Computing the Threshold

Threshold as a break-even point

A threshold is where encoding stops hurting and starts helping. The clearest place to see the arithmetic is the code-capacity model: errors strike only the data qubits (gates, measurement, and syndrome extraction are taken to be perfect), each qubit suffering an independent bit flip with probability $p$. We ask for the largest $p$ below which a code, decoded optimally, makes the logical error rate smaller than the physical rate — and below which growing the code drives it to zero.

The three-qubit majority vote

Take the distance-3 repetition code $|0\rangle \to |000\rangle$, $|1\rangle \to |111\rangle$, which corrects a single bit flip by majority vote. A logical error occurs when two or three of the three qubits flip, since the majority then points the wrong way. With independent flips of probability $p$,

For small $p$, $p_L \approx 3p^2 \ll p$: the code helps. But as $p$ grows the cubic catches up.

Solving for the crossover

The break-even condition is $p_L = p$ — the encoded qubit is no better than a bare one:

Factoring, $(2p - 1)(p - 1) = 0$, so $p = \tfrac12$ or $p = 1$. The physically meaningful crossing is

Below $p = \tfrac12$ we have $p_L < p$, and concatenating the repetition code (re-encoding each qubit again) squares the suppression at every level — so logical errors vanish in the limit of many levels. Above $p = \tfrac12$ the inequality flips and encoding amplifies the error. The threshold of the repetition code against independent bit flips is exactly one half, which is also the point where a single bit carries no information.

Try it

Find the repetition-code threshold yourself. Define $p_L(p) = 3p^2 - 2p^3$ and locate the nonzero crossing of $p_L(p) = p$ by solving the quadratic $2p^2 - 3p + 1 = 0$ for the root in $(0,1)$. Return that root as a plain number. (Compute it — do not hardcode $0.5$.)

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