Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 342 / 2

The Knill--Laflamme condition for code projector $P$ and errors $\{E_a\}$ isFor Pauli errors of weight at most $t$, the product $E_a^{†}{E}_b$ has weight at most $2t$. If $d\ge 2t+1$, such a product cannot be a nontrivial logical Pauli, because every such operator has weight at least the distance of a stabilizer code $d$. It is therefore either a stabilizer, acting as a scalar on the code, or it anticommutes with some stabilizer and maps the code to an orthogonal syndrome space. These are exactly the two possibilities required by the displayed condition, so every error on at most $t$ qubits is correctable.

The six displayed generators of the Steane code are independent and commuting, so the common positive eigenspace has dimensionit encodes one logical qubit. The binary columns of the underlying Hamming parity-check matrix are all nonzero and distinct. Hence no weight-one or weight-two Pauli lies in the stabilizer normalizer outside the stabilizer. On the other hand,is logical, and multiplying it by $X_1{X}_2{X}_3{X}_4$ gives the weight-three representativeThe analogous statement holds for $\overline{Z}$, so

Now include both $E_1=X_5$ and $E_2=X_6{X}_7$ in the proposed error set. Their product iswhich is non-scalar on the code and violates the Knill--Laflamme condition. Since multiplying by a logical operator does not change commutation with stabilizers, $X_5$ and $X_6{X}_7$ have the same error syndrome. A decoder cannot know which occurred and may apply a correction that leaves a logical $\overline{X}$ error.

All plaquette stabilizers commute and square to one. Every term inis minimized by eigenvalue $+1$, so any ground state satisfiesfor every plaquette.

The R-edge operator $X_{e_R}$ commutes with every X-type stabilizer. It overlaps the two endpoint R plaquettes in an odd number of vertices and every other plaquette evenly, so it anticommutes only with the two endpoint Z-type stabilizers. It flips those two eigenvalues to $-1$, creating two $r_x$ particles. Multiplying by an adjacent R-edge operator toggles the shared endpoint twice, annihilating that excitation there while creating one at the new endpoint. Repetition forms a string operator in a topological code whose two excitations move farther apart.

An $r_x$ string is a product of X operators, while a $g_z$ string is a product of Z operators. A process that carries $r_x$ once around $g_z$ can be represented by a closed red X string crossing the green Z string once. At the crossing qubit,while all other factors commute. Reversing the order of the two string operators therefore multiplies the state by $-1$. The full braid phase isso $r_x$ and $g_z$ are mutual semions.

A green X, Y, or Z string may end on the bottom green boundary without leaving a violated green plaquette beyond the lattice. The top vertex is likewise a zero-length green boundary where such strings can terminate. Thus both locations absorb $g_x,g_y,g_z$: they exhibit anyon condensation at a boundary.

A string joining these two green condensers preserves every stabilizer but cannot be reduced to stabilizers, so it is logical. In the pictured lattice the shortest such path contains nine qubits. No shorter nontrivial string connects equivalent condensing boundaries, henceFor example, the product of X operators along any shortest green path from the bottom boundary to the top corner is a logical $\overline{X}$, and the product of Z operators along a corresponding path is a logical $\overline{Z}$. They may be chosen to overlap on an odd number of vertices, so they anticommute as required for one encoded qubit.