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A linear map $\Phi$ between matrix algebras is completely positive when $\Phi \otimes {\mathrm{id}}_r$ is positive for every ancillary dimension $r$. In finite dimensions this is equivalent to positivity of its Choi matrix and to the existence of a Kraus representation.

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.

The operators $K_j$ in a Kraus representation are Kraus operators. For a measurement outcome $i$, its associated Kraus operators determine both the outcome probability and the conditional post-measurement state.

A quantum operation is a linear, completely positive, trace-nonincreasing map on operators; its output trace records the probability of a selected outcome. A deterministic process that sends every density matrix to another normalized density matrix is a quantum channel, satisfying linearity, complete positivity and trace preservation. Linearity preserves probabilistic mixtures. Complete positivity means that ${\mathrm{id}}_R\otimes \mathcal{E}$ remains positive for every finite-dimensional reference system $R$, so the process also acts consistently on entangled inputs. Trace preservation maintains total probability. Conditioning a selected branch by dividing by its output trace is generally nonlinear; it is a separate state-update step.

The Stinespring dilation of a deterministic operation on a system $A$ iswhere $U$ is unitary on system plus environment, the initial environment is fixed and independent of the input, and the environment is discarded. A mixed environment can be purified by adding an extra environment system. For different input and output dimensions, use an isometry $V:A\to B\otimes E$ instead of this same-system unitary form.

Choose an orthonormal environment basis $\{|\alpha ⟩\}$. Taking the partial trace gives the Kraus representationUnitarity and completeness of the environment basis giveConversely, suppose the Kraus operators obey this completeness condition. Define $V|\psi ⟩={\sum }_{\alpha }{K}_{\alpha }|\psi ⟩\otimes |\alpha ⟩$. Then $V^{†}V=I$, so $V$ is an isometry. For equal input/output system dimensions, extend the orthonormal images of the subspace $A\otimes |0⟩$ to an orthonormal basis of $A\otimes E$. Mapping a completed input basis to that output basis supplies a unitary $U$ with $U(|\psi ⟩|0⟩)=V|\psi ⟩$. Tracing the environment recovers the given operator sum. This proves both directions of the equivalence.

It also follows directly that every completely positive map in finite dimensions has an operator sum. Form its unnormalized Choi matrix $J=(\mathrm{id}\otimes \mathcal{E})|\Omega ⟩⟨\Omega |$, with $|\Omega ⟩={\sum }_i|i⟩|i⟩$. Complete positivity makes $J\ge 0$. Resolve $J={\sum }_{\alpha }|v_{\alpha }⟩⟨v_{\alpha }|$ and reshape each vector as $|v_{\alpha }⟩={\sum }_i|i⟩K_{\alpha }|i⟩$. Comparing the coefficients of $|i⟩⟨j|$ gives $\mathcal{E}(|i⟩⟨j|)={\sum }_{\alpha }{K}_{\alpha }|i⟩⟨j|K_{\alpha }^{†}$, hence the same formula for every operator by linearity. Trace preservation imposes the completeness condition just derived.

Finally, linearity of the operator sum is immediate from matrix multiplication. For any positive operator $B$ on reference plus system,because each summand is a positive congruence. This proves complete positivity for every reference dimension. Cyclicity of the trace givesKraus form is trace preserving exactly when ${\sum }_{\alpha }{K}_{\alpha }^{†}{K}_{\alpha }=I$. Without this normalization, the final printed assertion would be false: a single operator $K=I/2$ gives output trace one quarter on every input density matrix. Thus the completeness condition is an essential part of the physical Kraus representation.

A linear map $\Phi$ is a completely positive map if ${\mathrm{id}}_R\otimes \Phi$ sends every positive operator to a positive operator for every finite-dimensional auxiliary system $R$. Positivity of $\Phi$ alone tests only inputs without an auxiliary system and is weaker.

For the Kraus representation, let $\Omega \ge 0$ be any operator on the auxiliary system and the input system. ThenFor any vector $v$, its expectation is ${\sum }_k⟨(I_R\otimes A_k^{†})v,\Omega (I_R\otimes A_k^{†})v⟩\ge 0$. Every amplification is therefore positive, proving the Kraus-form map is completely positive. No normalization condition on the $A_k$ is needed for this positivity proof. To make it a deterministic quantum channel, one additionally requires ${\sum }_k{A}_k^{†}{A}_k=I$; a physical outcome branch instead obeys ${\sum }_k{A}_k^{†}{A}_k\le I$.

Transposition is positive on one system: for any positive $\rho$ and any $v$, $v^{†}{\rho }^{T}v$ is the complex conjugate of ${\bar{v}}^{†}\rho \bar{v}$, hence is real and nonnegative. It fails complete positivity already for a qubit. Apply partial transposition to the second subsystem of the Bell state $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$. The resulting operator isOn the antisymmetric vector $(|01⟩-|10⟩)/\sqrt{2}$, its eigenvalue is $-1/2$. Thus the partial transpose is not positive on this entangled input, andThe same witness embeds into any larger input dimension; in the exceptional one-dimensional case transposition is simply the identity.

Here a quantum operation is interpreted as a deterministic quantum channel, so it is completely positive and matrix trace preserving. Write its Kraus representation with ${\sum }_k{A}_k^{†}{A}_k=I$ and define the isometry in its Stinespring dilationThen $V^{†}V=I$ and $\Phi (\rho )={\mathrm{Tr}}_E(V\rho V^{†})$. Acting on $B$ gives ${\sigma }_{A{B}^{\prime }E}=(I_A\otimes V){\rho }_{AB}(I_A\otimes V^{†})$. An isometry preserves the nonzero eigenvalues of a density operator. Hence $S(A{B}^{\prime }E)=S(AB)$, $S(B^{\prime }E)=S(B)$, and $S(A)$ is unchanged. It follows that $I(A:B^{\prime }E{)}_{\sigma }=I(A:B{)}_{\rho }$.

Now discard $E$ and apply part 1. This proves data processing for quantum mutual information:The matrix trace-preserving convention matters. Postselection can increase conditional quantum mutual information: take a uniform classical bit $A$, copy it to $B$ with probability $\epsilon$, and otherwise put $B$ in an erasure state independent of $A$. Initially $I(A:B)=\epsilon$ bits. Projecting onto the nonerased sector and conditioning on success gives a perfectly correlated bit pair with quantum mutual information one. That success branch is matrix trace decreasing, and its renormalized action is not a deterministic channel. Retaining the full outcome flag and averaging all branches restores the ordinary channel inequality.

The printed normalization has its factors reversed. For the displayed map $\Phi (X)={\sum }_k{A}_{k}X{A}_k^{†}$, the condition ${\sum }_k{A}_k{A}_k^{†}=I$ says $\Phi (I)=I$, which is unitality. Trace preservation instead requiresThese are distinct trace-preserving and unital Kraus conditions. To disprove the printed equivalence, take $0<\gamma <1$ andThey obey ${\sum }_j{B}_j{B}_j^{†}=I$, butThus their completely positive map is not trace preserving. They are the adjoints of the operators of an amplitude damping channel. That channel is trace preserving but sends $I$ to $\mathrm{diag}(1+\gamma ,1-\gamma )$, so it cannot satisfy the printed normalization in any Kraus representation either. The corrected characterization is proved next.

Work in finite-dimensional input and output Hilbert spaces, with a linear map $\Phi$ on operators. Suppose first that $\Phi (X)={\sum }_k{A}_{k}X{A}_k^{†}$ and ${\sum }_k{A}_k^{†}{A}_k=I$. For every auxiliary dimension $m$ and every positive operator $W$ on the extended space,Each summand is positive because conjugation preserves positivity. This is exactly the definition of a completely positive map. Cyclicity of the trace givesso the map is also trace preserving.

Conversely, suppose $\Phi$ is completely positive and trace preserving, and let $|i⟩$ be an orthonormal input basis. Use the unnormalized entangled vector $|\Omega ⟩={\sum }_i|i⟩\otimes |i⟩$, with the reference factor first. Complete positivity makes the corresponding unnormalized Choi matrix positive:Its spectral decomposition can be written $J_{\Phi }={\sum }_k|v_k⟩⟨v_k|$, absorbing the nonnegative eigenvalues into the vectors. Define linear operators $A_k$ uniquely byExpanding the outer products and comparing their $(i,j)$ reference blocks yieldsThe matrix units span all operators, so linearity proves the Kraus representation $\Phi (X)={\sum }_k{A}_{k}X{A}_k^{†}$ for every $X$. Finally, trace preservation givesfor all $X$. Taking matrix units as test operators shows that the matrix in parentheses is zero. This proves both directions of the corrected CPT characterization, including the normalization rather than assuming it.

For an input quantum state ensemble $\{p_x,{\rho }_x\}$, write ${\sigma }_x=\Phi ({\rho }_x)$ and $\bar{\sigma }={\sum }_x{p}_x{\sigma }_x$. Its output Holevo quantity isThe Holevo capacity is the optimized one-use quantitywhere the supremum ranges over finite input ensembles and $S$ is the Von Neumann entropy in bits.

For $n$ uses of a memoryless quantum channel, the map is ${\Phi }^{\otimes n}$. An $n$-use code consists of a finite message set ${\mathcal{M}}_n$ of size $M_n$, an encoding state ${\rho }_m^{(n)}$ on the $n$ inputs for each message, and a decoding POVM $\{D_m^{(n)}:m\in {\mathcal{M}}_n\}$ on the output space, with $D_m^{(n)}\ge 0$ and ${\sum }_m{D}_m^{(n)}=I$. The encoding states may be entangled across uses. The measurement outcome is the receiver's decoded message, with conditional probabilityFor uniformly distributed messages the average error and rate areA rate $R$ is achievable if there is a sequence of such codes with $P_e^{(n)}\to 0$ and ${\mathrm{liminf}}_n{R}_n\ge R$. The unassisted classical capacity of a quantum channel is the supremum of achievable rates; the sender and receiver do not share prior entanglement.

The operational content of the Holevo-Schumacher-Westmoreland theorem is that ${\chi }^{\ast }(\Phi )$ is the capacity for product-state input encodings with collective output measurements. Allowing arbitrary entangled block inputs gives the regularized formulaThe limit equals the supremum because superadditivity of Holevo capacity follows by taking product ensembles for two blocks and using additivity of the Von Neumann entropy on product states. The coding theorem makes every rate below the one-block Holevo quantity achievable through many independently encoded blocks, with collective decoding.

The converse can be seen explicitly from the Holevo bound and Fano's inequality. For a uniform transmitted message $M$ and decoded message $\hat{M}$,The first inequality is obtained by subtracting the Fano bound on $H(M\mid \hat{M})$ from $H(M)={\log }_2{M}_n$. Dividing by $n$ and letting the average error vanish bounds every achievable rate by the regularized expression. The block-coding achievability just described supplies the reverse bound.

Under the assumed additivity of Holevo capacity across tensor powers, ${\chi }^{\ast }({\Phi }^{\otimes n})=n{\chi }^{\ast }(\Phi )$ for every $n$. ThusThe stated conditional assertion is true: under this additivity assumption, entangled input states cannot increase the asymptotic unassisted classical rate beyond the rate already achievable with product inputs. The decoder may still use collective measurements across channel outputs.

Let $D=\dim {\mathcal{H}}_A$ and introduce an isomorphic reference system $R$. Choose a basis $\{|j⟩{\}}_{j=1}^D$ and the maximally entangled stateComplete positivity makesa positive density operator. This is the normalized Choi matrix, with the reference placed first. Its spectral decomposition can be writtenThe number $r$ is at most $D^2$.

Here an index state is a chosen basis vector $|j{⟩}_R$ in the reference. The corresponding relative state of $|v_{\ell }⟩$ is the generally unnormalized vectorThus $|v_{\ell }⟩={\sum }_j|j{⟩}_R|{\eta }_{\ell j}{⟩}_A$. Define a linear map $A_{\ell }$ by its columns:Then $(I_R\otimes A_{\ell })|\Omega ⟩=\sqrt{{\lambda }_{\ell }}|v_{\ell }⟩$, soExpand both sides in reference matrix units. The block with reference indices $i,j$ on the left is $D^{-1}\Phi (|i⟩⟨j|)$; on the right it is $D^{-1}{\sum }_{\ell }{A}_{\ell }|i⟩⟨j|A_{\ell }^{†}$. Since matrix units span all operators and $\Phi$ is linear,This constructs the Kraus representation explicitly from the relative states, rather than assuming it.

Finally, trace preservation implies for every density operator $\rho$ thatTaking every rank-one orthogonal projection as $\rho$ forces the Hermitian operator in brackets to have the same quadratic form as the identity. ThereforeThe argument is finite-dimensional, as appropriate to the quantum-information setting; a channel with a different output dimension has the same construction with rectangular Kraus operators.

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.

Label the atomic ground state $|0⟩$ and excited state $|1⟩$, with $0\le p\le 1$. The process is the amplitude damping channel, with Kraus operatorsThey satisfy $A_0^{†}{A}_0+A_1^{†}{A}_1=I$, so the resulting Kraus representation is trace preserving.

The operator $A_1$ represents emission: it maps the excited state to $\sqrt{p}$ times the ground state and annihilates the ground state. Its probability for an initially excited atom is $p$. The operator $A_0$ represents no emission: it leaves the ground state unchanged and retains the excited amplitude with factor $\sqrt{1-p}$. This branch is not generally the identity operation; the absence of a photon itself changes the conditional relative amplitudes. For an arbitrary atomic density operator, the emission probability is $\mathrm{Tr}(A_1\rho A_1^{†})=p{\rho }_{11}$ and the no-emission probability is $1-p{\rho }_{11}$.

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.

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.

Set $K_y=|y⟩⟨y|$. The map is in Kraus representation:For any ancillary Hilbert space $R$ and positive operator $M$ on $R\otimes \mathcal{H}$,This verifies complete positivity directly, not merely positivity on unextended states. Cyclicity of the trace gives $\mathrm{Tr}\Lambda (X)=\mathrm{Tr}[X{\sum }_y{K}_y^{†}{K}_y]=\mathrm{Tr}X$. Thus $\Lambda$ is a CPTP map, the completely dephasing channel in the given basis.

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.

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.

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.

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 every ancillary system,when $X\ge 0$, proving complete positivity. Cyclicity of trace givesThis equals $\mathrm{Tr}\rho$ for every $\rho$ exactly whenthe trace-preserving condition in the Kraus representation.

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.