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In the unsquared quantum fidelity convention, the fidelity between a purification and its channel output is the square root of entanglement fidelity. Tracing out the purifying reference leaves the input and output marginals. Monotonicity of quantum fidelity under partial trace therefore gives the bound. Preserving an input marginal is a weaker requirement than preserving its correlations with a reference.

The Kraus formula for entanglement fidelity applies to any input, including mixed qubit states. For the amplitude damping channel, the diagonal Kraus trace is $[1+s_z+\sqrt{1-p}(1-s_z)]/2$ and the jump Kraus trace has squared modulus $p(s_x^2+s_y^2)/4$. Summing them gives the formula. Ground-state inputs have unit entanglement fidelity; excited-state inputs have fidelity $1-p$.

If $\{K_a\}$ is any Kraus representation of a quantum channel $\Lambda$, thenFor a purification of a density operator, each branch overlap is $⟨{\Psi }_{\rho }|(I\otimes K_a)|{\Psi }_{\rho }⟩=\mathrm{Tr}(\rho K_a)$. Summing the squared branch overlaps gives entanglement fidelity.

In the unsquared quantum fidelity convention, operation fidelity compares a purification of a density operator with its image under a quantum channel acting on the source subsystem:Its square is entanglement fidelity. Thus this convention is distinct from the squared quantity $F_e$; the two should not be installed as synonyms.

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.

Choose an orthonormal basis of the joint $d^2$-dimensional Hilbert space whose first vector is the given purification of a density operator $|\rho {⟩}_{QR}$. Apply rank-one dephasing to ${\sigma }_{QR}$ in this basis, and denote the resulting diagonal probabilities by $p_1,\dots ,p_{d^2}$. Their first entry isThe preceding entropy increase under nonselective projective measurement and entropy bound with one prescribed probability yieldTherefore the quantum Fano inequality isThe quantity $f^2$ is the entanglement fidelity of $\mathcal{N}$ on ${\rho }_Q$; it is already a squared overlap, so it is $f^2$, rather than $f$, that enters the binary entropy. The argument is an instance of the entropy bound from overlap with a pure state in dimension $d^2$. At $f^2=1$ the output is the original pure state and its Von Neumann entropy is zero. For $d=1$ the system is trivial and the same zero-entropy conclusion holds without evaluating ${\log }_2(d^2-1)$.

Choose Kraus representationsThe composite quantum channel has Kraus operators $T_{ba}=D_b{C}_a$. Each has matrix rank at most $k$, because it factors through the $k$-dimensional space $K$, andThe operation fidelity uses the unsquared quantum fidelity on a purification of a density operator, so its square is entanglement fidelity. The Kraus formula for entanglement fidelity givesIndeed each overlap $⟨{\Psi }_{\rho }|(I_R\otimes T_{ba})|{\Psi }_{\rho }⟩$ equals $\mathrm{Tr}({\rho }_Q{T}_{ba})$.

For any one of these Kraus operators $T$, let ${\Pi }_T$ be the orthogonal projection onto its image, of matrix rank $r_T\le k$. Since ${\Pi }_{T}T=T$, the Cauchy-Schwarz inequality for the Hilbert-Schmidt inner product gives the rank bound for a weighted operator trace:The last step uses part (i) at $r_T$, followed by nonnegativity of the eigenvalues of the density operator ${\rho }_Q$. Summing over the Kraus operators and using their completeness relation proves the finite-dimensional quantum compression converse:No invertibility of ${\rho }_Q$, or restriction to an isometric decoder, was used.

A compression-decompression scheme consists of quantum channels ${\mathcal{E}}_n:\mathcal{D}({\mathcal{H}}^{\otimes n})\to \mathcal{D}({\mathcal{K}}_n)$ and ${\mathcal{D}}_n:\mathcal{D}({\mathcal{K}}_n)\to \mathcal{D}({\mathcal{H}}^{\otimes n})$, where ${\mathrm{limsup}}_n{n}^{-1}{\log }_2\dim {\mathcal{K}}_n\le R$. For the emitted pure-state ensemble, reliability means that the mean squared quantum fidelity tends to one:The average is taken over the actual source distribution, rather than the worst possible input vector.

A standard stronger formulation of reliable quantum source compression requireswhere entanglement fidelity tests preservation of a purification and its reference-system correlations. It implies the mean-fidelity condition for every pure-state ensemble of ${\pi }^{\otimes n}$. Schumacher compression achieves this at any rate $R>S(\pi )$: choose $0<\epsilon <R-S(\pi )$, encode the quantum typical subspace, and map atypical outcomes to a fallback state. The composed channel has a Kraus term $P_{\epsilon }^{(n)}$, so its entanglement fidelity is at least $[\mathrm{Tr}({\pi }^{\otimes n}{P}_{\epsilon }^{(n)}){]}^2\to 1$.

Embed the input qubit as the span of $|0⟩,|1⟩$ in the three-dimensional output. Because the input is pure, its only purification has a trivial reference and entanglement fidelity reduces to its survival probability:Since $|e⟩$ is orthogonal to the embedded qubit subspace,

For a purification $|{\Psi }_{\rho }{⟩}_{RA}$ of $\rho$, the entanglement fidelity isUsing a Kraus representation $\Lambda (X)={\sum }_k{A}_{k}X{A}_k^{†}$ givesIn a Schmidt decomposition of the purification,Therefore

For a quantum channel on a $d\ge 2$ dimensional system, purify the input to $|\Psi {⟩}_{QR}$ and set $\sigma =(\mathcal{N}\otimes \mathrm{id})(|\Psi ⟩⟨\Psi |)$. Its entanglement fidelity $F_e=⟨\Psi |\sigma |\Psi ⟩$ satisfies $S(\sigma )\le h(F_e)+(1-F_e){\log }_2(d^2-1)$. This is the entropy bound from overlap with a pure state in dimension $d^2$. For a trivial one-dimensional system the output entropy is zero.

A quantum source-compression scheme has encoding and decoding quantum channels with compressed space ${\mathcal{K}}_n$ and rate ${\mathrm{limsup}}_n{n}^{-1}{\log }_2\dim {\mathcal{K}}_n$. For a pure-state source ensemble, average reliability means that the mean squared quantum fidelity after decoding tends to one. A stronger coherent criterion is $F_e({\pi }^{\otimes n},{\mathcal{D}}_n\circ {\mathcal{E}}_n)\to 1$, where entanglement fidelity also tests a purification reference. Encoding the quantum typical subspace achieves both criteria above $S(\pi )$. Schumacher's original quantum coding paper establishes quantum source coding.