## Friday, October 19, 2012

### How empty is the black hole interior?

I've exchanged a dozen of e-mails with Joe Polchinski, the most well-known physicist in the original team that proposed the firewalls. We haven't converged and Joe ultimately decided he didn't have time to continue and recommended me to write a paper instead (which he wouldn't read, I guess). However, he started to listen to what my resolution actually is and I could see his actual objection to it which seems flawed to me, as I discuss below.

Recall that $$\heartsuit$$ represents the (near) maximum entanglement and the firewall folks demonstrate that because $$R\heartsuit R'$$, the following things hold:$A\heartsuit B, \quad R_B \heartsuit B.$ The degrees of freedom $${\mathcal O}(r_s)$$ outside the black hole at $$t=0$$ when the black hole gets old are maximally entangled with some part $$R_B$$ of the early Hawking radiation (because $$R\heartsuit R'$$) as well as with the degrees of freedom inside the black hole $$A$$ which are "mirror symmetrically" located in the other Rindler wedge from (infalling) Alice's viewpoint.

But because a system can't be maximally entangled with two other systems, there is a paradox and one of the assumptions has to be invalid. AMPS continue by saying that what has to fail is the "emptiness of the black hole" assumption. They make another step and say that all field modes inside the old black hole are hugely excited so an infalling observer gets burned once she crosses the event horizon.

My answer is that the resolution is that $$A$$ and $$R_B$$ aren't really "two other systems"; $$A$$ is a heavily transformed subset of degrees of freedom in $$R_B$$ so $$B$$ is only near-maximally entangled with one system, not two, and everything is fine. I believe that this has been the very point of the black hole complementarity from the beginning.

Yup, I took this picture in Santa Barbara, not far from the KITP.

Now, Joe's objection is the following:
Even Alice must have a state, a pure state or a mixed state, ready to make predictions. Up to $$t=0$$, she describes the early Hawking radiation in the same way as Bob (who stays outside). For example, this wave function may imply that $$N_b=5$$ for an occupation number measured outside the black hole; the state may be an $$N_b=5$$ eigenstate.

It also means that the state is an eigenstate of a (complicated) observable at a later time which evolved from $$N_b$$. On the other hand, this observable doesn't commute with $$N_a$$, the occupation number for a field mode moderately inside the black hole, in the $$A$$ region. The state can't be an $$N_b$$ eigenstate and an $$N_a$$ eigenstate at the same moment because they're generically non-commuting operators, and consequently, it can't be true that $$N_a=0$$ which is needed for the emptiness of the old black hole interior from the viewpoint of an infalling observer.

I don't think so. What is true and what is not about Joe's statements above?

First, let us ask: Will her "later" state be an eigenstate of $$N_a$$ or $$N_b$$? Here, the answer is clear. We assumed the state to be an $$N_b=5$$ eigenstate so by the dynamical equations, the state will remain an eigenstate of an observable that evolved from $$N_b$$ via Heisenberg's equations. For this quantity (probably involving an undoable measurement in practice), the measured value is sharply determined.

To predict the value of another quantity such as $$N_a$$, we need to decompose the state into $$N_a$$ eigenstates and the squared absolute values of the probability amplitudes determine the probabilities of different results. Joe is right that in principle, because the operators $$N_a$$, $$N_b$$, and their commutator are "generic", there will inevitably be a nonzero probability for $$N_a\neq 0$$.

However, Joe isn't right when he suggests that this means a problem. Indeed, the probability of $$N_a\neq 0$$ will be nonzero. Nevertheless, this probability may still be tiny. In other words, the probability that a particular mode will be seen as $$N_a=0$$ may still approach 100 percent so in the classical or semiclassical approximation, quantum gravity will continue to respect the equivalence principle which implies that an observer falling into an old black hole sees no radiation (not even the Unruh one which he would see if he were not freely falling).

And I think that this is what happens. The probability of $$N_a\neq 0$$ is tiny but nonzero. We may try to be somewhat more quantitative.

Choose a basis of the $$\exp(S)$$-dimensional space of the black hole microstates so that the basis vectors are $$N_a$$ eigenstates. Now, my point is that the number of $$N_a=0$$ basis vectors can be and almost certainly is greater (and probably much greater) than the number of $$N_a=1$$ or higher eigenvalue eigenstates. We know how it would work if $$N_a$$ were counting the occupation number from a non-freely-falling, "static" observer's viewpoint.

In that case, the probability of having a greater number of particles (by one) would be suppressed by a factor similar to the Boltzmann factor $$\exp(-\beta E_n)$$ where $$E_n$$ is the energy of the mode and $$\beta$$ is the inverse black hole temperature, comparable to the Schwarzschild radius. In that case, we could prove that the probability of having a higher number of particles (by one) in some spherical harmonic $$Y_{LM}$$ would be suppressed by something like $$e^L$$; I am a bit sketchy here. This is just a description of the Unruh radiation that a "static" observer would experience right outside the black hole.

Things are harder for the freely infalling observer. Classically, she shouldn't see any radiation – because of the equivalence principle – so the higher values of $$N_a$$ should be even more suppressed. The suppression should become "total" for macroscopic black holes.

At the same time, however, the working of the low-energy effective field theory means that in the relevant Hilbert space, the creation operator increasing $$N_a$$ must relate states pretty much in a one-to-one fashion. So how it could be true that the number of $$N_a=1$$ eigenstates in the basis is (much) lower than the number of the $$N_a=0$$ eigenstates?

Before you conclude that my scenario is shown mathematically impossible, don't forget about one thing. If you create a quantum in a quantum field (in the black hole interior) and increase $$N_a$$, you also change the total mass/energy of the black hole from $$M$$ to $$M+E_a$$. So the $$N_a=0$$ states of a lighter black hole are in one-to-one correspondence with the $$N_a=1$$ states of a heavier black hole.

In other words (or using an annihilation operator), the $$N_a=1$$ states of the black hole of a given mass are in one-to-one correspondence with the $$N_a=0$$ states of a lighter black hole whose mass is $$M-E_a$$. But a smaller black hole has a smaller entropy, and therefore a smaller total number of all microstates. The ratio of the number of states is approximately equal to$\frac{ \exp(S_{\rm larger})} {\exp(S_{\rm smaller})} \sim \frac{\exp(M^2)}{\exp[(M-E_a)^a]}\sim\exp(2ME_a G)$ where I restored Newton's constant in the final result. All purely numerical factors are ignored. This result is still the quasi-Boltzmannian $$\exp(C\beta E_a)$$ with some unknown numerical constant $$C$$. Well, the calculation was really more appropriate for a static observer but even for a freely infalling one, it should still be true that the action of a creation operator creates a larger and heavier black hole. In other words, the annihilation operators produce a lighter black hole with fewer states, and therefore the excited states are in one-to-one correspondence with the smaller number of states of a lighter black hole.

Even if the calculation above is wrong despite the tolerance for errors in the numerical factors (if the parameteric dependence is different, and I hope it is), I think it's true that a fixed-mass black hole has fewer eigenstates with $$N_a=1$$ than those with $$N_a=0$$, so it's more likely that we will see empty modes. This likelihood is becoming overwhelming for modes that are sufficiently localized on the event horizon (those proportional to $$Y_{LM}$$ with a larger $$L$$). It means that if you pick a generic state such as the $$N_b=5$$ eigenstate and decompose it to the $$N_a$$ eigenstate basis, most of the terms will still correspond to $$N_a=0$$ which means that there will be a near-certainty that you will measure $$N_a=0$$ even though $$N_a,N_b$$ refuse to commute.

The fact that $$N_a\neq 0$$ may happen shouldn't be shocking. Look at a younger black hole and you will see that the interior can't be quite empty. It takes $${\mathcal O}(r_s)$$ of proper time to suck "most of the material" of the star from which the black hole was created but because some of the material recoils etc., there's a nonzero amount of material inside the black hole at later times, too.

Joe neglects the fact that $$N_a=0$$ is only an approximately valid statement and uses the strict $$N_a=0$$ to derive a paradox. I think that $$N_a=0$$ is just approximate and in fact, it's an interesting challenge to use the laws of quantum gravity – or specific laws in a formulation of string theory – to derive the percentage of states that have $$N_a=1$$, for example. Classically, almost all microstates must correspond to an empty interior (as seen by an infalling observer) because the highest-entropy, dominant microstates are those that (because of the second law of thermodynamics) appear "later" once the black hole is sufficiently stabilized, almost perfectly spherical, and after it has consumed the star material and its echoes. The reason behind $$N_a\approx 0$$ is therefore "entropic".

I don't know what the exact parametric dependence is so most of the formulae above were just "proofs of a concept" but I do think it shows that Polchinski et al. have overlooked a loophole that is arguably more plausible than all the loopholes they have discussed. The loophole says that the emptiness of the black hole interior simply isn't perfect but it is very good for large black holes and localized modes (certainly no deadly firewalls!). The equivalence principle at long distance scales, unitarity, and other assumptions of quantum mechanics and low-energy effective field theory may be preserved when complementarity is allowed to do its job and declare the information inside and outside the black hole as "not quite independent information".

And that's the memo.