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

Every element of the $n$-qubit Pauli group squares to either $1$ or $-1$. If a stabilizer generator $S_j$ had $S_j^2=-1$, closure of the group would put $-1$ in $S$, contrary to the definition. Thus $S_j^2=1$. Since Pauli operators are unitary,and $S_j^{-1}=S_j$ proves $S_j=S_j^{†}$.

If two stabilizers $g,h$ anticommute and a nonzero vector $|\psi ⟩$ belonged to the codespace, thena contradiction. Hence a nonzero stabilizer code requires $S$ to be Abelian. Its centralizer of a stabilizer group is

Measure each generator $S_j$. If a Pauli error $E$ has occurred, its outcome is $(-1{)}^{s_j}$, whereThe bit vector $s=(s_1,\dots ,s_m)$ is the error syndrome. Choose one representative $E_s$ with this syndrome. Another Pauli $E$ has the same syndrome exactly when $E_s^{†}E$ commutes with every $S_j$, namely when $E_s^{†}E\in C(S)$. ThereforeThis stabilizer-syndrome coset is all the syndrome reveals: multiplication by a stabilizer changes nothing on the code, while multiplication by an element of $C(S)\setminus S$ can change the logical state without changing any syndrome bit.

Fix the reference bit-flip chainand assign to each lattice edge $e=v{v}^{\prime }$ the bond signwhere $n_e(E)=1$ 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 occupationFor $N$ edges, its independent bit-flip probability isbecause $e^{\beta J}=\sqrt{(1-p)/p}$. Summing over products of stars therefore gives the Surface-code decoding as a random-bond Ising model identitywhere $r$ 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.

Choose both circular ends of the cylinder to be electric, or rough, boundaries. In the convention of part 1, retain plaquette generatorswith their boundary truncations, and use star generatorsonly at vertices not lying on an electric boundary. Omitting the endpoint star checks allows a $Z$ string to terminate at either boundary, which is precisely anyon condensation at a boundary for $e$.

Relative homology now has one nontrivial class represented by a primal $Z$ path $\gamma$ joining the two boundaries. The dual absolute homology has one class represented by an $X$ loop $\tilde{\gamma }$ around the cylinder. Their strings intersect once and anticommute, so they form one logical pair $\bar{Z},\bar{X}$. Every other closed or boundary-ending string is a product of stabilizers, so the code has exactly one logical qubit.

In the Random-bond Ising model mapping, there is no Ising spin for an omitted boundary star. An edge $e$ joining an interior vertex $v$ to an electric boundary is toggled by only $A_v$, so its bond factor ${\eta }_e{\sigma }_v{\sigma }_{v^{\prime }}$ is replaced by the boundary-field factor ${\eta }_e{\sigma }_v$. Equivalently, attach exterior spins fixed to $+1$ and retain the same bond formula. Edges wholly on a boundary contribute fixed constants. The bulk Hamiltonian is therefore supplemented bywith the signs still determined by $E_s{\bar{X}}^q$. This is the required boundary modification.