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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.

Use the Hadamard gate identity $H=(X+Z)/\sqrt{2}$ on the first data wire. The $X_1$ branch is corrected as above and has syndrome $01$. The $Z_1$ branch changes the encoded state to $\alpha |000⟩-\beta |111⟩$, with syndrome $00$. No bit-flip recovery is triggered, and decoding leaves $Z|\psi ⟩$ on the top wire.

By linearity of the full quantum circuit, its output for $H_1$ is thereforewhere the four-bit strings are ordered as wires $2,3,4,5$. The two syndrome states are orthogonal. Taking their partial trace removes the cross terms, so the reduced density matrix isThe top output undergoes a completely dephasing channel in the computational basis. The bit-flip component is corrected, but the undetected phase component destroys logical coherence after the syndrome is ignored.

Let $B={\sum }_k{g}_k{Z}_k$. In the Zurek spin-bath model, $H=-Z_D\otimes B$, so unitary time evolution with $\hslash =1$ is $U(t)=e^{it{Z}_D\otimes B}$. The bath Hamiltonian terms commute, and device states $|0⟩,|1⟩$ have $Z_D$ eigenvalues $+1,-1$. Hencewhere the two normalized conditional bath states areTake each bath factor normalized, $|{\alpha }_k{|}^2+|{\beta }_k{|}^2=1$, and $|a{|}^2+|b{|}^2=1$. This entails no restriction: if only the product is initially normalized, divide each nonzero factor by its norm; the product of these norms is one.

Taking the partial trace over the bath gives the reduced density matrixThe orientation of this conditional environment overlap fixes the sign of the phase in the upper-right entry. Factorizing the overlap yields the decoherence factorThe populations are conserved because $[H,Z_D]=0$. Only phase coherence can be reduced. In particular,Quantum decoherence here results from distinguishable conditional bath states, although the complete system remains in a pure state under unitary time evolution.

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 the Schmidt decomposition of the bipartite pure state:The two reduced density matrices, obtained by partial trace, areTheir nonzero eigenvalues are identical, even if the two ambient dimensions differ. Zero eigenvalues contribute no Von Neumann entropy, soThis common Von Neumann entropy is the entanglement entropy of the bipartite pure state.

Let $R=X(l)$ and $C$ be its complement, with Hilbert-space dimension $d_C$. Use Haar twirling conditional expectation to defineThis is an operator supported on $R$. The normalization $1/d_C$ is essential: the partial trace alone would not fix an operator already supported on $R$.

Subtract the integrand from $A_X(t)$ and use the operator norm triangle inequality. Since $A-UA{U}^{†}=[A,U]U^{†}$ and every unitary operator has norm one, the assumed Lieb-Robinson bound, applied to a unitary supported on the entire complement, givesThe factor $\min (|X|,|C|)$ was bounded by $|X|$; there is no sum over sites and hence no volume-dependent prefactor.

ChooseFor $t,l\ge 0$, $e^{2kst}-1\le 2kst{e}^{2kst}$, and thereforeThus $A_{X(l)}(t)=A_R(t)$ is the desired Lieb-Robinson localization by Haar twirling. The displayed choice assumes the usual positive constants $\mu ,s$; if the interaction bound has $s=0$, the error is identically zero and any positive $v$ works. For negative times the same argument uses $|t|$.

The useful object is the Choi state of the entanglement-breaking channel. Put $d_A=\dim {\mathcal{H}}_A$, introduce a reference $R$ with the same computational basis as $A$, and use the normalized maximally entangled stateThe entanglement-breaking channel makes $C_{RB}$ a separable quantum state. Its partial trace is ${\mathrm{Tr}}_B{C}_{RB}=I_R/d_A$, because $\mathcal{E}$ is a quantum channel. Write a separable positive operator decompositionDiscard any term with $\mathrm{Tr}{\beta }_y=0$: a positive operator with zero trace is zero. DefineEach ${\tau }_y$ is a density operator. Each $E_y$ is a positive operator, and taking the matrix transpose of the partial trace identity givesThus the $E_y$ form a POVM. In this reference-first convention, the Choi reconstruction formula isThe factor $d_A$ is essential because we used a normalized Choi state. A measurement channel storing the outcome $y$, followed by the prescribed quantum state preparation of ${\tau }_y$, has precisely this action on every linear operator $L$. The desired factorization is therefore

The coefficient matrix in the computational basis isFor a bipartite pure state, the partial traces give ${\psi }_A=M{M}^{†}$ and ${\psi }_B=M^T\overline{M}$. Therefore the requested reduced density matrices areThe columns of $M$ are orthogonal. Their norms are $\sqrt{2/3}$ and $\sqrt{1/3}$, so normalizing them gives the orthonormal setWith $|v_0⟩=|0⟩$ and $|v_1⟩=|1⟩$, a Schmidt decomposition isExpanding this expression reproduces the minus sign in the $|01⟩$ coefficient. The two reduced density matrices both have eigenvalues $2/3$ and $1/3$, consistent with these Schmidt coefficients.

For any normalized pure state on finite-dimensional Hilbert spaces ${\mathcal{H}}_A\otimes {\mathcal{H}}_B$, the Schmidt decomposition givesThe $u_j$ and $v_j$ form orthonormal sets in their respective spaces. The positive numbers $\sqrt{p_j}$ are the Schmidt coefficients, and $r$ is the Schmidt rank. Taking a partial trace gives the reduced density matricesThus the nonzero eigenvalues of the two reduced density matrices agree.

One construction is a singular value decomposition of the coefficient matrix $M$ in $|\psi ⟩={\sum }_{a,b}{M}_{ab}|a⟩|b⟩$. If $M=U\Sigma V^{†}$, the columns of $U$ give $u_j$, the complex conjugates of the columns of $V$ give $v_j$, and the singular values give $\sqrt{p_j}$. This also explains why the Schmidt coefficients are real and nonnegative.

Apply the measure-and-prepare channel locally to an arbitrary bipartite density operator ${\rho }_{AR}$. The resulting state iswhereTo see that this is the correct output, the partial trace is cyclic for operators acting only on the traced subsystem, so the displayed expression equals ${\mathrm{Tr}}_A[(E_a\otimes I){\rho }_{AR}]$. Put $q_a=\mathrm{Tr}{R}_a$. The POVM completeness relation implies ${\sum }_a{q}_a=1$, and terms with $q_a=0$ vanish. Thereforeis a separable quantum state. Thus the channel is an entanglement-breaking channel, for every reference-system dimension.

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.

With $|\Omega {⟩}_{RA}=n^{-1/2}{\sum }_i|i{⟩}_R|i{⟩}_A$,The partial trace over $R$ isIts norm is $\mathrm{Tr}{\rho }_A=1$. The positive square root of an operator $\sqrt{{\rho }_A}$ is unique, proving the claim.

The pure state $|\Psi {⟩}_{AB}$ is entangled when it cannot be written as a product state $|a{⟩}_A|b{⟩}_B$, equivalently when its Schmidt rank exceeds one. Its subsystem states are the reduced density matriceswhere the partial trace is characterized by $\mathrm{Tr}(M_A{\rho }_A)=⟨\Psi |M_A\otimes I_B|\Psi ⟩$ for every observable $M_A$.

Let Bob's measurement have Kraus operators $M_b$ satisfying ${\sum }_b{M}_b^{†}{M}_b=I_B$. If Alice does not learn Bob's random outcome, her state after the measurement isThis no-communication theorem means that Bob can change Alice's conditional state after she learns his outcome, but cannot change any local outcome distribution available to her alone. Standard nonrelativistic quantum mechanics is therefore operationally compatible with the prohibition of superluminal signalling in special relativity, despite its nonlocal conditional-state updates.

An exact nondisturbing state readout would destroy this protection if it reported the globally collapsed state on an absolute-time slice. For example, Alice and Bob may share Bell states. At a prearranged time Bob encodes a bit by measuring his qubit in either the computational or Hadamard basis. The usual projection postulate assigns Alice respectively one of $|0⟩,|1⟩$ or one of $|+⟩,|-⟩$. A device returning the exact pure state lets Alice identify which basis Bob chose without waiting for his outcome, producing a superluminal signal. Equivalently, Bob may choose whether to measure, and the device distinguishes the resulting proper pure state from the original improper maximally mixed local state.

A causal alternative is a quantum state readout device located at a spacetime point $x$ that reports the local quantum state under objective collapse: take reduced states on spacelike hypersurfaces through $x$ and let those hypersurfaces approach the past light cone of $x$. The result includes localized collapses in $J^{-}(x)$ and excludes spacelike-separated collapses. This assumes a fixed Minkowski spacetime, localized collapse events, ordinary local unitary dynamics between them, and outputs that may control only operations in their causal future. The postulate is logically consistent because it adds a classical record of this local state without changing the state or any standard measurement probability. It also cannot support an indirect signalling algorithm. Inductively through the algorithm's events, every readout at $x$ depends only on operations, collapses, and earlier readouts in $J^{-}(x)$; any operation selected from that output lies in the readout's future light cone. Composing readouts, unitary evolutions, and measurements therefore never carries a controllable dependence outside a future light cone, so relativistic causality is preserved.

Let a projective measurement have mutually orthogonal projections $P_i$ satisfying ${\sum }_i{P}_i=I$. On a density operator $\rho$, the Born rule and the Lüders rule giveconditional on outcome $i$. On the first part of a bipartite quantum system, replace $P_i$ by $P_i\otimes I_2$.

The subsystems are isolated when there is no interaction term coupling them. Their Hamiltonian operator has the formso their subsequent unitary time evolution factorizes as $U_1\otimes U_2$.

The reduced density matrix of subsystem 1 iswhere the partial trace is characterized byfor every local observable $M_1$. Thus ${\rho }_1$ contains exactly the statistics accessible by measurements on subsystem 1.

Suppose a projective measurement $\{Q_j\}$ is performed on subsystem 2 and its outcome is not communicated. The resulting nonselective state isFor every $M_1$, the cyclic property of the trace and ${\sum }_j{Q}_j^2=I_2$ giveHence ${\rho }_1^{\prime }={\rho }_1$. No local projective measurement on subsystem 1 can reveal whether the remote unreported measurement occurred. This is quantum no-signalling, which prevents a choice made at a spacelike separation from transmitting information faster than light and makes the measurement formalism compatible with relativistic causality.

The proposed nondisturbing device would violate no information without disturbance. In an ordinary quantum instrument, let $M_{i\alpha }$ be the Kraus operators associated with classical output $i$. If every pure state $|\psi ⟩$ remains unchanged even conditional on the displayed outcome, every nonzero $M_{i\alpha }|\psi ⟩$ must be parallel to $|\psi ⟩$. A linear operator for which every vector is an eigenvector is a scalar multiple of the identity, so $M_{i\alpha }=c_{i\alpha }I$. Its output probabilityis independent of the state. It cannot equal $⟨\psi |P_i|\psi ⟩$ for arbitrary projectors. Equivalently, repeated nondisturbing samples would permit quantum state tomography of one specimen and then quantum cloning, contradicting ordinary quantum theory.

Such devices would not necessarily enable superluminal signalling. One consistent operational extension could make every sequence of outputs depend only on the local reduced density matrix and local settings. Since an unreported remote measurement leaves that matrix unchanged, all local device statistics would remain unchanged too. Other extensions could add nonlocal outcome-dependent rules and permit signalling, but that behavior is additional to the device specification rather than forced by it.

Extend the quantum conditional entropy from normalized states to positive operators bySince $\mathrm{Tr}{X}_{AB}=\mathrm{Tr}{X}_B$, the two terms involving $\log t$ cancel under $X\mapsto tX$, so $F(tX)=tF(X)$. Thus $F$ is positively homogeneous, and part (b) extends its concavity from states to the positive cone.

Apply part (c) with $X={\sigma }_{AB}$ and $Y={\rho }_{AB}$. Differentiating the matrix logarithm under the trace givesThe inequality from part (c), after moving $F(\rho )$ to the left, becomesThis is the data-processing inequality for quantum relative entropy under partial trace. Tensoring each output with the appropriate maximally mixed state does not change either side, so it also proves data processing under normalized partial traces. Singular $\sigma$ follows by approximation on its support.

Apply the assumed data-processing inequality for quantum relative entropy to the normalized partial trace over $C$, with the two input statesThe channel sends them to ${\rho }_{AB}\otimes I_C/d_C$ and $I_A/d_A\otimes {\rho }_B\otimes I_C/d_C$. Additivity over the common maximally mixed factor reduces data processing toExpanding the Umegaki relative entropy in terms of Von Neumann entropy givesAfter cancelling $\log d_A$, this is precisely the Strong subadditivity of Von Neumann entropy

Let ${\rho }_{n,j}^{\prime }$ be the $j$th one-system marginal of ${\rho }_n^{\prime }$, and put ${\epsilon }_n=\Vert {\rho }_n^{\prime }-{\rho }^{\otimes n}{\Vert }_1$. Contractivity of trace distance under partial trace givesuniformly in $j$. Because the one-system state space is a compact space, continuity of $f(\cdot \Vert \sigma )$ implies uniform continuity. Hence there is a function $\delta (\epsilon )\to 0$ such thatfor every $j$.

Multipartite superadditivity and additivity now implyThereforewhich proves the required lower asymptotic semicontinuity of quantum relative entropy argument.

For a bipartite density operator ${\rho }_{SA}$, the reduced density matrix of $S$ isIt is characterized byfor every observable $M_S$ on $S$. Equivalently, in any orthonormal basis $\{|a_k⟩\}$ of $A$, the partial trace isand the result is independent of that basis.

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.