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

Solution

Fix the reference bit-flip chain

E_{s, q} = E_{s} \overset{ˉ}{X}^{q}

and assign to each lattice edge

e = v v^{'}

the bond sign

η_{v v^{'}} (s, q) = (- 1)^{n_{e} (E_{s, q})},

where

n_{e} (E) = 1

E

contains

X_{e}

and is zero otherwise. A product of star stabilizers is specified by Ising spins

σ_{v}

: choose the star at

v

when

σ_{v} = - 1

. The resulting error has edge occupation

n_{e} = \frac{1 - η _{v v^{'}} σ _{v} σ _{v^{'}}}{2} .

For

N

edges, its independent bit-flip probability is

p^{∣ E ∣} (1 - p)^{N - ∣ E ∣} = [p (1 - p)]^{N /2} exp (β J \sum_{v v^{'}} η_{v v^{'}} σ_{v} σ_{v^{'}}),

because

e^{β J} = (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} \overset{ˉ}{X}^{q} S_{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.

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