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In a purified Stinespring dilation, coherent information satisfies $I_c=S(RE)-S(E)=S(\rho )-I(R:E)$. Nonnegativity of quantum mutual information implies $I_c\le S(\rho )$. Equality means that the reference and environment are uncorrelated, so the environment learns nothing about the purified input reference.

For a channel on $B$, use a Stinespring dilation $B\to B^{\prime }E$. Isometry invariance gives $I(A:B)=I(A:B^{\prime }E)$, so the loss on discarding $E$ is $I(A:B)-I(A:B^{\prime })=I(A:E\mid B^{\prime })$. This quantum conditional mutual information is nonnegative by Strong subadditivity of Von Neumann entropy. The identity holds for arbitrary mixed inputs and identifies the correlations lost to the discarded environment.

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.

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.

For the Kraus operators of the measurement in quantum mechanics, the Born rule givesThe density matrix $A_i\rho A_i^{†}$ is the unnormalized state of that outcome. It is positive, its trace is $p_i$, and completeness gives ${\sum }_i{p}_i=\mathrm{Tr}\rho =1$. There is no conditional state to assign to an event with $p_i=0$.

For a projective measurement, each $P_i$ is an orthogonal projection, satisfying $P_i=P_i^{†}=P_i^2$. Completeness becomes ${\sum }_i{P}_i=I$. Fix $v$ in the range of $P_j$. ThenSince $P_{j}v=v$, the $j$th summand already equals $\Vert v{\Vert }^2$. All other summands are nonnegative, so $P_{i}v=0$ for $i\ne j$. Consequently $P_i{P}_j=0$, and taking adjoints also gives $P_j{P}_i=0$. This proves the stronger completeness forces orthogonality of measurement projections:

For the unitary dilation of a measurement instrument, take a quantum ancilla with orthonormal states $|0⟩,|1⟩,\dots ,|n⟩$, initially in $|0⟩$. DefineThe completeness equation impliesso $V$ is an isometry. Its action on the input subspace can be extended to a unitary operator $U$ on the enlarged system:Choose orthonormal bases in the two complementary subspaces and map one to the other. Their dimensions agree; the extra unused quantum ancilla level also ensures this extension is possible for an infinite-dimensional system. After the coupling, the joint density matrix is

Now perform a projective measurement on the larger system, withThese are pairwise orthogonal and sum to the full identity. The unused $|0⟩$ level has zero weight in $\Omega$, so merging it into outcome one adds no probability. For every $i$,Its trace is $p_i$, and taking the partial trace over the quantum ancilla after normalization gives exactly ${\rho }_i$. Hence unitary coupling to an quantum ancilla, a projective measurement, and discarding the quantum ancilla reproduce both the probabilities and the conditional states of the general measurement. This is the instrument version of a Stinespring dilation, rather than a construction reproducing only the outcome probabilities.

Take a normalized environment quantum stateThe Controlled-Z gate is symmetric between the two qubits, so it can be written with the environment as the controlling system:Expanding the joint density operator after the interaction givesIn the partial trace, the two off-diagonal environment operators have trace zero, while each diagonal projector has trace one. HenceEquivalently, the Kraus operators are $K_0=\sqrt{1-ϵ}I$ and $K_1=\sqrt{ϵ}Z$, and $K_0^{†}{K}_0+K_1^{†}{K}_1=I$. This explicit Stinespring dilation describes the system's dephasing by discarding information in the environment. A fresh environment in this state at each storage interval also realizes the memoryless iteration used in part (a).

The channel acts only on the input system $Q$; the reference $R$ is untouched. WriteThe coherent information isThe reference marginal remains ${\rho }_R$, whose nonzero eigenvalues agree with those of $\rho$. Unlike a classical information quantity, coherent information can be negative.

The lower diagram uses a Stinespring dilation: append an environment in a fixed pure state, apply a joint unitary operator, and discard the environment by partial trace. Before discarding it, the global $RBE$ state is pure.

A deterministic quantum operation here is a CPTP map. By Stinespring dilation, there is an isometry $V:B\to B^{\prime }E$ with $\Lambda (\tau )={\mathrm{Tr}}_E(V\tau V^{†})$. Let ${\omega }_{A{B}^{\prime }E}=(I\otimes V){\rho }_{AB}(I\otimes V^{†})$. An isometry preserves all nonzero eigenvalues, soThe quantum mutual information is $I(A:B)=S(A)+S(B)-S(AB)$. The difference after discarding the environment iswhere the last inequality is Strong subadditivity of Von Neumann entropy, in the form $S(A{B}^{\prime })+S(B^{\prime }E)\ge S(B^{\prime })+S(A{B}^{\prime }E)$. Thus data processing for quantum mutual information givesSubsystem $A$ is unchanged; its prime simply labels the output. Keeping the environment would preserve the mutual information, while tracing it out cannot increase it.

Take a Stinespring dilation $V:B⟶B^{\prime }E$ of the local quantum channel and define ${\tau }_{A{B}^{\prime }E}=(I_A\otimes V){\rho }_{AB}(I_A\otimes V^{†})$. Tracing out $E$ gives the prescribed output, with $A^{\prime }=A$. An isometry preserves the nonzero eigenvalues of a density operator; therefore $S(B^{\prime }E{)}_{\tau }=S(B{)}_{\rho }$, $S(A{B}^{\prime }E{)}_{\tau }=S(AB{)}_{\rho }$, and $S(A{)}_{\tau }=S(A{)}_{\rho }$.

Consequently $I(A:B^{\prime }E{)}_{\tau }=I(A:B{)}_{\rho }$. The loss of quantum mutual information isThe last inequality is Strong subadditivity of Von Neumann entropy, or nonnegativity of quantum conditional mutual information. HenceThis mutual-information loss as conditional mutual information shows exactly which correlations are discarded into the environment. No purity assumption on the original state is needed.

Choose a Stinespring dilation $V:A⟶B_{1}E$ of the quantum channel. Applying it to the input purification gives the pure vectorIts $R{B}_1$ marginal is the output state in the definition of coherent information. Complementary subsystems of a pure state have the same nonzero eigenvalues, by the Schmidt decomposition. Thus $S(B_1)=S(RE)$ and $S(R{B}_1)=S(E)$. It follows thatHere $S(R\mid E)$ denotes quantum conditional entropy, evaluated on the complementary output $RE$. This is coherent information as an environment conditional entropy. Equivalently $I_c=-S(R\mid B_1)$; the environment expression has the positive sign, while the receiving-system expression has the negative sign.

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.

Use two results: the isometric Stinespring dilation of a quantum channel, and Strong subadditivity of Von Neumann entropy. Let $V:B\to B^{\prime }E$ dilate the given operation, and defineAn linear isometry of Hilbert spaces preserves the nonzero eigenvalues, so $S(B^{\prime }E{)}_{\omega }=S(B{)}_{\rho }$ and $S(A{B}^{\prime }E{)}_{\omega }=S(AB{)}_{\rho }$. Subtracting the two coherent information expressions givesThe last inequality is Strong subadditivity of Von Neumann entropy, in the form $S(A{B}^{\prime })+S(B^{\prime }E)\ge S(B^{\prime })+S(A{B}^{\prime }E)$. Thus the data-processing inequality for coherent information isThe lost coherent information is precisely the quantum conditional mutual information between the reference $A$ and discarded environment $E$, conditional on the retained output $B^{\prime }$.

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.

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 $V={\sum }_j{A}_j\otimes |j{⟩}_E$.

For the pure state ${\sigma }_{RQE}$ induced by a purification and Stinespring dilation, the coherent information is

An open-boundary matrix product state with physical dimension $q$ and bond dimension $\chi$ iswhere the $A^i$ are $\chi \times \chi$ matrices and $|\ell ⟩,|r⟩$ are boundary vectors. For periodic boundary conditions, replace the boundary contraction by the matrix trace $\mathrm{Tr}(A^{i_1}\cdots A^{i_N})$.

The same matrices define the matrix product state transfer mapWhen ${\sum }_i(A^i{)}^{†}{A}^i=I$, the Stinespring dilationis an isometry. Repeatedly applying $V$ stores each Kraus label in a fresh physical register:Contracting the remaining virtual system with $⟨\ell |$ gives an MPS, while tracing over all recorded labels gives repeated application of the completely positive map $\mathcal{E}$. The MPS is therefore a coherent unravelling of the channel. Equivalently, retaining the Kraus-label registers realizes a purification of its output. A non-normalized MPS tensor gives the same construction with a general completely positive map; an appropriate canonical gauge normalizes the transfer map on its support.

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 Stinespring dilation isometry isApplying it successively to fresh environment systems givesContracting the final system with a boundary vector turns every amplitude into a boundary contraction of the matrices $K_{\alpha }$. This is a matrix product state with bond dimension at most the system dimension.

For an injective MPS, the fundamental gauge freedom istogether with the inverse transformation of boundary vectors; an overall phase is also immaterial. Its channel converges to a unique fixed state exactly when eigenvalue one is simple and there are no other peripheral eigenvalues, equivalently when it is a primitive quantum channel. A unique fixed state without convergence only requires the eigenvalue-one eigenspace itself to be one-dimensional.