with P.

That epsilonP acts on a wavefunction to generate the infinitessimal difference between f(x) and f(x + epsilon) is rather obvious. Nothing wrong with writing does the obvious: it is good strategy. What is not good strategy is to write down something which—however mathematically correct—has nothing to do with anything, which is what you did. You could just admit that, which you refuse to do, but you tacitly have by correcting the post.

Now since f(x + epsilon) is, to first order, f(x) + epsilonPf(x) (leaving aside i and h-bar), one can exponentiate P to get the generator of finite translations. That is true and easy to say and understand. Your writing down the translation operator and then taking a to be epsilon is getting everything backwards: it is because P is the generator of infinitesimal translations that the exponentiation is the finite translation operator, not the other way around.

So you pompously ask whether I see that your irrelevant equation is correct. Yes, and I see that it is irrelevant. Which was my point.I wrote: "Look: why are you even writing down f(X)? Which function of the X operator do you have in mind? " and I wrote to Sabine "I am not questioning its truth, I am questioning its relevance. I have provided what I take to be a quite standard account of P as the generator of the translations without ever mentioning some operator f(X). I still can't make out what f(X) has to do with anything here." And I wrote to Dark star: "One more time: I am not questioning the accuracy of the equation, but its relevance." How much clearer can I be?

As for your little footnote: no, it is not "perfectly accurate to say that x is an eigenvalue". To be an eigenvalue there has to be an eigenfunction and there are no eigenfunctions of the operator X. We pretend that the "delta function" is a function, but it isn't: it is a distribution. Why you insist on ignoring this is beyond me. I will give you the benefit of the doubt and assume you understand it.

My post went on to note this obscure sentence of yours"Two particle locations, x and x+\epsilon, are thus on the same orbit of the symmetry (in this case there is only one orbit)." go back and read my post if you don't see how weird what you write is. In fact, go back a read my post to see what a clear exposition is. So if you concede that that sentence is ill constructed and want to explain it, let's do that before going on. If you are just going to pretend that you did not make a hash of that whole first example, and learn something from that, then what's the point?

Dark Star

ReplyDeleteSo you, along with Chai tea, along with BHG do not understand that X has no eigenstates, that the "delta function" is a distribution and not a function, and that therefore it cannot be differentiated. But you also boast of your deep understanding. OK: then you do your little exercise. Put in for P what it is, namely d/dx (omitting i and h-bar) and try to solve your equation. And when you get to P|x> do write back and say what that equals. Inquiring minds want to know.

As everyone except Tim knows:

ReplyDelete$\psi(x)=\bra{x}\ket{\psi}$

And right on cue, there's that legendary whataboutism once again in response to Chai Tea. Never mind that this is just a distraction from Tim's... progress towards understanding the rudiments of symmetries in quantum mechanics, let's engage (if only to save others some agony). |3> is the state in which there's a 100% probability of finding the particle at x=3. There is no question such a state is an element of the Hilbert space, even in the most mathematical (axiomatic) sense. Under time evolution, in the absence of potentials, this state evolves to a gaussian probability distribution centered around x=3. The fact that its inner product with another X-eigenstate is a distribution, rather than a function, should tell you that distributions are part of the basic description of quantum mechanics, not that there's some magical gobbledegook that the physics textbooks are lying to you about (and this is even a basis-dependent statement... in the P-eigenstate basis the inner product is (p|x) = exp(ipx) ). But screw the words, there is no conflict whatsoever with any physical prediction: the fact that it is a distribution is just an artifact of trying to demand that a function with support on a set of measure zero square-integrates to one. For any physical question the wave function is effectively smeared over the resolution of the measuring apparatus and so you will get the same answer for any physical question by replacing the distribution by its smeared version, which is an ordinary function. But I guess you would not like such states not to be part of the Hilbert space because of your anti-distributional prejudice. Again with the modifications of quantum mechanics -- spill the beans, Tim, tell us how it is. Teach us some foundations!

ReplyDeleteThank you, Bee, for the succinct summary of Tim's approach. Here we seem to be in stage b. It's just incredible.

Ah, using dark star's way of circumventing HTML traps, and switching X's and x's to conform to everyone else's convention,

ReplyDeletelet me write that everyone here but Tim can make sense of

psi(x)=(x|psi) (*)

And yes, I know (and so does everyone else here) that (x|x')=\delta(x-x') and so |x) is not properly normalized. But that doesn't make (*) nonsense.

And again, you tend to interpret what others write in the least charitable way possible: when I wrote (again, switching conventions to agree with everyone else's)

"x here is the eigenvalue of the operator X: X|x)=x|x)"

I meant, of course, that in that eigenvalue equation, X|x)=x|x), x is the eigenvalue and X the operator, not that the operator X has only one eigenvalue.

And now to address your "serious point": if you think that in order to calculate the effect of the operator P on the state |psi) you have to take a derivative of a delta function when making use of (*), you are mistaken. You just need to evaluate (x|P|psi), which you can do, e.g., by inserting the identity \int dp |p)(p|.

Tim,

ReplyDeleteNice try at revisionist history, but the written record is clear for all to see. I wrote an equation, stating explicitly that x and p were operators in this equation. You then wrote " Every x in that paragraph is not an operator, it is a variable", which is a false statement, as others have pointed out. Now you deny it. Sabine has already nicely summarized your mode of operating, and I have nothing to add in this regard.

Now you write "the "delta function" is a distribution and not a function, and that therefore it cannot be differentiated.". The definitive text in QM is probably Dirac's famous treatise, and if you open it you will find derivatives of delta functions all over the place. So is Dirac also a poor confused soul, his sorry state due to his lack of studying "foundations"? Give me a break.

Since you are still struggling to understand my elementary example, I will repeat it with a little bit of extra commentary to help you out. No one but you seems to be confused about it, and everything I originally said is completely correct. Please try hard to focus on what I actually say here. (signs and i's are glossed over as usual)

We want to establish that if P|\psi> = 0 then \psi(x+a) = \psi(x). Of course the "obvious" way of getting this is to use that in the position basis P = -i d/dx, but the whole point here is that we want to proceed from first principles so that we can generalize to more complicated examples (as I already said). So if you would just follow along the payoff will come soon enough. I will now state a series of mathematical facts. If you disagree with any of them please state your precise mathematical disagreement and I will clarify as needed.

1) Starting from the canonical commutation relations [X,P] = i we would like to define a translation generator. We do this at the operator level, not restricting to the position basis. The *definition* of a translation generator, let's call it D, is that it should obey [\epsilon D,f(X)] = f(X+\epsilon) - f(X) for any function f(X). This is a nice way of writing the definition, because the right hand side shows very explicitly that the argument of any function f is getting shifted by an infinitesimal displacement, but you can of course write this in other ways as you wish. Now, it should be obvious that in fact D= P fits the bill, so P is the translation operator. Not clear? Just ask

2) Let |x> be the state that obeys f(X)|x> = f(x)|x> for any f(X). You could just use f(X)=X if you like, but I find it helpful to take a general f. Then (1+\epsilon P)|x> = |x+\epsilon>. Proof: f(X) (1+\epsilon P)|x> = (f(X) + \epsilon P f(X) + [f(X),\epsilon P] )|x> = (1+\epsilon P )f(X+\epsilon) |x> = f(x+\epsilon) (1+\epsilon P)|x>, where I dropped terms of order \epsilon^2. So indeed, this state is an eigenstate of f(X) with eigenvalue x+\epsilon, which establishes the result. Questions?

3) Suppose the state |\psi> obeys P|\psi> = 0. Then \psi(x+epsilon) = = = = \psi(x). We have now obtained the desired result, \psi(x+\epsilon) = \psi(x). Questions?

Do you see how we started from the fundamental starting point, namely the canonical operator commutation relations, and systematically arrived at the final result. Do you see how the fact that P pushes us along an orbit in which x is translated implies that the wavefunction \psi is constant on the orbit (if it is annihilated by P)? Those are the lessons I want you to absorb, since they will make it technically easy to turn to the case of gauge transformations; the mathematical structure will be the same.

Once again, I am just repeating what I originally said. Please voice your confusions in terms of precise requests for clarification and I will be happy to oblige.

Perhaps going back to the basic issue: whether asymptotically AdS or otherwise, what is the Cauchy surface foliation in a spacetime with a blackhole that evaporates? Do you all agree on that much?

ReplyDeleteBHG, DarkStar, Chai Tea, Sabine, and anyone else

ReplyDeleteWe have reached critical mass. Here is the explosion. We will see if anything survives.

I started on this exchange months ago under one assumption: someone who reads Sabine's blog might know something that bears on my paper. I did that because I did not want the paper to overlook a relevant argument I got some useful feedback and interesting papers to read from comments. And there was the suggestion, initially raised by Dark Star and then taken over by BHG, that AdS/CFT refutes the claim in the paper. I have never pretended to be an expert on AdS/CFT. I have been trying to get information from people who confidently claim to understand AdS/CFT. What has happened since then?

This happened: BHG took over the task of trying to explain both AdS/CFT and its bearing on my paper. In the early days I started to repeat, in every post, the question: "Can you give me a clean and precise statement of exactly what AdS/CFT says?" This is the first question to ask in a properly conducted discussion. After try after try after try, I finally got this from BHG "AdS/CFT is a work in progress". Post of 9:22 AM, June 18, 2017.

One rational thing for me to do, at that point, would be to just write "Call me back when there is an actual thesis to discuss. Then we can investigate 1) whether there are any good reasons to think that the thesis is true and 2) whether it actually has any bearing at all on the argument in my paper". I could have, and probably should have, just ended this whole thing then.

But at least since June 18—for six months—I have been being charitable. Very charitable. Despite the fact that BGH has admitted that he cannot say exactly what AdS/CFT claims, I have proceeded on the assumption that enough has been worked out for it to have some bearing on my paper. And after six months there is, to be quite direct about it, not a shred of evidence that that is true.

One thing that is supposed to follow is that the final state, after the evaporation, is a pure state, because it is in the CFT and then by a duality it is in the AdS gravity theory. And the main point of my paper is that the only thing that one can properly mean by "the state after the evaporation" is "the state on a Cauchy surface that lies to the future of the evaporation event”. If you mean less than that, for example if you mean “The state on some edgeless space-like hypersurface after the evaporation event”, then there is not, and never has been, any reason to expect that state to be pure, or for it to be the product of a unitary evolution from an earlier Cauchy surface, or for it to “preserve all the information” in an earlier Cauchy surface. I never claimed this as a novel observation. Just the opposite, I credit Wald for making it. And it is, in any case, obviously correct. Indeed, what seems to get a lot of physicists’ backs up about my paper is that it is simply works out some straightforward consequences of a this observation. “How dare you, a mere philosopher, suggest that we have for decades overlooked such a trivial observation!” That is about the sum and total of the negative reactions I have gotten about this paper. There are also physicists who have said “Yes, of course that’s right. What has everyone been going on about all these years, with black hole complementarity, and firewalls, and fuzzballs, and ERP = ER? What in the world has gone wrong with foundational physics?”

What is novel in the paper? As far as I know, it is the first time in print that the exact structure of foliations of the space-time into Cauchy surfaces has been identified as such—under the supposition that the Evaporation Event is indeed an event and the space-time is globally hyperbolic. If you remove the Evaporation Event, so there are no Cauchy surfaces at all, then the discussion has to proceed in the slightly (but not substantially) different way.

Con't.

But if you are going to do that, then don’t call Sigma 1 a Cauchy surface in the first place. One thing is clear: Sigma 2_out, which is what every introduction to the supposed paradox is concerned with, is not Cauchy. Everyone who thinks there is a problem is either tacitly assuming it is Cauchy, or has never explicitly asked themselves if it Cauchy, or has some secret argument about why one should expect the state on it to preserve information even though it isn’t Cauchy.

ReplyDeleteAbout the foliation into Cauchy surfaces: although this is the first discussion I know of it in print, Christi Stoica worked this out independently and made a nifty GIF of it. He never bothered to publish. I have been pointing this out in public for a quarter century, and did not publish. I just got fed up when Sabine posted about the problem and did not point this out. What I found out from e-mails with her was that she had never considered that foliation. And that it refutes a claim that she made with Lee Smolin in a published paper. So the “but everybody knows this” meme that Sabine pushed in her original post is, to say the least, ironic.

So where have we gotten in all this time? The relevant claim has been pared down to this: First, we are taking AdS/CFT for granted, not because it has been proven but because those are the terms of the discussion. Second, although there is still no clean statement of what AdS/CFT says, it is supposed to establish enough of a duality to the CFT to conclude that the spectrum of the energy is discrete and (up to some set of global symmetries) non-degenerate. Third, and this is the critical point, BHG is under the impression that if my solution were correct, the AdS energies (by which I mean the ADM mass or the appropriate analog in AdS) must be degenerate. But at this point, the only argument to this effect that I can identify cuts just as strongly against the claim that the eigenstates of ADM mass are non-degenerate in plain vanilla AdS, quite apart from any evaporating black holes. And an argument that kills AdS/CFT cannot be used to critique my argument by reference to AdS/CFT. That is where the discussion is.

Where is the discussion going? On Dec. 23 I wrote:

”As long as we are getting conceptual issues out in the open, let's at least start to talk about the elephant in the room: diffeomorphism invariance. Diffeomorphisms are regarded as symmetries of a gravitational theory so states related by a diffeomorphism lie on the same gauge orbit and represent the same solution. And as a consequence, foliations by different sets of Cauchy surfaces are also regarded as yielding the same solution, up to a gauge symmetry. But what that means is that the picture presented in your last message is conceptually confused”

It looks like the trajectory of the discussion it headed in this direction. And one thing is for sure: diffeomorphism invariance is a quite subtle and difficult issue. It tripped up Einstein, and has been tripping up discussions of quantum gravity as well. Unlike the points I make in my paper, which are quite easy to grasp, coming to terms with diffeomorphism invariance and its implications is a tricky business. Given the level of discussion here, this conversation will simply not be up to the task.

If you think you understand diffeomorphism invariance and its implications, then start with a couple of questions. First, did you ever read Einstein’s original paper that introduces the “hole argument”? Second, have you read the papers of Earman and Norton that reformulate and generalize that argument? (If you don’t know who Earman and Norton are, that is not a good sign.) If the answers to these questions are “no”, then you have no grounding and no standing to comment on these issues except to ask questions. So if this is where this is going, and the answer is “no”, then you have to immediately adopt the attitude that I know a lot more about all this than you do. If you are psychologically incapable of that, just stop now.

Con't

Con't

ReplyDeleteI am just fed up with the arrogant tone of all of you. If you want to make assertions confidently, and even with sharp rhetoric, I am fine with that, but you better be able to back it up. And also to admit when you have made a mistake, whether trivial or important. BHG, Dark Star, Chai Tea and Sabine have all been asking if I really don’t understand that BHG’s commutation equation is correct even after I have written to each of you individually that my issue is not about correctness but relevance. Every single one of you has “conveniently” overlooked (or more accurately: maliciously ignored) my saying this. BGH has not even attempted to explain what relevance that equation has for the topic at hand, but has just rewritten the presentation in an improved way that omits it. Without acknowledging that is what he has done.

(I can be super-charitable at this point and suggest that maybe somehow you have been tacitly thinking in Heisenberg rather than SchrÃ¶dinger picture, in which case the equation might have some relevance. But since the conclusion of the argument is stated in SchrÃ¶dinger picture, this is really a bad way to proceed. BHG quite pompously brings up the Hamitonian in this context, as if the Hamiltonian were of the form f(X). The potential term is, but the Hamiltonian obviously isn’t. And I already mentioned the potential term. But enough if this.)

Now for a confluence of various reasons, which I can explain if anyone is interested, I am drawing a line. It is very simple. If you want to post anything on this thread from here on and expect any response from me, then the first thing you have to do is reveal your actual name and who you are. I have done this from the beginning, and those of you who have not are taking the luxury of anonymity to be obnoxious, to ignore what I have written, to make false claims about me, and to make technically incorrect statements, knowing that none of that can actually come back to you in your everyday life. This has had a poisonous effect all along, and it is worthwhile to see if something useful can come from eliminating that poison.

The most useful feedback I have gotten—mentions of papers I did not know—came from people who post under their own name. That is a correlation worthy of note.

BHG has at least taken the time and effort to engage. He has had the integrity, if he thinks there is something wrong with my paper, to try to explicate what it is. And we have gotten far enough to see that, if there is anything there at all, it is peculiar to AdS and the argument cannot be extended in any obvious way to asymptotically flat space-times. We have been sharp with each other, but I don’t mind that. And I have had the impression that essentially we are just talking to each other. But now, all of a sudden, Dark Star and Chai Tea and even Sabine—who says that she is not even reading the exchange—have to chime in with the false claim that I denied the truth of what BHG wrote, willfully ignoring what I did write, namely that it was irrelevant. And then have the gall to say that I have not been charitable.

So from here on in, this is the rule. Sabine is of course free to comment: she makes her arguments in the open rather than under the cover of a screen name. Anybody else interested: either reveal who you are or don’t expect a response from me. You won’t get one. I myself would prefer that Sabine not even post submissions to this thread that fail this condition, but that is up to her. If anyone thinks they can argue a convincing case why they should be allowed to do this anonymously, you can try to make it and I will consider it. But short of that, I am done. Either things get serious, and people take responsibility for what they post, or it is officially beneath my notice.

Tim,

ReplyDeleteThe claim Lee and I made in our paper was that if you remove the singularity you already know that everything that enters the black hole horizon must come out eventually. This means that a (quasi-)remnant is the most plausible explanation for what happens. We then address commonly made objections to remnants to demonstrate that this option has been discarded without any good reason. I don't see how the slicing matters for that.

Having said that, the point of my blogpost, if you recall that, was that the question isn't answerable by logic alone. We don't have a theory of quantum gravity and so all discussions about the issue eventually come down to questions of taste. You may be the only person in existence who likes the idea that the singularity remains there. Fine. I don't have a particular problem with that. You have not so far explained how the loss of unitarity on the outside doesn't make a problem. I'm not saying that because it worries me, but because that's the obvious objection you'd be expected to answer.

Best,

B.

bhg, dark star, Chai tea

ReplyDeleteLike Tim, I am not a big fan of pseudonymity. It is a hugely unbalanced situation if some of us post under our real name while others don't. If you don't want your real name to appear on the page here you can send me a note and I'll pass it on to Tim, so then everyone knows who is who. My email is hossi[at]fias.uni-frankfurt.de

I'm not sure why Tim thinks he can set the terms of the debate, but this is Sabine's space we've been running all over and so: I would be thrilled to return to pseudonymous mostly silent observation, I was wasting much less time that way. I chimed in only when things reached the point of utter absurdity, thinking that the BHG might enjoy a reprieve.

ReplyDeleteTim, we have nothing whatsoever to gain by convincing you -- BHG and I (and certainly also Chai Tea if s/he was around at the time) understood the flaws in your argument very quickly, especially because as you acknowledge it is not a new one. I would venture that all three of us have encountered, thought seriously about and frankly debunked this argument before. Really it is the BHG who has been charitable, and your inability to recognize this exhausts my vocabulary of superlatives. Your very approach to the scientific process is deeply flawed and I sincerely hope that you are not representative of your discipline.

B.

ReplyDeleteThanks for the response. I do appreciate it.

The question of whether a breakdown of unitarity—or if you will having the transition from Sigma 1 to Sigma 2_out be pure-to-mixed, as Hawking originally proposed with his superscattering matrix—must cause any evident empirical problems has been completely settled. The GRW spontaneous collapse theory breaks unitarity literally all the time, at a massive scale, certainly much more massive than would arise from evaporating black holes. It is known that GRW makes slightly different predictions than standard QM, and there has been a very intensive investigation for years to see if any of those differences would show up in the lab, but so far the theory cannot be empirically ruled out. So there is certainly no quick argument from pure-to-mixed transitions to any empirical difficulties.

On this subject, several people I have discussed this with have been under the impression that there is a paper of Susskind, Banks and Peskin that shows how Hawking's superscattering matrix would yield some unacceptable consequences. But on close inspection, that paper is, well, kind of silly. In short, having no actual theory to analyze, they just make one up out of thin air and criticize it. That paper was out of SLAC. The really interesting thing is that in that paper they reference another SLAC and CERN paper, by Ellis, Hagelin, Nanolpoulos and Srednicki, which investigated the same question, and which they claim to radically improve upon. The Ellis et al paper is quite lovely, and starts out the right way: to check for the loss of purity, you have to be doing delicate interference experiments (of course), which are essentially the polar opposite of high-energy scattering. And in the nice paper of Ellis, et.al., the authors give thanks to one J. S. Bell.

The issue addressed in my paper is actually quite independent of the existence of the singularity: if you want to sew on some extra space-time to continue through the singularity, that's fine by me. It is not the singularity that gives rise to the problem but the event horizon. It is because of the event horizon that things that go into the black hole can't come out again. And in an evaporating black hole the event horizon does this peculiar thing of shrinking down to zero area and disappearing. If that can happened without any singularity then the problem is just the same.

My paper does spend more time than usual on what happens at the precise point where the event horizon disappears, and suggests that the space-time fails to be a manifold there. If (as Wald prefers) you delete what I call the Evaporation Event, then you can keep the manifold structure but you lose the global hyperbolicity—there are no Cauchy surfaces at all—and you gain a naked singularity. The choice there really is one that depends more on taste than argument.

So getting rid of the singularity is not the issue. What you would have to do is get rid of the event horizon. But if you get rid of the event horizon then, properly speaking, you get rid of the black hole altogether. And if there are no black holes then a fortiori there are no evaporating black holes, and so there can't be a problem.

One more note. When I said that I did not want to deal with screen names, I actually meant not that just I should know who is posting but that everyone should: the information should be public knowledge. We are not discussing matters here that require some sort of cover. The only thing it accomplishes, as far as I can see, is that it makes it much easier to write without carefully considering either content or tone of what you write, because you operate under the cover of darkness. If you want to make an argument, then stand behind it. Take the risk that you will be shown to be mistaken in public. You may be embarrassed if that happens, but so what? As Socrates said, the proper punishment for someone who is ignorant is to be taught. In the end you learn something.

Cheers,

Tim

Tim, Bee,

ReplyDeleteI think anonymity serves a useful purpose of separating ideas from personalities. However, I agree that it is distasteful to hurl insults under the veil of anonymity. I therefore pledge to reveal my real name next time I feel compelled to depart from neutral scientific dialogue, which I have been mostly successful in adhering to. Of course, Tim is at this point free to end the conversation, which I think would be a shame, since I think there will be a genuinely helpful back and forth once the current line of inquiry gets to the topic of diffeomorphisms.

Tim asks the question: what is the "relevance" of my equation [\eps P, f(X)]= f(X+\epsilon)-f(X) ? I am honestly confused about what kind of an answer he is looking for. I showed that if some operator P obeys this equation for all f(X), and if a state |\psi> obeys P|\psi> = 0, then it follows by simple manipulations which I provided that \psi(x+\epsilon) = \psi(x). So my argument is of the the form if p and q then r. The relevance of p is that it is needed in the argument to establish r. Perhaps the question concerns why I say "for all f(X)" rather than just using f(X)=X, which would also be sufficient to establish r. As I said, I think it is pedagogically useful to write the equation for a general f(X), since then you can see the expression f(X+\epsilon) appearing, which makes it clear to the eye the sense in which a translation is occurring. So I think this is a useful and helpful way to state the fact that P is the translation generator. I am honestly not sure what else to say, but am happy to say more if asked.

I put some careful thought into how to build up an argument starting with an example I thought we would quickly agree on. Perhaps its relevance to the ultimate questions of interest are not immediately apparent, but I assure you they will be.

Also, I think it is important to emphasize that I did provide a sharp refutation of Tim's scenario (as far as I can make sense of it) that stands fully intact. It is useful to summarize it:

ReplyDelete1) Tim asserts that the bulk obeys the ordinary rules of QM, and that we should take the standard Penrose diagram at face value. Good, so let us ask where this leads when we consider black hole formation and evaporation in AdS.

2) There is a well defined notion of time t at the boundary assuming standard asymptotically AdS boundary conditions. We therefore have a Hilbert space of physical states at each time t, and the wavefunction describing the process is \psi(t, ...) where ... denotes the relevant configuration space. In the bulk description, the ... correspond to the values of the metric and matter fields on a spatial manifold (more details of this were going to explained in the line of argument I was developing, but we may never get to this).

3) This wavefunction evolves according to Hamiltonian evolution, d\psi/dt = -i H\psi, where H is the boundary Hamiltonian of Regge and Teitelboim. The full H_bulk also includes volume terms, but these are constraints and so annihilate the physical state \psi, leaving just the boundary part of H_bulk.

(2) and (3) just correspond to following the standard rules of QM

4) Now consider this wavefunction at very late times t, long after the black hole has evaporated. In the Penrose diagram, all Cauchy surfaces are disconnected at this time (i.e. this property holds for all Cauchy surfaces that hit the bulk at such a time). The wavefunction \psi is therefore a function of metric and matter fields on a disconnected surface. Note that there is no statement here about the full spacetime being disconnected; the point is the wavefunctions at a time t are associated with fields on a spatial manifold, and it is this manifold that is disconnected.

5) In every example we know of, if we have a disconnected Cauchy surface then the Hilbert space is a tensor product, with one factor for each connected component of the surface. How could it be otherwise? The constraints of QG that define physical states do not include any mixing between degrees of freedom on one component or another.

6) If we have a tensor product Hilbert space, and each space is nontrivial, then it is clear that the Hamiltonian will have a huge degeneracy corresponding the freedom to change the "state behind the horizon" while changing nothing in the region outside.

7) This by itself is not necessarily problematic. But now suppose there is a dual CFT. The CFT has a Hamiltonian H_CFT, and part of AdS/CFT duality is the statement that the spectrum of H_CFT is the same as that of H_gravity. Then we have a contradiction, since we know that H_CFT has no large degeneracy in its spectrum.

So we have concluded that there is a clash between taking the Penrose diagram seriously and making minimal assumptions about the properties of a CFT dual.

As far as I know, and I would love to be corrected on this, the only objections Tim has raised is that he will not commit to the bulk having a physical Hilbert space, and even it exists he will not commit to it being a tensor product at late times. I don't understand these objections since I claim they follow from assuming the usual rules of QM and taking the Penrose diagram seriously, which was supposed to be the full point.

My view is that the Penrose diagram is deeply misleading, but in which precise way I do not know. This is of course a major topic of current research.

BHG

ReplyDeleteI have had to give this some thought. Here is my proposal.

You have earned special consideration due to the time and effort you have put into this exchange. It has not been nearly as productive as it might have been, in large part because of the attitude expressed by Dark Star, namely that you and s/he and Chai Tea, and any other mildly competent physicist has already considered the argument put forward in my paper and has "debunked" it, and the only thing standing in the way of my seeing this is that I am dim, and don't know what is in a first-year introduction to quantum theory. Although you suppress it more, this is also the tone that pervades your posts. You might usefully compare the tone of the exchange above between myself and Sabine. I use—and enjoy using—sharp rhetoric. But I assume you are aware that I know the fundamentals of basic non-relativistic quantum mechanics perfectly well, and that the sorts of insults that have been hurled at me on this topic are nonsense. If not, then why have you continued this so long? And because of the testosterone-filled (that is sharp-rhetoric right there, and I mean it as such) tendency to fall back on things like the "I suggest you do some actual calculations..." ploy, I want to fight back on the same ground. For example, if I wanted to continue on about the difference between a distribution and a function—the sort of technical mathematics that suits this tone—I could do it. Although ultimately to no useful end, because the atmosphere has become that of a fight rather than a conversation.

If you don't think that you have at least been forced to think more precisely about things and express them more clearly in virtue of this exchange—if you think this has all been for my benefit and none for yours, and that you have learned nothing as Dark Star asserts—then let's just quit.

I recently had an interaction with a physicist that was one of the most unpleasant in my life. This person actually informed me—not based on anything at all that I had said—that I did not know what a microcanonical ensemble is. When I said that indeed I did, he informed yet again, in no uncertain terms, that I did not. And then, with a flourish, wrote down a partition function as if the mere sight of the formula would scare me under the table. It was a display of an attempt at the sort of bullying that physicists often resort to. (Somewhat ironically, because it is not unusual for them not to have the fine details of the mathematics under control.) Fortunately, I was with a physicist friend who knew that the whole display was just empty posturing and bullying because he knew perfectly well that I know what a microcanonical ensemble is. l am, as I say, just fed up with these idiotic pissing matches that are grounded, in the end, in nothing but ego. I do not deny that I have done this as well. But it is out of a pent-up frustration at a pervasive dismissive tone *that is not backed up by showing that I have actually missed anything important*.

So here is my proposal. We have gotten far enough to see that the argument you think there is against my paper can only be made in AdS and turns crucially on the claim that the energy eigenstates in AdS must be non-degenerate (up to symmetry). But I still cannot see that even if true that tells at all against anything in my paper. As I mentioned above, as far as I can see the most recent arguments you have been giving are not so much arguments against my position but arguments against AdS/CFT itself. So I will continue discussing that with you even though you will not reveal who you are, if you are interested. But that's it. I won't get any further into general discussions of symmetry or diffeomorphisms with anyone using a pseudonym. Because while I acknowledge that I am no expert on AdS/CFT, I am an expert on this, and will really not tolerate this attitude any more.

It's your call.

Tim,

ReplyDeleteI have been scratching my head trying to understand why you responded to my quantum mechanics example with "This whole paragraph is really confused." I believe I have now identified the main source of confusion. You wrote:

"What the symmetry of the space means is that if you take any solution and operate on it with a symmetry operator, you get another solution."

and

" A "gauge orbit" is a set of solutions that can be generated from one another by the action of the symmetry operators."

These are not the standard definitions of "symmetry" and "gauge orbit", and differ from my usage (which I claim is standard). Regarding "symmetry", this is usually discussed in terms of the action, S(\phi). A transformation, \phi --> \phi + \delta \phi is a symmetry if S(\phi+\delta \phi) = S(\phi). This is a much stronger condition than taking solutions to solutions, and this extra strength buys you the key consequence of a symmetry. Namely, we can then apply Noether's theorem and obtain conserved currents and charges. So it is really key that a symmetry preserves the action even if \phi is not a solution. Your statement is not wrong, but it leaves out the most important aspect of a symmetry.

For "gauge orbit", this has nothing to do with solutions per se; it is a statement about the configuration space of the theory. In E&M, the configuration space is the space of gauge fields, A_i(x). Two gauge field configurations are defined to be on the same gauge orbit if they are related by a gauge transformation. In some other places you seem to be thinking of points on the gauge orbit as being different QM states or (equivalently) wavefunctions. If that were the case, then indeed my claim that "if the symmetry generator annihilates a state then the wavefunction takes the same value on all points of an orbit" would be meaningless. But when we stick to the definition that points on the gauge orbit are different gauge field configurations related by a gauge transformation, then the statement makes perfect sense, and is indeed true, as I explicitly demonstrated. Similarly, for single particle quantum mechanics, points on the symmetry orbit are not different states but different points in configuration space,i.e. different values of x. So the result I was deriving is a statement about a single wavefunction possessing some property (namely independence of x), not a statement regarding two different wavefunctions.

I hope that helps.

BHG

ReplyDeleteI will make one comment about the symmetries and then let that drop. I think we can spend some useful time going though your enumerated points, and maybe that can get us to a conclusion.

My comment repeats something I said, and I will try not to make this soundl contentious so there is no issue of debating it. The first example you go through, using the P operator as the infinitessimal generator of translations, concerns a physical symmetry, namely the homogeneity of Euclidean space (in the non-Relativistic setting). That symmetry persists in Minkowski space but fails in a generic solution to the EFEs. This has interesting consequences, of course, for the application of Noether's theorem.

The second example, if we begin with classical EM, is of a completely different character. The gauge symmetry is not based on a physical symmetry at all: it arises from the existence of multiple distinct mathematical representations of the same physical situation, i.e. the same distribution of E and B fields. Because the origin of this symmetry is mathematical rather than physical, as in the first example, the morals one draws from it are completely different.

So the first task is not to just attend to the similarities of case 1 and case 2, but to consider just as carefully their differences.

Next, if you move to QED, you have the additional complication that the Aharonov-Bohm effect demonstrates that A and phi have some sort of different status than they do in classical EM: in some sense they are more "physical". This already gets quite delicate.

But even if you can sort out that new wrinkle, it is still a job to understand diffeomorphism invariance. Is it more like translation invariance, based in a physical symmetry; more like classical gauge invariance, purely mathematical; or something else altogether? This is treacherous territory, and would require extreme care.

BHG:

ReplyDeleteOK, back to the main topic.

Let's start with 2) and not move on until it is sorted out. You write:

2) There is a well defined notion of time t at the boundary assuming standard asymptotically AdS boundary conditions. We therefore have a Hilbert space of physical states at each time t, and the wavefunction describing the process is \psi(t, ...) where ... denotes the relevant configuration space. In the bulk description, the ... correspond to the values of the metric and matter fields on a spatial manifold (more details of this were going to explained in the line of argument I was developing, but we may never get to this).

I don't see how this can be correct, or at the least it is misleading. Let me say why.

The time t at the boundary can be used to assign a time to any Cauchy surface. That is just the time assigned to the boundary of the surface at the boundary. But of course, there will be infinitely many such Cauchy surfaces that are assigned the same time by this procedure. Any pair of Cauchy surfaces that overlap at the boundary will. Some of these Cauchy surfaces will "cut high" through the bulk, lying to the future of the Evaporation event, others will "cut low", through the dust that will eventually form the black hole. So there is no such thing as *the* bulk state at time t: there are infinitely many such states. And a fortiori there are infinitely many time-parameterized sets of states that have exactly one state for each value of t. But a generic such time-ordered set is not a foliation of the space-time. So let's impose some more conditions. If Slice Ss time parameter precedes Slice S', then every event on S lies in the past Domain of Dependence of S'. And every event lies on some slice in the set. These conditions ensure that every such set is a foliation of the space-time.

But still, there are infinitely many such time-ordered parameterized sets. So even if we impose these extra conditions, there is no such thing as “the state of the bulk” that corresponds to a particular boundary time.

At this juncture there are two choices.

A) Somehow pick exactly one of these sets, i.e. a particular foliation of the space-time. Then each value of the time-parameter t does indeed index a unique bulk state, namely the state with the intrinsic geometry and external curvature of the member of that foliation indexed by the given time parameter. So you can ask what the physical conditions on that particular Cauchy surface are: whether there there are photons, or electrons and if so where they are.

B) Declare all of the Cauchy surfaces that share the same time parameter in these infinitely many sets are to be somehow regarded as “equivalent” or “the same” or “gauge-equivalent” to each other. Say that they all somehow represent the same state.

The challenge for A is to explain how the unique foliation is chosen. Having done that, one can unproblematically talk about “the bulk state at t”. For each t there will be a unique three-metric, matter field, etc.

The challenge for B is to understand what sort of equivalence exists between the set of states that are all indexed by the same value of t. One can call this “gauge-equivalence”, but a word is just a word and does not magically carry any consequences. Indeed, it is hard to see what physical similarities there are at all between the members of this equivalence class except that they were all Cauchy surfaces contained in the same 4-dimensional solution, and all asymptotically overlap. The “state” that these are representations of must be the entire bulk state including the complete formation and evaporation of the black hole. This will be true at time t1 and t2 and t3, etc. So the sequential appearance of events in time has nothing to do with the lapse of boundary time. In fact, the whole point of even having a boundary time becomes quite obscure: why not just declare all Cauchy slices through the 4-D solution to be “gauge equivalent” irrespective of their asymptotics?

So which is it: A or B?

bhg,

ReplyDeleteI think you could save yourself much breath by noting that Tim's scenario is a type of remnant solution that violates the BH-entropy. If you assume the AdS/CFT correspondence is correct then this will not work.

As I mentioned earlier, however, you (not you personally, but you people in the field in general) basically assume the BH-entropy by allowing only states that can be expanded around the boundary, so not much of a surprise that the # of states scales with the surface. Can't say I find this terribly convincing. What's missing in the argument - as you alluded to earlier - is to show that all other states aren't possible for one reason or the other. This may or may not be related to what Tim is saying, but in any case that's what's bugging me.

Best,

B.

Tim,

ReplyDeleteRegarding who has set the tone of this exchange, no surprise, but I see it very differently than you do. There's little point in rehashing all of this, but I would like to state my perspective once and then move on. First, note that Sabine in her initial posting complained about your "condescending tone" and more recently listed her complaints about your mode of operating. So I am frankly surprised that you bring up your interactions with her as a model of how things should proceed, since I suspect she would not agree.

Let's look back at the two exchanges that elicited the most fireworks, namely the issue of Gauss' law and my QM example. Here is the beginning of the latter, taken directly from your post

BHG: "1) Let's start with single particle non-relativistic quantum mechanics and discuss the symmetry of translation invariance. There is a symmetry generator p, the momentum. THat p is the generator of translations is the statement that [\epsilon p, f(x)] = f(x+\epsilon) - f(x). ( Here x and p are operators of course). \epsilon is infinitesimal, and a finite translation is obtained by exponentiation. Two particle locations, x and x+\epsilon, are thus on the same orbit of the symmetry (in this case there is only one orbit). Now, suppose the system is in a state described by wavefunction \psi(x), and suppose that the symmetry generator p annihilates the wavefunction, p \psi(x) = 0. It follows that \psi(x+\epsilon ) = \psi(x). That is, it follows that the wavefunction takes the same value on all points of the symmetry orbit."

Tim: "This whole paragraph is really confused. You say that in what you write x and p are operators. Well, no. Every x in that paragraph is not an operator, it is a variable. p is an operator, x isn't. So you have to stop right now and reverse your attitude"

Two points here: First, telling someone that they have to "reverse their attitude" is unlikely to foster constructive debate. Second, your negation of my statement that x and p are operators in that equation is both incorrect and much less constructive than writing something like "please elaborate on how to think of ... as an operator equation". Instead, your remark amounts to an accusation that I don't know elementary facts about operators in QM, which is insulting. It would require the forbearance of Job not to respond to this with insults of my own, and I regrettably succumbed to this.

cont

cont

ReplyDeleteRegarding Gauss' law we had the exchange:

Tim: "I don't think you are taking the time now to even read what you write. For example, you wrote, and I quote,"

BHG : "In electrodynamics the operator that generates gauge transformation has exactly the same structure: a volume integral that vanishes by Gauss' law plus a surface integral that yields the charge. I take it you disagree with this as well?"

Tim: And just now:

BHG: "I clearly stated and used precisely what Gauss' law is, which the statement that the divergence of the electric field equals the charge density."

Tim: "Do you really not want to correct something in one or both of these claims? Trying to make sense of what you even have in mind here is beyond my capacity, and your additional claim that everything is clearly stated and precise just makes the situation worse. I am going to be very explicit and brutally honest about what I have just quoted, since that seems to be what it might take to get you to attend to what you are claiming.

First: yes, I certainly disagree with what you asked about, because it is plainly wrong."

As I see it, the trouble here is that you had flipped in your mind the usual definitions of Gauss' law and Gauss' theorem. This can happen, and there's no foul there. But rather than check a book or ask me for clarification you went on to give a pretty insulting lecture about how I don't understand calculus. In fact, rereading what I wrote I stand by my statement that it is clear and correct, although the first point was perhaps to concise for the target audience, but this is exactly what could have been remedied by a simple question.

I admit to at times becoming exasperated and resorting to insults. As I said, I regret this and pledge not to repeat it. That said, all this is now water under the bridge as far as I am concerned.

Bee,

ReplyDeleteMy impression is that Tim's scenario is not really a remnant scenario but is rather the one that Hawking originally proposed. That is, he is not arguing for the presence of some long lived object that remains in the spacetime, accessible to observers. Instead, the claim is that the black hole evaporates completely into radiation, but that the radiation is not in a pure state. His point, as I understand it, is just to argue that this original scenario has been discarded (by some) without good reason (in his view).

Now, major figures like Unruh and Wald take a similar stance, so it is clearly not "dumb". And I agree that if we put AdS/CFT aside there is no very convincing argument against it. It's hard to make much progress without more information about the equations of quantum gravity.

AdS/CFT is really the only viable framework at present for quantum gravity, in the sense of being a setup that both reproduces Einstein gravity in the low energy limit and has a non-perturbative formulation. My point all along has been that Hawking's original scenario, applied to black holes in AdS, seems incompatible with what we know about AdS/CFT. Now, even if you agree with this, you can counter that maybe AdS/CFT is irrelevant to how quantum gravity works in our universe, or maybe AdS/CFT contains some internal inconsistency yet to be discovered. No argument about that. I am just saying that in the one case where we do have a formulation of QG that lets us address the question at hand, it seems that Hawking's proposal is not realized as far as we can tell.

Finally, I am surprised by your statement " so not much of a surprise that the # of states scales with the surface. Can't say I find this terribly convincing. ". As you surely know, the black hole state counting in AdS/CFT goes way beyond showing that the # of states scales with the area. First, it gets the coefficient correct, but more importantly there are now large collections of computations that successfully reproduce an infinite series of corrections to the area law due to quantum gravity effects in the bulk. These are *highly* nontrivial computations in which much more comes out than was put in. This is the kind of thing that impresses folks like Unruh and Wald -- it's clear that AdS/CFT must contain a good deal of truth.

Tim,

ReplyDeleteRegarding your message "I will make one comment about the symmetries and then let that drop" I have no disagreements with what you write there. My choice of examples regarding NR QM, E&M, and diff invariance in GR was not to imply that these are the same. The point of the NR QM example was to establish a mathematical lemma in a familiar context, so that this lemma could then be applied in the different context of gauge and diff transformations. My purpose in labeling these examples as 1, 2, 3 was precisely to make clear that there are distinct issues at play in each.

I think it is very useful pedagogically to proceed via the route: 1) global spacetime symmetry 2) gauge symmetry in E&M 3) diff invariance in GR. I have thought deeply about the meaning of diff invariance in quantum gravity. I also have thought (less) about the "hole argument". My perspective is that the hole argument is interesting historically and philosophically, but I haven't run across a discussion that seems likely to me to be helpful in terms of formulating the equations of quantum gravity, but I would be very happy to be shown otherwise.

I will make comments on your other message later

Tim,

ReplyDeleteI now turn to your questions regarding my point (2). This is indeed a key point and your questions are clearly stated so I am happy to oblige. This is in fact the point I was hoping to eventually clarify in my commentary starting with the QM particle example.

Let's put aside black holes and consider the simpler process of planet formation in AdS from a cloud of particles. I want to carefully distinguish three levels of description of this process. i) as a classical solution of the equations of GR coupled to matter ii) as a semi-classical process corresponding to treating the metric classically but the matter quantum mechanically; this is the limit in which Newton's constant is taken to be small while the planet mass is large, so that MG_N is fixed iii) the metric and matter are both treated quantum mechanically. My claim is going to be that it is only at level (iii) that it makes sense to say that the bulk state depends only on the boundary time. In the following I refer to AdS, but in fact everything holds equally well for asymptotically flat spacetime.

Level (i): we have a classical spacetime, and in this spacetime we can consider a 1-parameter family of Cauchy surfaces, labelled by \alpha, that all asymptote to the AdS boundary at a specified time t. So for each \alpha we have a spatial metric, g_{ij}(x;\alpha), and matter fields \phi(x;\alpha). These Cauchy surfaces do not cover the full spacetime, since they are necessarily spacelike and attached to a fixed time at the boundary. I doubt we will have any disagreement about anything here

Level (ii) We have the same spacetime as in level (i), but now have quantum fields/particles in this spacetime. On each Cauchy surface we have a quantum state. Focusing now on the 1-parameter family of Cauchy surfaces defined above, our description of the planet formation process restricted to these Cauchy surfaces is described by a 1-parameter family of quantum states, \phi(\phi(x);\alpha). That is, for each \alpha we have a quantum state, which we can use to compute the probability for a certain particle to be at a certain location on the \alpha Cauchy surface. The state changes with \alpha, and indeed we can write a Schrodinger equation that governs this change: d\psi/d\alpha = iH_\alpha \psi. H_\alpha is obtained explicitly by integrating the matter Hamiltonian weighted against a lapse function over the \alpha Cauchy surface. The lapse function appearing in H_\alpha goes to zero at the boundary, since we are not evolving the surface there. For "early" \alpha the quantum state \psi(\phi(x);\alpha) describes a cloud of particles, while for "late" \alpha it describes a planet, assuming that the planet formation process occurs within the region covered by our family of Cauchy surfaces.

Level (iii) Now we get to the meat. The metric is now treated quantum mechanically, so the wavefunction takes the form \Psi[ g_{ij}(x),\phi(x)], where I write \Psi rather than \psi to emphasize that the metric is now an argument of the wavefunction. The wavefunction \Psi assigns a complex number to each choice of spatial metric g_{ij}(x) and field configuration \phi(x). The wavefunction is also annihilated by the local (i.e. one for each x) constraint operators, H_t(x) \Psi = H_i(x) \Psi = 0. A key point is that this wavefunction does not depend on a parameter \alpha, since its domain includes all possible spatial metrics g_{ij}(x). For a given t, there is thus a single wavefunction \Psi, not a 1-parameter family of wavefunctions as in level (ii). So for instance, we can choose the pair (g_{ij},\phi) to correspond to a cloud of particles, or to a planet, and the single wavefunction \Psi assigns a complex number to each choice.

ReplyDeleteAn obvious question is how level (iii) reduces to level (ii) in the limit of small G_N. The point is that \Psi [g_{ij},\phi] will be peaked around certain correlated values of g_{ij} and \phi. So let g_{ij}(x;\alpha} be the 1-parameter family of spatial metrics in the level (i) or (ii) description. I claim that if we define a \psi via \psi(\phi(x);\alpha) = \Psi[g_{ij}(x;\alpha),\phi(x)) then indeed this is the same \psi that was discussed in level (ii), provided that we are indeed in the semiclassical regime. [There is a slight technical lie here: we really should not fix g_{ij} precisely but rather localize it to a narrow wavepacket, but let me ignore this for now]. I can point you to references where this is shown very explicitly.

So our level (iii) description indeed naturally reduces to level (ii) and level (i) in the appropriate limit. Now, there is much more to be said here about how to work with and interpret the level (iii) wavefunction, and how it evolves with the time t, but perhaps it is best to pause here.

BHG

ReplyDeleteUnfortunately, we do have a disagreement even on point 1, or rather on point 1 together with your claim that "In the following I refer to AdS, but in fact everything holds equally well for asymptotically flat spacetime."

The difference, of course, is that in an asymptotically flat space-time, there really is no "boundary" properly speaking at all. The "boundary points" Scri + and Scri- and spatial and past and future timelike infinity are all mathematical fictions. So choose a Cauchy surface, and then take the collection of all Cauchy surfaces that approach it asymptotically. Union all of those surfaces together. That union will be the entire space-time. (save the fictional boundary points). In other words, in the Penrose diagram of an asymptotically flat space-time, every event not on the boundary is spacelike separated from Spatial Infinity.

There is a quantifier inversion issue here. If you choose a single such foliation that limits to t, then there will be events left off it. But if you choose any single event, there will be a foliation that limits to t that includes it.

So is the idea that you choose a single foliation first and then work only with it, disregarding the features of other foliations? If so, is there any procedure for choosing the unique foliation?

I want to keep track of places where it does not seem clear that everything that works in AdS works for asymptotically flat space-times..

Sabine and BHG,

ReplyDeleteSince you mentioned my solution, let me save some time here. BHG writes:

"My impression is that Tim's scenario is not really a remnant scenario but is rather the one that Hawking originally proposed. That is, he is not arguing for the presence of some long lived object that remains in the spacetime, accessible to observers. Instead, the claim is that the black hole evaporates completely into radiation, but that the radiation is not in a pure state. His point, as I understand it, is just to argue that this original scenario has been discarded (by some) without good reason (in his view)."

I'm a little surprised at this, because it is the very point I was most explicit about in the paper, and indeed is the point of the paper. So to be clear.

It is not a "remnant solution" if the remnant is supposed to be accessible to observers who remain outside the event horizon.

Is it Hawking's solution? Yes and no. Yes in that the evolution from Sigma 1 to Sigma 2_out, which is the only evolution that Hawking considers, is pure-to-mixed, loses information, and would be implemented in scattering theory by a superscattering matrix. Again, the Susskind, Banks and Peskin objections to this strike me as entirely insubstantial, and I am told that there is a paper by Wald that argues to the same conclusion.

But no, in the sense that if you asked Hawking whether unitarity fails, he would just answer unequivocally "yes", whereas I answer unequivocally "no, not in the sense that matters". The sense that matters as far as the basic principles of QM are concerned is not that the evolution from Sigma 1 to any old later surface be unitary, but that the evolution from Sigma 1 to any other *Cauchy* surface be unitary. Since Sigma 2_out is not Cauchy, the evolution to it is neither here nor there. But Sigma 2_Out U Sigma 2_In is Cauchy, and unitary, and information-preserving.

Hawking's blind spot was not seeing that the interior of the black hole is still space-like separated—and hence available for information storage—from the observer who has watched the black hole evaporate to completion. At that point, the whole event horizon is in his past, but none of the interior of the black hole is. So the information can be stored there, with no loss. The unitarity of Quantum Theory is not broken where it should not be broken.

And to jump ahead a bit, BHG's argument against this solution breaks down at point 5 in his exposition. You might as well write:" In every example we know of, if any Cauchy surface is disconnected, than the space-time itself is disconnected." That is true if you are dealing with a manifold, and that the Hilbert space for disconnected space-times is indeed a tensor product. That's because situations such as is suggested here cannot arise on a manifold. But this one does not arise on a manifold.I am exactly proposing a situation in which the Cauchy surface disconnects but the space-time doesn't. And that means that the usual argument for the solution space to be a tensor product dissolves. As far as I can tell.

Tim, BHG,

ReplyDeleteWhether or not you want to call it a 'remnant' solution doesn't really matter. I used this term because it violates the BH-entropy. It has to because there is no mechanism to get information in the radiation. If you have a mechanism to get information into the radiation, then this scenario would be indistinguishable from any other information-release case.

Regarding the question whether or not the thing entirely evaporates, I believe Tim's point is that the question is ill-posed. Be that as it may, if you try to construct a qft for an outside observer, he/she will see a potential loss of unitarity, which is the case Hawking originally considered with all the debate that followed that and so on.

Best,

B.

bhg,

ReplyDeleteRegarding your other comments. You write

"it's clear that AdS/CFT must contain a good deal of truth. "Truth... about what? The only thing your comment highlights is that it's internally consistent. That doesn't tell you it's actually realized in nature. You have "successfully reproduce[d] an infinite series of corrections". Great, but how do you know these corrections are actually correct in the sense of describing observations? No need to answer that question, we all know you have no way of telling.

This is just to say that I am not among the people who are impressed. Best,

B.

Bee,

ReplyDelete""it's clear that AdS/CFT must contain a good deal of truth. "

Truth... about what? "

I am speaking about "truth" rather than "Truth", if you know what I mean. AdS/CFT provides what seems to be a mathematically consistent quantum theory that yields standard Einstein gravity (in AdS) at low energies. It thus contains the elements needed to formulate and resolve Hawking's original "paradox". Once this is fleshed out we can ask whether the resolution critically depends on being in AdS. This seems to me to be a very sensible way to proceed, especially since the alternative is to throw up one's hands. But all of this has been said many time before...

Tim is absolutely right in pointing out that the position operator doesn't have any eigenvectors that live in the hilbert space L^2 of square integrable functions. The famous eigenvalue eigenvector equation that one often see in qm textbooks for physicists is only valid in a rigged hilbert space, where the eigenvector in question is a distribution. Every mathematical physicist that works on the foundations of qm knows this. That's because they don't follow the tradition of Dirac that physicist follow,they follow von Neumann's tradition. Heck, von Neumann had to prove the general spectral theorem for unbounded self adjoint operators precisely to make sense of operators like the position operator that has continuous spectrum but not eigenvectors... just calculate the norm of exp ipp_0, the supposed eigenvector of the momentum operator with eigenvalue p_0, and you will see it is not an element of L^2.

ReplyDeleteTim,

ReplyDeleteFair point about the distinction between AdS and asymp. flat spacetime. Rather than deal with this, I suggest we restrict to AdS for the time being.

My post is meant to explain the precise meaning associated with there being a state at boundary time t. Do you agree with that content?

Once we reach agreement on the meaning of there being a state at time t, we can proceed more quickly if you will explicitly commit to one of the follows options (which cover all cases). Here I refer to a very late time t, such that all Cauchy surfaces are disconnnected. At such a time t, which of the following holds:

A) The Hilbert space of physical states is a tensor product

B) The Hilbert space of physical states is not a tensor product

C) There is no Hilbert space of physical states

If you support (A) or (B) then I can address that. If you advance (C) then I will have to ask you for more details about what you mean. How do you make predictions in this theory? - I don't see how Born's rule could apply, etc

Tim,

ReplyDeleteActually, I don't see how your proposal differs in any respect from what Hawking originally claimed. You write

"But no, in the sense that if you asked Hawking whether unitarity fails, he would just answer unequivocally "yes", whereas I answer unequivocally "no, not in the sense that matters".

But look at what Hawking actually writes in his famous "Breakdown of predictability .. " paper from 1975, in particular the discussion around p. 2463. He says that you do get unitary evolution from the initial state to a final state in the tensor product of H_2 and H_3, where H_2 is the state on the "hidden surface" (i.e. behind the horizon) and H_3 is the state on the "final surface" (i.e. outside the horizon). He is just saying that there is not unitary evolution from the initial state to a state on H_3. Said another way, there is not a unitary S-matrix. This is what I have always understood by his argument.

BHG-

ReplyDeleteGreat! We have already made more progress in the last day than we had for literally months!

Let me make a couple of points about Hawking's 1976 paper. First, it really helps to know that that is what you have been thinking of. That paper is very problematic in terms of how Hawking states things and even what he is interested in, but you are quite right that that particular passage is relevant to one of the main points in my paper, and I should have referenced it. It is just that one also has to then go into the context to see that the overall impression one gets from the paper is more misleading than enlightening, in my opinion. This takes us straight into foundational issues.

The first equation on p. 2463 is indeed important, but please note that the entire discussion there references three Hilbert spaces and the tensor product space of them. These are all *kinematic* Hilbert spaces, not *solution* spaces. This is important in the sequel. The possible *physical* states will certainly be a *subspace* of the product space H_1 X H_2 X H_3 that the scattering matrix is defined over!

Also note that Hawking is using that equation as a way of specifying the dynamics of the theory. There is no mention of a Hamiltonian, etc. The implications of the Hamiltonian are all encoded in the S matrix, which directly gives the function from an initial state in the asymptotic past (xi_C) to a *complete* final state in the asymptotic future (the state in H_2 X H_3). [NB: it is *this* state that I call the "final state". That disagrees with Hawking's use of words. For him, the state in H_3 alone is the "final state. Please be careful about this!) Let's assume that the initial state xi_C is pure (Hawking will eventually loosen this, for good reason). That is the state on Sigma 1. Then quantum mechanics says that the state on H_2 X H_3 will also be pure and will contain all the information encoded in xi_C. Now please pay close attention to the paragraph that follows the equation.

Hawking says that "even given xi_C, one "cannot determine the final state [i.e. the state of Sigma 2_Out] but only" the reduced density matrix for Sigma 2_Out that results from tracing over Sigma 2_In! But what does he mean by "the final state"? If the wave function is complete, and if the final state (Note: here I use "final state" in my sense, not his!) is the pure state in H_2 X H_3, and if that in an *entangled* state (as generically it will be), then the reduced density matrix is the only thing you can mean by "the final state" {Note: here I use "final state" in his sense, i.e. the state on Sigma 2_Out). If the wave function is complete and never collapses, then you can indeed determine the final state on Sigma 2_Out from the initial state on Sigma_1: it is the very density matrix that he calculates!

Con't.

Con't.

ReplyDeleteNow maybe Hawking has in mind a collapse theory, in which the evolution of the wave function is not always deterministic and unitary. If that is so, then the whole discussion goes differently. But in that case the supposed tension between black hole evaporation and unitarity etc. never comes up at all: in a collapse theory QM is not unitary and not deterministic and not predictable and not retrodictable ever in any circumstances, black hole or no. If you think that, then there was never any prima facie tension between black hole evaporation and QM in the first place. In a collapse theory, information is lost at every collapse, so there can't be any "information loss paradox": information loss is omnipresent. So if Hawking does have in mind a collapse theory, then we are in a different ballpark altogether. What is sure is 1) no contemporary presentation of the so-called paradox assumes a collapse theory: just the opposite, they assume determinism, predictability and unitarity of the wave function dynamics as central postulates of QM and 2) The contemporary presentations want the evolution from the state on Sigma 1 to the state on Sigma 2_Out to be unitary, to preserve information, and to be pure-to-pure. My whole point is that condition 2) does not at all follow from condition 1), and there are no grounds to expect, or reason to demand, 2).

Here is something that Hawking says (on p.2462) that I don't think one can make sense of: "Of course if the observer measures the wavefunctions of all the particles that are emitted in a particular case he can a posteriori determine the internal state of the black hole, but it will have disappeared by that time." This sentence is a complete disaster. I can articulate why, if I need to, but my point is that however Hawking is thinking about what the problem is here, it is not correct. This is the sense in which I don't think that I am just channelling or repeating Hawking in my paper. If you want to say that I am starting from him and correcting him, fine. That is actually how the paper is written, but I could have gone into more detail.

The main point is, however you want to carve that up, is there any cogent objection to my solution of the "paradox"? If you want to say (which you seem to) that there is no objection at all *until you take into account AdS/CFT*, then we again have the puzzle of why anyone thought there was a problem in the first place! The so-called paradox existed and was considered a big deal long before AdS/CFT. On your account, why was that?

I will answer your last post directly in another post.

BHG

ReplyDeleteIn direct response:

Take C off the table. My whole point in bringing up C was in service of making clear the distinction between the kinematic space, which contains state that are not"physical" and the solution space, which contains only the "physical" states. The issue of the latter not being a Hilbert space never had to do with the inner product, which is what you need for Born's rule, but with the condition and the space be complete. But if you come up with an incomplete solutions space you can always complete it. It seems like I have nit been able to get this point across even yet, so let me try one last time.

The standard way that Quantum Theory, and classical theory as well!, proceeds is to first define a kinematical space and then impose conditions to pick out the physical solutions from the non-physical solutions in the kinematical space. In classical mechanics, the kinematical space is the space of all continuous functions from points in phase space to a time parameter t, i.e. parameterized trajectories through phase space. Not all of these trajectories correspond to evolutions that could physically occur. Indeed, if we use the phase space here, some of these functions are not even conceptually coherent!. For example, consider all the states in phase space with P= 0.Now take a trajectory through phase space that lies entirely in that subspace, so (as it were) the positions of all the particles are constantly changing position while their momenta are identically zero. This is not a physical possibility for a very basic reason. So to get to the solution space we have to clean out all conceptually incoherent states. But even having done that (which essentially corresponds to take all the parameterized trajectories through *configuration* space as basic, and then matches each of these with a unique trajectory through phase space), one then imposes dynamical laws to cut down to the physical solutions.This set of trajectories forms the "solution space").

So: as long as we don't confuse a kinematic space with a solution space, I can say that right not it appears that the solution space for Quantum gravity is not a product space. So I will go with B.

Now what?

Note, by the way, that indexing by your variable t seems not to play any role at all. For appropriate t, the entire evolution from, say, dust to Hawking radiation on Sigma 2_Out is given by varying alpha, not t. I really don't see any important role t is playing.

Tim,

ReplyDeleteI've tried to tell you that ever since you contacted me, but I'll try it once again. The problem is that if you want to describe the creation and subsequent evaporation of a black hole in a quantum field theory, it doesn't seem to be unitary (according to Hawking's calculation). That's the evolution from the asymptotic 'in' states to the asymptotic 'out' states. Yes, that discards the part of the wave-function that's behind the horizon. Yes, that's the origin of the problem. Yes, we can discuss whether such asymptotic states are well-defined, but they are arguably common use. No, the inconsistency is not that the math is broken, the inconsistency is between the seemingly non-unitary *outside* evolution and the unitarity of qft. Nobody thinks it's non-unitary if you take into account the inside.

If you just want to accept this, fine, then you have to explain why we don't have to worry about this lack of unitarity. Or, more concretely, how to deal with the production and decay of black holes in qft, hence Hawking's talk about the S-matrix and so on. What do you propose how it works? For all I can tell, you're saying nothing about it. No, I am not saying that there is necessarily something wrong with denouncing unitarity. As I have tried to say several times, we have no reason to think that there is only one way to resolve this problem. Math alone won't do.

As for bhg, for all I can tell he's just saying that this option isn't compatible with AdS/CFT, which shouldn't come as a surprise because, as I said above, if you don't get out information in the radiation you have to throw away the microcanonical interpretation of the BH-entropy, which is widely believed to hold in AdS/CFT.

Sabine,

ReplyDeleteYes, you have been saying that for quite a while. And thereby ignoring the main point of my paper.

You write:

"That's the evolution from the asymptotic 'in' states to the asymptotic 'out' states. Yes, that discards the part of the wave-function that's behind the horizon. Yes, that's the origin of the problem. Yes, we can discuss whether such asymptotic states are well-defined, but they are arguably common use. No, the inconsistency is not that the math is broken, the inconsistency is between the seemingly non-unitary *outside* evolution and the unitarity of qft. Nobody thinks it's non-unitary if you take into account the inside."

Simple question: what does the phrase " the unitarity of qft." mean? If it means that the evolution from the state on a Cauchy surface to a state on a non-Cauchy surface is supposed to be unitary, then that is just not a feature of qft and never has been, black hole or no black hole. Wald's example, replicated in my paper, shows this for surfaces in Minkowski space-time. So that can't be what you mean by the "unitarity of qft". If you want a unitary and deterministic and information preserving evolution in qft, then it has to be from Cauchy to Cauchy. Nothing else will do. This is not controversial, is it?

So if that is not controversial, and the fact that Sigma 2_out is not Cauchy is not controversial (it isn't), then why is there any conflict at all between the non-unitary "outside" evolution and the unitarity of qft? It is just the opposite: the unitarity of qft *implies* the non-unitarity of the "outside" evolution! We would really have a problem if the "outside" evolution *were* unitary! In fact, we would have just the problem that the "solutions" to the non-problem display: breakdown of the monogamy of entanglement, for example!

The information does not get out in the Hawking radiation. It can't get out in the Hawking radiation because that is only entangled with the anti-Hoawking radiation, not with the infalling matter. More than that, it would violate the no-cloning theorem if the information *did* get out with the Hawking radiation. The "solutions" are trying to violate fundamental quantum-theoretic principles: no wonder they don't work. They better not.

So the only "puzzle" left is to answer Hawking's question: where is the information after the black hole evaporates? Answer: inside the black hole, i.e. in the interior region of the event horizon. That sounds impossible at first hearing, but it isn't, as I explain.

If you mean something else by "the unitarity of qft" please specify. Otherwise, where is the problem?

Sabine again,

ReplyDeleteTo your other question: "If you just want to accept this, fine, then you have to explain why we don't have to worry about this lack of unitarity. Or, more concretely, how to deal with the production and decay of black holes in qft, hence Hawking's talk about the S-matrix and so on. What do you propose how it works? For all I can tell, you're saying nothing about it."

I think that Hawking's proposal to use a superscattering matrix if you plan to use S-matrix-like theory is perfectly correct, just as he says. Of course, the whole virtue-and-drawback of an S-matrix or SS-matrix is that you just gloss over what actually happens in the interaction region! So if you are interested in the fine details of how black holes form and evaporate, the SS-matrix won't help. For that we need a Hamiltonian, or something analogous. I.e. you need a complete theory of quantum gravity. Of course! But there is still no information loss paradox to deal with.

The Hamiltonian should give you a unitary and information-preserving evolution from Sigma 1 to Sigma 2_In U Sigma 2_Out, and so the evolution to Sigma 2_Out alone is gotten by tracing over Sigma 2_in, just as Hawking does. That gives you the superscattering matrix. Now all we need is the Hamiltonian! But there is no "information loss paradox" either standing in the way or giving any clues about how to arrive at that.

Tim,

ReplyDeleteI specified what I mean. Or rather, what people in the field usually mean. I have done so many times. Evolution from asymptotic in-states to asymptotic out-states. You are correct to say that this is in some cases (such as for black holes) not an evolution from a Cauchy to a Cauchy surface. Yes, as I said equally many times, this explains the lack of unitarity.

For all I can tell, your point is merely to say that this isn't something anyone should be worried about. Which, as I have now repeated like a dozen times, is all fine with me. I have also, rather unsuccessfully it seems, tried to explain to you why no one gives a shit if you say there's no need to worry because that doesn't tell them just what happens when you produce a black hole (say, in some particle collision) and it subsequently decays. Can you tell us what happens? How does your modified non-unitarity qft for the asymptotic states look like? Where is your theory, Tim? What's your proposal? You're just saying "oh, I don't think it's a problem" but you haven't offered any workable scenario. Where's the scattering matrix? What does it look like? How do you deal with that non-unitary process if you combine it, say, with qcd? How does that work? Where's the math, Tim? Where's the substance?

It seems that

ReplyDelete1. The emphasis on S-matrix or CFT in AdS/CFT obscures the local nature of physical laws.

2. It ought to be possible to verify via local calculations, that if such-and-such conditions are satisfied locally, then globally unitarity will obtain. Conversely it should be possible to pin-point where/when exactly potential "leaks" in unitarity might be. Otherwise these debates will be endless (but maybe that is the nature of the Universe?)

Tim,

ReplyDeleteRe Hawking: my interpretation of what he is saying in that paper seems to be different than yours, but I agree that the wording is sometimes cryptic, and I don't think I want to invest time in a close reading. Rather, let me point out a much clearer and more concise statement that appears in his subsequent paper on the topic "Unpredictability of Quantum Gravity". The passage is:

"The situation changed, however, when it was realized that black holes evaporate by emitting particles with a thermal spectrum [1]. Suppose that one started from an initial pure quantum state which could be described in terms of a complete set of commuting observables on a space-like surface in the past. The same quantum state could also be described in terms of observables in the future only in this case one had to have two sets of observables, observables at infinity which described the outgoing particles and observables inside the black hole which described what felt through the event horizon. The system would still be in a pure quantum state but an observer at infinity could measure only part of the state; he could not even in principle measure what fell into the hole. Such an observer would have to describe his observation by a mixed state which was obtained by summing with equal probability over all the possible black hole states. One could still claim that the system was in a pure quantum state though this would be rather metaphysical because it could be measured only by an angel and not by a human observer. "

This seems to me to be exactly the same thing as you are saying.

As to why people thought there was a "problem" to resolve pre AdS/CFT, here my comments essentially align with Sabine's. The issue at hand is not unitarity per se, but rather unitarity of the S-matrix. As we all know, the principle of a unitary S-matrix has been key in formulating the wildly successful Standard Model. In fact, one can think of relativistic QFT as the general solution to the problem of constructing a unitary, Lorentz invariant S-matrix that obeys cluster decomposition. So if you give up this principle, what do you do? As far as I know, no one has written down a consistent quantum theory that reproduces the successes of the aforementioned framework while allowing for a bit of non-unitarity of the S-matrix. The simplest attempt fails, as Banks, Peskin, Susskind showed. Of course, the BPS argument is far from conclusive since they made the specific assumption of Markovian evolution of the density matrix, but once you give this up you're basically swimming in a sea of infinite possibilities. In this light, pre AdS/CFT the phrase "black hole information paradox" is indeed somewhat misleading, I will grant you that.

Tim,

ReplyDeleteBack to the main topic. First of all (the easy part) you asked about the role of t versus \alpha. Keeping in mind that we are in AdS, and considering the Penrose diagram for black hole formation/evaporation, it's clear that for a given t only a subset of points of the full spacetime are spacelike separated to the points at time t at the boundary. As we evolve t we "scan" over different regions of the spacetime that are so spacelike separated. So the wavefunction as a function of t will contain information about different regions of the spacetime. This is key because I want to eventually take t sufficiently large such that there are no connected Cauchy surfaces at this time, and I want to analyze the state at that time.

Now, as to the rest of your message, I am afraid I don't follow this at all. Perhaps you could point to a reference where "solution space" and "kinematical space" are used in this manner. Here is how I would phrase things in the simplest context of the canonical formulation of a single classical particle in 1 spatial dimension moving in some smooth potential. One has a phase space, which is R^2, and points are labelled as (x,p). On this phase space is a Hamiltonian function H(x,p). Points in phase space evolve in time according to Hamilton's equations: dx/dt = dH/dp, dp/dt = -dH/dx. (technically I assumed here that (x,p) are canonical coordinates so that the symplectic form is just dx dp, but let's not get into that). That's it.

Now, one could talk of a "solution space" as the space of all solutions to Hamilton's equations. But note that this space is equivalent to the phase space, because if you start at time t=0 at any point (x,p) then this uniquely labels a solution of Hamilton's equation with this boundary condition. This idea forms the basis of what is called "covariant phase space"

You seem to want to say, independent of Hamilton's equations, that there is some other restriction based on what are "physical possibilities", for example disallowing trajectories that involve a change in x without corresponding change in p. But where is this coming from? In the canonical formulation, x and p are treated as completely independent variables (unless there are constraints, which is not the case for the system at hand). There is no relation between their evolution beyond what comes out of Hamilton's equations. Indeed, when you derive Hamilton's equation from a variational principle you consider variations over all paths in phase space, and you would get the wrong answer if you first tossed out "unphysical possibilities" however that is defined.

Finally, I don't even understand the wording of sentences like " In classical mechanics, the kinematical space is the space of all continuous functions from points in phase space to a time parameter t, i.e. parameterized trajectories through phase space." In the part of the sentence before the" i.e." you speak of functions from phase space to to y, while a parametrized trajectory is a function from t to phase space. So the two halves of the sentence on opposite sides of the "i.e." mean opposite things.

Dear Sabine

ReplyDeleteAs with BHG, maybe I can now make progress with you after all these months. After a prelude, I will explain

1) What I take the situation to be.

2) What I understand the significance of the "information loss paradox" was supposed to be.

3) What the exact form of my "solution", or better to say debunking, of the paradox is.

4) What happens physically when a black hole evaporates.

5) Why the S-matrix approach is a distraction.

6) How sociological factors created the situation.

7) Why, as you ask, anybody should give a shit.

Prelude

This will be a long post. That will certainly put you off, and even make you suspicious. You recently wrote:

“This sometimes works because your argumentative style is so unnecessarily verbose that few make the effort of digging through it. “

The phrase “unnecessarily verbose” brought to mind the scene in Amadeus where the Emperor, after hearing The Abduction from the Seraglio, complains that there are “too many notes”. Physicists often have this reaction to my work, together with the judgment that I am “arrogant”. I think what they actually have in mind is “condescending”, and they think that because I am going through things very deliberately, step-by-step, small inference by small inference by small inference, constantly pointing out the exact meanings of terms being used. Their immediate reaction is “Yes, I know all this, why are you treating me like a child?” And the answer is that there are mistakes in the usual presentation, mistakes in what they think they know, and the mistakes turn on ambiguities, tacit premises that need to be brought to light, invalid argument structure, etc. I am trying to make the situation so explicit that no one can deny any claim I make. And the reaction, if they can’t point to an error but still don’t like the conclusion, is to just wave the whole thing away. “Too many notes”.

There is another reason my discussions are so long. I am, in fact, trying to both deal with a particular problem and to explain certain foundational issues to people who have literally never taken a class or read a text about foundations. The irony is that while I am falsely accused of not having taken an introductory quantum mechanics course, the truth is that probably none of the people posting here has ever taken an introductory foundations of physics course. Because foundations simply is not taught. Where is it you think you ever actually learned about foundations of physics? By the fleeting remarks in the standard intro physics texts, written by people who also never studied it? How did you come by this knowledge? Do great physicists at least understand foundations? Hawking’s most famous remark on foundational questions is “When I hear about SchrÃ¶dinger’s cat, I reach for my gun”. What is the appropriate response to such a statement, except “Fuck you, Hawking”?

As you know, Travis Norsen and Hans Westman and Travis Meyers and I have spent months trying to straighten Gerard ‘t Hooft out about Bell’s theorem and the meaning of “superdeterimism”. For an expert in foundations, this is literally Foundations 101. But not understanding it has led ‘t Hooft on a wild goose chase for decades. And my claim is that the whole physics community working on quantum gravity has been equally misled by the “information loss paradox”. The only chance of making this case is to make the case airtight, which means being very deliberate and careful. If you take the time to read carefully, to “make the effort to dig through it”, I would not have to fall into repetition, which is indeed boring.

I have broken this down into 7 sections so we can deal with them separately. If you have an objection, locate it in a section. If a section seems fine to you, just take that one off the table.

2) The Significance of the “Paradox”

ReplyDeleteA paradox arises when starting from principles that are generally and widely accepted (the “doxa”), one can derive either a contradiction or an unacceptable consequence. A contradiction is obviously the most powerful here result: it signals that something generally and widely accepted is false, although it does not indicate which of the premises is the false one. If the conclusion is not a contradiction but just unacceptable for some other reason, one can always “bite the bullet” and accept the unacceptable. These are really the same case though: if the conclusion is unacceptable, then the denial of the conclusion is widely and generally accepted, so one can just reframe the argument as a derivation from widely accepted premises, now including the denial of the bad consequence, to a contradiction.

In cases of extreme desperation, one can, instead of abandoning one of the premises, attack the very logic itself used in the derivation. This desperate gambit sometimes goes by the name “quantum logic”. We will not consider it.

The reason that the “information loss paradox” has been considered so important is that it is exactly presented as being of this form. One starts with very general and widely accepted principles—some from GR and some from QM—and in the setting of black hole evaporation derives a contradiction. If correct, then it shows that the tension between GR and QM is not merely a problem of not having found the right way to combine them mathematically, but rather that they simply cannot be combined in any way that retains both of their core commitments. Something’s gotta give. It could be GR and it could be QM and it could even be both, but one of our most successful fundamental theories will have to be abandoned or modified in a radical way. In short, the particular setting of black hole evaporation puts GR and QM in direct logical conflict. So understanding how to resolve the conflict in that setting will give guidance about which theory has to be modified, and therefore how to proceed.

The conflict is supposed to go like this: According to the basic principles of QM, the evolution from the state on Sigma 1 to the state on Sigma 2_Out must be unitary, deterministic, and information-preserving. But according to the basic principles of GR, that evolution cannot be unitary (in one sense of “unitary”! Warning!) or information preserving. So you have to choose sides here. Save QM and abandon GR or save GR and abandon QM. If you can show that one or the other of these options can yield a fruitful way to proceed, you have pointed a path forward.

3) The debunking solution

ReplyDeleteThe point of my paper is that this is just a false dilemma. The choice offered is a false choice from the get-go. It is true that the basic principles of “standard” QM imply—in the appropriate sense!—that the evolution of a quantum system is deterministic and unitary and information-preserving. Now by “standard” quantum theory I here mean a theory that is committed to two principles: a) The completeness of the wavefunction, in the sense that all of the physical properties of a system are reflected, somehow, in the wavefunction. That is, no “hidden variables”. And b) No collapse of the wavefunction. The wavefunction always evolves deterministically and linearly and unitarily (in the right sense of “unitary”: preserving the inner product!). That evolution “preserves information” in the sense that later wavefunctions can be predicted from earlier ones, and earlier ones retrodicted from later ones.

It is true that any approach to quantum theory that is committed to a) and b) is, willy-nilly, going to be a Many Worlds theory. I don’t think that physicists generally appreciate this: that is Foundation 101. And that any Many Worlds theory has to confront the Measurement Problem, AKA The Problem of SchrÃ¶dinger’s Cat, is also Foundations 101. Stephen “I reach for my gun” Hawking certainly appears to be completely unaware of this, and his discussions always proceed as if he had solved the Measurement Problem in a non-Many-Worlds way, although he won’t breath a word how. So it is impossible to make any clear sense out of what he writes. I will try to get through this discussion without having to go into these foundational issues, because they are not relevant to my discussion of the Information Loss Paradox as it is standardly presented. I will take a) and b) as terms of the discussion.

Now the debunking solution is simple. It notes that the only sense in which QM or QFT (by which I mean theories like QED and Chromodynamics, which are much, much more than just scattering matrices!) are committed to determinism and unitarity and linearity and information-preserving, the only sense from the very beginning, has been for Cauchy-to-Cauchy evolution. That is Wald’s point. And that Sigma 1 to Sigma 2_out, what you call the “outside evolution” is not Cauchy-to-Cauchy is trivial. So no principle of QFT has ever implied that the “external evolution” should have the properties that everyone is demanding that it have! Even more, the basic principles imply—taking the Penrose diagram given by Hawking seriously— that that evolution can’t be unitary, and linear and information preserving.

Furthermore, again taking the Penrose diagram seriously, there is absolutely nothing that prevents a theory from having exactly the properties that QM or QFT does demand: unitarity and linearity and information preservation on all Cauchy-to-Cauchy evolutions. So the basic principles of GR and the basic principles of QM are not in any conflict at all in this case. There isn’t, and never was, any paradox.

4) What Happens Physically?

ReplyDeleteAccording to the solution I am advocating, here is what happens. First, enough matter and energy, of whatever kind, get crushed together for a trapped surface and an event horizon to form. That event horizon appears, from spatial infinity, to be radiating Hawking radiation. It therefore shrinks. The event horizon gets smaller and smaller until eventually it shrinks to nothing and disappears. When it disappears there is a flash of light: all of the light caught on the event horizon gets released. There is no “remnant” or anything else “left over”. But from the perspective of someone who has seen the evaporation run to completion, from that location in space-time in the future of the Evaporation Event, the interior of the black hole still exists and contains information about what went in to form the event horizon in the first place. This answers Hawking’s “Where’s the information” puzzle.

5) The irrelevance of the S-Matrix

We are here interested in fundamental physics, that is, physical principles and laws that govern all of physical reality, all interactions, all of space-time. If we have such laws, then they obviously cover all possible experimental conditions. In particular, they cover “scattering experiments”, whose initial states and final states asymptotically approach free states. A fundamental theory will determine all the details in such a setting: how the particles interact with each other and scatter, etc. A fundamental theory will imply a scattering matrix.

But the converse is certainly not true. Just knowing how scattering works in this ideal limit does not tell use what a fundamental theory should, namely what is going on between the initial state and the final state. Indeed, the scattering ansatz is clearly inappropriate for most physical interactions. So our job is not just to get the scattering right, it is to get a complete theory.

But there is one conceptual/linguistic point that should be emphasized. Suppose that the scenario presented in my paper is correct. Then there will, of course, be a mapping from incoming free states of matter that will form a black hole to outgoing free states of Hawking radiation. That mapping will carry pure states to mixed state: improper mixtures, to be exact. That mapping will not be “unitary” in the sense of “preserving inner products”. But it will be unitary in the sense of “preserving probability”, that is, the sum of the probabilities of all of the possible outcome states will be 1. Of course.

Now when I say that the S-matrix is irrelevant, I of course do not mean there isn’t one. Just as Hawking said, it will be a superscattering matrix. What is the exact form of it? Well that question obviously requires a full theory of quantum gravity, which I never even hinted I have! If that is your complaint about the paper, then your expectations were never vaguely reasonable.

6) Speculations on Sociology

ReplyDeleteIf the debunking argument is right, then there never was a paradox, there never was an even prima facie conflict between GR and QM, there never was any reason to think that focusing on the question “How does the information get out” would be a good idea. In fact, it is a positively bad idea, since the information does not get out. But then why has the “Information Loss Paradox” caused so much spilled ink over the last 40 years?

That, I think, is due to sociology. Once the situation is presented as a death-match between QM and GR, as a place where one or the other theory will be forced to abandon a basic principle, the partisans of each theory face off against each other. The Relativists, including Hawking and Wald and Thorne, take the side labeled “Information is lost”, meaning that information on Sigma 1 does not make it to Sigma 2_out. The high-energy folks, such as Susskind and Banks and Preskill, take the side labeled “Information is not lost”, meaning that the information does get out to Sigma 2_Out. Bets are made. Tempers flare. Each side feels that it is defending its own pet theory from the other, because one or the other theory will have to concede a major defeat. All of this goes on for 40 years.

The debunker says that the whole situation is confused. Nobody has to lose. Nobody’s pet theory has to change. Just be clear about what the theory really entails, and everybody is OK. All of this argument has been a waste of time.

The debunker also says that, since you made the bet, it is Hawking and Thorne that win. Information does not get out to Sigma 2_Out. It is not somehow transported across the event horizon. But so what? It doesn’t have to be.

7) Why Give A Shit?

Finally, why should the debunking be considered important at all, given that by itself it does not provide any actual theory of quantum gravity, any actual detailed superscattering matrix, etc.?

I would have thought that the answer is obvious. Because pretty much all of the quantum gravity community, who do not understand this, seem to have been wasting their time. What is the whole “firewall” proposal about? Supposedly, solving the information loss paradox. (Not by explaining how the information gets out, but explaining why it never gets in in the first place.) What is ER = EPR all about? Trying to solve the information loss paradox by smuggling the information out through wormholes. What is the “Page time” all about? What it would take for the evolution from Sigma 1 to Sigma 2_Out to be pure-to-pure rather than pure-to-mixed. What was “black hole complementarity” all about? God knows, it is such a mess, but information loss is in there somewhere.

More pointedly, why did Hawking concede his bet and Thorne not concede? Because Thorne correctly sees what the situation is and Hawking does not. That’s what the debunker says. So if you give a shit about any of that, you should give a shit about this analysis.

Even more directly, the quantum gravity community has been wasting precious resources for 40 years trying to solve a non-problem. And worse than that, trying to come up with a solution which, if it worked, would create a fundamental problem, because the solution would have to violate some fundamental principle. And until this “paradox” is properly diagnosed, that waste of time and effort will go on.

So I do not come bearing the solution to quantum gravity. But if I’m right, I do come bearing information that will help get the quantum gravity community to get their heads screwed on straight. If I’m wrong in what I argue, then I’m wrong. Find the error and point it out. I am doing my best to facilitate that. But if I’m right, then I think you really ought to give a shit.

Tim,

ReplyDeleteYour diagnosis is a good starting point.

Let me add to this. You are missing that no one thinks the singularity is anything but a signal that GR breaks down. If you remove the singularity, you end up at pretty much the same point as Lee and I - with a remnant as the "obvious" solution. It's the obvious "conservative" solution, as we put it, meaning it's what you get if you just keep on using qft and gr without any twiddles and thumbs, and no, there is no problem with that. If you want to know what happens with the remnant though, you'll need to have a theory of quantum gravity. Same is the case in your scenario. Even if you want to keep the singularity you'll need quantum gravity to figure out what happens in the vicinity.

Next, you will notice however that there is nothing that prevents you (or rather theoretical physicists) from making changes to the physics in the vicinity of the *horizon* (not the singularity), provided that's not in conflict with anything we know. There are thus many "non-conservative" ways of doing this and none of them is "wrong" in a mathematical way.

Thus, the problem can't be solved based on logic alone, so why discuss it to begin with. Here by "problem" I am not referring to a paradox (I think we all agree that there isn't one) but to answering the question "what happens". Math alone will just not give a unique answer, and there is no reason to think so. I hope your great philosophy courses have taught you that.

Having said that, you could have concluded from your own diagnosis that your paper would be entirely ignored which is what I tried to warn you about well ahead of time. This would have been avoidable, to some extent at least. I consider this a missed opportunity. You could have made a difference. For reasons I can't fathom, you chose not to use this opportunity.

Though I think that it's an exaggeration to say that all of the 40 years were wasted. We did learn a lot about quantum field theory in curved space, renormalization of the stress-energy, and yes also about the relations between gravity and gauge-theories via AdS/CFT. But yes, most of it is a waste of time. Good luck convincing anyone of it. Best,

B.

BHG-

ReplyDeleteJust quickly on Hawking, and then let's leave him alone, because we won't get anything out of dissecting things further.

Even if you can read my solution into Hawking's words (and we would have to check what he says about the post-evaporation situation to verify this), the last sentence is completely contrary to anything I would ever say: " One could still claim that the system was in a pure quantum state though this would be rather metaphysical because it could be measured only by an angel and not by a human observer. " I have no idea what this is supposed to mean. As we know 1) no quantum state on an individual system can ever be measured by any human observer (if I hand you an electron and you want to know what spin state it is in you are out of luck). 2) the only states that really matter for information preservation are universal states, i.e. states on complete Cauchy surfaces, and these can never be accessed by any human observer, and 3) there are, and can be, no human observers "at infinity" because, first, there is no such place and second even if there were there will never be any humans there. My point in saying all this is that there is no real content to Hawking's dismissal of the claim that the universal state is a pure state as "rather metaphysical" because it cannot be observed by a human. It is not at all metaphysical—it is a straightforward physical claim. If you want to restrict yourself to things that some living, breathing human being could access then the whole information paradox will never get off the ground. The paradox is about the evolution of the universal quantum state, not about what anyone could measure, even in principle. So let's just let Hawking go.

On the kinematical space/solution space distinction. First, yes, I messed up that sentence: it should be a function from t into the phase space. Sorry.

Second, I characterized a solution to a classical problem as a (continuous) function from t into the phase space (i.e. a t-parameterized path through phase space, or a continuous t-monotonic path through Phase-spece X R) because that is what we want: just such a function. Call the space of all such functions the "mathematically complete space". Now I want to make the following distinctions within the mathematically complete space. Some functions are what I will call kinematically incoherent. For example, the function that keeps the xs fixed through all time but does not have p = 0 for all time. The kinematically incoherent functions are easy to characterize. Take a set of N point particles in R^3, and consider all of the possible continuous motions of all of the particles. Each such motion corresponds to a trajectory through phase space. But not every trajectory though phase space can be so generated. If it can't, it is kinematically incoherent.

So starting with the mathematically complete space, eliminate all the conceptually incoherent ones. What remains are the kinematically possible trajectories.As I understand it, you claim that if I use a variational principle on the kinematically possible trajectories rather than the mathematically complete space I will get some wrong answers to some problems. I find this claim highly dubitable. Do you have any citation of such a proof? Nothing hangs on it, but I am just interested.

A little googling has shown me that the term "kinematical possibility" is more often used by philosophers, but the term "kinematical constraint" is certainly used when discussing variational principles and virtual work. (See, e.g. Lanczos). A kinematical possibility is just a motion that satisfies all the kinematical constraints, and the kinematical space is the set of all kinematically possible motions. A kinematically incoherent trajectory can never meet all of the kinematical constraints because it does not even correspond to a motion!

Con't

A trajectory can be kinematically possible but not dynamically possible. The dynamics of the theory are laws that posit relations between the state of the system at one time and its state at other times. Only trajectories that meet this extra constraint are dynamically, i.e. physically, possible. The set of all such trajectories is the solution space.

ReplyDeleteI hope at least it is clear what I mean. I want to start with the whole time-dependence of the successive states built into the solution. There is really no need to start with a mathematical space that contains conceptually incoherent trajectories. But if we start with the space of kinematic possibilities, we can regard the dynamics as an extra constraint that picks out the solutions from it.

A solution, then is not a point in phase space but a t-parametized trajectory through phase space. It may be (or may not be) that every point in phase space is the initial point of some solution. I hope you now see that this claim of yours is not well-formulated:

"Now, one could talk of a "solution space" as the space of all solutions to Hamilton's equations. But note that this space is equivalent to the phase space, because if you start at time t=0 at any point (x,p) then this uniquely labels a solution of Hamilton's equation with this boundary condition. This idea forms the basis of what is called "covariant phase space""

Yes, the solution space is the space of all solutions to Hamilton's equations! But such a solution is not, as you seem to suggest, a point in phase space, it is a t-parameterized trajectory through phase space, and not every such trajectory is a solution.

There is a little irony in your saying that one needs the kinematically possible non-solutions—and even the kinematically incoherent non-solutions!—for the variational principle to pick out the solutions. That is, you are remarking on the mathematical necessity of having the non-physical trajectories in order to determine the physical ones. Earlier, you kept saying that the Hilbert space contains only solutions, and the non-solutions are just garbage. You are now taking the opposite tack.

Anyway, in non-relativistic qm a solution is a time-indexed function into the Hilbert space. Not every such function solves the Hamiltonian. Again, from the set of all time-indexed functions into the Hilbert space the Hamiltonian selects only the solutions.

Thinking about it this way smoothes the path to quantum gravity. A solution to the quantum gravitational equations will be a full 4-dimensional space-time. We start with a kinematical Hilbert space, which is all of the mathematically possible representations of a space-time and its contents. And from this large Hilbert space we want to select some subset of solutions. In the case of Wheeler-deWitt the condition for being chosen is being annihilated by the Hamiltonian. The set of all such states in the original space forms the solution space, That's how I understand the term.

Sabine,

ReplyDeleteWell, I am glad that we seems to be coming to a bit of agreement! Let me try to expand the circle of convergence.

I certainly agree that there is some new physics needed if you want to cure the singularity. My paper just isn't about that problem. If you can completely cure the singularity and extend the space-time there, that alone does not touch the supposed paradox, which arises from the horizon structure. So what does the paper do?

One thing is this: if you want to keep a globally hyperbolic spacetime and solve the problem in this "conservative" way, then you will have to have a non-manifold point at the Evaporation Event. So going that way, we can at least focus attention on the physics of non-manifolds. Of course, maybe at some fundamental scale space-time isn't a manifold anywhere, but playing around with this case will give some insight into what that might mean.

On the other hand, if you remove the Evaporation Event then you no longer have a globally hyperbolic space-time, and you need to deal with a bare singularity. There is also a lot to chew on there.

Finally, if you think there is some untapped potential in the paper, please feel free to just write your own, drawing on anything useful here. At the moment, I am just trying to understand what the supposed problem was supposed to be. BHG claims that AdS creates a problem for this debunking solution, but I still can't make out the argument. Somebody will learn something by the time we are done.

Tim,

ReplyDeleteI am not motivated to add further papers to the pile on a questions that, as I explained above, I consider to be impossible to make headway on without experimental input.

I also didn't mean that there is anything to expand on in your paper (at least not that I see) but that it wasn't hard to see the way you presented it would just make everyone in the field say you didn't get the point. To avoid yet another misunderstanding, note that I did not say you didn't get the point, but that your paper clearly raises the impression by not acknowledging the S-matrix issue that launched the whole discussion.

All you have to say about this (here, not in the paper) is:

"Well that question obviously requires a full theory of quantum gravity, which I never even hinted I have! If that is your complaint about the paper, then your expectations were never vaguely reasonable."Well yes, obviously, and obviously everyone knows that. That's exactly the reason why no one studies this case or the case with remnants, because you cannot calculate anything and thus not write papers about it. (Or maybe you can, if you're a philosopher, but not if you're a physicist.) That's opposed to AdS/CFT which is great because you can calculate things and publish them, nevermind that the results might have nothing to do with physics.

Best,

B.

Tim,

ReplyDeleteI do want to say one more thing about Hawking, because I think he gets the situation just right in the passage I quoted, and I don't agree with your criticism of it. To avoid issues about infinity, let's imagine performing experiments in a large sealed lab. In the context of non-gravitational physics, I think we all agree that we can prepare a pure state, let it evolve in time, and it remains a pure state. Furthermore, this claim can be verified experimentally: by preparing a large number of identical copies of the initial state (i.e. repeating the experiment many times) we can, by standard protocols, verify whether or not the final state is indeed pure. Now imagine that our experiment involves black hole formation/evaporation. According to Hawking, the evolution is from pure states to pure states, but the radiation alone is in a mixed state, with the purification of the state lying behind the horizon. Key point: this scenario is experimentally indistinguishable to one in which the initial pure state evolves to a mixed radiation state period. Since we cannot in principle measure the state inside the black hole, there is no conceivable experiment that can distinguish whether the evolution is fundamentally from pure to pure or from pure to mixed. Hence his rhetorical flourish about angels and metaphysics. In all other examples of effective pure to mixed evolution (i.e. due to photons escaping out the window of our lab and heading to Andromeda) we can remove the issue by working in a sealed lab, but not so (according to Hawking) in the case of black holes. That's a big difference.

So if in AdS/CFT, the CFT reflects the full state of S2_in + S2_out in the bulk then naturally, the CFT will exhibit unitarity; if the AdS/CFT claim is that the CFT reflects only S2_out, then there is indeed another paradox.

ReplyDeleteIn particular, the claim seems to be that the state in S2_out and the CFT together can be "relaxed" to the vacuum. This "relaxation" doesn't touch S2_in in the bulk. The vacuum is non-degenerate in the CFT; but S2_in causes the bulk vacuum to be degenerate to a high degree, and therefore AdS/CFT rules out S2_in. (Basically S2_in doesn't leave a trace in the CFT, so at late times the CFT maps only to S2_out.)

Tim,

ReplyDeleteYour usage of terms like "solution space" and "kinematical space" are quite different from what appears in any of the physics literature I am aware of, and it will be important to arrive at some agreement on terminology. I will lay that out shortly, but first let me address your specific questions

1) You ask "As I understand it, you claim that if I use a variational principle on the kinematically possible trajectories rather than the mathematically complete space I will get some wrong answers to some problems. I find this claim highly dubitable. Do you have any citation of such a proof?"

The proof is as follows (for more details just google "phase space variational principle" or look in a classical mechanics book for a derivation of Hamilton's equation from a variational principle). Let's just consider a single particle in a potential. The action in phase space is S = \int ( p dx/dt - H(x,p))dt. To get the equations of motion we do the following. We consider all trajectories x(t), p(t), assuming no relation between the two functions. We impose the boundary condition x(t_1) = x_1 and x(t_2) =x_2, but with no boundary condition on p. Then we demand that the action be stationary under all such variations. Doing so, you find that the trajectory we are varying around must obey Hamilton's equations. Now, if you had imposed some restrictions on the class of trajectories (corresponding to omitting some set of "unphysical possibilities") you will clearly get something weaker than Hamilton's equations. In particular, if some variation (\delta x(t), \delta p(t)) is disallowed by this criterion, then clearly you are throwing out the part of Hamilton's equations projected onto this variation. Hamilton's equation only arise by demanding S be stationary under *all* variations that obey the boundary conditions.

2) Tim: " A solution, then is not a point in phase space but a t-parametized trajectory through phase space. It may be (or may not be) that every point in phase space is the initial point of some solution. I hope you now see that this claim of yours is not well-formulated:"

BHG: "Now, one could talk of a "solution space" as the space of all solutions to Hamilton's equations. But note that this space is equivalent to the phase space, because if you start at time t=0 at any point (x,p) then this uniquely labels a solution of Hamilton's equation with this boundary condition. This idea forms the basis of what is called "covariant phase space""

Tim: Yes, the solution space is the space of all solutions to Hamilton's equations! But such a solution is not, as you seem to suggest, a point in phase space, it is a t-parameterized trajectory through phase space, and not every such trajectory is a solution."

Actually, my claim is well formulated, and it is a mathematically rigorous statement that you can read about in the beautiful and accessible paper by Crnkovic and WItten "Covariant description of canonical formalism ...". Again, the statement is that the space of solution to Hamilton's equations is *mathematically* the same space as the phase space; any solution is uniquely labelled by a point in phase space corresponding to its data at some initial time, and any point in phase space uniquely specifies a solution.

3) "There is a little irony in your saying that one needs the kinematically possible non-solutions—and even the kinematically incoherent non-solutions!—for the variational principle to pick out the solutions. That is, you are remarking on the mathematical necessity of having the non-physical trajectories in order to determine the physical ones. Earlier, you kept saying that the Hilbert space contains only solutions, and the non-solutions are just garbage. You are now taking the opposite tack."

ReplyDeleteNo, I am being entirely consistent. I.e. in E&M states that don't obey the Gauss law constraint are indeed unphysical garbage. Now, in the usual form of the path integral you integrate over all trajectories, even those that don't obey the Gauss' law constraint. But the integration over A_t precisely produces a delta function that enforces Gauss' law, so in fact only trajectories that obey Gauss' law contribute, and hence you can reformulate things such that only such trajectories appear from the beginning. The situation here should not be confused with the discussion above about single particle mechanics, where there is no constraint (no analog of Gauss' law). There you need to integrate over all trajectories period. The claims about some states being unphysical garbage is specific to theories in which there are phase space constraints; the claim being that states that don't obey the constraints are garbage. Part of the confusion here, and this is what I'll get to later, is we need to be careful to distinguish between solutions of the equations of motion versus solutions of the constraints. These are *very*different notions.

Next, when I get a chance, I will write a comment to standardize nomenclature, which is important to avoid confusion.

BHG-

ReplyDeleteI'll be brief so we don't get side-tracked. What you say is "a big difference" is a difference, but not big—indeed of measure zero—in importance.What we are discussing are the principles that are validated (or not) by theories, e.g. "information preservation" (backward determinism). None of this has to do with what an actual observer could do. With a positive cosmological constant, Alice and Bob, on different galaxies, will eventually lose the ability even in principle to communicate, and so can't verify what has happened in each other's labs. And no universal state can ever, even in principle be measured even once, much less could one manufacture a huge ensemble of identically prepared universes to do tomography on even if it could. Talk of what can be verified by observation has no place in this conversation at all: it is an analysis of the physical properties of a world according to a theory. Whether that world contains observers at all, and what they could in principle observe, is neither here nor there.

Talk of measurement here, or "what one could tell"from, e.g. the state of the Hawking radiation, is just metaphorical talk about what the state of the Hawking radiation plus the laws of physics imply about other times and places. If you try to take it literally then the whole discussion never gets off the ground.

BHG

ReplyDeleteRegarding your point 2. In the case you describe there is a 1-to-1 correspondence between solutions and points in phase space because the initial value problem is completely unconstrained. That is, you are using a Hamiltonian without constraints. If there are constraints, then not every point in the phase space may yield a solution. But more important, I am trying to get you to think a different way about all this: instead of thinking of there being an initial value and then laws that determine the time evolution, think of starting with a collection of full 4-dimensional models and then cutting these down to the physical possibilities by some criterion, e.g. that it minimize the action. In such a case the "space of solutions" is certainly not in one-to-one correspondence with the space of kinematic states. It is only by conceptualizing things this way that we can understand what is going in with GR. GR, the Hamiltonian does not generate later states from earlier ones: it annihilates the solutions. This is a disaster if you think of the Hamiltonian as generating the future behavior. It is perfectly OK if what is being annihilated is a full 4-d solution. Then it is just a way to mathematically pick out the solutions.

BHG

ReplyDeleteYour argument in point 1 is an argument by "clearly". What you characterize as clearly true is not at all clearly true! I am using a variational principle to identify—among a set of trajectories—ones that locally minimize the action. That is a purely local matter in the relevant mathematical space, i.e. the space of all trajectories through phase space. There can be a perfectly well-defined mathematical fact about what action is ascribed to the kinematically incoherent trajectories but 1) it better not be that any such trajectory locally minimizes the action, or the theory will have a solution that literally makes no physical sense and 2) if these kinematically incoherent trajectories are nowhere near the minimizing paths in the space of trajectories, then deleting them will have no effect at all on the solutions. And the solutions are all you care about. So what you portray as clear is not clear in the least. Indeed, for classical mechanics, it is false.

If you don't impose the restriction to kinematically coherent trajectories, then the entire theory fails completely. This is trivial to see. Let the potential be 0, so we are in the free theory. If the initial location of the particle is x_i and the final location is x_f, then what we want to get is just inertial motion from x_i to x_f: p should be constant such that p(t_f - t_i) = x_f - x_i. dx/dt is constant = p/m. So the action is just the integral of p^2/m. In other words, the solution minimizes the time integral of the kinetic energy. If the particle went from x_i to x_f by any other continuous trajectory—speeding up in some places and slowing down in others—then the action would be increased because the higher action in the speeded-up parts would overbalance the lower action in the slowed-down parts. And that is because the kinetic energy is quadratic in the velocity.

Good. Now lets allow in the kinematically incoherent trajectories. Then you are screwed. Take any trajectory—which you are allowing— that starts at (x_i,0) and ends at (x_f,0) and has any continuous function you like for the x part and constant p=0 for the p part. The action is clearly 0, because of p = 0! So the physical solution is certainly not the global minimum! Is it even the local minimum? Obviously not. We are going to compare the action along the inertial trajectory dx/dt = C to a trajectory where the particle starts out going slower and then speeds up to get to x_f on time. So for the primed trajectory, dx'/dt = C + epsilon f(t), where Int(f(t)dt = 0, so the average velocity remains the same and the particle gets to x_f on time. Let's bound f(t): |f(t)| < B.

Now: consider the following epsilon-indexed set of trajectories: (x'(t),p'(t)) = (Ct +epsilonf(t), Cm( 1- epsilon(2B/C-B))). When epsilon = 0, we get the inertial trajectory, but for arbitrary bounded f(t), the action of every trajectory with epsilon > 0 is strictly less than the action on the inertial trajectory. Of course, every one of these trajectories is kinematically incoherent, but you are insisting on minimizing our the full set! So the situation is just the opposite as you claimed: if you don't restrict your variation to the kinematically coherent trajectories, you don't even get the right answer! Some nearby kinematically incoherent trajectory will have a lower action

Tim,

ReplyDeleteTim: "Your argument in point 1 is an argument by "clearly".... if you don't restrict your variation to the kinematically coherent trajectories, you don't even get the right answer! Some nearby kinematically incoherent trajectory will have a lower action"

You are just making a simple computational error here. I will explain precisely where that is if you want me to, but I would appreciate it if you would save me the effort by consulting a reference on the phase space variational principle (pretty much any mechanics textbook will do) since I promise that you will find it works exactly as I stated. I am not trying to pull a fast one.

BHG

ReplyDeleteLeave the computational error aside, and this isn't really important but look: if a trajectory locally minimizes the action in the mathematically complete set of trajectories, including the incoherent ones, then it obviously minimizes the action in any subset of that complete set that it is part of, and the deletion of any set of trajectories from the mathematically complete set can't change that, So your claim that leaving them out would give the wrong result in the classical case just cannot be right. Right?

BHG

ReplyDeleteLooking over our last exchanges, I got myself into unnecessary complications because of your remark about treating x and p as independent variables. So what I should have said (which is equivalent to what I did say, but simpler) is that the set of kinematical possibilities is just the set of all time-parameterized curves in *configuration* space, not *phase* space. Each of those will, of course, correspond to a time-parametized trajectory through *phase* space, and it is over these that we want to minimize the action.

I think you have gotten confused by this: when solving a one-dimensional problem this way, you do minimize the action over a set of paths, but those are not usually depicted as paths in phase space: they are paths in the x-t place, and more particularly functions from t to x. You minimize over all such functions, which is the same as minimizing over all t-parameterized paths in the configuration space. You are not minimizing over all paths in phase space. The fact that p is proportional to dx/dt is critical for the proof. In that sense, p and x are not really treated as independent of each other. I can't vary x(t) without varying x'(t) and hence p(t).

BHG

ReplyDeleteNever mind about the phase space variational principle. I am still puzzled about it—I had to get straight what the exact form of the kinetic energy term is—and I seem to have even simpler examples of why the proper solution is not minimizing, and even seems not to be stationary—but this will obviously take us away from what we wanted to talk about. So just let it go. I'll get it cleared up myself.

Tim,

ReplyDeleteOK, just as long as we're clear that what I said about the phase space variational principle is correct and indeed standard.

However, this does raise a more serious problem that has me worried. In another recent message you wrote: ", I am trying to get you to think a different way about all this: instead of thinking of there being an initial value and then laws that determine the time evolution, .... " and " . It is only by conceptualizing things this way that we can understand what is going in with GR. GR, the Hamiltonian does not generate later states from earlier ones: it annihilates the solutions."

The trouble is that the claim you are basing this on, " the Hamiltonian does not generate later states from earlier ones: it annihilates the solutions." is simply unambiguously false. In a space with an asymptotic structure like AdS, which is what we are discussing, it is just a flat out mathematically wrong statement no matter how you slice it, and I have tried to explain this over and over. The Hamiltonian does not annihilate the state. The Hamiltonian acting on a physical state is a boundary operator that acts nontrivially on the state, in particular evolving it with respect to boundary time according to the Schrodinger equation. Now, to try to convince you of this I have

1) offered intuitive arguments

2) directed you to the original literature (Regge and Teitelboim)

3) pointed to statement by experts in GR and quantum gravity (Marolf, Horowitz)

What I cannot do is give you a line by line derivation. The reason I can't do is twofold. First, it is impractical to do so in a blog comment. But more importantly, the technical knowledge required to follow this argument involves ideas that build on but are an order of magnitude (or more) sophisticated than the single particle phase space variational principle stuff that you are clearly still trying to absorb. So what am I supposed to do? All I can think of apart from repeating (1,2,3) is to ask that you put some trust in me that I really do know the in and outs of this subject

Unfortunately, this is not a minor point. It is absolutely vital to understand that the quantum state in AdS is a function of boundary time that evolves according to the boundary Hamiltonian. In particular, the wavefunction at a given time does not contain information about the full spacetime, only about the points in the spacetime that are spacelike connected to the boundary points at time t (in the semiclassical regime at least). Again, I can give you intuitive arguments for this, but you must understand that I understand this very well on a deep technical level, and I am only providing less rigorous intuitive arguments in an effort to speak your language.

BHG

ReplyDeleteLet's try to avoid the technical details and think conceptually about what is going on. You seem to believe that the solution to all of this lies in some sophisticated math, but it does not. Let me make now one basic point that should have been recognized long ago, and which is implicit in what I have already written, but whose significance has not been appreciated (including by me!). The big point is this:

AdS is not globally hyperbolic!

The implications are clear: AdS does not contain any Cauchy surfaces. Therefore, my argument, which runs on being careful about your Cauchy surfaces, has no precise analog in AdS. Both of us have been ridiculously blind to this simple point: me by thinking that I can just adapt my argument from the asymptotically flat setting to the AdS setting, and you by thinking that you can just adapt your argument from the AdS setting to the asymptotically flat setting. This has led to a bunch of confusions on both our parts. The main one is the one I think that you acknowledged: in the globally flat case, the entire space-time (save the fictitious boundary) is spacelike separated from spacelike infinity. This is not true for any point along the time-like boundary of AdS. Or, the other way to put this is: AdS has a timelike boundary and the asymptotically flat space-time does not. That means that there are deep and fundamental mathematical differences between dealing with AdS and dealing with the Penrose diagram that I discuss in my paper. Those differences are have been bubbling up here, unrecognized by both of us.

Now there are two ways to proceed from this point. One is for you to concede that the difference between AdS and any asymptotically flat (or, more generally any globally hyperbolic) space-time is absolutely critical to the argument you are trying to make, and that no matter how it comes out it has no direct bearing on my paper. In a space-time like AdS, with a timelike boundary, of course there can be no question about "information preservation": information on any edgeless spacelike hypersurface (i.e. the state on such a surface) is never enough to determine the state on the whole space-time, and hence on every other edgeless spacelike hypersurface.

Here is another way to put this. In order for there to be an "information Loss Paradox", there has to be at least some prima facie reason to think that information won't be lost. What I point out in my paper is that "information loss" is really just code for "failure of determinism in the past direction" and that the only situation in which one has any right to expect such determinism is for Cauchy-to-Cauchy evolution. But let me now re-express that in a way that will be useful and eliminates talk of Cauchy surfaces: you only have a right to expect determinism for the evolution from the state on one surface to the state on another surface that lies in the domain of dependence of the surface on which the "initial" data lie. Since the entire space-time lies in the domain of dependence of a Cauchy surface, Cauchy-to-anything evolution ought to be deterministic, and therefore Cauchy-to-Cauchy ought to be bideterministic: both predictable and information-preserving.

Con't.

Con't

ReplyDeleteSince AdS has no Cauchy surfaces, the first question is what sort of evolution one has any right to expect to preserve information in the first place. And here you have two choices.

1) You are only interested in some sort of initial data from which the state on the *whole spacetime* follows. It looks like this is a perfectly fine thing to ask for in the case I discuss: data on Sigma I, for example, should do the trick. But then when we move to AdS you are screwed. Exactly because of the timelike boundary, there just are no surfaces that have the relevant property. So just pack up and go home. AdS is not the right setting to even ask this question.

2) You are interested in some sort of initial data that determine what the ultimate outcome of the black hole evaporation is, even if it does not determine the entire space-time structure. Then we can proceed even in AdS. Take a spacelike hyoersurface that cuts "below" the formation of the black hole. Take another spacelike hypersurface that runs "above" the evaporation, *but limits to the same point on the boundary as the first surface*. The second can lie in the domain of dependence of the first, *but only if they limit to the same point on the boundary*.In this case, as I mentioned before, the relevant evolution that takes you from the first surface to the last is what you called alpha evolution, not t evolution along the boundary. To put it bluntly, the boundary is far enough away from the black hole that what happens in one place has no influence in the other. So if you really want to discuss black hole evaporation and information in AdS, pick an appropriate point on the boundary so that the whole of the evaporation process (or at least most of it, if there are remnants) is at space like separation from that point, and leave the time evolution at the boundary out of the discussion. Focus instead on alpha evolution.

This has been the conceptual source of a lot of our disagreements. The Hamiltonian in AdS essentially needs to have a boundary term exactly because it is not globally hyperbolic: data can disappear into and unexpectedly emerge from the boundary. To that extent, the boundary is like a naked singularity. One can stipulate some boundary conditions to get that under control, but the boundary conditions will not be a consequence of the Hamiltonian.

It is because I am only used to thinking about the Hamiltonian in a globally hyperbolic setting, where there are no timelike boundaries to deal with, that we have constantly been at loggerheads over the Hamiltonian. You insist that the Hamiltonian has two bits: a bulk term and a boundary term. And in AdS we see why that has to be the case: the bulk term governs what we may call the alpha evolution from one spacelike hypersurface to another in its domain of dependence. This evolution does not depend on what is happening at the boundary, but only because the surfaces limit to the same point on the boundary. If you want to evolve to cover the whole AdS space-time, then you need the timelike boundary conditions at the timelike boundary. That is what your surface term in the Hamiltonian is all about.

So let's start by making this distinction. There is t evolution at the boundary of AdS, and there is also alpha evolution from one spacelike hypersurface to another in the bulk. My concern has always been alpha evolution. In the case I discuss, there is no t-evolution in this sense because there is no timelike boundary. Insofar as your considerations depend logically on t evolution, whatever they are they won't export to globally hyperbolic spacetimes.

Do you agree or disagree? If we agree on this, we have made tremendous progress. If not, we are not going to make any, and need to sort out the problem.

im,

ReplyDeleteYes, I am well aware that AdS is not globally hyperbolic and that this makes it different than asymptotically flat spacetime. In AdS we impose boundary conditions at the timelike boundary, and for generic matter fields there is a unique boundary condition that preserves the symmetry of AdS, so we use this. Once the boundary conditions have been chosen, the theory is "effectively" globally hyperbolic in the sense that data on a spacelike surface that cuts across the full spacetime uniquely determines the future evolution in time. So while calling such a surface a "Cauchy surface" is indeed a slight misnomer, there is really not much change in the physical content. We could call it an "initial value surface" if you prefer. In particular, one certainly has unitary evolution in AdS in the absence of black holes, there is no question about that (and of course I clam that this extends to the black hole case as well, taking the final state to consist just of radiation).

Next, I want to emphasize that the need for a boundary term in the Hamiltonian is not a consequence of this lack of global hyperbolicity. Indeed, note that the Regge/Teitelboim paper was written in the context of asymptotically flat spacetime, although the idea is much more general. Rather, the boundary term in the Hamiltonian arises due to the existence of an asymptotic structure. Fixing the asymptotic structure imparts a physical meaning to the t coordinate at infinity, and so there is a physical sense in which things can evolve with respect to it, unlike the case of a closed universe, where any given t is just an arbitrary time coordinate (this is my attempt to offer an intuitive argument).

Finally, and I think I have said this before, my arguments will only apply directly to black holes in AdS, as there is so far no analog of AdS/CFT in asymptotically flat spacetime. I am quite confident that in AdS/CFT all the information comes out in the Hawking radiation, but the only argument I can give at present does use properties of AdS in an important way. My *belief* is that this is just a temporary crutch, and that once the mechanism is better understood this can be extended to other spacetimes, including the expanding universe that we live in (again, I note that just as we don't live in AdS, neither do we live in asymptotically flat spacetime). But you are welcome to be skeptical about this.

I do want to ultimately focus on t evolution rather than \alpha evolution. One trouble with \alpha evolution is that it is only an approximate concept that makes sense in the semiclassical regime, while t evolution is well defined in general. To make a clean argument I want to consider t evolution to the far future, long after the black hole has evaporated.

BHG

ReplyDeleteThen let me ask this question. Fix a boundary time t_0. There is some state psi(t_0) which is supposed to correspond to (or even *be*) a state in the CFT. But that state is also supposed to be, or correspond to, a state of the bulk. But as we have seen, unless you are assuming or have fixed some foliation of the bulk, no state that can be determined locally at the boundary can be assigned to a particular hypersurface in the bulk. Because many, many different spacelike hypersurfaces in the bulk will overlap at the boundary. In the asymptotically free case, the union of all those spacelike hypersurfaces is the entire physical spacetime, so any state at spacelike infinity must characterize the entire space-time. In the case of AdS it is not so extreme, but it is still quite substantial: relative to some fixed boundary time there will be many, many hyper surfaces that meet the boundary at that time. The union of them all would be a huge volume in the AdS, a volume that could include the entire formation/evaporation of the black hole. So what part of the bulk, exactly, do you take that boundary state to represent? Or does it somehow represent the whole 4 volume? If not that, what principle can be used to discriminate particular events or surfaces in the bulk as represented in that boundary state?

Dear Professor Tim,

ReplyDelete1. What does unitarity of quantum mechanics mean in a non-globally-hyperbolic spacetime?

2. The AdS boundary CFT is said to have unitarity. Is this unitarity a case of the AdS boundary being globally hyperbolic, or does there too apply the extended definition of unitarity that you have to give when answering (1.) above?

3. Is it still true for the AdS spacetime and AdS blackhole that to any observer for whom the evaporation event lies in the past, the event horizon is neither in the past nor in the future of such an observer?

Answer to these three questions will greatly help mere mortals follow the dialog.

Thanks in advance!

Tim,

ReplyDeleteI am happy with how the discussion is progressing. I first want to clarify one point, which will also help explain how these boundary terms in the Hamiltonian come about. When we talk of a spacetime as being asymptotically this or that, we mean that as we go to infinity we require the metric to approach a specified metric (e.g. AdS or Minkowski space) with a specified accuracy, by requiring that the deviations die off sufficiently rapidly at large r. We then try to write down a Hamiltonian that generates time evolution compatible with these boundary conditions. For a general classical system the Hamiltonian is required to be a function on phase space such that the equations of motion (which have typically already been specified) take the form dx/dt = {x,H}, dp/dt = {p,H}, where { , } is the Poisson bracket, and (x,p) are the canonical coordinates on phase space. In a field theory like GR, H is the integral over a spacelike surface. When you compute the Poisson bracket you need to do an integration by parts, and the corresponding boundary term must vanish for the bracket to be defined. It is the vanishing of this boundary term that forces H itself to have an explicit boundary term whose role is to cancel this other boundary term. This procedure is only possible if admissible boundary conditions have been chosen, otherwise there might not exist any boundary term in H that can do the job. So the form of asymptotic falloff conditions and the form (and existence of) H are tightly linked. This is the content of Regge/Teitelboim.

The reason I bring this up now is that in the asymptotically flat case is if the "points" at spatial infinity at a certain time t are connected by spacelike surfaces to the entire spacetime, how can I say that H generates evolution with respect to t? It seems like the entire spacetime is already captured by surfaces connected to a single choice of t, so how can things change with t? The point is that we have to remember the asymptotic fall off conditions. I am only allowed to consider a subset of the possible spacelike surfaces that asymptote to a given t. If the surface approaches infinity while "bending too rapidly" the metric will violate the asymptotic conditions. So in fact, at a given time t the set of surfaces allowed only cover part of the spacetime; the covered region changes with t, making evolution in t sensible.

cont

Coming back now to AdS, let's consider a spacetime that is well in the semi-classical regime. At a given time t_0 on the boundary, we can ask for the region of the spacetime covered by spacelike surfaces that asymptote to t_0 while obeying the falloff conditions. The claim is roughly that the wavefunction \psi(t_0) will describe this region (sometimes called the Wheeler- deWitt patch). The more precise story, which is quite appealing, is the following. Recall that the wavefunction takes the form \psi[ g_{ij}, \phi; t_0], where g_{ij} is the spatial metric and \phi denotes matter. Now, we can ask for the value of \psi on any choice of g_{ij} and \phi we choose. So how can it be that \psi only contains information about a limited spacetime region? Here's how it works. In the semi-classical regime, we can divide the space of (g_{ij},\phi) into those for which |\psi|^2 is of order 1, and those for which |\psi|^2 is exponentially small. One can show that the (g_{ij},\phi) in the former category are precisely those for which, in the classical spacetime, a surface can be found on which (g_{ij}, \phi) take the specified values, and the surface asymptotes to t_0 in the allowed way. As t_0 evolves, we can therefore piece together the full classical spacetime.

ReplyDeleteThe CFT also has a wavefunction \psi(t), and the claim of AdS/CFT is that the \psi(t) in the bulk and in the boundary are the same wavefunction, just expressed in terms of two very different sets of variables, i.e. bulk variables or CFT variables.

As to where the argument is going, we are going to ask about \psi(t) in the far future, piecing together what we know about its structure from the bulk and from the CFT.

Dear Arun,

ReplyDelete1) Even in a non-globally-hyperbolic space-time, one can say that if the state on a hypersurface should determine the state on any later hypersurface in its future domain of dependence. The future domain of dependence of a surface is the set of all events e whose past light-cones are completely cut through by the surface: no timelike curve can get from past timelike infinity to e without intersecting the hypersurface. And the same mutatis mutandis for the past domain of dependence.

In these terms, a Cauchy surface is just a spacelike hypersurface whose domain of dependence (past and future) is the whole spacetime.

Suppose that there is a quantum state on some hypersurface. Then (assuming no collapses) the quantum dynamics can be used to calculate a quantum state on any surface in its domain of dependence. This will generally involve tracing out over degrees of freedom outside the domain of dependence, so this can be a pure-to-mixed evolution. But if the laws used to evolve the original state to the later surface are unitary, that restricts the sorts of evolutions there can be.

2) As far As I understand it, the CFT space is globally hyperbolic. That means that specifying the state on the boundary at one time determines it at later times.This does seem to be in direct conflict with the apparent need to fix the boundary conditions—in addition to the initial state—to evolve the AdS gravity theory. So I think that's a great question. Maybe BHG can answer it.

3) Yes, that will still be true. So that part of my account is unchanged.

BHG

ReplyDeleteI'm not sure what you mean by this sentence:

"As t_0 evolves, we can therefore piece together the full classical spacetime. "

If I am following, these procedures work in the semi-classical regime, but the black hole evaporation is not in that regime, is it? Here is another way to put it: fix a time t_E such that the entire creation and subsequent evaporation lie in the Wheeler-deWitt patch of t_E. Either you can somehow make sense of the space-time structure of the solution in that patch or you can't. If you can. then worrying about the states at other times seems pointless: we will already know what happens when a Black Hole evaporates.And if you can't. because you are out of the semi-classical regime, then how will looking at other moments of time, whose Wheeler-deWitt patches are just as obscure, help the situation? We can't piece together the whole spacetime if we can't get a result for that critical region. So we reach a dilemma:

If one can make sense of the space-time from before the formation of the event horizon to after the evaporation of the black hole using a single Wheeler-deWitt patch, then knowing the state at boundary times other than T_E is unnecessary. If one cannot make sense of the space-time structure using just the state at t_E, then knowing the state at other boundary times won't help. So either way, the evolution in boundary time is beside the point.

What boundary conditions on AdS are viable for classical fields is answered in:

ReplyDeletehttps://arxiv.org/pdf/hep-th/0402184.pdf

Dynamics in Non-Globally-Hyperbolic Static Spacetimes III: Anti-de Sitter Spacetime

Akihiro Ishibashi, Robert M. Wald

Arun

ReplyDeleteThanks for the citation. That looks like it is exactly in the right direction, but the "static" is worrisome. Our example is as far from static as one can get! but maybe this can be helpful.

Tim,

ReplyDeleteAbout boundary conditions in AdS, all I will say is that there is a huge literature on this (going back to Witten's first AdS/CFT paper), and a basic entry in the AdS/CFT dictionary is the relation between boundary conditions in the bulk and the corresponding modification of the CFT action.

Now, as to your main question, let me first note that in my discussion of piecing together the classical spacetime I specifically stipulated that the spacetime was semiclassical. I was just trying to explain how a classical picture properly emerges from the wavefunction in this case. As you correctly point out, in the case of an evaporating black hole the semclassical approximation fails near the singularity. So in this case, we are not able to explicitly compute the wavefunction \psi[g_{ij, \phi,t], and indeed this probably doesn't even make sense, since the metric description presumably breaks down near the singularity. If we were able to compute the correct version of \psi[g_{ij}, \phi,t] in this case, we could indeed use it to see how information does or does not escape from the black hole. But we can't so we are going to have to make some assumptions; what I have provided is the framework for describing the quantum state and its evolution, but that only takes us so far. The challenge is this: can we tell a story about how \psi[g_{ij},\phi,t] "looks" and evolves as t goes from early to late times in a way that respects the structure of the standard Penrose diagram, and is also consistent with AdS/CFT. My claim is that this is not possible, but we are going to have to go through this very carefully, and I suspect we will both learn something in the process.

First of all, on the CFT side the structure and evolution of \psi[\phi_CFT, t] (here \phi_CFT stands for the CFT field variables) looks fairly benign in the case of black hole formation and evaporation. During the collapse phase the CFT is in some out of equilibrium states. During the main process of black hole formation and decay the CFT is thermalizing, Finally the CFT settles down to some pure state that is approximately thermal. So this seems all very familiar.

Now we have to face the more difficult question of evolution in the bulk...

BHG

ReplyDeleteIt is nice that we are making progress, and I don't want to slow it down, but maybe we can keep a couple of discussions going in parallel.

I am getting conflicting impressions from what you have written. Here is a conceptual problem, and it cuts to the heart of one puzzle about AdS/CFT: how can a duality hold between physics in different dimensions? And not just that, but the lower dimensional theory has even less structure (only conformal structure) and so would seem to have even fewer degrees of physical freedom to correspond to the degrees of freedom in the bulk theory.

But more to the point. As I understand it, if I fix a moment at the boundary (a CFT Cauchy surface), what corresponds to that is some state in the bulk? Well, there are infinitely many bulk Cauchy surfaces that will cut the boundary along that CFT Cauchy surface, and no obvious physical characteristic that would single any particular one out. So here is a suggestion: the Cauchy surface at the boundary will correspond to the collection of all the points in the bulk that are space-like separated from the boundary cut. That would form the "Wheeler-DeWitt patch", if I am following. And the whole of the black hole formation and evaporation process can lie in one such patch. So if we understood what is going on in that single patch we would be done.

Note that on this picture an N-2 dimensional locus in AdS corresponds to an N dimensional locus in the bulk, so the mismatch in dimension is even greater. And as the Cauchy surface evolves in time in the CFT, there will be a succession of Wheeler-DeWitt patches that correspond, with successive Wheeler DeWitt patches largely overlapping. That is rather different from the picture you presented in which as the boundary time increases there is a single corresponding wavefunction \psi[g_{ij, \phi,t] in the bulk. Am I to understand that that wavefunction is defined over all of the Cauchy surfaces in the Wheeler-DeWitt patch? And that a successive state on the boundary will have a wavefunction that corresponds to a slight "higher" Wheeler-DEWitt patch in the bulk? And that where the two patches overlap the two wavefunctions will be compatible? (e.g. that the probability for the existence of a neutron star at some particular place in the bulk will be the same in the two patches, assuming they overlap at that location?).

If this is right, then there is yet more redundancy (and hence fewer independent degrees of freedom) in the CFT. And it is even more surprising than any 1-1 map between degrees of freedom in the bulk and on the surface could exist.

Maybe this citation also might help: Unruh, Wald, 2017 "Information Loss" (https://arxiv.org/abs/1703.02140v1):

ReplyDelete"The one sentence version of AdS/CFT argument against information loss is that since the conformal field theory - being an ordinary quantum field theory in a fixed classical spacetime - presumably does not admit pure state to mixed state evolution, such evolution must also not be possible in quantum gravity, including when black holes form and evaporate."

"The AdS/CFT correspondence is a conjecture. Our difficulty in assessing the validity of the AdS/CFT argument against information loss is not so much that this conjecture has not been proven, but rather that it has not been formulated in sufficient detail and with sufficient precision to make a clear argument. ..." and the rest of the paragraph.

Tim,

ReplyDeleteRegarding the apparent mismatch between degrees of freedom I will say the following. In the classical limit the number of degrees of freedom is infinite on both sides, so there is no well defined comparison there. At finite hbar, and hence at finite Planck length, we can roughly say that in the bulk there is one degree of freedom per Planck size cell, and this is a large number if the characteristic length scale of AdS is large in Planck units. The CFT is defined in one dimension lower, but the degrees of freedom are NxN matrices, and the duality dictionary says that N is going to infinity in the classical limit. So the degrees of freedom that might seem to be missing in the CFT are hiding in the NxN matrix structure. There is a simpler (but less interesting) analog of AdS/CFT called the "c=1 matrix model" where this picture can be verified in explicit detail: a function describing the distribution of eigenvalues of the matrices can be interpreted as a function in a new emergent spatial dimension.

To the rest of your message, I am not sure I understand the question since you are referring to two "rather different" pictures that I have presented, whereas I see only one picture. To avoid confusion, let me emphasize that the description of wavefunctions and Wheeler-DeWitt patches can be given independently of AdS/CFT -- i.e. everything here was understood well prior to AdS/CFT, although you have to dig in the literature a bit, since the vast majority of papers on Wheeler-DeWitt concern quantum cosmology, where there is no asymptotic structure and hence no t evolution of the type under discussion here. To recap, at any given bounday time t the wavefunction \psi{g_{ij},\phi,t] assigns a complex number to any choice of spatial metric g_{ij} and matter field configuration \phi. In the semiclassical regime, one can isolate the space of (g_{ij},\phi) for which the wavefunction is of order 1 rather than being exponentially small. The claim is that the space of these (g_{ij},\phi) is the same as the space of (g_{ij},\phi) that can be obtained as slices in a classical solution of the Einstein equations such that the slice asymptotes to time t at the boundary. By studying how the space of such (g_{ij},\phi) evolves with t, one can in principle reconstruct the full classical spacetime. As we go from t to t + delta t, the wavefunction indeed describes in this way a new Wheeler-DeWitt which overlaps with the old, but contains a new region "above" the old. Again, this is under the assumption that the full spacetime has a good semiclassical description.

BHG

ReplyDeleteLet me try to clarify. As I understand: associated with each Cauchy surface of the boundary theory, there is a Wheeler-Dewitt patch in the bulk, which is the set of all bulk events at spacelike separation from every point of the Cauchy surface, i.e. such that there is no timelike curve connecting it to the Cauchy surface. This would make sense because the Wheeler-DeWitt patch of the asymptotically flat spacetime, taking spacelike infinity to be the boundary point, would be the whole spacetime.

If that is right, then if we compare the Wheeler Dewitt patch for some Cauchy surface on the boundary at t0 to the patch associated with the Cauchy surface at a nearby time t1, these patches will overlap. And the wave function for each time will give the probability of a certain collection of fields on a spacelike hypersurface of a given intrinsic geometry and extrinsic curvature that intersects the boundary at the Cauchy surface. Is that right?

If all that is right, then when two Wheeler-DeWitt patches, associated with different boundary times, overlap, that creates constraints for them to be compatible in the overlap region, assigning probabilities to local events. So that translates into constraints among the states in the CFT. It seems like a hard thing to get a formal handle on.

Oh: and regarding the matching of physical degrees for freedom in the AdS theory and the CFT theory: just cardinality is usuallynnot enough. Penrose has one way of counting infinite degrees of freedom that have continuity constraints, and it is much more discerning than just cardinality. Similarly, one can find a 1-1 map from a one-dimensional space to a two-dimensional space, but not a continuous function. I assume this is not well understood in this case.

Tim,

ReplyDeleteRegarding WdW patches: yes this is correct, except where you write "...probability of a certain collection of fields on a spacelike hypersurface of a given intrinsic geometry and extrinsic curvature". The wavefunction does not depend on the extrinsic curvature: the extrinsic curvature is basically the canonical momentum conjugate to the instrinsic metric, and of course wavefunctions depend on only one of x or p, but not both.

Next: indeed the wavefunctions must obey constraints in order for this piecing together of WdW patches to make sense. What are these constraints: they are precisely the physical state constraints H_i \psi= H_t \psi =0 ! This is closely related to why I say that wavefunctions that do not obey the physical state conditions are just "garbage". It's not that they don't simply correspond to spacetimes that fail to obey the Einstein equations, rather they simply have no consistent spacetime interpretation at all. In the CFT there are no constraints of this sort since the states are "already physical" --- the state space of the CFT is in one-to-one correspondence with physical states in the bulk. The unphysical bulk wavefunctions simply have no counterpart in the CFT, which is again perfectly fine as these wavefunctions have no physical meaning attached to them.

Regarding matching degrees of freedom, this will become relevant so let me say a bit more about what we do and do not understand. The simplest case concerns bulk states consisting of a diffuse collection of particles, sufficiently widely spaced that their interactions are negligible. For such states there is a completely precise, explicit and very well understood map between states in the bulk and states in the CFT. As the density of particles in the bulk gets higher so that interaction energies are small but non-negligible the map between states gets more complicated but is still understood. WHen the interaction energies become of order 1, but no black holes are present, the map is understood in principle but is hard to make explicit. When black holes are present the CFT description of particles outside the horizon is understood in principle, but when the particles are inside the horizon there are basic conceptual confusions and the situation is unclear. This last fact is why people can argue about whether the inside of a black hole even makes sense, or is instead replaced by a fuzzball/firewall.

Now, it may at this point be useful to give a shortened version of my argument that proceeds along the following lines. We can think about forming a black hole by collapsing an initially diffuse cloud of quanta. This black hole will eventually decay into a final diffuse cloud of quanta. My argument would be to say that when we view this evolution in the CFT we are ending up with a state which I claimed above is of the type for which we understand the bulk-boundary map well; in particular it is a diffuse cloud of quanta in the external AdS region that is not entangled with anything else. This is just a sketch, and it may or may not be more useful to first go through a more careful discussion of the full time evolution of the wavefunction.

BHG

ReplyDeleteSince we are making progress, I don't want to slow you down. But please verify the following:

1) Under the AdS/CFT duality, the state on a Cauchy surface of the CFT corresponds to the state on the Wheeler-DeWitt patch associated with that Cauchy surface.

2) The Wheeler DeWitt patch associated with a Cauchy surface is the set of bulk events that are spacelike separated from all the events in the Cauchy surface.

3) Due to the overlap of successive Wheeler-DeWitt patches in the bulk, and the demand that all the patches be consistent with one another, constraints in one patch may be inherited by all later patches.

If we agree on these points, then I think I can see how the argument will turn out.

The citation of Unruh & Wald given by Reimond above is to be read. In particular footnote 7 is to be understood (can't say that I do).

ReplyDeleteAny event on the AdS boundary that is in the future of the evaporation event still has the black hole event horizon neither in its past nor in its future, I presume, and the Cauchy surface of the CFT on which that event lies hence includes in its Wheeler-DeWitt patch the event horizon?

ReplyDeleteArun and Reimond and BHG,

ReplyDeleteI have to say that although I knew of the Wald and Unruh paper and looked at it long ago, my attention was not fixed on AdS/CFT at that point and this is indeed a relevant paper for this discussion. The point about the gluing theorems was raised to me yesterday as a general objection to AdS/CFT, and we may well get some insight into pushing on that point. Also, note that their frustration not the AdS/CFT has not be *proven* but that it had not even been *formulated* with enough precision is exactly parallel to my own experience asking BHG what AdS/CFT says, and eventually being told the it is a "work in progress" (as I mentioned above).

In theory, then, the topic could well become not whether AdS/CFT is somehow in conflict with the solution outlined in my paper, but whether AdS/CFT is tenable at all.The status of the "gluing theorems" would be a prime example of this argument. But for the moment it think it is best to allow AdS/CFT to remain, and see what bearing it has (or does not have) on information loss. If it doesn't have any bearing at all, then it is not relevant to my paper in any case.

Tim,

ReplyDeleteI essentially agree with your points, 1,2,3 aside from some technicalities (some of which I already mentioned) that are probably best left aside for now. In 3, I am not exactly sure what you mean by "inherited". The wavefunction at each time has to be annihilated by the constraints H_i(x) and H_t(x). The wavefunction at one time is related to that at another time by evolution under the Hamiltonian.

Proceeding, with my argument, I would like to give an account of the evolution of the bulk wavefunction that respects as best as possible the standard Penrose diagram. Of course, since there are singularities in the bulk we have to make some assumptions since we don't have a complete theory of quantum gravity in the bulk. In the following I ignore anything to do with the CFT, and essentially try to present the usual case for info loss.

We start at early time with a diffuse cloud of particles that are arranged to undergo collapse to a black hole. At sufficiently early time we know the WDW wavefunction according to the rules we have described. In particular, the wavefunction at early times is a function of connected spatial metrics g_{ij}. As time evolves, we know from the Penrose diagram that our WDW patch starts to include points inside the horizon and regions near the singularity. And some of the points in the WDW patch lie on spatial surfaces that are necessarily disconnected. So at these intermediate times the WDW wavefunction is a function of both connected and disconnected spatial metrics g_{ij}. Since there is no way to smoothly transform a connected surface into a disconnected one, we need to assume that the correct laws of physics at the singularity specify a rule for doing so. As we evolve further to very late times, the WDW patch includes only disconnected spatial surfaces, and so the wavefunction depends only on these.

Let's think more about this late time wavefunction. The full wavefunction is that of a pure state, but the density matrix describing the outer component corresponds to some diffuse cloud of quanta, and it is strongly entangled with degree of freedom inside the horizon. By consider different initial states we would end up with different final pure states, almost all of which look approximately the same when restricted to the exterior region. These different states are elements of some Hilbert space of physical states. In particular, as part of taking the Penrose diagram seriously I claim that the Hilbert space is expected to be a tensor product, with one factor describing the wavefunction on each connected component of the complete surface. There is no mechanism for it to be otherwise: the kinematical Hilbert space is given by all possible (normalizable etc) functions of the metric and matter fields, and since we can choose these independently on the two components we have a tensor product. All that is left is to impose the constraints, but these are local expressions H_i(x) and H_t(x) and don't couple together the two components. So we arrive at the conclusion that the physical Hilbert space at very late times is a tensor product, and the wavefunction is a particular pure state in this tensor product.

This scenario, which seems to follow almost inevitably from reasonable assumptions about physics in the bulk implies information loss to observers who can only make measurements in the late time exterior region. I believe that the above is essentially just a restatement of Hawking's original argument, but phrased in the language of a WDW wavefunction.

BHG-

ReplyDeleteSo that was just what I have been expecting. And that was also exactly why I put in point 3. It is what shows that your argument has missed out something.

At first I was puzzled about why the AdS geometry should even be relevantly different from the plain vanilla asymptoticly flat space-time of the original Penrose diagram. That was why I was saying that we could just fix a particular boundary time, call it T0, such that the Wheeler-DeWitt patch at T0 essentially is the standard diagram (except for some differences out at the boundary, which require boundary conditions, but that is tangential to the argument.) In this single WDW patch, alpha evolution takes us from your initial to your final state, at least if the AdS is big enough. And I think I have explained why there is no argument that the solution space for just that patch will be a product space. Unlike the case of a pair of disconnected space-times, any change in the interior of the black hole will (according to your account of AdS) force a change in the amount of Hawking radiation and hence in the exterior state. (So it is here that the non-degenerate spectrum of the Hamiltonian comes into play). So at T0 we can't argue that the solutions space is a product space.

But now for my point 3. Let's take a slightly later time, T1, whose WDW patch largely overlaps that of T0. In particular, there will be Cauchy surfaces in the WDW of T1 that overlap a Cauchy surface in T0 everywhere but some region that starts very, very far from the region where the black hole is evaporating. And by "passing on the constraint", what I meant was that by the same argument that the space of solutions, confined to the WDW of T0 is not a product space, we conclude that the same must be true for the space of solutions restricted to the WDW patch of T1. Because we can make the difference between T0 and T1 as small as we like.

But once we have established that, we are done. If for some small enough delta t we can prove that the solution space restricted to the WDW patch of T is a product space iff the WDW patch of T + delta t is, then that feature must hold (or not hold) for all t. Even in the very late periods, you cant make a change in the disconnected bit without forcing a change in the external region. And essentially for the same reason I gave: make the interior region slightly more massive and you make the exterior region slightly more full of Hamming radiation, and that difference will propagate froward in time forever, until you reach whatever exterior part of the Cauchy slice you like.

The only way around this that I can see is to say that you can make changes to the interior region without at all changing the exterior. But any argument to that effect is just an argument against AdS/CFT itself. And without AdS/CFT we have no reason not to just accept that the solution space is a product space.

Tim,

ReplyDeleteIn case it isn't clear, what I am trying to do here is to get you to commit to a particular scenario in the bulk which I can then address. So my post, which did not mention the CFT side, is trying to lay out the possibilities, but it will ultimately be up to you to say what precisely you have in mind in terms of the evolution of the wavefunction and the structure of the state and Hilbert space at late times.

One issue that we need to get straight concerns terminology, because you seem to attach a nonstandard meaning to "solution space". At any given boundary time t there is a "kinematical space" of wavefunctions \psi[g_{ij},\phi;t] that is the space of (suitably normalizable) functionals of the spatial metric and matter fields. Then we have the constraint equations H_i(x)\psi = H_t(x) \psi =0. The "solution space" (as normally defined) is the subspace of the kinematical space that solves these equations. So in particular, this is a statement about the space of wavefunctions at some particular time t, and has nothing to do with evolution in t (indeed we haven't even mentioned the boundary Hamiltonian here, which is what evolves \psi in t). Also, It is this solution space which must be endowed with a suitable inner product such that it becomes a Hilbert space -- the physical Hilbert space.

With that in mind, I can't parse your argument, since it relies on statements like "the solution space restricted to the WDW patch of T" which make no sense to me if the solution space is used as above. I can't really tell, but it seems like you are using "solution space" here to mean the space of solutions of H\psi = i d\psi/dt. That is definitely not what I am referring to, and whether that space is a tensor product or not is neither here nor there; for sure, it is not a Hilbert space since we clearly don't want to try to write down an inner product on the space of such wavefunctions.

So may I ask that you restate your objection while adhering to the above definition of "solution space" (which we may as well call the physical Hilbert space)? Again, I am happy to address any scenario you want to defend, but I don't think you should be quite so quick to give up on the very late time Hilbert space space being a tensor product, since the alternative will be even more bizarre, as we can discuss.

As a useful diagnostic question, suppose we considered E&M on the background geometry of the evaporating black hole spacetime (i.e. consider quantum E&M on a fixed background spacetime). This theory has constraints, namely Gauss' law. Do you agree or disagree that in this case the Hilbert space at very late time is a tensor product?

BHG:

ReplyDeleteSo I am puzzled here. What I mean by the "solution space" is the space of solutions to the complete set of dynamical equations over the complete space-time. There only two possibilities. One that any solution in just a WDW patch can be extended uniquely to a global solution over the whole space-time, and the other is that it can't. If it can't then that can happen two different ways: either the solution in the WDW patch can be extended to a global solution in more than one way, or it can't be extended at all, that is, in the one case there are two global solutions that agree in the WDW patch (and there is a breakdown of global determinism), in the other there is a solution to the WDW patch that is not the restriction of any global solution to that patch.

Now I have inferred, from what you have said, that the third case in not possible: there are never two distinct global solutions that agree on any WDW patch. So the only case to worry about is the second one: there are solutions on just the patch that are not the restriction of any global solution to the patch. I don't have any a priori reason to believe or disbelieve in this possibility. But if it can happen, then those inextendible solutions are of no interest to us: they can never arise in practice anyway. So we are justified in only paying attention to WDW patch solutions which can be extendible to a global solution.

So if it is correct that every global solution is unique paired with a solution of every patch, then there is no practical difference between the too. And if there are solutions in a patch that cannot be extended to the whole space-time, ignore those: they are irrelevant to the issue.

In case that nature of my objection is still not clear, let me try to say what I think you are doing and why it won't work.

In the original case of asymptotically flat space-time, I explained why you can't vary the interior state without varying the exterior state. What you are trying to do is take a time late enough in the AdS that the WDW patch is disconnected: no continuous curve lying in the patch connects the interior and exterior regions. The you want to argue that the solution space for just that patch must be a product space. But that is not a relevant point. Take the set of global solutions over the whole spacetime. Now take each such solution and restrict it to the late patch, which is disconnected. Does that set of restricted solutions form a product space? No.

So really, all you have shown is that the set of local solutions—solutions to the fundamental equations just in the given WDW patch—is much larger than the set of global solutions restricted to the patch. I can accept that. But since I am only interested in global solutions, it is not a relevant point. The set of global solutions, restricted to the WDW, is not a product space even if the set of local solutions, restricted to that subspace, is.

Tim,

ReplyDeleteGood, this clarifies matters, and I believe I can now summarize the situation as follows. Possibility 1 is that the Hilbert space of the theory at a given boundary time t can be understood entirely by considering the space of solutions of the constraint equations at that time, without reference to other times. I gather that you agree that in this case the very late time Hilbert space will be a tensor product, simply because all the surfaces in the WDW patch are disconnected, and the constraints do not link different components in any way. Possibility 2 is that to determine the Hilbert space at time t you have to understand the space of global solutions to the constraints *and* the Schrodinger equation, and then restrict to the desired time t, and doing so will give some new conditions beyond what is obtained in (1).

If I am characterizing this accurately, then I just want to point out that (2) is a departure from how things normally work in QM, and quite a radical one in my view. Ordinarily, if we are given a quantum system, the Hilbert space at time t is the space of suitably normalizable wavefunctions that obey any constraints that are present. I don't normally need to specify any details of the Hamiltonian, and I certainly don't need to first study the space of solution of H\psi = id\psi/dt to determine the Hilbert space. This is a good thing, for several reasons. First, as a practical matter, determining the space of fully time dependent solutions is computationally intractable unless the Hamiltonian is very simple. But more importantly, the Hamiltonian might be time dependent, and could change in the far future in such a way as to render inadmissible some otherwise healthy global solution. So to determine the Hilbert space now, under this viewpoint, I need to know what the Hamiltonian looks like in the arbitrarily distant future.

So I would classify (2) as a scenario in which the usual rules of QM break down. Again, the question is whether some late time observers in the exterior region need to know about the global history of the universe including its future evolution to do physics, or whether they only need to consider the WDW patch at time t to do physics at time t. In the latter case, and with the Hilbert space being a tensor product, there is no problem: the observers can set up initial data and do experiments without needing to know about whether some black hole formed and evaporated in the past, or what this data will evolve to in the arbitrarily distant future. But under (2), one does need to consider such things.

Note that I am not saying that your possibility (2) in necessarily wrong: my point is just that to make black hole evaporation compatible with AdS/CFT something radical needs to occur (which is why it's a paradox), and I claim that that is exactly what is happening here. Finally, I have not heard anyone else argue for this sort of scenario before, but please let me know if it has appeared elsewhere.

BHG

ReplyDeleteExcellent! Yes, you characterize what I am saying perfectly. The only dispute we have is whether this is "radical". Everything I am saying just follows straightforwardly from things you have acknowledged. You are just as committed to these consequences as I am. I am not proposing any innovation to quantum theory at all, just working out what it entails in this particular situation.

Just to make this clearer, let me explain how I see the situation and why quantum theory looks a little different in this setting. There are two steps that take us away from the sorts of situations we are used to when we first learn the theory.

Step 1 is trying to apply quantum theory to GR, where there is not only not a fixed background space-time, but there are no canonical foliations of a generic space-time. In SR, the Lorentz frames provide a set of canonical foliations, and in cosmology we often deal with space-time that are so symmetrical that there is a single canonical foliation given by slices of constant curvature. But in the generic GR solution there just are not conditions that pick out some preferred class of foliations. Choosing a foliation is therefore a very arbitrary kind of thing, just as choosing a gauge in classical EM is. And—following out this analogy—just as we gather together all of the different possible choices of gauge in EM into a "gauge orbit" and regard all of these different choices as just different ways of expressing the same physical state of affairs, so in this case the choice of foliation is treated as "choosing a gauge", and all of the solutions that derive from these different choices are regarded as physically identical. The difference is that different choices of foliation carve up the space-time differently, so there will not be a 1-to-1 correspondence between a particular slice in one foliation with a particular slice in another. Hence what are physically equivalent are not given by slice-to-slice mappings from one foliation to another, but rather one the totality of all slices in one foliation to the totality of all slices in another. That is, we do not map slice-to-slice in the different foliations, but rather whole-space-time to whole-space-time.

Since the "dynamics" is usually presented as equations relating earlier slices to later slices, this situation is often expressed by saying that the dynamics in Quantum GR is "pure gauge", and that any two Cauchy slices taken from the same 4-D space-time are "gauge equivalent" or "lie in the same gauge orbit". I'm sure you are familiar with these locutions.

So the first big conceptual warning sign that has to be put up is that this sense of "gauge orbit" and "gauge equivalent" is very. very different from the sense of those terms in classical EM. In that case, two states *on a Cauchy surface* are gauge equivalent means that the physical situation on that surface—i.e. the E and B fields—are exactly the same. A and phi may be different, but they are not physical degrees of freedom. But if we say that the states on any two Cauchy slices taken from a solution to the EFEs are "gauge equivalent" we mean something very,very different. The physical situations on those Cauchy slices may be completely different: one contains a diffuse dust, say, and the other a star. The physical situations are very different. The sense in which they are "gauge equivalent" is just that they are both Cauchy slices from the same complete solution. Because they are both Cauchy, there is a clear sense in which they both "contain the same information": the complete solution can be recovered from each of them.

Con't

Since we usually think of the dynamics of a theory as specifying the evolution "from one time to another time", i.e. from one Cauchy surfaces to another Cauchy surface, if we decide to call all the Cauchy surfaces "gauge equivalent" then we have rendered the dynamics "pure gauge". From this point of view, the dynamics never takes us out of a "gauge orbit": rather the dynamics defines what the "gauge orbit" is.

ReplyDeleteAnd now we take the last step: If we call all of these states, defined on the various Cauchy surfaces "the same state" because they are "gauge equivalent", then in this sense "the state never changes"! In this sense, the state is "stationary", and the Hamiltonian, which generates the temporal evolution, should give no evolution at all! That is, H applied to a solution ought to yield 0. Or to put it the usual way, the Hamiltonian annihilates the solutions: they are all eigenstates with eigenvalue zero. The eigenstates are massively degenerate, of course, since every Cauchy slice through every solution the to EFEs is now an eigenstate of the Hamiltonian with Eigenvalue 0!

This situation is sometimes called the "problem of time", but it is no real problem at all. You just have to keep careful track of how the terms "gauge equivalent" and "gauge orbit" are being used. They just don't mean the same thing as in classical EM, where we first learned them. And we have to start thinking of the state on a Cauchy slice not as representing just what is going on on that slice, but as a means of representing the full 4-dimensional space-time.Now we can make sense of the WDW equation have the form of H|psi> = 0. That is step one in our conceptual analysis.

Since you ask whether this is written anywhere, Step One is indeed written. I wrote it, back in 2002, in response to a paper on the hole argument by John Earman. The response is called "Thoroughly Muddled McTaggart" and can be found here:

https://quod.lib.umich.edu/p/phimp/3521354.0002.004/1

Earman's paper, called "Thoroughly Modern McTaggart", is here:

https://quod.lib.umich.edu/p/phimp/3521354.0002.003/1

So this is an example of a discussion in the foundations literature that I'm sure no physicist would be aware of. This is the literature that arising exactly from thinking about the consequences of diffeomorphism invariance.

Next up: Step Two

Con't.

Now: Step 2.

ReplyDeleteStep 1 takes care of the standard Penrose diagram for asymptotically flat space-time. That space-time is globally hyperbolic, and fixing a solution on any one Cauchy surface fixes it everywhere. But AdS is not globally hyperbolic, so there are other difficulties.

We consider AdS as being covered by a sequence of WDW patches, with the patches overlapping. Each patch is the events spacelike separated from a set of points on the boundary. In a globally hyperbolic spacetime such a patch would be the whole spacetime. but because that is not the case here we need a collection of patches to cover it.

Now you want to look at such a WDW patch that is "late enough" not to constitute a connected space-time, with the part disconnected from the boundary being in the interior of the event horizon. I hope the following is obvious: if a WDW patch contains any event on the Event Horizon then the interior of the Event Horizon will not be disconnected: the event on the Horizon can be connects both to the interior and to the boundary. So the earliest WDW patch with the property you want is the earliest patch where the lower boundary of the patch runs above the Evaporation Event. For such a WDW patch and every successive patch, the patch itself is disconnected and the entirety of the interior lies in the patch as the disconnected part.

So my point is that where 2 WDW patches overlap they have to agree, and all of these WDW patches overlap on the entire interior and so have to agree on the interior. And they also have to agree with all of the earlier patches that have parts in the interior. As time goes on the the boundary time, the interior state is "frozen". So as far as that particular solution is concerned, you can't change the state of the interior at all any remain in the gauge orbit of that solution.

Now let's look at an early boundary time, before the black hole evaporates. You want to say that the boundary state then fixes a unique interior state. So those early WDW patches fix a unique interior state of the Black Hole. determined by the boundary state. That interior state will continue to be uniquely associated with that boundary condition, and hence with the state outside the event horizon as the sequence of WDW patches sweeps over the formation of the black hole and the evaporation. So the interior state will be uniquely associated with an exterior state. And the solution space is not a product space.QED

Tim,

ReplyDeleteI agree with your general comments about time in GR, although I remark that, as you know, in AdS slices associated with different *boundary* times are not gauge equivalent in any mathematical or physical sense.

I think I can see where our precise difference in perspective lies. In your discussion you consider the evolution of a particular state (or class of states) corresponding to collapsing matter, and then you analyze this state at very late times. You conclude that this space of states does not have the structure of a tensor product. I agree with this statement, but want to point out that this results from the imposition of a constraint, namely that the late time state did in fact evolve from one of these initial states. This is a constraint that links the data on the two disconnected components, and so one indeed does not have a tensor product. However, my claim is that this is not a statement about the full late time Hilbert space failing to be a tensor product, but rather the failure of the subspace obeying this constraint to be such. States not in this subspace are perfectly valid states, it is just that when evolved backwards in time they won't correspond to a collapsing ball of matter in a single connected slice; for example, the two disconnected components might remain disconnected into the far past.

In particular, I would like to focus on the perspective of an observer who was born somewhere in the very late time region. Under what I would call the "normal rules" of QM this observer would reason as follows. The universe at some boundary time t consists of the component I can directly access plus some unknown other number of disconnected components corresponding to all the black hole evaporation events that have occurred, and the wavefunction is in some complicated entangled state. So what I will therefore do is to measure the quantum state in my component, and having done that once I can proceed to do physics while completely ignoring the existence of these other disconnected components. Namely, I can prepare states in my own component subject only to the gravitational constraint equations in my own component, and there are no other constraints due to the existence of other components.

If the situation just described does not hold, I call this a radical breakdown of the laws of QM. But translated into mathematical terms, what i have described above is the statement that the Hilbert space is a tensor product, with measurement and state preparation referring to operators acting within one component of the tensor product. I say that it would be extremely weird if there turns out to be some constraint that prevents the observer from preparing a state that, when evolved back in time, does not correspond to the collapse of a particular ball of matter. The only constraints should be the usual GR constraints acting in the observer's own component.

cont

To summarize, whether or not the late time Hilbert space is a tensor product leads to observable consequences for an observer confined to the very late time region. The lack of a tensor product structure implies constraints on the class of states that the observer can prepare, but these constraints cannot be understood as arising from the known laws of physics that govern the observers own component; instead they are some additional "global laws" of a sort that have never appeared before in physics.

ReplyDeleteRemember, I view the black hole formation paradox as deserving the moniker of "paradox", so it's not so much that I am disputing the logic leading to your claim, as I am pointing out that you end up in a radical place. I was actually expecting you to take a different tack. Namely, I was expecting you to agree that the late time Hilbert space is a tensor product, since this makes the bulk physics look standard. And I was expecting you to counter my argument that this conflicts with the non-tensor product nature of the CFT by saying that this is evidence for the breakdown of AdS/CFT. Instead, irrespective of AdS/CFT I am arguing that your proposal implies, in my eyes, a radical departure for physics in the bulk.

BHG

ReplyDeleteExcellent! We are on the same page now! Just to reaffirm:

If by "solution space" one means "space of global solutions for the entire AdS space-time", and accepting your own characterization of AdS/CFT, we agree that the solution space in not a tensor product.

Note: this discussion is being conducted under the assumption that AdS/CFT, as you characterize it, is correct. I remain skeptical about that, but that is neither here nor there, and I won't press my skepticism.

We also agree about this consequence: the state on Sigma_2 out is an (improper) mixture, due to entanglement with Sigma_2 In.

We also agree that in some very forced and attenuated and really metaphorical sense, that improper mixed state is "available" to the outside observer. That is, under the fiction that the outside observer had access to an infinite collection of Sigma_2 Out regions, and was assured that they all were identically prepared, and could make global measurements on Sigma_2 Out, then the outside observer could, in principle, nail down the state of Sigma_2 Out to arbitrary accuracy as a mixed state.

We also agree that such a fictional observer could not determine via retrodiction what the initial state was. For that, one would need the state on Sigma_2 In. That's where the information about how the Black Hole formed is kept. Obviously.

And here's is a brand new point: even if you give me the state of Sigma 2_Out as a density matrix, and the state of sigma 2_ In as a density matrix, one can't retrodict: one needs the entanglement as well. Measurements confined to either surface alone cannot even in principle get the phase information about the entanglement.

And here's a really novel point: even the complete state on Sigma 2_In U Sigma_2 Out including the entanglement may not be enough to retrodict the initial state on Sigma_1! This points to a breakdown of the usual claims about retrodictablilty in GR due to the presence of the Evaporation Event. There is essentially an irreversable function of addition or superposition when figuring out what will come out of the future light cone of the Evaporation Event. At least that is how I would do it. We can come back to this later if we want.

But here is where I completely disagree. There is no sense in which this is a radical change in anything *from any practical perspective*. No actual observer ever has had or ever will have access to the state on any Cauchy surface, connected or disconnected. Obviously. So all of this talk about "an observer who was born somewhere in the very late time region" is all just metaphorical or empty. If you mean a real, physical observer it is empty, since no real physical observer could obtain the relevant information about the quantum state even in principle. And if it is just metaphorical, then let's just drop the pretense that this has anything at all to do with actual observation, or any consequences for actual predictive abilities.

Con't

Con't.

ReplyDeleteMy own take is that what I am proposing is a breakdown of the standard laws of GR, rather than of QM. The addition of a non-manifold point obviously requires some modification of GR. And if we leave out the Evaporation Event and leave a bare singularity, that also requires a modification. But in neither case does that modification really have to do with the question "where is the information about what collapsed to form the black hole?'. In either case, the answer to that question is "It is inside the event horizon". So there is no puzzle about that.

If we are agreed about all this, then as far as my paper goes I think we are done. AdS/CFT does not change the situation with respect to what is in the paper at all. The point about the breakdown of perfect retrodictablility *due to the non-manifold geometry of the Evaporation Event* is something I did not put in that draft but had planned to put in in any case, because it comes up even apart from AdS/CFT. So as long as I make note of that, I think I can report that AdS/CFT does not actually affect the analysis at all.

If you leave the Evaporation Event out, and retain a 4-manifold (as Wald is inclined to do), then again you get a similar result, but because the space-time is not globally hyperbolic (even when asymptotically flat) this is also not really a shock. In that case, no one should have ever called Sigma 1 a Cauchy surface in the first place. On my preferred solution, Sigma 1 still is Cauchy.

If we agree on all the factual claims here, then the rest is just subjective judgments about how "radical" something is. I prefer radical (in some sense) and coherent over both radical and incoherent, such as "black hole complementarity" or radical and hand-wavy/unrigorous as in ER = EPR, or radical and physically unmotivated as in firewalls, In truth, adding the Evaporation Event in is really rather boring as regards supposed the "information loss" problem as usually formulated, and just a bit intriguing about getting the physics of the Evaporation Event right. I have some ideas about how to do that, but they are embryonic at the moment.

So once we settled down to work we could make real progress! Do you have any outstanding puzzles or objections?

Tim says: "...all of this talk about 'an observer who was born somewhere in the very late time region' is all just metaphorical or empty."

ReplyDeleteMaybe I can help BHG with this paper by Susskind et al. "Complexity Equals Action" (https://arxiv.org/abs/1509.07876), since there the observer or measurement plays no essential role other than just tracing out density matrices or cutting through tensor networks...

Tim,

ReplyDeleteYour post raises several issues, but I just will just address one point here, to stay focused. Once again, the claim I am defending is that there exists no complete scenario that is not radical in some way. At present, we are starting from the assumption that the Penrose diagram is an accurate depiction of the physics and then seeing where that leads. AdS/CFT is actually not playing much of a role in this, so we can put it aside for the moment.

You write "... we agree that the solution space in not a tensor product.". To be clear, I am in agreement that this solution space is not a tensor product provided one imposes the restriction that the initial state represents a collapsing ball of matter in a single connected AdS space. I am not agreeing to any version of this statement that refers to all possible solutions with no restrictions imposed.

Now, what I want to press you on is the experience of a real observer born in the late time region. I am of course in agreement that this single observer cannot determine the density matrix of the outer region. However, I do think that this observer can carry out measurement that collapse the wavefunction according to the usual rules of QM. In particular, although a spatial slice in AdS has infinite volume, an observer can shoot a light ray towards the boundary and it will return in finite time. So in this way, the observer can make measurements on his full AdS component. Before he makes any measurements, we agree that the system is in a pure state, but it involves entanglement between \Sigma_in and \Sigma_out. You say that the late time Hilbert space is not a tensor product, and the question at hand is whether or not this is a radical claim.

Here is an analogy. Suppose an observer is born inside a sealed laboratory. The photons etc. in the lab are entangled with those in another sealed laboratory such that the full quantum state is pure. First consider the case in which the Hilbert space is a tensor product. Then our observer can proceed as follows. He makes a careful measurement of all the photons etc in his lab, and thereby collapses the wavefunction to a product state |psi_lab 1> x |\psi_\lab 2>. Having done that, he can now do physics in his lab while completely ignoring the existence of the other lab -- the wavefunction is no longer entangled, and will remain so. I call this the "normal situation". Instead, let us now consider the case that the Hilbert space is not a tensor product. For example, suppose there is a proton in lab 1 and a neutron in lab 2, and we impose the constraint that together their spins must be in the singlet state. This is a constraint on the space of allowed states that holds for all time. We could implement this by adding to the Hamiltonian a term that energetically favors the singlet state, and then in the limit that the corresponding coupling goes to infinity we obtain the desired constraint. What does this imply for our observer? Without getting into details one thing is clear: our observer will find himself unable to measure the spin state of the proton. For if he did, say measuring the proton to be spin up, then the combined state would no longer be a singlet, since the latter assigns equal probability for the proton to be spin up or down. So our observer finds himself in a weird world, in which he cannot carry out certain measurements that we normally say are feasible.

Now I want to apply all this to the situation in which lab 1 is our late time AdS component, and lab 2 is the region inside the horizon. You are saying that the Hilbert space is not a tensor product, so there must be some constraint that enforces this. My claim is that any such constrain will lead to a situation as above in which our observer is unable to carry out certain measurements that he would be able to carry out were the Hilbert space a tensor product. I call this radical.

BHG,

ReplyDeleteI am very happy to focus down on the point that you are raising. I have to say that this post has taken me completely by surprise, but that is a good thing: you have had some considerations in the back of your mind all this time that I was totally unaware of. So let me try to get clear on just how you are thinking about this.

The operative sentence in your post is "However, I do think that this observer can carry out measurement that collapse the wavefunction according to the usual rules of QM". What i need to understand better is what you think the "usual rules of QM" are.

An anecdote will help here. When I first read about the Information Loss Paradox back in (I think) the 80s, so before AdS/CFT, it was in the New York Times' Science Times. And I just could not figure out what in the world was going on. In that article, it said that there was a conflict between QR and QM because GR implied that information would disappear and it is a fundamental principle that QM preserves information because the evolution of the wavefunction is always unitary and deterministic and time-reversible. And I just shook my head: after all these years of being told that the great innovation of QM is fundamental indeterminism, I was now being told just the opposite!

So I finally figured out that what they had to have in mind was a no-collapse theory: either Many Worlds or (unlikely what they were considering) pilot wave theory. A collapse theory, like GRW, loses information all the time in every circumstance and is neither predictable nor retrodictable ever. I figured that collapses were just off the table in this context. So can you just say a few words about what you think "the usual rules of QM" are, and particularly how it solves the Measurement Problem? Only in light of that can I answer your question.

Tim,

ReplyDeleteSure, I can say more. However, I will try to prevent the discussion from becoming too entangled in interpretations of QM rather than issues specific to black hole evaporation. When I speak of "the usual rules of QM" what I have in mind here is that the late time observer can treat his state, which consists of some diffuse cloud of quanta, exactly as he would have had the state not arisen from black hole evaporation but instead from some standard non-gravitational process. As far as I know, all "interpretations of QM" (as opposed to modifications thereof) at least agree on the rules for making predictions. Essentially, one first determines the state by measuring a suitable number of observables. Here I do not wish to get into issues of whether the observer has thereby "collapsed the wavefunction" (although I admittedly did use those words), or placed himself in one of "many worlds", or whatever. All I want to say is that this observer has now determined the state; the state can then be evolved forward in time via Hamiltonian evolution, and probabilities are computed via the Born rule. I.e. what's in the textbooks. So what I am asking you to address is whether the late time observer can do physics according to this textbook recipe, or are there some novel issues present that he needs to take into account?

BHG

ReplyDeleteNo, of course not. The late time observer is in the same situation as anyone is whose local state is entangled with distant states, i.e. everyone always. Whether the other entangled part is just far away or behind and event horizon or on a disconnected part of a Cauchy surface makes no practical difference. We handle that all the time.

Tim,

ReplyDeleteI don't understand what your position is. You just wrote that "The late time observer is in the same situation as anyone is whose local state is entangled with distant states,". And you agreed that this observer can do physics as the textbooks say: he can measure a suitable number of observables, thereby collapsing the wavefunction to the form \psi_tot = \psi_{observer} x \psi_{other}, and henceforth disregard the existence of \psi_other as far as making predictions in his own "lab" are concerned. That is how the situation would be if the observer were in a sealed lab whose contents were entangled with degrees of freedom in some other distant lab.

But on the other hand, you have also told me that the late time physical Hilbert space is not a tensor product. If that's the case, then the above situation cannot obtain, since then the Hilbert space, by definition, is not spanned by a basis of product states as above. I.e. it's then not possible to say that the state is \psi_tot = \psi_{observer} x \psi_{other} after the observer has made his measurements.

So there is an incompatibility in these claims.

BHG

ReplyDeleteLet me describe the situation in more detail, and then we can decide if you want to go on in this direction or not.

When the issue of information conservation is raised, and the associated issues of predictability and retrodictability, the only natural venue is the universal quantum state, the universal wavefunction. The assumption is that there is such a universal wavefunction, which always evolves unitarily and predictably. Anything less than that and there will certainly be room to lose or gain information.

But if ultimately there is only one wavefunction,the universal wavefunction, that poses an interesting foundational question: since no one ever does know or could know what the universal wave function is, how do we use such a theory to make predictions? How is a wavefunction even ever ascribed to a subsystem, such as a beam of electrons in a lab? What is the relation between the universal wavefunction and these little wave functions assigned to subspaces or subsystems?

This particular question has a clean answer in Bohmian mechanics, given by the so-called conditional wavefunction. This is something that can only be defined because the theory postulates actual particles that always have definite positions. I can explain how that works if you like. There really is no standard account of this. So all I meant is that first figure out how you deal with the plain vanilla case; I hand you an electron in an unknown state, maybe a pure state, maybe it is entangled with something else. You take it into the lab, use a "preparation procedure", and then ascribe a pure state to it. That is what we do, and for predictive purposes it works fine. But that whole procedure ought to be just part of the universal state evolving without collapse.

As I say, I know how to tell this story for Bohm's theory. I have no idea how to tell it for "standard QM", because "standard QM" just is not a well-posed exact theory. If you feel like you can reconcile the unitary evolution of the universal state with the way we portray preparation practices go right ahead. Whatever story you tell will work in the same way here. So if you think that there is any essential difference once the evaporating black hole is brought in, please say what it is.

Tim´s “…lose or gain information” is the pivotal hint.

ReplyDeleteTo gain (Shannon) information one needs probability and thus a bit of randomness. If all would be predetermined, if all is already written into the initial conditions of this universe there would be nothing new, no real probabilities.

The new thing about QM was that for the first time real QM randomness and thus probabilities entered in physics with the measurement. Peculiar was that this was observer dependent.

But SR told us that we have to formulate observer-independent laws. So, the task is to find observer-independent reductions of the state. And we should not be concerned about the anthropocentric view about our ignorance, but the mechanism that evolves the universe.

To gain (Shannon) information it is necessary to lose (unitary) information by state reductions.

Tim´s “universal quantum state, the universal wavefunction” let´s call the entire state ( = state vector, not density matrix). It is a product of smaller states. Since it is a product state it is not an entangled state. Each smaller state can evolve into an entangled state with itself or other smaller states (subsystems). Collapses on the smaller states do not affect the rest of the product – this is the very essence of a product state.

Tim´s 2 sentences “But that whole procedure ought to be just part of the universal state evolving without collapse” and “… reconcile the unitary evolution of the universal state with the way we portray preparation practices …” – well, with an exclusively unitary evolution this simply is not possible. Collapses in the smaller states are needed.

The problem with Bohmian mechanics is that it wants to reinstall complete determinism (no Shannon information anymore) and it is hard or even impossible to formulate it in a relativistic way (https://arxiv.org/abs/1307.1714). Also to try to marry GRW with SR is painful (https://arxiv.org/abs/1708.09371) and GRW, CSL, Lindblad they all violate unitarity and this is unacceptable for QFT. Further Everettians could solve the “measurement problem” but then they run into the “measure problem”.

Observer-independent triggered reductions like the Diosi-Penrose model are in principle capable to serve the job, but they need to be formulated such, that between measurements the unitarity is conserved. (i.e. the SchrÃ¶dinger-Newton equation is not good enough).

I think that BHG's and Tim's concerns about lab measurement in the face of a universal wave function are either addressed by cluster decomposition, or else we have a severe hole in our physics, and more fundamental than any incompability of General Relativity with Quantum Mechanics.

ReplyDeleteArun,

ReplyDeletewith cluster decomposition I guess you are referring to the following:

In QM if two observables commute they are simultaneously diagonalizable and hence measurable. Since no signal can be send between spacelike separated points, a measurement at one of these points cannot influence the measurement at the other point. By the way, these two facts from QM and SR are essential for the spin–statistics theorem.

Nevertheless, in EPR-like experiments there is a correlation between spacelike separated measurements, but this correlation cannot be used for superluminal signaling.

Tim,

ReplyDeleteThe question I am asking is independent of the interpretational issues surrounding QM. I want to consider a non-philosophically inclined observer who just wants a set of rules to follow to make physically verifiable predictions, and it our job to supply him with these. Let's consider two situations

1) There is a large sealed lab full of quanta in a pure state. At some moment a partition is inserted, dividing the lab in two, although the full state remains highly entangled. The lab on one side of the partition is carried off to the far side of the moon. At a subsequent date an observer is born in the lab that remains on earth, and he wants a set of rules to follow. He may or may not know (or care) about the existence of the other lab. He has with him a book "Introduction to QM", which instructs him to proceed as follows: first carefully measure all the positions, spins, etc. of the particles in your lab. You now have a quantum state \psi_lab, which is an eigenstate of the operators you measured. You can then evolve forward in time via the Hamiltonian of your lab and then make predictions based on Born's rule. Now, I think we would all agree that these rules will work in the same sense that QM always "works", and furthermore these rules make no reference to the existence of some other lab. However, it's also clear that the success of this protocol critically relies on the fact that the full Hilbert space is a tensor product, since otherwise (by definition) the tensor product states \psi_lab x \psi_other will not be in the Hilbert space (see my earlier example of the singlet state constraint).

2) In AdS, a black hole forms and evaporates, and an observer is born in the very late time AdS region. The question is, can the observer proceed via the same protocol as in (1)?

As I see it, you have made two incompatible claims regarding this question. Claim 1 is that, yes, the situation is exactly the same in the two cases insofar as the basic rules of QM go. Claim 2 is that the late time physical Hilbert space in the black hole case is not a tensor product; but this implies that the protocol cannot succeed as I explained above.

So that is the question, and it has nothing to do with retrodiction or issues of that sort, but is instead a practically minded question about what rules our observer should follow to make predictions.

BHG

ReplyDeleteGood, so let's start to take this question apart. And it is not at all a "philosophically inclined" question: it is a central question of physics, one that is never addressed in physics texts. Let me start by pointing out some problematic aspects of the question as you pose it.

The main question before us is two-fold:

1) As a practical matter, how do we assign wavefunctions (or density matrices) to subsystems of the universe? These are, of course, the wavefunctions (or density matrices) that are actually used to make predictions and test quantum theory.

2) How does quantum theory itself account for the practical success of the techniques used in the answer to question 1.

Now you have already made an assumption about the answer to 1) which is clearly inaccurate. You write: "He has with him a book "Introduction to QM", which instructs him to proceed as follows: first carefully measure all the positions, spins, etc. of the particles in your lab.You now have a quantum state \psi_lab, which is an eigenstate of the operators you measured." Just at a purely day-to-day practical level, this is clearly impossible. I assume that |psi_lab> is supposed to be vector in the Hilbert space associated with the lab. But then you would have to "measure" every particle in the lab, including all of the particles that constitute the equipment in the lab! No "Introduction to QM" book will give you a clue about how such self-measurements are to be carried out. Measurements of measurement equipment land us immediately into Wigner's friend scenarios: Will the measurement of the post-measurement state of a second measuring device + target system reveal an entangled state of the pair or a product state? If there are no collapses, then it ought to be an entangled state, which is not an eigenstate of the measurement device being in a definite outcome state. Now put a twist on this and have the second measurement be a self-measurement: one where you are trying to determine if *you yourself* are in an entangled state with some target system you have just measured or rather in a product state. David Albert discusses some of the curious features of such self-measurements in any no-collapse theory in Quantum Mechanics and Experience, but I guarantee that no Introduction to QM book even mentions them. So your whole set-up presumes something that is both conceptually and practically extremely problematic: the possibility of complete self-measurement.

Con't

Con't

ReplyDeleteHere is a second point: what you call "measurement" is not really a measurement in the normal sense of that term: it is a state preparation. It is simply impossible to measure—in the sense of determine by a laboratory procedure—the spin state of a particle presented to you. If you could do that then you could superluminally signal, which the no-Bell-telephone theorems (and the ETCRs) guarantee you cannot do. So although you speak of "measuring" the state of the lab, what you really mean is preparing a state of the lab, and knowing in what state it has been prepared. Now state preparation procedures are just different beasts than "measurements".

As I said, nothing you can do will reveal the spin state of a particle whose spin has been prepared in an unknown eigenstate of spin. But we nonetheless can do a state preparation that will yield a beam of particles in a known state. For example, set up a Stern-Gerlach magnet in the z-direction. shoot a beam of electrons (in whatever initial state you like) through it. Do nothing to the part of the beam directed upward, and put the beam directed downward through a magnetic field that causes it to precess the z-spin through 180°. Now recombine the beams. you now have a beam of pure z-up electrons.

Now this very preparation procedure appears to lose information. After all, it seems like you get a z-up electron out no matter what sort of an electron you feed in, so one cannot retrodict from the final state what the initial state was. If that were so, then the state preparation procedure would already violate unitarity. But we are presuming that unitarity is not violated for the entire lab system. So that question—where is the information about the initial state of the electron hiding in the final state of the electron + preparation equipment?—is rather similar to the question we are asking in the evaporating black hole scenario.

So in a no-collapse theory the information cannot get lost. Which means that the end state has to be different for an electron that was initially z-up and an initial state that was initially z-down. I recommend trying to chase the information down in this example.I'll let you chew on that for a bit.

I hope I have convinced you that the scenario you mention contains a lot more difficulties than you present it as having. I would be delighted to work through these difficulties with you, if you are interested.

As a lurker (apart from one post in the first week of this saga) who has slogged through all 597 messages so far, I feel I’ve earned the right to make a small intervention in the debate. (Authors should be grateful to faithful readers, as well as vice-versa!) So, BHG and Tim, please consider the following, which I hope might lead your debate back onto its recent positive track, as opposed to morphing into a debate about how to think about measurement in QM in general. Which I think will likely turn frustrating for both of you.

ReplyDeleteLet’s go back to BHG’s posts where he argued that if Tim’s proposal were correct, there would be weird constraints on what late-time observers could measure (or on what results they could get – it wasn’t clear to me precisely what the concern was). Let me cite one claim in particular that I think is important:

“I say that it would be extremely weird if there turns out to be some constraint that prevents the observer from preparing a state that, when evolved back in time, does not correspond to the collapse of a particular ball of matter.” (post of Jan. 27)

On the contrary: if we are assuming determinism (which, I understand, has been a background assumption of the whole information loss paradox – unitary evolution), then no observer, ever, anywhere, can “prepare a state” such that when evolved backwards, it produces anything other than the actual past! Consider the situation in classical Newtonian physics, which we’ll pretend was fully deterministic for the sake of the argument. I go around the world feeling like I can do all sorts of things. But can I choose to do something (say, eat a hot dog for lunch) which is such that, combined with the rest of the actual state of the world (surrounding me and my hot dog), evolved backwards in time it yields a past different from the actual past? Not if we hold fixed the assumption that the past was the way it actually was! And this is why, for ages, thinkers have worried that determinism and free choice are incompatible. Do I

noticemy lack of freedom? Can I do any experiment to prove to myself that I am fully constrained by past states of the world? No. This lack of freedom is invisible to me.Coming back to BHG’s late-time observers in ADS, and applying a charitable reading of what he meant when talking about collapsing the wavefunction (apparent collapse, for an observer in a “branch” of the universal wavefunction), the concern seems to be that the constraints on the solution space that Tim’s proposal involves would prevent the late-time observer from doing some state-preparation type measurement such that, if evolved backwards unitarily in time, it leads to a universe without a certain collapsing ball of matter. But I don’t see why this would be the case at all. The late-time observer can do state-preparation type measurements such that,

if the post-preparation state of Sigma-2_out were the entire universal quantum state, the backward evolution in time would not yield that certain collapsing ball of matter. But the state the observer “prepares” is NOT the universal quantum state; it is a tiny fragment of a much bigger and more complicated state, one that involves (among other things), entanglement with stuff behind an event horizon. That full, bigger state, when evolved backwards in time, is (of course) going to yield that collapsing ball of matter (on the branch-line traced back from our observer! On yet other branches of the universal wave function, maybe there never was any black hole...)CONT.

CONT.

ReplyDeleteMy point is: the late-time observer in ADS is not going to be aware of the global constraints that lead the global wavefunction to have a unique past, any more than classical-me is aware that he couldn’t have had a chicken sandwich for lunch when he chooses the hot dog. Whether Tim’s proposal is radical or not, I can’t see that is radical in the sense of being a proposal that entails things will

observablylook very different to late-time physicists.BHG offered an analogy using two labs and some constraints that would (he claimed) make it impossible for a physicists in one lab to even measure the spin of a proton. But I confess I could not understand his train of thought here. Maybe if that could be made clearer, the debate could get back onto the path of progress.

Thanks to anyone who made it this far!

Superdeterminism is definitely a tenable CHOICE ;-)

ReplyDelete