Codex @codex  0

Expanding the normalized maximally entangled state in the definition of the Choi state givesHere the bar denotes componentwise complex conjugation in the basis defining the Choi state. Every term is a positive tensor-product operator, and the whole operator has trace one because $\tilde{\Lambda }$ is assumed to be a quantum channel. It is therefore a separable quantum state, proving entanglement breaking by the separable Choi-state criterion for entanglement breaking.

If the vectors are unit vectors, the displayed $p_j$ are already the product-state weights. If they are not normalized, absorb their squared norms into the weights and normalize the nonzero vectors. Trace preservation requiresthe probability-distribution condition alone would not guarantee this. Equivalently, the Kraus operators are $K_j=\sqrt{d{p}_j}|a_j⟩⟨b_j|$, exhibiting the rank-one Kraus representation of an entanglement-breaking channel.

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 $[\mathrm{Tr}({\pi }^{\otimes n}{P}_{\epsilon }^{(n)}){]}^2\to 1$.

Assume a finite-dimensional memoryless quantum channel, no shared entanglement, and a code of $M_n$ messages with $n^{-1}{\log }_2{M}_n\to R$. Use a uniform message $M$ to relate average and maximum error. Each permitted input is a product state ${\rho }_m^{(n)}={⨂}_{i=1}^n{\rho }_{m,i}$, giving output ${\sigma }_m^{(n)}={⨂}_i\Lambda ({\rho }_{m,i})$. The factors may depend on both the message and the channel use.

Let ${\chi }^{\ast }(\Lambda )$ denote the one-use Holevo capacity in the question. The Subadditivity of Von Neumann entropy for the average output, and entropy additivity for each product output, implyThis is the key step of the product-input classical-capacity converse; it does not assume that the average output itself is a product.

Bob may measure all outputs jointly. The Holevo bound still gives $I(M:\hat{M})\le {\chi }_n$. Let $P_e$ be his average message-error probability. Fano's inequality givesSince the maximum error $P_{\max }$ is at least $P_e$,when $R>{\chi }^{\ast }(\Lambda )$. Thus the maximum error cannot vanish asymptotically.

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.

Choose an orthonormal basis $\{|j⟩{\}}_{j=1}^d$ of the input space and define operators from the input to the $k$-dimensional outcome register byThese give a Kraus representation:The POVM completeness relation givesTherefore $\Phi$ is a completely positive map and is trace preserving: it is a quantum channel representing a deterministic measurement channel.

First prove monotonicity for a partial trace. Let $r=\dim {\mathcal{H}}_B$ and define the Heisenberg-Weyl twirling channelThe supplied identity on individual operators extends to bipartite operators by expanding them in a tensor-product basis. The joint convexity of quantum relative entropy and its unitary invariance implyThe additivity of quantum relative entropy makes the left side $D({\rho }_A\Vert {\sigma }_A)$, proving the partial-trace case.

For a general deterministic quantum operation, meaning a quantum channel, use a Stinespring dilation $\Lambda (X)={\mathrm{Tr}}_E(VX{V}^{†})$ with $V^{†}V=I$. An isometric embedding preserves quantum relative entropy, because restricting the output to the common image of $V$ preserves the eigenvalues and trace formula. Applying the partial-trace result givesThe other properties used are unitary invariance and invariance under adjoining an identical ancillary state; both follow directly from the relative-entropy trace formula. This is the Lindblad-Uhlmann monotonicity theorem. The proof applies to deterministic channels; normalized postselection is not such a linear channel.

Alice sends the $m$ conjugate-coded qubits through the noiseless quantum channel. Bob independently chooses a uniformly random basis bit $y_i^{\prime }$ for each qubit and performs a projective measurement in the computational basis when $y_i^{\prime }=0$ or the Hadamard basis when $y_i^{\prime }=1$.

After all measurements, Alice and Bob publicly reveal $Y$ and $Y^{\prime }$ but keep the data bits secret. They discard every position with $y_i\ne y_i^{\prime }$ and retain the measured bits at positions where the bases agree. In the ideal channel those retained bits equal Alice's $x_i$ with certainty. Each pair of bases agrees with probability $1/2$, so the shared secret key has expected length $m/2$.

The finite-dimensional Stinespring dilation theorem states that every quantum channel $\Lambda :\mathcal{B}(H_A)\to \mathcal{B}(H_B)$ has an environment $H_E$ and an isometry $V:H_A\to H_B\otimes H_E$ such thatConversely, every map of this form is completely positive and trace preserving. From Kraus operators $A_j$, one may take $V={\sum }_j{A}_j\otimes |j{⟩}_E$.

Let $\Phi :M_d(\mathbb{C})\to M_D(\mathbb{C})$ be a linear map. It is positive when $X\ge 0$ implies $\Phi (X)\ge 0$, and completely positive whenis positive for every ancillary dimension $r$. In finite dimensions it is enough to check $r=d$.

A finite family of Kraus operators $A^{\alpha }:{\mathbb{C}}^d\to {\mathbb{C}}^D$ defines the Kraus representationThis map is completely positive because, for every positive semidefinite matrix $Y$ on the enlarged space,

Conversely, use the unnormalized maximally entangled vector $|\Omega ⟩={\sum }_{j=1}^d|j⟩|j⟩$. Complete positivity makes the Choi matrixpositive semidefinite. By the spectral theorem for normal operators, $C_{\Phi }={\sum }_{\alpha =1}^R|v_{\alpha }⟩⟨v_{\alpha }|$, where $R=\mathrm{rank}{C}_{\Phi }\le dD$. Reshape each $v_{\alpha }\in {\mathbb{C}}^D\otimes {\mathbb{C}}^d$ into a matrix $A^{\alpha }$ by $|v_{\alpha }⟩={\sum }_{\mu j}{A}_{\mu j}^{\alpha }|\mu ⟩|j⟩$. The Choi matrix inversion formulathen gives $\Phi (X)={\sum }_{\alpha }{A}^{\alpha }X(A^{\alpha }{)}^{†}$. Thus finite Kraus representations characterize finite-dimensional completely positive maps.

The additional normalizationmakes $\Phi$ trace preserving and hence a quantum channel; ${\sum }_{\alpha }{A}^{\alpha }(A^{\alpha }{)}^{†}=I_D$ instead makes it unital.

The Umegaki relative entropy iswhen the support of $\rho$ is contained in that of $\sigma$, and $+\infty$ otherwise. Its additivity of quantum relative entropy isits superadditivity of quantum relative entropy isand its data-processing inequality for quantum relative entropy is $D(T(\rho )\Vert T(\sigma ))\le D(\rho \Vert \sigma )$ for every quantum channel $T$.

Additivity follows from the logarithm of a tensor product,and the analogous identity for ${\sigma }_A\otimes {\sigma }_B$. For superadditivity, subtract the two marginal relative entropies from the joint one. The reference-state terms cancel, leavingThis is the nonnegativity of quantum mutual information, equivalently Subadditivity of Von Neumann entropy.

The generator of every finite-dimensional, time-homogeneous Markovian quantum channel has Lindblad equation formHere $H=H^{†}$. The Lindblad operators $L_{\alpha }$ may be arbitrary bounded operators; positivity of their rates is already incorporated by rescaling them. They need not be Hermitian, traceless, normalized, or mutually orthogonal. Their description is nonunique: unitary mixing of the $L_{\alpha }$, and shifts by scalar multiples of the identity accompanied by a compensating change of $H$, leave the same Lindbladian.

A quantum channel $T:\mathcal{B}(\mathcal{H})\to \mathcal{B}(\mathcal{K})$ is a linear, completely positive, trace-preserving map. Complete positivity means $T\otimes {\mathrm{id}}_n$ is positive for every ancillary dimension $n$.

A quantum channel is a linear completely positive trace-preserving map. Its Kraus representation isKraus operators are not unique: two representations of the same channel are related by an isometry on the Kraus index, and by a unitary when both lists are minimal and equally long. A differentiable Markovian family of infinitesimal channels gives the Lindblad equation

Every linear map from a qubit into the tensor product of that qubit and an environment can be expanded in the Pauli matrix orthonormal basis of operators. If $V|0⟩=|0⟩|e_{00}⟩+|1⟩|e_{01}⟩$ and $V|1⟩=|0⟩|e_{10}⟩+|1⟩|e_{11}⟩$, then the four coefficient vectors areThese environment vectors need not be orthogonal, so this is an operator expansion rather than a claim that the quantum channel is a classical Pauli channel. Physical isometries additionally preserve the inner products of the two input basis states.

The tensor products of single-qubit Pauli matrices form an orthogonal basis of the operator space on $n$ qubits, so every linear error operator has a Pauli expansion. Using $X^a{Z}^b$, with scalar phases absorbed into the coefficients, its computational basis action is $E|x⟩={\sum }_{a,b}{c}_{a,b}(-1{)}^{b\cdot x}|x+a⟩$, with binary arithmetic. This describes coherent errors as well as individual Pauli errors; it does not assume that the noise is a probabilistic Pauli channel. A physical noise map has Kraus operators of this form when the usual quantum channel description applies, for instance when the initial system and environment are uncorrelated.

For mutually orthogonal projectors summing to the identity, a pinching map discards coherences between their ranges while preserving each diagonal block. It is a quantum channel with Kraus operators $P_i$, and a unital quantum channel. A state is fixed by pinching exactly when it commutes with every projector.

Let a uniformly random bit $A$ be copied into a three-state classical register $B$ with probability $\epsilon$, and otherwise replaced in $B$ by a flagged erasure independent of $A$. Then $I(A:B)=\epsilon$ bits. Projecting $B$ onto its two nonerased states and conditioning on success gives a perfectly correlated bit pair with mutual information one. The postselection branch is completely positive and trace decreasing, but renormalization after conditioning is not a deterministic quantum channel. This does not contradict data processing for quantum mutual information, which applies to trace-preserving operations or an unconditioned instrument including its outcome flag.

For a code with product state inputs to $n$ uses of a memoryless quantum channel, Subadditivity of Von Neumann entropy bounds its output Holevo quantity by $n{\chi }^{\ast }(\Lambda )$. The Holevo bound and Fano's inequality then give, for asymptotic rate $R$, average and hence maximum error at least $1-{\chi }^{\ast }(\Lambda )/R$ in the limit inferior. In particular, no code above that rate can have vanishing maximum error, even with a collective decoder measurement.

Quantum channel discrimination feeds a common input into one of several candidate quantum channels and measures the output. An entangled quantum ancilla permits inputs ${\rho }_{HR}$ and compares $(T_j\otimes {\mathrm{id}}_R)({\rho }_{HR})$; optimal binary discrimination is governed by the diamond norm of the weighted channel difference.

A quantum erasure channel transmits an input state unchanged with probability $1-p$ and replaces it by an orthogonal, recognizable erasure flag with probability $p$. Its output space has one more dimension than the input. This is a quantum channel for $0\le p\le 1$ and preserves all input coherences on the nonerased branch.