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Let the quantum channel send $B$ to $B^{\prime }$. From its Kraus representation, choose a Stinespring representation $V:B\to B^{\prime }E$ and formThe channel output is ${\sigma }_{A{B}^{\prime }}={\mathrm{Tr}}_E{\sigma }_{A{B}^{\prime }E}$. Isometric invariance of Von Neumann entropy givesTherefore the quantum mutual information satisfies $I(A:B{)}_{\rho }=I(A:B^{\prime }E{)}_{\sigma }$. Expanding its loss after discarding the environment givesThe final expression is the quantum conditional mutual information $I(A:E\mid B^{\prime })$. The Strong subadditivity of Von Neumann entropy states exactly that it is nonnegative. HenceThis proves data processing for quantum mutual information for any mixed input and any local quantum channel, not just the atomic channel in earlier parts. The correlations lost are exactly the conditional correlations with the discarded environment.

Independent applications of the phase-flip channel are a classical mixture of the eight physical Pauli Z gate error patterns. A specified weight-$k$ pattern has probability ${ϵ}^k(1-ϵ{)}^{3-k}$. The recovery from the preceding part succeeds on weight zero and one, and yields the logical Pauli X gate on weight two and three. There are three weight-two patterns and one weight-three pattern, so the logical failure probability isAveraging the corrected branches gives the complete decoded quantum channelIn particular, the leading logical error is $3{ϵ}^2$, rather than an error of first order in $ϵ$. For the stated range,Thus the failure probability is smaller than the physical error probability. The decoded channel is a bit-flip channel; it is not $D_{p_L}$, whose error operator would be $Z$. This distinction follows from the logical codeword exchange in the phase-flip repetition code.

Three standard objectives are state preparation, observable optimization and process implementation. All optimizations are over admissible control inputs and a specified final time $T$.

For state preparation, steer an initial density operator to a target ${\rho }_d$. An exact goal is ${\rho }_f(T)={\rho }_d$, or one may minimize the squared Hilbert-Schmidt distanceFor a pure target ${\rho }_d=|{\psi }_d⟩⟨{\psi }_d|$, maximizing $⟨{\psi }_d|{\rho }_f(T)|{\psi }_d⟩$ is a useful alternative. The target must be physically attainable: Hamiltonian engineering alone preserves the initial spectrum of the density operator.

For an observable objective, choose a Hermitian operator $O$ and maximize its expected final value:Population transfer is a special case with $O$ the projector onto the desired level or subspace; suppressing an unwanted population reverses the sign or minimizes this quantity.

For coherent process engineering, implement a target unitary operator $V$ on every input, rather than merely arranging one state transfer. If the propagator is $U_f(T)$, a phase-insensitive error isThe error vanishes precisely when $U_f(T)=e^{i\varphi }V$: equality in the Hilbert-Schmidt inner product Cauchy-Schwarz inequality bound makes the two operators proportional. For an open-system process, the analogous objective compares the implemented quantum channel with the target channel. These objectives can be supplemented by pulse-energy, amplitude, bandwidth and duration constraints in quantum optimal control.

Implement the quantum channel by a unitary dilation with a pure initial environment. The final state on $RBE$ is pure. The Schmidt decomposition across either bipartition gives equal complementary Von Neumann entropies, so $S(RB)=S(E)$ and $S(B)=S(RE)$. By the Subadditivity of Von Neumann entropy,The reference is untouched and purifies the input, hence $S(R)=S(\rho )$. Therefore $\boxed{I_c(\Phi ,\rho )\le S(\rho )}$.

After the first quantum channel, denote the joint reference-output state by ${\omega }_{R{B}_1}$. Dilate the second CPTP map as an linear isometry $B_1\to B_2{E}_2$, with the resulting state denoted $\tau$. Linear isometries preserve nonzero eigenvalues, both globally and on the transformed output, givingTracing out $E_2$ implements the second channel, soThe Strong subadditivity of Von Neumann entropy for $R,B_2,E_2$ gives $S(R{B}_2)+S(B_2{E}_2)\ge S(B_2)+S(R{B}_2{E}_2)$. Rearranging proves the data-processing inequality for coherent information:

They need a pre-shared entanglement resource that permits deterministic production of a Bell pair. In particular, one shared Bell pair and two classical bits suffice for exact quantum teleportation. This condition is stronger than merely having some entanglement.

Let Alice's input be $|\psi {⟩}_C=\alpha |0⟩+\beta |1⟩$, and let her system $A$ and Bob's system $B$ share $|{\Phi }^{+}{⟩}_{AB}=(|00⟩+|11⟩)/\sqrt{2}$. Use the Bell states $|{\Phi }^{\pm }⟩=(|00⟩\pm |11⟩)/\sqrt{2}$ and $|{\Psi }^{\pm }⟩=(|01⟩\pm |10⟩)/\sqrt{2}$. Direct expansion givesAlice performs a projective measurement in this Bell basis and sends its two-bit outcome. Each outcome has probability $1/4$. Bob applies respectively $I,Z,X,ZX$, recovering $|\psi ⟩$ exactly, without either party knowing $\alpha ,\beta$. Before he receives the outcome, his averaged state is $I/2$, so the protocol does not communicate without the classical message. Alice's measurement consumes the input and the shared entanglement.

For necessity, exact transmission of every pure qubit means the induced quantum channel is the identity: pure-state projectors span the operator space, and a physical channel is linear. Alice can therefore prepare a Bell pair locally and transmit one half by the proposed protocol, leaving a Bell pair shared with Bob. The protocol would itself be a deterministic local operations and classical communication conversion of their resource into a Bell pair. This proves the exact teleportation resource criterion. With no shared entanglement, it is impossible: local operations and classical communication preserve separable quantum states, whereas the resulting Bell pair would be entangled.

Iterating the Bloch-vector map of amplitude damping givesThereforeThe repeated amplitude damping limit describes irreversible relaxation of the excited state $|1⟩$ into the ground state $|0⟩$, with energy transferred to the environment and no compensating thermal excitation. At $p=1$, the ground state is reached in one action. At the endpoint $p=0$, the quantum channel is the identity and the initial Bloch vector remains unchanged; the ground-state conclusion requires nonzero damping.

In the computational basis, writeThe Kraus representation of the amplitude damping channel givesReading off its Bloch vector gives the Bloch-vector map of amplitude damping:Here $0\le p\le 1$. The shift in $s_z$ shows that this quantum channel is not unital unless $p=0$; its fixed ground state lies at the north pole, rather than the center of the Bloch sphere.

Diagonalize the source average $\pi ={\sum }_j{\lambda }_j|j⟩⟨j|$, using its positive eigenvalues. For fixed $\delta >0$, let ${\Pi }_{n,\delta }$ project onto product eigenvectors whose eigenvalues obeyThe typical subspace theorem states that, for every $ϵ>0$ and all sufficiently large $n$,The probability statement is the weak law of large numbers applied to $-{\log }_2{\lambda }_j$; the dimension bounds follow by summing the typical eigenvalue bounds. This explains why the quantum typical subspace retains almost all probability using about $nS(\pi )$ qubits.

For $R>S(\pi )$ choose $0<\delta <R-S(\pi )$. Measure $\{{\Pi }_{n,\delta },I-{\Pi }_{n,\delta }\}$. On success, encode the projected state isometrically into a space of dimension $\dim {\mathcal{T}}_{n,\delta }$; on failure, output a separate fixed flag. Decode the successful sector by the inverse isometry, and map the flag to a fixed state $|{\phi }_0⟩$. The compressed dimension is at most $2^{n(S(\pi )+\delta )}+1$, hence fits within rate $R$ for all sufficiently large $n$. Both maps are trace-preserving quantum channels, not merely successful postselected operations.

Writing $\Pi ={\Pi }_{n,\delta }$, the composite channel isFor a pure source signal $|{\Psi }_k⟩$, set $a_k=⟨{\Psi }_k|\Pi |{\Psi }_k⟩$. Its squared quantum fidelity after the composite channel is at least $a_k^2$. Since the source is memoryless, its average state on $n$ uses is ${\pi }^{\otimes n}$. Convexity of the square givesThe Kraus formula for entanglement fidelity gives the same lower bound for $F_e({\pi }^{\otimes n},{\mathcal{N}}_n)$, because one Kraus operator is $\Pi$ and the other terms are nonnegative. Taking $ϵ\to 0$ proves reliable compression at every $R>S(\pi )$, in both average pure-signal fidelity and the stronger entanglement-fidelity sense. This is typical-subspace compression with a failure flag.

For a finite-dimensional memoryless quantum channel $\Lambda$, the Holevo-Schumacher-Westmoreland theorem identifies its product-state classical capacity with the optimized output Holevo quantity:With product codewords and a collective measurement on the outputs, every rate below this value is achievable with vanishing error; no larger rate can be reliable under that product-input restriction. For a fixed classical-quantum output alphabet the corresponding optimized entropy difference gives its classical coding capacity. If entangled inputs across channel uses are also allowed, the general unassisted capacity is the regularized value $C(\Lambda )={\lim }_{n\to \infty }{n}^{-1}{\chi }^{\ast }({\Lambda }^{\otimes n})$. The requested calculation below concerns the single-use optimized product-input expression, so no unproved additivity assumption is needed.

Take a Stinespring dilation $V:B⟶B^{\prime }E$ of the local quantum channel and define ${\tau }_{A{B}^{\prime }E}=(I_A\otimes V){\rho }_{AB}(I_A\otimes V^{†})$. Tracing out $E$ gives the prescribed output, with $A^{\prime }=A$. An isometry preserves the nonzero eigenvalues of a density operator; therefore $S(B^{\prime }E{)}_{\tau }=S(B{)}_{\rho }$, $S(A{B}^{\prime }E{)}_{\tau }=S(AB{)}_{\rho }$, and $S(A{)}_{\tau }=S(A{)}_{\rho }$.

Consequently $I(A:B^{\prime }E{)}_{\tau }=I(A:B{)}_{\rho }$. The loss of quantum mutual information isThe last inequality is Strong subadditivity of Von Neumann entropy, or nonnegativity of quantum conditional mutual information. HenceThis mutual-information loss as conditional mutual information shows exactly which correlations are discarded into the environment. No purity assumption on the original state is needed.

Dilate the second quantum channel by an isometry $W:B_1⟶B_2{E}_2$, and set ${\tau }_{R{B}_2{E}_2}=(I_R\otimes W){\sigma }_{R{B}_1}(I_R\otimes W^{†})$. Its $R{B}_2$ marginal is the final output. Isometry invariance of Von Neumann entropy gives $S(B_1{)}_{\sigma }=S(B_2{E}_2{)}_{\tau }$ and $S(R{B}_1{)}_{\sigma }=S(R{B}_2{E}_2{)}_{\tau }$. The difference between the two values of coherent information is thereforeNonnegativity follows from Strong subadditivity of Von Neumann entropy. This proves the data-processing inequality for coherent information:The dilated state here need not be pure. The proof uses only isometry invariance and strong subadditivity, so it applies even when the first channel has already entangled the input with its own discarded environment.

Choose a Stinespring dilation $V:A⟶B_{1}E$ of the quantum channel. Applying it to the input purification gives the pure vectorIts $R{B}_1$ marginal is the output state in the definition of coherent information. Complementary subsystems of a pure state have the same nonzero eigenvalues, by the Schmidt decomposition. Thus $S(B_1)=S(RE)$ and $S(R{B}_1)=S(E)$. It follows thatHere $S(R\mid E)$ denotes quantum conditional entropy, evaluated on the complementary output $RE$. This is coherent information as an environment conditional entropy. Equivalently $I_c=-S(R\mid B_1)$; the environment expression has the positive sign, while the receiving-system expression has the negative sign.

Both output states have the same reference marginal, and $S(R)=S(\rho )$. Expand quantum mutual information to obtain the mutual information and coherent information identityThe input-entropy term is identical in the two expressions. Subtract them and use the preceding data-processing inequality for coherent information:Thus the final quantum channel cannot increase the reference-output quantum mutual information, as required by data processing for quantum mutual information.

In the state-picture convention, a Stinespring representation of a completely positive map $\mathcal{N}:\mathcal{L}({\mathcal{H}}_Q)\to \mathcal{L}({\mathcal{H}}_{Q^{\prime }})$ consists of an auxiliary Hilbert space ${\mathcal{H}}_E$ and a linear map $V:{\mathcal{H}}_Q\to {\mathcal{H}}_{Q^{\prime }}\otimes {\mathcal{H}}_E$ such thatThe partial trace discards the environment. For example, from a Kraus representation $\mathcal{N}(X)={\sum }_j{K}_{j}X{K}_j^{†}$, take $V={\sum }_j{K}_j\otimes |j{⟩}_E$. The adjoint, or observable-picture, form is ${\mathcal{N}}^{†}(Y)=V^{†}(Y\otimes I_E)V$.

A general completely positive map does not require $V$ to be an linear isometry of Hilbert spaces. If $\mathcal{N}$ is trace preserving, then $V^{†}V={\sum }_j{K}_j^{†}{K}_j=I$, so $V$ is an linear isometry of Hilbert spaces. This is the Stinespring dilation of a quantum channel, which will be used in the data-processing proof.

Use two results: the isometric Stinespring dilation of a quantum channel, and Strong subadditivity of Von Neumann entropy. Let $V:B\to B^{\prime }E$ dilate the given operation, and defineAn linear isometry of Hilbert spaces preserves the nonzero eigenvalues, so $S(B^{\prime }E{)}_{\omega }=S(B{)}_{\rho }$ and $S(A{B}^{\prime }E{)}_{\omega }=S(AB{)}_{\rho }$. Subtracting the two coherent information expressions givesThe last inequality is Strong subadditivity of Von Neumann entropy, in the form $S(A{B}^{\prime })+S(B^{\prime }E)\ge S(B^{\prime })+S(A{B}^{\prime }E)$. Thus the data-processing inequality for coherent information isThe lost coherent information is precisely the quantum conditional mutual information between the reference $A$ and discarded environment $E$, conditional on the retained output $B^{\prime }$.

Write $X$ and $Y$ for the classical quantum registers $\tilde{X}$ and $\tilde{Y}$. By the quantum mutual information formula $I(U:V)=S(U)+S(V)-S(UV)$, expanding the right-hand side of the quantum mutual information balance identity givesAll Von Neumann entropies here are evaluated in $\omega$.

We use three facts: quantum mutual information is nonnegative by nonnegativity of quantum relative entropy; a local quantum channel cannot increase it by data processing for quantum mutual information; and the quantum mutual information of a classical-quantum state is its ensemble's Holevo quantity. Write $\chi (\mathcal{T})$ for the Holevo capacity, the supremum of the output Holevo quantity over finite input ensembles.

The $XB$ marginal is an output ensemble for $\mathcal{E}$, with inputs $\rho (x{)}_A={\mathrm{Tr}}_C\rho (x{)}_{AC}$. Consequently $I(X:B{)}_{\omega }\le \chi (\mathcal{E})$. Also ${\omega }_{XBD}$ is obtained from ${\sigma }_{XYD}$ by a quantum channel on $XY$ which retains $X$ and prepares $B$ from $Y$. HenceTo bound the latter even for entangled states $\rho (x{)}_{AC}$, exhibit the conditional input ensemble after a local measurement. Setwhen $q(y|x)>0$; zero-weight outcomes can be omitted. The numerator is a positive operator, its trace is $q(y|x)$, and the $q(y|x)$ sum to one. ThusThis is a classical-quantum state with an output ensemble for $\mathcal{N}$, so $I(XY:D{)}_{\sigma }\le \chi (\mathcal{N})$. Combining these bounds with $I(B:D{)}_{\omega }\ge 0$ yieldsThe last step takes the supremum over all input ensembles on $AC$. The left-hand channel direction is $B\leftarrow A$, as established in part (i); the printed $A\leftarrow B$ in the last inequality is a typographical reversal. Independent product ensembles also give the reverse inequality, so this proves Holevo-capacity additivity for entanglement-breaking channels.

The useful object is the Choi state of the entanglement-breaking channel. Put $d_A=\dim {\mathcal{H}}_A$, introduce a reference $R$ with the same computational basis as $A$, and use the normalized maximally entangled stateThe entanglement-breaking channel makes $C_{RB}$ a separable quantum state. Its partial trace is ${\mathrm{Tr}}_B{C}_{RB}=I_R/d_A$, because $\mathcal{E}$ is a quantum channel. Write a separable positive operator decompositionDiscard any term with $\mathrm{Tr}{\beta }_y=0$: a positive operator with zero trace is zero. DefineEach ${\tau }_y$ is a density operator. Each $E_y$ is a positive operator, and taking the matrix transpose of the partial trace identity givesThus the $E_y$ form a POVM. In this reference-first convention, the Choi reconstruction formula isThe factor $d_A$ is essential because we used a normalized Choi state. A measurement channel storing the outcome $y$, followed by the prescribed quantum state preparation of ${\tau }_y$, has precisely this action on every linear operator $L$. The desired factorization is therefore

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.