 Five-qubit concatenation threshold

Assume independent physical-error probability $p$, correction of zero or one errors, and failure on every block with at least two errors. The failure map is $f(p)=1-(1-p{)}^5-5p(1-p{)}^4$. Factoring $f(p)-p$ gives $p(1-p)(4p^3-11p^2+9p-1)$. The unique interior threshold is the root of $4p^3-11p^2+9p-1=0$. Below it concatenation decreases the failure probability toward zero. Two levels give $f(f(p))=1000p^4+O(p^5)$, compared with $10p^2+O(p^3)$ at one level.

// Target: quantum-information-theory.bigb