What does this equation mean?

Hayden and Preskill’s own result, sharpened into an explicit, efficient decoding circuit by Yoshida and Kitaev, is that once scrambling of this kind has run for about $t_*$ , collecting only a handful more qubits of radiation than were thrown in — not a number that grows with the hole’s size — suppresses $\epsilon_L$ exponentially in that small handful [ 15 ] . The mechanism behind that exponential suppression is worth seeing exactly rather than taking on faith, and Page’s own formula for the average entanglement entropy of a subsystem of a random pure state supplies it. For an m -dimensional subsystem embedded in an mn -dimensional Haar-random pure state, $\langle S_{m,n}\rangle = \sum_{k=n+1}^{mn}\frac1k \;-\; \frac{m-1}{2n}$. is exact [ 4 ] . Fixing m=2 (a single qubit) and letting n — the effective size of everything else the qubit could be entangled with — grow gives $\langle$ $S_{2,4}$$\rangle$ = 0.5095 nats, $\langle$ $S_{2,16}$$\rangle$ = 0.6465 nats, and $\langle$ $S_{2,64}$$\rangle$ = 0.6814 nats, closing in on the ceiling $\ln$ 2 = 0.6931 nats. The residual gap to that ceiling shrinks by a factor of 3.94 when n quadruples from 4 to 16 , and by 3.98 when it quadruples again from 16 to 64 — converging, as more radiation-sized dimension is added two qubits at a time, on a clean factor of four per doubling of collected qubits, or two per single qubit. This is an exact, hand-computable corroboration of the folklore “each extra collected qubit roughly halves what is left to recover,” not a re-derivation of Hayden and Preskill’s original bound, and it is why $t_L(\epsilon)$ depends on $\epsilon$ only logarithmically: driving $\epsilon$ from 10^{-1} to 10^{-9} costs on the order of thirty extra qubits of radiation, a negligible addition to $t_*$ for any astrophysically sized hole. None of this requires the hole’s dynamics to be literally Haar-random for an unbounded time; Harrow and Low showed that circuits of only polynomial depth already approximate the Haar distribution’s first two moments closely enough for arguments like this one to apply [ 14 ] .