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A completely positive map with ${\sum }_{\alpha }{K}_{\alpha }^{†}{K}_{\alpha }\le I$ is a trace-nonincreasing quantum operation, representing an outcome branch with its probability retained in the trace. A deterministic quantum channel has equality and is trace preserving. Dividing a branch output by its trace gives a conditional state but generally produces a nonlinear update; that normalization is not part of the linear quantum operation.

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 $1+\gamma$.