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

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.