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The Schumacher compression theorem concerns a memoryless quantum information source with average density matrix $\rho$ on a finite-dimensional Hilbert space. For $n$ uses the average state is ${\rho }_n={\rho }^{\otimes n}$. Compression consists of an encoding quantum channel into a space of dimension $D_n$ and a decoding channel back to the original block space, with asymptotic rate $n^{-1}{\log }_2{D}_n$ qubits per source use. The optimal reliable rate is $S(\rho )$: every larger rate is achievable, and no smaller rate is reliable. This holds for average pure-source fidelity; the direct construction also preserves a purification reference, and the converse for that stronger entanglement fidelity criterion holds with fidelity tending to zero below the threshold. The encoding and decoding are deterministic channels; no free classical transmission of the source label is allowed.

Diagonalize $\rho ={\sum }_j{\lambda }_j|j⟩⟨j|$ and omit zero-probability eigenvalues when sampling. Define the quantum typical subspace projector $P_{n,\delta }$ by retaining product eigenvectors withThe sampled random variable $-{\log }_2{\lambda }_j$ has expectation $S=S(\rho )$. The weak law of large numbers gives $w_n=\mathrm{Tr}({\rho }_n{P}_{n,\delta })\to 1$. Every retained eigenvalue is at least $2^{-n(S+\delta )}$, and the total probability is at most one, soThis proves the two typical subspace theorem estimates used in the direct part.

Write $P=P_{n,\delta }$. Encode its range isometrically into a compressed space and reserve one orthogonal failure flag. Decode the range back isometrically and the failure flag to a fixed pure state $|\phi ⟩$. This typical-subspace compression with a failure flag is a physical channel whose composite action isIndeed its Kraus operators are $P$ and $|\phi ⟩⟨e_a|$, where the $e_a$ form an orthonormal basis of $\ker P$, and their adjoint products sum to $I$. The Kraus formula for entanglement fidelity givesFor any pure-state ensemble of ${\rho }_n$, the average squared quantum fidelity is at least this number: apply convexity of $|z{|}^2$ separately to the amplitudes $⟨{\psi }_x|K_{\alpha }|{\psi }_x⟩$ with their source probabilities. Thus average fidelity also tends to one. The compressed dimension is at most $2^{n(S+\delta )}+1$, and its rate is at most $S+\delta +o(1)$. Choose $\delta <R-S$ for any prescribed $R>S$. One may also take ${\delta }_n=n^{-1/4}$: the finite spectral information variance and Chebyshev's inequality still give $w_n\to 1$, with rate tending to $S$. A pure source, $S=0$, can simply be re-prepared exactly without sending any qubits.

For the coherent converse, we prove the finite-dimensional quantum compression converse. If the compressed space has dimension $D$, the state of reference plus compressed system has a pure-state decomposition with Schmidt rank at most $D$. Applying decoder Kraus operators locally cannot increase the Schmidt rank of those vectors, so the final reference–output state has Schmidt number at most $D$. Let $|{\Psi }_{{\rho }_n}⟩$ purify the source, with squared Schmidt coefficients equal to the decreasing eigenvalues ${\mu }_1\ge {\mu }_2\ge \cdots$ of ${\rho }_n$. For any normalized vector $|v⟩$ of Schmidt rank at most $D$, its reference support lies in a rank-at-most-$D$ projector $Q$. HenceThe last bound follows by maximizing ${\sum }_j{\mu }_j⟨j|Q|j⟩$ subject to $0\le ⟨j|Q|j⟩\le 1$ and their sum at most $D$. Average over a Schmidt-rank-bounded decomposition of the final state to obtain $F_e\le {\sum }_{j=1}^D{\mu }_j$.

At rate $R<S$, choose $0<\delta <S-R$. The total source probability outside the quantum typical subspace tends to zero, while every eigenvalue inside it is at most $2^{-n(S-\delta )}$. Therefore for $D_n\le 2^{nR}$,This argument allows arbitrary encoding and decoding channels, not only subspace projection.

For completeness, average pure-source reliability also forces the same rate, even though it is a weaker criterion than entanglement fidelity. Here is a separate converse for that formulation. Let the emitted block pure states be ${\psi }_x$, with probabilities $p_x$, and let ${\sigma }_x$ be the decoded states. If their mean squared fidelity tends to one, then their mean trace distance ${\bar{\delta }}_n$ tends to zero, since $\frac{1}{2}\Vert {\sigma }_x-{\psi }_x{\Vert }_1\le \sqrt{1-⟨{\psi }_x|{\sigma }_x|{\psi }_x⟩}$. This inequality follows by choosing a purification with that overlap and using contraction of trace distance under a partial trace.

We need only an elementary entropy continuity from Jordan decomposition bound, derivable from Question 4. If $\delta =\frac{1}{2}\Vert \sigma -\tau {\Vert }_1$, write the positive and negative parts of $\sigma -\tau$ as $\delta {\omega }_{+},\delta {\omega }_{-}$. The common density matrixand the two mixture bounds giveHere $g$ is increasing and concave, $g(0)=0$. Apply this in block dimension $d^n$, both to the average states and to individual outputs, using convexity of trace distance and concavity of $g$. Since the original ensemble is pure, its Holevo quantity is $S({\rho }_n)=nS$. The decoded ensemble hasThe Holevo quantity under a quantum channel cannot increase. To specify the entropy ingredient, Strong subadditivity of Von Neumann entropy states $S(XB)+S(BE)\ge S(B)+S(XBE)$. Attach the classical source label $X$, dilate the decoder isometrically to $BE$, and trace out $E$. The displayed inequality says $I(X:B)\le I(X:BE)$; the isometry preserves the input mutual information. Since $I(X:B)$ of a labelled ensemble is its Holevo quantity, this proves the needed decoder monotonicity. Before decoding, the system has dimension $D_n$, so its Holevo quantity is at most ${\log }_2{D}_n$. Thus ${\log }_2{D}_n\ge nS-o(n)$ and the rate cannot be below $S$. Both operational fidelity formulations therefore have the claimed noiseless coding threshold.

Write $\overline{\rho }={\sum }_i{p}_i{\rho }_i$. The Holevo quantity iswhere $S$ is Von Neumann entropy, with a consistent logarithm base. Attach an orthogonal classical label register $X$ to form the classical-quantum stateThe entropy of an orthogonal quantum mixture givesHere $H(p)$ is the Shannon entropy of the classical label. Hence $I(X:B)=\chi (\mathcal{E})$, where $I$ denotes quantum mutual information.

Let $\Phi :B\to C$ be the proposed quantum channel. Using its Kraus representation, define an linear isometry $V={\sum }_{\ell }{A}_{\ell }\otimes |\ell {⟩}_E$, since $V^{†}V=I$. This is a Stinespring representation. Apply it to $B$, obtaining a state ${\eta }_{XCE}$; tracing out $E$ applies $\Phi$ to every ensemble member. An linear isometry preserves nonzero eigenvalues and therefore Von Neumann entropy, soTheir difference isThe Strong subadditivity of Von Neumann entropy states that this expression is nonnegative for every tripartite state. ConsequentlyThus the Holevo quantity under a quantum channel decreases because discarding the dilation environment cannot improve correlations with the classical label. The probabilities remain unchanged throughout.

Encode the ensemble label in an orthogonal classical register:The classical-quantum state has $S(XB)=H(p)+{\sum }_x{p}_{x}S({\rho }_x)$, $S(X)=H(p)$ and $S(B)=S(\bar{\rho })$, where $\bar{\rho }={\sum }_x{p}_x{\rho }_x$. HenceApplying ${\mathrm{id}}_X\otimes \Lambda$ changes the conditional states to the output ensemble. The same identity gives $I(X:B^{\prime })=\chi ({\mathcal{E}}^{\prime })$. Data processing for quantum mutual information, proved in Question 3, therefore yields the Holevo quantity under a quantum channel:This concerns the information available in the whole ensemble, not the change in the entropy of each individual state.