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The passage around this formula

It is worth being explicit about which part of that conditional is load-bearing. Neither $t_L$ nor $t_G$ is arbitrary on its own — each is anchored to a cited, checkable piece of theory, and the crossover tolerance $\delta^\ast$ is computed, not chosen after the fact to produce a headline. What is a choice, and should be named as one, is treating $\epsilon$ and $\delta$ as commensurate just because both are dimensionless numbers between zero and one. A fidelity and a normalized angular tolerance measure different things, and no law of physics says a “demanding” qubit fidelity and a “demanding” compass tolerance sit at the same point on their respective scales. The reversal at $\delta$ $\approx$…

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