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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.

In a quantum-trajectory unravelling, a pure state evolves between jumps underand a jump of type $\alpha$ sends $|\varphi ⟩$ to $L_{\alpha }|\varphi ⟩/\Vert L_{\alpha }|\varphi ⟩\Vert$. Averaging $|{\varphi }_t⟩⟨{\varphi }_t|$ over the stochastic jump records recovers $\rho (t)$.

Equivalently, over a short interval $dt$ use the operators in a Kraus representationand the Stinespring dilation $V={\sum }_{\mu }{K}_{\mu }\otimes |\mu {⟩}_E$. Iterating with fresh environment systems produces a pure system-environment history whose partial trace is the Lindblad evolution.

If the fixed point is pure, stationarity requires $L_{\alpha }|{\psi }_{\ast }⟩={\ell }_{\alpha }|{\psi }_{\ast }⟩$ for every $\alpha$, together with preservation of its ray by the adjusted effective Hamiltonian. After shifting the jumps one may take it to be a dark state of a Lindblad equation. Once a trajectory reaches that ray, neither no-jump evolution nor a jump takes it away; in the example of part (c), a jump transfers any excited component directly into the singlet.

The Kraus representation isThe second identity is precisely trace preservation. Different Kraus families related by an isometry represent the same channel.

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

An entanglement-breaking channel admits a Kraus representation with rank-one Kraus operators $K_j=|a_j⟩⟨b_j|$. Its Choi state is a sum of positive tensor-product operators, hence separable. Conversely, spectral decompositions of each prepared state and each POVM effect in a measure-and-prepare channel produce rank-one Kraus operators. Trace preservation imposes ${\sum }_j\Vert a_j{\Vert }^2|b_j⟩⟨b_j|=I$.

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.

In finite dimensions, a completely positive map has the form $\mathcal{N}(X)={\mathrm{Tr}}_E(VX{V}^{†})$ for $V:{\mathcal{H}}_Q\to {\mathcal{H}}_{Q\prime }\otimes {\mathcal{H}}_E$. From a Kraus representation take $V={\sum }_j{K}_j\otimes |j⟩$. The observable-picture adjoint is ${\mathcal{N}}^{†}(Y)=V^{†}(Y\otimes I_E)V$. A general completely positive map need not have isometric $V$; it is a linear isometry of Hilbert spaces precisely when the map is trace preserving.

For a Kraus representation $\Phi (X)={\sum }_k{A}_{k}X{A}_k^{†}$, trace preservation is equivalent to the first identity, by cyclicity of the trace. The second says $\Phi (I)=I$, namely unitality, and is a distinct constraint. For $0<\gamma <1$, the amplitude damping channel has operators $A_0=\mathrm{diag}(1,\sqrt{1-\gamma })$ and $A_1=\sqrt{\gamma }|0⟩⟨1|$ and is trace-preserving but not unital. Taking the adjoints as Kraus operators instead gives a unital completely positive map whose output trace on $|0⟩⟨0|$ is $1+\gamma$.