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A zero-temperature energy-relaxation channel transfers the excited population of a qubit to its ground state. Its Kraus operators are $A_0=\mathrm{diag}(1,\sqrt{1-p})$ and $A_1=\sqrt{p}|0⟩⟨1|$ for $0\le p\le 1$. They satisfy ${\sum }_j{A}_j^{†}{A}_j=I$. The excited population is multiplied by $1-p$ and coherences by $\sqrt{1-p}$. For nonzero $p$, it is not a unital quantum channel.

For a bipartite pure state, entanglement entropy is the Von Neumann entropy of either reduced density matrix. When Alice makes a local measurement with pure conditional branches, Bob's reductions average to his original reduction. Concavity of Von Neumann entropy therefore gives ${\sum }_r{p}_{r}S({\rho }_{B,r})\le S({\rho }_B)$. When Bob measures, apply concavity to Alice's reductions. In each pure branch the two reduced entropies coincide by the Schmidt decomposition. Refining measurement outcomes to individual Kraus operators and iterating proves the inequality for arbitrary finite-round LOCC. Discarded local systems can be measured and their outcomes refined as well, so discarding does not evade the bound.

Exact singlet verification accepts the spin singlet state with certainty, rejects every orthogonal state, and preserves the singlet on acceptance. Each nonzero accepting Kraus operator is then a scalar multiple of the singlet projector, whose operator Schmidt rank is four. The local Kraus rank bound from an entangled resource excludes a single shared Bell pair, which has Schmidt rank two. Two shared Bell pairs suffice by a Bell-state nondemolition measurement followed by reporting whether its label is the singlet. This protocol may disturb triplet coherence. The rank bound concerns the resource Schmidt rank, and is not an entropy lower bound of two bits for arbitrary resource states.

Complex conjugation sends a complex number $z=a+ib$ to its complex conjugate $\overline{z}=a-ib$. It is an involution and a field automorphism of $\mathbb{C}$, with fixed field $\mathbb{R}$. Applied componentwise to a complex vector, it produces the conjugated vector appearing in a Choi state for a rank-one Kraus operator.

The matrix $C$ in the Knill--Laflamme condition is positive semidefinite, since $v^{†}Cv$ is the squared norm of ${\sum }_a{v}_a{E}_a|\psi ⟩$ for any normalized code state. Diagonalizing it gives linear combinations $F_j$ with $P{F}_j^{†}{F}_{l}P=c_j{\delta }_{jl}P$. For $c_j>0$, $V_j=F_{j}P/\sqrt{c_j}$ is an isometry of the code into an error-image subspace, and these subspaces are orthogonal. Measuring their projectors and applying $V_j^{†}$ recovers the logical state. Complete the operation arbitrarily on the orthogonal complement to obtain a quantum channel. This construction corrects every noise map whose Kraus operators lie in the specified error span, including coherent superpositions of error operators.

Let $|{\Phi }_d⟩=d^{-1/2}{\sum }_{j=0}^{d-1}|j,j⟩$ and $1\le m\le d$. It can be converted deterministically by LOCC to $|{\Phi }_m⟩$. For each $m$-element subset $S$ of the local basis labels, let ${\Pi }_S$ be its projector and let Alice's Kraus operator be $K_S={\Pi }_S/\sqrt{\left(\genfrac{}{}{0ex}{}{d-1}{m-1}\right)}$. Every label occurs in $\left(\genfrac{}{}{0ex}{}{d-1}{m-1}\right)$ subsets, so ${\sum }_S{K}_S^{†}{K}_S=I$. Each outcome has probability $m/[d(\genfrac{}{}{0ex}{}{d-1}{m-1})]=1/\left(\genfrac{}{}{0ex}{}{d}{m}\right)$ and leaves the maximally entangled state $m^{-1/2}{\sum }_{j\in S}|j,j⟩$. Alice reports $S$, and both parties relabel these $m$ basis states to $0,\dots ,m-1$ by local unitary operations.

For $d=6$, $m=4$, only three outcomes are needed: partition the six labels into disjoint pairs $T_0,T_1,T_2$ and use $K_r=(I-{\Pi }_{T_r})/\sqrt{2}$. Each label is retained twice, proving completeness, and each outcome has probability $1/3$. Every branch leaves four equal Schmidt coefficients, hence two independent Bell pairs after binary relabelling.

For nonnegative $u,v$ with $u^2+v^2=1$, Alice applies Kraus operators $M_0=\mathrm{diag}(u,v)$ and $M_1=\mathrm{diag}(v,u)$ to one half of a Bell state. Their squared effects sum to the identity, and both outcomes have probability one half. The first branch is the target; the second becomes the target after both parties apply Pauli X gates. Sending the outcome by classical communication makes this exact deterministic entanglement dilution, rather than a postselected conversion.

Share a Bell state meter pair, apply local system-to-meter CNOT gates, and measure each meter in the computational basis. The parity of the local meter records measures system $Z_A{Z}_B$ while preserving all coherence within either parity sector. The complete tuple of local records has Kraus operator $P_{u\oplus v}/\sqrt{2}$. Each local record is uniform, and the nonlocal result requires local operations and classical communication to compare records. A local Hadamard gate conjugation supplies the corresponding $X_A{X}_B$ measurement.

An entanglement monotone quantifies entanglement without increasing on average under local operations and classical communication. The average is over recorded measurement in quantum mechanics outcomes: individual successful branches can have more entanglement, provided their probabilities compensate for that increase. This distinction makes an entanglement monotone useful for bounding probabilistic state conversion. A pure-state entanglement monotone need only satisfy this inequality on pure input states and pure conditional branches, with local measurement outcomes refined to individual Kraus operators.

Let three parties share the normalized quantum stateFor $n$ copies, group each party's $n$ qubits together. Across any nontrivial division of these three parties, the Schmidt decomposition has one nonzero coefficient for every binary string of length $n$, so its Schmidt rank is $2^n$. Similarly, $m$ copies of the standard GHZ state have Schmidt rank $2^m$ across the same division.

Every recorded LOCC branch acts by a product of local Kraus operators. Schmidt-rank contraction under product operators therefore rules out an exact successful conversion when $m<n$, even if success need only have nonzero probability. This is a rank obstruction for the full target, not merely a bound on its entanglement entropy.

Approximation changes the question. In the expansion of $|{\Psi }_p{⟩}^{\otimes n}$, squared coefficients have the binary-product distribution with parameter $p$. A quantum typical subspace retains weight tending to one while using only $2^{n{h}_2(p)+o(n)}$ strings, where $h_2$ is binary entropy. The normalized truncation approaches the full target in quantum fidelity, although its Schmidt rank can be far smaller than $2^n$. Consequently, the exact finite-copy rank argument does not forbid an asymptotic entanglement dilution protocol with vanishing approximation error.

If encoding and decoding quantum channels factor through a $k$-dimensional space, their composite Kraus operators all have matrix rank at most $k$. The rank bound for a weighted operator trace and the Kraus formula for entanglement fidelity implyThus high operation fidelity requires the leading $k$ eigenvalues of the source density operator to carry almost all its probability. The bound applies to arbitrary completely positive trace-preserving encoders and decoders.

A generalized quantum measurement can be described by Kraus operators $M_{i\alpha }$, where $i$ labels the recorded outcome and $\alpha$ labels alternatives unresolved by that record. Completeness requiresFor an input density operator $\rho$,when $p_i>0$. The one-operator-per-outcome form is the special case with one $\alpha$ for each $i$.

The POVM effect $E_i={\sum }_{\alpha }{M}_{i\alpha }^{†}{M}_{i\alpha }$ determines the outcome probability, while the whole outcome map determines the post-measurement state. Keeping only $i$ can mix states that a refined record $(i,\alpha )$ would distinguish. Forgetting every outcome gives the trace-preserving quantum channel $\rho \mapsto {\sum }_{i,\alpha }{M}_{i\alpha }\rho M_{i\alpha }^{†}$. This is the distinction between a positive operator-valued measure and a quantum instrument. ^[ref]

For example, two unresolved operators $M_{00}=|0⟩⟨0|$ and $M_{01}=|0⟩⟨1|$ give the sole recorded outcome with certainty and reset every input qubit to $|0⟩$. A single operator with effect $I$ would instead act as an linear isometry; it cannot describe that reset on the same two-dimensional output.

The operators $K_j$ in a Kraus representation are Kraus operators. For a measurement outcome $i$, its associated Kraus operators determine both the outcome probability and the conditional post-measurement state.

If two parties share a pure resource of Schmidt rank $R$, a fully refined branch of their local operations has Kraus operator $K_{\mu \nu }={\sum }_{j=1}^R\sqrt{p_j}{A}_{\mu j}\otimes B_{\nu j}$. Hence its operator Schmidt rank is at most $R$. Local ancillas, locally adaptive readouts and refined discarded environments are included in the branch operators. Shared classical randomness only mixes such branches. Classical comparison of records does not increase their operator Schmidt rank.

Matrices $A^i$ acting on a finite-dimensional virtual space define a completely positive mapThe same matrices form the local tensor of a matrix product state. To interpret $\mathcal{E}$ as a quantum channel, require the additional trace normalization ${\sum }_i(A^i{)}^{†}{A}^i=I$.

Under this condition,is an isometry, because $W^{†}W={\sum }_i(A^i{)}^{†}{A}^i=I$. Keeping the physical label $i$ preserves a coherent record of each Kraus operator. Tracing it out gives ${\mathrm{Tr}}_{\mathrm{phys}}(WX{W}^{†})=\mathcal{E}(X)$. This is the finite-dimensional Stinespring representation of the quantum channel. ^[ref]

Starting from $|L⟩$ and applying this isometry repeatedly produces amplitudes $A^{i_n}\cdots A^{i_1}|L⟩$ for the physical chain and remaining virtual system. A final boundary contraction with $⟨R|$ gives the matrix product state amplitudes $⟨R|A^{i_n}\cdots A^{i_1}|L⟩$. Tracing all physical labels instead gives ${\mathcal{E}}^n(|L⟩⟨L|)$.

Without the stated normalization, the matrices still define a completely positive map and a matrix product state tensor. They do not automatically describe a deterministic quantum channel; normalization or conditioning must be specified.

Measurement backaction is the effect of a measurement in quantum mechanics on later predictions for the measured system. A quantum instrument with one Kraus operator $K_m$ per recorded outcome hasThe conditional density operator ${\rho }_m$ exists only when $p_m>0$. Its nonlinear normalization expresses conditioning; the unnormalized map $\rho \mapsto K_m\rho K_m^{†}$ is linear. More generally, an unresolved outcome can contain several Kraus operators: replace the numerator by ${\sum }_a{K}_{ma}\rho K_{ma}^{†}$ and take its trace for $p_m$. ^[ref].

Keeping the outcome gives a selective quantum measurement. Discarding it gives a nonselective quantum measurement with quantum channel $\rho \mapsto {\sum }_m{p}_m{\rho }_m$. The latter describes the whole measured population without its labels, whereas each ${\rho }_m$ describes a selected subpopulation. Neither description requires the measured quantity to have possessed its eventual value before the interaction.

For a concrete test, prepare the Pauli X eigenstate $|+⟩=(|0⟩+|1⟩)/\sqrt{2}$. Its $X=+1$ probability is one. An ideal $Z$ projective measurement gives zero or one with equal probability, leaving $|0⟩$ or $|1⟩$ under the Lüders rule. Discarding the result leaves the maximally mixed state $I/2$. A subsequent $X$ measurement now gives either sign with probability one half, even if the $Z$ result was kept. This observable change in statistics is backaction. Measuring $X$ on the original $|+⟩$ instead leaves it unchanged: disturbance depends on both the preparation and the instrument. Measurement-based quantum feedback uses the recorded outcome to choose a subsequent control operation, so both parts of the instrument matter.

A measurement channel writes a POVM's outcome into an orthogonal classical register:For any input orthonormal basis $|j⟩$, the Kraus operators $K_{a,j}=|a⟩⟨j|\sqrt{E_a}$ realize this map and satisfy ${\sum }_{a,j}{K}_{a,j}^{†}{K}_{a,j}=I$. It is therefore a trace-preserving completely positive map.

If a common recovery quantum channel with Kraus operators $R_{\mu }$ restores every pure code state after each error $E_a$, then each vector $R_{\mu }{E}_a|\psi ⟩$ is parallel to $|\psi ⟩$. Linearity on basis vectors and their pairwise superpositions forces $R_{\mu }{E}_{a}P=c_{\mu a}P$, with the scalar independent of the logical state. Insert ${\sum }_{\mu }{R}_{\mu }^{†}{R}_{\mu }=I$ between two error operators to obtain the displayed Knill--Laflamme condition. This proves necessity without presuming an orthogonal syndrome description of recovery.

Write a source and target in Schmidt decomposition with squared coefficients $(1-s,s)$ and $(1-t,t)$, where $0<s,t\le 1/2$. If $s\ge t$, Nielsen's pure-state conversion theorem allows deterministic conversion. If $s<t$, the Schmidt-tail entanglement monotone gives $p\le s/t$. Alice attains this bound with a local two-outcome measurement in quantum mechanics whose success Kraus operator is $K_s=\mathrm{diag}(\sqrt{s(1-t)/(t(1-s))},1)$ and whose failure operator is $K_f=\mathrm{diag}(\sqrt{1-s(1-t)/(t(1-s))},0)$. They obey $K_s^{†}{K}_s+K_f^{†}{K}_f=I$. The successful unnormalized state is $\sqrt{s/t}$ times the target, and failure leaves a product state. A classical message tells Bob which branch occurred. Known local changes of Schmidt bases allow the same protocol for arbitrary two-qubit pure states with these coefficients.

Use deterministic two-qubit entanglement dilution, retaining the specified complex probability amplitudes. Alice implements a quantum instrument on $A$ with Kraus operatorsThe Kraus representation completeness condition holds even for complex $\alpha ,\beta$:To realize this using the permitted quantum gates and a basic measurement, Alice initializes an ancilla qubit $C$ to $|0⟩$. Conditional on her computational-basis state being $j$, she applies to $C$ the unitary matrixEach has orthonormal columns. Therefore $W=|0⟩⟨0{|}_A\otimes V_0+|1⟩⟨1{|}_A\otimes V_1$ is a unitary gate, withA computational-basis measurement of $C$ implements the desired quantum instrument. Its two unnormalized branches on $AB$ areEach has probability $1/2$. Alice sends her outcome $m$ as one classical bit. For $m=0$, both parties keep their qubits unchanged. For $m=1$, both apply a Pauli X gate, givingThus both outcomes produce the exact target using one classical bit and local operations. The construction includes $\alpha =0$ or $\beta =0$ and arbitrary relative quantum phases; it requires no quantum channel between the parties.