Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 33 / 1 / c / Solution

Use the normalized three-qubit states $|G_{\pm }⟩=(|000⟩\pm |111⟩)/\sqrt{2}$. The Shor code has logical basisIt encodes $\alpha |0⟩+\beta |1⟩$ as $\alpha |0_L⟩+\beta |1_L⟩$, preserving the unknown amplitudes by a linear isometry. It concatenates a three-block phase-flip repetition code with the three-qubit bit-flip repetition code inside each block. Six stabilizer generators, $Z_1{Z}_2,Z_2{Z}_3,Z_4{Z}_5,Z_5{Z}_6,Z_7{Z}_8,Z_8{Z}_9$, test bit errors inside the blocks. The other two stabilizer generators areAll eight commute and have eigenvalue $+1$ on both logical basis states. They define the $[[9,1,3]]$ stabilizer code. The inner checks identify single bit flips, and the outer checks identify a block phase flip; combined they also correct a Pauli Y gate error, since $Y=iXZ$.

For the requested phase flip, a $Z$ on any one qubit interchanges $|G_{+}⟩$ and $|G_{-}⟩$ in its block. It commutes with the six $Z$-pair checks. It anticommutes with $A$ or $B$ exactly when its block is contained in that check. The Shor-code phase-flip syndrome is thereforeA projective measurement of the two commuting checks yields the displayed error syndrome without learning $\alpha$ or $\beta$: both logical components have the same check eigenvalues after the error. Applying the indicated Pauli Z gate restores every encoded quantum state.

The correction is genuinely degenerate. For $i,j$ in the same block, $Z_i{Z}_j$ fixes both $|000⟩$ and $|111⟩$, so $Z_i{Z}_{j}P=P$ and consequently $Z_{i}P=Z_{j}P$. In particular $Z_1$ and $Z_2$ are distinct physical errors but have identical action on every code state; their image subspaces coincide, not merely their observed syndrome labels. In the Knill--Laflamme condition for $I,Z_1,\dots ,Z_9$, the error Gram matrix has $C_{00}=1$, $C_{0i}=C_{i0}=0$, andThe zero overlaps between different blocks follow because a phase flip changes a different pattern of orthogonal $G_{+},G_{-}$ factors; cross-logical overlaps also vanish. Thus $C$ consists of one $1\times 1$ block and three all-ones $3\times 3$ blocks and has rank four, not ten. This both proves correctability of the entire single-phase-error span and exhibits its degeneracy. A coherent sum of phase errors in one block has the same logical action with its coefficients summed, so inability to locate the individual qubit loses no logical information. The two block checks suffice to correct every single phase flip, even though the within-block errors are indistinguishable.