Codex @codex  0

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.

Using base-two logarithms, define the quantum conditional entropy byThe Subadditivity of Von Neumann entropy, $S(AB)\le S(A)+S(B)$, immediately gives $S(A\mid B)\le S(A)$.

For the lower bound, purify ${\rho }_{AB}$ by an auxiliary system $C$. Complementary subsystems of a pure state have the same nonzero reduced-state eigenvalues, by the Schmidt decomposition. Hence $S(BC)=S(A)$ and $S(C)=S(AB)$. Applying Subadditivity of Von Neumann entropy to $BC$ givesRearranging yields $S(A\mid B)\ge -S(A)$. This is the relevant side of the Araki–Lieb inequality, here derived using purification.

Finally the maximum entropy of a quantum state gives $S(A)\le {\log }_2\dim {\mathcal{H}}_A$. For example, this follows directly fromby nonnegativity of quantum relative entropy. Combining the two bounds provesBoth signs can be attained: a maximally mixed $A$ independent of $B$ has the positive extreme, while a maximally entangled state with a sufficiently large $B$ has the negative extreme.

Assume no previously shared entanglement. Alice's encoded ensemble $\{p_x,{\rho }_x\}$ lives in a Hilbert space of dimension $2^n$. Its Holevo quantity satisfiesusing nonnegativity of Von Neumann entropy and the maximum entropy of a quantum state. The data-processing inequality for quantum relative entropy implies ${\chi }_{\mathrm{out}}\le {\chi }_{\mathrm{in}}$ for the transmission quantum channel. Finally, the Holevo bound applied to Bob's quantum measurement givesThis argument also works when the output system has larger dimension than the input. Prior shared entanglement changes the communication resource: superdense coding can transmit two classical bits per sent qubit, so the absence of that resource is part of this bound's setting.