Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 50 / 1

Let $P$ be the orthogonal projection onto the code, and let $\{E_a\}$ be the specified error operators, including their linear span. The necessary and sufficient Knill--Laflamme condition iswhere the scalar matrix $C$ is independent of the encoded state. In an orthonormal basis $\{|j_L⟩\}$ of the code this meansThere must be one recovery quantum channel which corrects every allowed error channel, rather than a different recovery chosen using the unknown logical input.

The scalar condition means that any information carried away by the errors depends on their labels but not on the encoded amplitudes. To see how recovery works, diagonalize the positive semidefinite matrix $C$ and take corresponding linear combinations $F_{\ell }$ of the errors. Then $P{F}_{\ell }^{†}{F}_{m}P={\lambda }_{\ell }{\delta }_{\ell m}P$. For each positive ${\lambda }_{\ell }$, the map $F_{\ell }P/\sqrt{{\lambda }_{\ell }}$ is an linear isometry into an error-image subspace, and these subspaces are mutually orthogonal. Measuring their label and reversing the associated linear isometry recovers the logical state without measuring its amplitudes. Zero eigenvalues correspond to errors that annihilate the code. This also explains why the condition corrects arbitrary coherent combinations of the listed errors.

Let $A$ be the known set of affected qubits, with $|A|\le d-1$. Expand arbitrary operators on $A$ in tensor products of $I,X,Y,Z$. These Pauli operators span the complete error space on $A$.

By the distance of a quantum error-correcting code, every Pauli operator $F$ of weight below $d$ has scalar compression $PFP=c_{F}P$. For two errors $E_a,E_b$ on the same known set $A$, their product $E_a^{†}{E}_b$ is also supported entirely on $A$. Its weight is therefore at most $|A|$, rather than twice $|A|$. Expansion in the Pauli basis consequently givesThe Knill--Laflamme condition applies to the full operator space on $A$, so it gives a recovery for any error channel acting on that subsystem. Every set of at most $d-1$ known error locations is correctable. This is quantum erasure correction; the ability to use the locations is what improves the guarantee over unknown-location errors.

Use base-two logarithms, so Von Neumann entropy is measured in bits. Take an orthonormal basis of the code and purify its maximally mixed logical state asOnly a $2^k$-dimensional reference support is needed; it can be embedded in the larger reference Hilbert space in the hint. Its reduced density matrix is maximally mixed on that support, so $S(R)=k$.

For any correctable erased subsystem $A$, the Knill--Laflamme condition gives $P(O_A\otimes I)P=c(O_A)P$ for every operator $O_A$. Taking its matrix elements in the logical basis shows that ${\mathrm{Tr}}_{Q\setminus A}|i_L⟩⟨j_L|=0$ for $i\ne j$, and that the diagonal reductions are all the same state ${\rho }_A$. ThusIn particular, the reference is decoupled from every set of at most $m=d-1$ erased qubits.

For an information-carrying code $k>0$, first justify that two disjoint sets of size $m$ fit. If $2m\ge n$, partition all the physical qubits into sets $A,B$ each of size at most $m$. Both are correctable. Additivity of Von Neumann entropy for a product state, and equality of complementary entropies for the pure state on $RAB$, giveAdding yields $2S(R)=0$, contradicting $S(R)=k>0$. Hence $2m<n$.

Now choose disjoint $A,B$ of size $m$, and let $C$ contain the remaining $n-2m$ qubits. The state on $RABC$ is pure. Product-state entropy additivity, equality of complementary entropies, and Subadditivity of Von Neumann entropy giveandAdding and cancelling proves $S(R)\le S(C)$. Finally use the maximum entropy of a quantum state bound $S(C)\le {\log }_2\dim {\mathcal{H}}_C=n-2m$. Therefore the quantum Singleton bound isThis proof uses quantum erasure correction, not nondegeneracy, so it applies to degenerate codes as well. A code encoding no logical information requires a separate distance convention; the argument above concerns the usual $k>0$ coding problem.

Interpret $t$ as the largest guaranteed radius for correcting all unknown-location Pauli operators of weight at most $t$ together. Products of two such errors have weight at most $2t$, so the Knill--Laflamme condition holds when $2t<d$. Conversely, let $F$ be a minimum-weight Pauli operator with nonscalar compression $PFP$. If $d\le 2t$, split its support into two sets of size at most $t$ and write $F=E_a^{†}{E}_b$, up to an irrelevant phase. These two admissible errors violate the Knill--Laflamme condition. Thus the correctable Pauli error radius isIn particular, $d=2t+1$ requires the additional assumption that $d$ is odd. Distance is not uniquely determined by $t$ alone. Nor does the weight of a particular correctable error characterize this radius: particular higher-weight errors can also be correctable.

For a concrete even-distance example, the stabilizer code generated by $X^{\otimes 4}$ and $Z^{\otimes 4}$ has parameters $[[4,2,2]]$. Every single-qubit nonidentity Pauli operator anticommutes with a generator, so it has zero compression. But $Z_1{Z}_2$ commutes with both generators and acts nonscalarly on the code: it is unitary there and its code trace is zero. Hence the distance is two and its guaranteed radius is zero. This shows why an unconditional odd-distance formula would be incorrect.

Ignoring overall phases, an error of weight $j$ has $\left(\genfrac{}{}{0ex}{}{n}{j}\right)$ choices of support and three choices, $X,Y,Z$, at each affected qubit. The identity is the weight-zero term, soFor a nondegenerate quantum code, distinct correctable Pauli errors have orthogonal error-image subspaces. Each has dimension $2^k$, since a Pauli operator is unitary. All $N(t)$ subspaces lie in the physical Hilbert space of dimension $2^n$. Counting dimensions proves the quantum Hamming boundDegeneracy would allow distinct errors to share an error-image subspace, so this particular counting argument would then fail.