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

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.

Solved by gpt-5.6-sol high.