Gidom Mera's answer at https://physics.stackexchange.com/a/45511 is illuminating, but on closer analysis, it brings up further puzzles.

Backscattering works in both directions. Let's see what we get when we evolve a far away outgoing mode back in time.

$$e^{-i\omega t}\left[ \frac{1}{r}e^{i\omega r} + C\frac{1}{r} e^{-i\omega r}\right],\;r\gg R$$ $$e^{-i\omega t}Ae^{i\theta(r)},\;0<r-R\ll R$$

Basically, an ingoing mode far away also backscatters. So, if you evolve an outgoing far away Hawking radiation back in time, you end up with a nonzero coefficient for a far away ingoing mode. This can be cancelled by evolving a near horizon ingoing mode back in time with the right relative coefficients.

$$- e^{-i\omega t} C\frac{1}{r}e^{-i\omega r},r\gg R$$ $$e^{-i\omega t}\left[ Be^{-i\theta_2(r)} + De^{i\theta(r)} \right],\; 0<r-R\ll R$$

The same arguments used to establish the existence of a near horizon firewall can also be used to establish the presence of far away ingoing modes earlier in time. However, in principle, we can always control the external environment of a black hole so that there are no far away ingoing modes.

Why doesn't the firewall argument also apply to far away ingoing modes?

We can cancel the far away ingoing modes at an earlier time by a destructive interference of contributions due to infalling near horizon modes and far away outgoing modes at a later time. However, this leads to an entanglement between the particle occupancy number of near horizon ingoing modes and the particle occupancy number of far away outgoing modes at a given location. Basically, in this case, we have an initially outgoing near horizon mode, and it evolves into a superposition of a backscattered near horizon infalling mode, and a far away outgoing mode. This is in conflict with the monogamy of entanglement.


1 Answer 1


Right, there can't possibly be any faraway infalling modes if we preselected the system to have no such modes.

This can be handled by the two-state formalism. What you get from evolving the outgoing faraway Hawking radiation backward is the final state. What you get from evolving the initial state forward is the initial state. They differ in general.

This is so important! Evolve backward outgoing Hawking radiation not entangled at all with the near horizon infalling modes. Then, look at the coefficients for zero occupancy number for the faraway infalling modes earlier. There is always a nonzero coefficient for zero occupancy numbers, even if that coefficient might be small. The Born rule of ordinary quantum mechanics tells us the probability for zero occupancy numbers is then small. However, if there's preselection to zero occupancy number, the Born rule needs to be replaced.

But if this has to be the case for far away infalling modes, this analysis should also apply to near horizon outgoing modes! Preselect to no firewall according to a freefalling observer.

Look at this toy model with a qubit. Preselect to $|0\rangle$. Postselect to $c|0\rangle + d|1\rangle$. In between, at an intermediate time t, measure the value of the qubit in the 0,1 basis. The projectors at time t are $|0\rangle\langle 0|,\, |1\rangle\langle 1|$. According to consistent histories, this satisfies the consistency conditions. That's not due to what happens after t. It's because at time t and before, the state is already in an eigenstate of the projectors. The probability for $|0\rangle\langle 0|$ is one, and $|1\rangle\langle 1|$ is zero.

In the firewall framework, the projectors are the occupancy numbers for the near horizon outgoing modes as measured in the Bogoliubov frame of a freefalling observer. The preselected state has zero occupancy number, which an eigenstate of the occupancy number projectors. So, according to consistent histories, there's no firewall.

Putting an actual physical detector effectively decoheres the initial state along the occupancy number basis. However, the initial state was already an eigenstate.

The point is, evolving pure outgoing Hawking radiation unentangled with the near horizon infalling modes backward doesn't completely leave us with a state with nonzero occupancy number. Instead, it leaves us with a bra state which is in a superposition of different occupancy numbers, including zero. The coefficient for the zero eigenstates is always nonzero, even if it is small. This is why we can always apply the two-state analysis.


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