Distillation Thresholds

A distillation protocol is only useful if iterating it reduces the error. Whether it does depends on where the input error sits relative to a threshold — the analogue, for magic states, of the fault-tolerance threshold for codes.

The recurrence and its fixed points

The $15$-to-$1$ protocol implements the leading-order map

Iterating distillation produces the sequence $\varepsilon_0, \varepsilon_1 = f(\varepsilon_0), \varepsilon_2 = f(\varepsilon_1), \dots$ Its long-run behavior is governed by the fixed points of $f$, the solutions of $f(\varepsilon) = \varepsilon$:

The trivial fixed point $\varepsilon = 0$ is the ideal magic state. The nontrivial fixed point

is the break-even point of this leading-order recurrence — the distillation threshold for the idealized $\varepsilon_{\text{out}} = 35\,\varepsilon_{\text{in}}^{3}$ map with perfect Clifford operations. It is the clean reference number for the $15$-to-$1$ protocol, not an exact, universal threshold: the true recurrence carries higher-order corrections ($35\,\varepsilon^3 + O(\varepsilon^4)$), and faulty circuit elements (see below) shift the usable value.

Stability: which way does the flow go?

Whether the iteration converges to $0$ or runs away depends on which side of $\varepsilon^{*}$ we start. Compare $f(\varepsilon)$ to $\varepsilon$:

• If $\varepsilon_0 < \varepsilon^{*}$, then $35\,\varepsilon_0^2 < 1$, so $f(\varepsilon_0) = \varepsilon_0\,(35\,\varepsilon_0^2) < \varepsilon_0$. Each round shrinks the error, and the sequence converges to $0$ — rapidly, since the convergence is cubic.
• If $\varepsilon_0 > \varepsilon^{*}$, then $35\,\varepsilon_0^2 > 1$, so $f(\varepsilon_0) > \varepsilon_0$. Each round increases the error: distillation diverges and is useless.

So $\varepsilon^{*}$ is an unstable fixed point separating a basin of attraction (toward perfect magic states) from a runaway regime. This is exactly the structure of a threshold.

Caveats: this is the ideal-Clifford threshold

The $\varepsilon^{*} = 1/\sqrt{35}$ figure assumes the distillation circuit itself is perfect — only the input magic states are noisy. In a real device the Clifford gates and measurements that implement the protocol are also faulty, which lowers the usable threshold and is the reason distillation is run on encoded (error-corrected) qubits. The more careful analyses of Reichardt and others track these circuit-level errors and give somewhat smaller, model-dependent thresholds.[^reichardt] The $1/\sqrt{35}$ value is the clean, protocol-level number every practitioner carries in their head.

Try it

Solve the fixed-point equation $35\,\varepsilon^{3} = \varepsilon$ for its nontrivial root and return the threshold $\varepsilon^{*}$. Derive it; the grader also checks that your value really is a fixed point of the map.

[^reichardt]: Reichardt, Quantum universality from magic states distillation applied to CSS codes, arXiv:quant-ph/0411036.

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