Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 342 / 2

The commuting termsare independent stabilizer generators. Since $J>0$, every ground state has $S_j=+1$. The resulting two-dimensional stabilizer subspace isthe phase-flip repetition code. A convenient pair of logical Pauli operators isIndeed, they commute with every stabilizer, anticommute with each other, and are not stabilizers.

For phase-flip errors $E_A={\prod }_{j\in A}{Z}_j$ and $E_B={\prod }_{j\in B}{Z}_j$, the operator entering the Knill--Laflamme condition is $E_A^{†}{E}_B=E_{A△B}$ and has weight at most $2t$. Every nonempty proper product of $Z_j$ anticommutes with some $S_j$ unless it is the full logical operator $\overline{X}$. The Knill--Laflamme conditions therefore hold whenever $2t<n$, and fail once two allowed errors can differ by $\overline{X}$. Thus

Each physical $X_j$ commutes with all stabilizers, and $X_j{X}_1$ is a product of stabilizers. Hence every $X_j$ acts on the code as the undetectable logical operator $\overline{Z}$. The code cannot detect, and therefore cannot correct, even a single bit flip.

A syndrome determines a phase-error set only up to multiplication by $\overline{X}=Z_1\cdots Z_n$: an error of weight $w$ and its complement of weight $n-w$ have the same error syndrome. Their probability ratio isFor $p<1/2$, maximum likelihood therefore chooses the representative of smaller weight. This is exactly majority-vote decoding of a repetition code: correction succeeds when fewer than half the qubits are flipped, with a random tie-break at $w=n/2$ when $n$ is even.

If $W\sim \mathrm{Bin}(n,p)$, then $\mathbb{E}W=np$ and $\mathrm{sd}(W)=\sqrt{np(1-p)}$. For every fixed $p<1/2$, the distance from the mean to the decision boundary in standard deviations isThus the logical failure probability $\mathrm{Pr}(W>n/2)$ tends to zero, in fact exponentially by a Chernoff bound. At $p=1/2$ the two representatives are equiprobable and decoding cannot improve with $n$. Hence

Label the physical qubits by $(i,j)$, with outer block $1\le i\le n$ and inner position $1\le j\le m$. The inner bit-flip repetition code has stabilizersand logical operators ${\overline{X}}_i={\prod }_{j=1}^m{X}_{i,j}$ and ${\overline{Z}}_i=Z_{i,1}$. Replacing each outer $X_i$ by ${\overline{X}}_i$ gives the outer stabilizers

This is the surface code on a chain of spheres. The $n+1$ pole-touching points are vertices, and the $m$ longitudes on sphere $i$ are its qubit-carrying links. At an interior touching point, the star operator is exactly $S_i^X$. Each face between adjacent longitudes has the two-edge plaquette operator $Z_{i,j}{Z}_{i,j+1}$; only $m-1$ of the $m$ face operators on each sphere are independent.

Using the conventions of part a, logical operators areThe first may be placed on any one sphere and is a dual equatorial cut crossing all $m$ longitude links. The second may use any fixed longitude and is a pole-to-pole path through all $n$ spheres. Multiplication by stabilizers deforms either representative without changing its logical action. Their minimum weights are $m$ and $n$, respectively, so the distance of a stabilizer code is

Under independent physical phase flips, an inner block suffers an effective phase flip of its encoded qubit exactly when it contains an odd number of errors. The binomial parity identity givesand hence the effective outer error probability isFor every $p<1/2$ and finite $m$, one has $q_m<1/2$, so outer majority-vote decoding succeeds with probability tending to one as $n\to \infty$. At $p=1/2$, one has $q_m=1/2$. The physical threshold is therefore still

For fixed $p\in (0,1/2)$, however, $q_m$ increases monotonically to $1/2$ as $m$ grows. Its distance from threshold isConsequently the majority-vote standardized separation is only of order $(1-2p{)}^m\sqrt{n}$, and the large-deviation exponent is of order $n(1-2p{)}^{2m}$. Thus larger $m$ leaves the threshold unchanged but makes the logical phase-error probability decay more slowly with $n$.