← Back to article

Equation 92 · The Bit Comes Back Before the Bearing

What does this equation mean?

t∗t_*

Read the formula alongside the article passage below. Each part has a deeper page with its role in the equation, the supporting passage and nearby citations.

the negligible addition to. Read the equation part by part below; each part has a contextual explanation and a link to its mathematical background.

Read it piece by piece

t∗t_*

Symbol t_*

the negligible addition to.

Understand this part →

How to interpret it

Read this expression with the definitions, units, and assumptions supplied by the article.

What the article says around this equation

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 S2,4S_{2,4}⟩\rangle = 0.5095 nats, ⟨\langle S2,16S_{2,16}⟩\rangle = 0.6465 nats, and ⟨\langle S2,64S_{2,64}⟩\rangle = 0.6814 nats, closing in on the ceiling ln⁡\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…
Read the full surrounding passage
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 S2,4S_{2,4}⟩\rangle = 0.5095 nats, ⟨\langle S2,16S_{2,16}⟩\rangle = 0.6465 nats, and ⟨\langle S2,64S_{2,64}⟩\rangle = 0.6814 nats, closing in on the ceiling ln⁡\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 tL(ϵ)t_L(\epsilon) depends on ϵ\epsilon only logarithmically: driving ϵ\epsilon from 10^{-1} to 10^{-9} costs on the order of thirty extra qubits of radiation, a negligible addition to t∗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 ] .

Read the equation in its article →

Sources cited in the surrounding passage

These citations give research context. Read each source to check which claims it supports.

Return to The Bit Comes Back Before the Bearing

Browse the mathematical compendium →