Fix the reference bit-flip chain
$$
E_{s,q}=E_s\bar X^q
$$
and assign to each lattice edge $e=vv'$ the bond sign
$$
\eta_{vv'}(s,q)=(-1)^{n_e(E_{s,q})},
$$
where $n_e(E)=1$ if $E$ contains $X_e$ and is zero otherwise. A product of star stabilizers is specified by Ising spins $\sigma_v$: choose the star at $v$ when $\sigma_v=-1$. The resulting error has edge occupation
$$
n_e=\frac{1-\eta_{vv'}\sigma_v\sigma_{v'}}2.
$$
For $N$ edges, its independent bit-flip probability is
$$
p^{|E|}(1-p)^{N-|E|}
=[p(1-p)]^{N/2}
\exp\left(\beta J\sum_{vv'}
\eta_{vv'}\sigma_v\sigma_{v'}\right),
$$
because $e^{\beta J}=\sqrt{(1-p)/p}$. Summing over products of stars therefore gives the <Surface-code decoding as a random-bond Ising model> identity
$$
\Pr(E\in E_s\bar X^qS_X)
=\frac{[p(1-p)]^{N/2}}{r}\,Z_{s,q},
$$
where $r$ is the number of Ising configurations representing the same stabilizer product, usually $r=2$ on a closed connected lattice because a global spin flip changes no bond. The class-independent prefactor cancels when the two logical classes $q=0,1$ are compared.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 342 / 2 / c / Solution

{c}

For independent bit flips, summing the probabilities of all error chains in one syndrome and logical class equals, up to a class-independent factor, the partition function of a random-bond <Ising model>. A reference error fixes bond signs and products of star stabilizers become Ising spin configurations.

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 Incoming links: Surface-code decoding as a random-bond Ising model