Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 35 1 Solution Created 2026-10-03 Updated 2026-10-14
The Pauli group includes scalar phases . For a nonzero stabilizer code, its Abelian subgroup must satisfy . Otherwise a common positive eigenvector would satisfy and vanish; for example is Abelian but yields no nonzero stabilizer code. We impose this standard admissibility condition.
Every element of such a stabilizer group is Hermitian and squares to : an imaginary-phase Pauli with square is excluded. The associated stabilizer code and its orthogonal projector areThe group multiplication proves , Hermiticity gives , and its range is exactly the common positive eigenspace. If has independent generators, . Every nonidentity element has trace zero, while , soThus it encodes logical qubits.
For the symplectic representation of a stabilizer code, label by , ignoring scalar phase. Products add labels, andThis is the nondegenerate alternating form on the binary symplectic space of Pauli labels. The image of is an r-dimensional isotropic subspace of a binary symplectic space, since the generators commute and no nontrivial scalar lies in . Consequently and . A binary generator matrix satisfies . Conversely an isotropic subspace can be lifted by choosing commuting Hermitian generators with independent labels; their products cannot equal , and the orthogonal projector formula supplies the stabilizer code. The labels of the centralizer of a stabilizer group are , with nontrivial logical actions represented by .
For an example, the three-qubit bit-flip repetition code has and . Its two binary rows have zero X part and Z parts . The errors have respective error syndromes , so all single bit flips can be corrected. This is not a stabilizer code correcting every single-qubit error: acts as a nontrivial logical phase error.
Let denote the Pauli centralizer. The exact Pauli error criterion for a stabilizer code isEquivalently, no difference of error labels lies in . The scalar phases in are essential: an error product is harmless even though it is not in .
We prove the criterion. For outside , some anticommutes with . Since , we have , hence . If , then . If , it restricts to a unitary operator on the stabilizer code, butno is scalar. A scalar restriction would have scalar coefficient zero by this trace equation, contradicting unitarity. Thus this third case is not scalar.
To justify why scalar compressions are exactly what correction requires, suppose one recovery corrects every . Dilate that recovery to an isometry including its auxiliary system. Correct recovery of every pure stabilizer code state gives . The auxiliary state is independent of : linearity on basis states and their superpositions forces their auxiliary vectors to coincide. Preservation of inner products then giveswhich is the Knill--Laflamme condition . The three cases above prove necessity of the boxed criterion.
For sufficiency, put errors in the same class when . They act identically on the stabilizer code up to scalar phase, so give the same error-image subspace. Under the criterion, different classes give mutually orthogonal subspaces , because their cross-compressions vanish. Measure which of these subspaces is occupied and apply the inverse of a representative error. This returns every stabilizer code state unchanged; extend the recovery on the orthogonal complement by preparing any fixed stabilizer code state. Linearity also corrects any valid noise channel whose Kraus operators are linear combinations of the specified errors: within each syndrome all amplitudes multiply the same logical state, and trace preservation normalizes the recovered state. This proves sufficiency, including degenerate error classes.