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A channel's Choi state is its normalized Choi matrix,It is a density operator and satisfies ${\mathrm{Tr}}_{\mathrm{out}}{C}_{\Lambda }=I_{\mathrm{in}}/d$. Some authors use an unnormalized Choi matrix of trace $d$, so reconstruction factors depend on the convention.

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.

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.

Use the unnormalized Choi matrixIt is positive because ${\Phi }_A$ is a completely positive map. Take its spectral decomposition, absorbing square roots of nonzero eigenvalues into the vectors:Expand $|v_k⟩={\sum }_{i,j}{v}_{k,ij}|i{⟩}_A|j{⟩}_B$ and define the Kraus operator $A_k$ by $(A_k{)}_{ij}=v_{k,ij}$. Equivalently,The preceding relative state identity then gives $(I\otimes ⟨{\varphi }_B^{\ast }|)v_k=A_k{\varphi }_A$. Inserting the spectral decomposition into the stated reconstruction identity yieldsThe conjugated index vector is required here; it is present in the PDF but missing in the converted TeX version of this identity.

Rank-one projectors span all operators by polarization. Explicitly, putting $P(w)=|w⟩⟨w|$ givesThus linearity extends the equality to every $X\in \mathcal{B}({\mathcal{H}}_A)$:Finally trace preservation and cyclicity of the trace giveFor pure-projector inputs this says the Hermitian operator ${\sum }_k{A}_k^{†}{A}_k-I$ has zero expectation on every vector, so it vanishes. ThereforeThis is the spectral Kraus decomposition. Using a normalized Choi state $J_{\Phi }/d$ instead would require the extra factor $\sqrt{d}$ when reshaping its eigenvectors into Kraus operators.

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.

Fix an orthonormal basis $\{|i⟩{\}}_{i=1}^d$ of the input Hilbert space, where $d=\dim \mathcal{H}$, and a reference copy $R\simeq \mathcal{H}$. Use the normalized maximally entangled vector $|{\Omega }_d⟩=d^{-1/2}{\sum }_i|i{⟩}_R|i{⟩}_{\mathcal{H}}$. The normalized Choi–Jamiołkowski state isComplete positivity makes $J\ge 0$, and trace preservation gives ${\mathrm{Tr}}_{\mathcal{K}}J=I_R/d$, hence $\mathrm{Tr}J=1$. It is therefore a genuine density operator. In the unnormalized Choi matrix convention the factor $1/d$ is omitted and the trace is $d$; specifying normalization distinguishes a Choi state from that matrix convention. The reference basis also fixes the transpose in the Choi reconstruction formula, $\Lambda (X)=d{\mathrm{Tr}}_R[(X^T\otimes I)J]$.

The map $\Lambda$ is a completely positive map when $\Lambda \otimes {\mathrm{id}}_m$ maps positive operators to positive operators for every $m$. In a basis $\{|i⟩\}$ define the unnormalized Choi matrixIf $\Lambda$ is completely positive, $J(\Lambda )=(\mathrm{id}\otimes \Lambda )(|\Omega ⟩⟨\Omega |)\ge 0$, where $|\Omega ⟩={\sum }_i|ii⟩$. Conversely, decompose $J={\sum }_k|a_k⟩⟨a_k|$ and reshape each vector into an operator $A_k$. The Choi reconstruction formula gives $\Lambda (X)={\sum }_k{A}_{k}X{A}_k^{†}$, which is completely positive. Thus

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 normalized Choi matrix of the trace-to-identity map is $I\otimes I/d$, while that of transposition is the flip operator $F/d$. Hence the Werner–Holevo channel hasthe normalized projector onto the antisymmetric subspace.

If the channel were random unitary, part (i) would express $C_T$ as a mixture of maximally entangled vectors. Every vector in such a mixture must lie in the support of $C_T$, hence in the antisymmetric subspace. Under vectorization, an antisymmetric vector corresponds to a skew-symmetric matrix $A$, while maximal entanglement requires $A{A}^{†}$ to be proportional to $I$. In odd dimension, $\det A=\det (-A^T)=-\det A$, so $\det A=0$; such an $A$ cannot be proportional to a unitary. Thus for odd $d\ge 3$, and in particular $d=3$, this unital channel is not random unitary, disproving the converse. The odd-dimensional qualification matters because antisymmetric maximally entangled vectors can exist in even dimension.

For $T(\rho )={\sum }_k{p}_k{U}_k\rho U_k^{†}$,Each $(U_k\otimes I)|\varphi ⟩$ is a maximally entangled state, so the Choi matrix is a convex combination of maximally entangled pure states.

If $T$ is entanglement breaking, applying it to one half of $|\varphi ⟩⟨\varphi |$ immediately shows that its Choi matrix is separable.

Conversely supposeis separable. The Choi reconstruction formula for the normalized convention isThe trace-preserving condition ${\mathrm{Tr}}_1{C}_T=I/d$ implies ${\sum }_k{M}_k=I$, so $\{M_k\}$ is a POVM. Part (i) now proves that $T$ is entanglement breaking. Thus

For a bipartite vector $|v⟩\in {\mathcal{H}}_R\otimes {\mathcal{H}}_A$ and a chosen orthonormal basis of $R$, the generally unnormalized relative state on $A$ is obtained by the partial inner product with one basis vector. It gives the expansion $|v⟩={\sum }_j|j⟩\otimes |{\eta }_j⟩$. The squared norm of a relative state is the probability of its corresponding basis outcome when $|v⟩$ is normalized. Relative states of eigenvectors of a Choi matrix construct the columns of its Kraus operators.

Use the unnormalized Choi matrix $J(\Phi )=(\Phi \otimes \mathrm{id})(|\tilde{\Psi }⟩⟨\tilde{\Psi }|)$, which has trace $d$ for a channel. A positive such matrix has a spectral decomposition with vectors $v_k=\sqrt{{\lambda }_k}{u}_k$. In paired bases define the Kraus operator $A_k$ by $(A_k{)}_{ij}=(v_k{)}_{ij}$, so $v_k=(A_k\otimes I){\sum }_j|j⟩|j⟩$. Partial contraction with a conjugate index vector gives $A_k|\varphi ⟩$, hence $\Phi (|\varphi ⟩⟨\varphi |)={\sum }_k{A}_k|\varphi ⟩⟨\varphi |A_k^{†}$. Rank-one projectors span the operator space, proving the Kraus representation on all inputs. Trace preservation is equivalent to ${\sum }_k{A}_k^{†}{A}_k=I$. This construction needs at most $\mathrm{rank}J$ nonzero Kraus operators.