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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.

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.

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.

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 binary linear codes $C\subseteq C^{\prime }\subseteq {\mathbb{F}}_2^N$, with dimensions $k,k^{\prime }$. For each coset $u+C$ in $C^{\prime }/C$, defineRepresentatives of one coset give the same vector, and distinct cosets have disjoint computational supports, so these vectors are orthonormal. Their span is the CSS code $\mathcal{X}$, encoding $\boxed{k^{\prime }-k\text{ logical qubits in }N\text{ physical qubits}}$.

For $a,b\in {\mathbb{F}}_2^N$, write $X^a={⨂}_j{X}_j^{a_j}$ and $Z^b={⨂}_j{Z}_j^{b_j}$. The code is stabilized by $X^c$ for $c\in C$, because it permutes the sum, and by $Z^z$ for $z\in (C^{\prime }{)}^{\perp }$, because $z\cdot (u+c)=0$. These Pauli operators commute: $c\cdot z=0$. There are $k+N-k^{\prime }$ independent stabilizer generators, so their common positive eigenspace has dimension $2^{N-(k+N-k^{\prime })}=2^{k^{\prime }-k}$; it is exactly the span above.

The commuting check measurements separate the two types of error. A bit error $X^a$ changes the signs of the $Z$ checks by $(-1{)}^{z\cdot a}$, giving the ordinary syndrome of $a$ for $C^{\prime }$. A phase error $Z^b$ changes the $X$ checks by $(-1{)}^{c\cdot b}$, giving the syndrome of $b$ for the dual code $C^{\perp }$. If $C^{\prime }$ corrects $t_X$ bit errors and $C^{\perp }$ corrects $t_Z$ bit errors, these checks identify the respective patterns, and inverse Pauli operations recover the state without revealing the logical coset. This is the nested-code CSS syndrome recovery procedure.

The sharper distances account for harmless stabilizer errors. A Pauli $X^a{Z}^b$ commutes with every check precisely when $a\in C^{\prime }$ and $b\in C^{\perp }$. Among these, it acts as a stabilizer, up to a phase, precisely when $a\in C$ and $b\in (C^{\prime }{)}^{\perp }$. To justify the latter assertion, $X^a$ with $a\in C^{\prime }\setminus C$ permutes distinct logical cosets, while $Z^b$ with $b\in C^{\perp }\setminus (C^{\prime }{)}^{\perp }$ has unequal signs on two such cosets. Neither can act as a scalar, and a combined operator has the same nontrivial permutation or phase distinction. For positive logical dimension setThe quantum code distance is $\boxed{d=\min (d_X,d_Z)}$: a nontrivial logical Pauli has at least one nonstabilizer component of weight at least the corresponding distance, while a pure $X$ or pure $Z$ logical operator attains the minimum. The weight of $X^a{Z}^b$ counts the union of its supports, so this argument also includes $Y$ errors.

Here is the correction proof for coherent errors, not only a classical decoding argument. Let $P$ be the code projector and let $E,F$ have bit-component weights at most $t_X$ and phase-component weights at most $t_Z$. If $2t_X<d_X$ and $2t_Z<d_Z$, the corresponding components of $E^{†}F$ have weights below $d_X,d_Z$. If this product anticommutes with a stabilizer $S$, then $P(E^{†}F)P=-P(E^{†}F)P=0$, using $SP=P$. If it commutes with all stabilizers, the weight bounds force its two components into $C$ and $(C^{\prime }{)}^{\perp }$, so its code compression is a scalar multiple of $P$. Thus the Knill--Laflamme condition holds for the entire error family.

To see directly why this gives a common recovery, diagonalize its positive Gram matrix $c_{EF}$ and form the corresponding error combinations $F_j$. Their restrictions obey $P{F}_j^{†}{F}_{l}P=c_j{\delta }_{jl}P$. For $c_j>0$, $F_{j}P/\sqrt{c_j}$ is an isometry onto a syndrome subspace, and the different images are orthogonal. Measure these subspaces and apply the inverse isometry; zero-$c_j$ combinations annihilate the code. This recovers every quantum channel whose Kraus operators lie in the error span. Arbitrary operators supported on at most $t$ qubits expand in Pauli operators with the same support. Consequently the CSS code corrects all errors on up to $\boxed{t=\lfloor (d-1)/2\rfloor }$ qubits, and more generally the asymmetric family just described. The sufficient classical bound is obtained by replacing $d_X,d_Z$ by the minimum distances of $C^{\prime }$ and $C^{\perp }$, respectively; degeneracy can improve it.

The Schmidt decomposition and local unitary operations put any entangled two-qubit pure state intoBoth Schmidt coefficients are nonzero. Alice performs a two-outcome generalized measurement with Kraus operatorsTheir effects sum to the identity, so this is a legitimate local measurement in quantum mechanics. On success,The normalized output is the Bell state ${\Phi }_{+}$. Bob's unitary matrix $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ maps it to the spin singlet state. Alice communicates her success flag; no shared quantum operation is needed. At $\lambda =1/2$ the filter is the identity and success is certain. This is the singlet target case of optimal stochastic conversion of a two-qubit pure state.

For dichotomic local outcomes $A_x,B_y\in \{-1,1\}$, the CHSH inequality isIt assumes a local hidden-variable theory with freely chosen settings independent of the hidden-variable distribution. For the spin singlet state, direct evaluation of its Pauli matrices correlations givesChoose $a_0=e_z$, $a_1=e_x$, $b_0=(e_z+e_x)/\sqrt{2}$, and $b_1=(e_z-e_x)/\sqrt{2}$. The four correlations are $-1/\sqrt{2},-1/\sqrt{2},-1/\sqrt{2},+1/\sqrt{2}$, respectively, giving

The success flag must be recorded and communicated before the subsequent independent choice of Bell-test settings. This conditioning is legitimate because setting-independent heralding preserves Bell locality: a putative local hidden-variable theory for the entire sequential experiment would, after successful filtering, still have local response functions with a setting-independent conditional hidden-variable distribution. More explicitly, if the hidden variable is $\xi$ and $h$ is the already determined success flag, conditioning changes its distribution to $\rho (\xi \mid h=1)$ but leavesClassical messages used in preparation can be included in $\xi$; none are exchanged during the spacelike Bell measurements. Such a conditioned model still obeys the CHSH inequality, contradicting the displayed singlet correlations. Therefore no local hidden-variable theory reproduces all sequential measurement correlations of any entangled two-qubit pure state. This argument does not discard outcomes on the basis of the later measurement settings.

The resource here is one copy of a two-qubit Werner state, with the physical range $1/2<F\le 1$. The impossibility must include successful conditional branches of local operations and classical communication; average monotonicity alone would not exclude probabilistic enhancement.

For a normalized two-qubit pure state $|\chi ⟩={\sum }_{i,j}{M}_{ij}|ij⟩$, define its concurrence as $C(\chi )=2|\det M|$. Its mixed-state concurrence of two qubits is the convex roof extensionCombining near-optimal decompositions proves convexity. A separable quantum state has zero concurrence, because it has a decomposition into product vectors with zero coefficient determinant.

First establish the concurrence of a two-qubit Werner state. A pure state with squared Schmidt coefficients $\lambda ,1-\lambda$ has $C=2\sqrt{\lambda (1-\lambda )}$. In this Schmidt basis a maximally entangled state has coefficient matrix $V/\sqrt{2}$ with $V$ unitary. Its overlap magnitude is at most $(\sqrt{\lambda }+\sqrt{1-\lambda })/\sqrt{2}$, because $|V_{ii}|\le 1$. Thus the singlet overlap satisfiesAveraging any decomposition of $W_F$ gives $C(W_F)\ge 2F-1$.

For the reverse bound use the orthonormal triplet vectorsEach is normalized, and direct evaluation of its two-by-two coefficient determinant gives $C({\chi }_{j,\pm })=2F-1$. Their equally weighted six-state mixture has cancelled singlet–triplet coherences and triplet weight $(1-F)/3$, so it is exactly $W_F$. Hence

Every fully specified record of an LOCC protocol gives a product Kraus operator $A\otimes B$ on the original pair. Local ancillas and discarded local outcomes can be included by refining the record. Write its success probability and normalized output asFor invertible $A,B$, the concurrence scaling under invertible local filters isHere is a derivation of that scaling. Each pure coefficient matrix transforms to $AM{B}^{\mathsf{T}}$, so its unnormalized determinant is multiplied by $\det A\det B$. If that pure component has success probability $p_j$, its normalized concurrence is multiplied by $|\det A\det B|/p_j$, while its mixture weight becomes $r_j{p}_j/p$. Every ensemble average therefore scales by $|\det A\det B|/p$. Invertibility provides the reverse correspondence for every decomposition of ${\rho }^{\prime }$, so taking both infima proves equality. If either local filter has rank one, the output is a separable quantum state and the desired bound is immediate.

It remains to show that the filter normalization cannot make this ratio exceed one. Use the Pauli matrices representationWrite the positive matrices $X=A^{†}A=a_{0}I+a\cdot \sigma$ and $Y=B^{†}B=b_{0}I+b\cdot \sigma$, so $a_0\ge |a|$, $b_0\ge |b|$. Taking traces givesFor completeness, squaring the middle comparison leaves $(a_0|b|-b_0|a|{)}^2\ge 0$, and both sides are nonnegative. This proves the single-copy Werner filtering normalization bound, and hence

Forgetting some record information or grouping several successful records mixes these branch outputs; convexity preserves the same bound. Any alleged final $W_{F^{\prime }}$ would therefore obey $2F^{\prime }-1\le 2F-1$, contradicting $F^{\prime }>F$. Thus even heralded single-copy LOCC cannot produce the specified improved Werner state. This proves that single-copy Werner filtering cannot increase concurrence. It does not forbid collective entanglement distillation from several noisy pairs; allowing additional shared entangled input copies would change the resource assumption.

Let $P$ be the orthogonal projector onto the quantum code $\mathcal{X}$, and choose an orthonormal basis $|i_L⟩$ for it. The necessary and sufficient Knill--Laflamme condition is that one matrix $C$, independent of the logical labels, satisfiesEquivalently, $⟨i_L|E_a^{†}{E}_b|j_L⟩=C_{ab}{\delta }_{ij}$. This makes the information in the error label independent of the encoded quantum state. It applies to the whole linear span of the errors, and hence to every noise quantum channel with Kraus operators in that span.

Here is a proof of both directions. For necessity of the Knill-Laflamme condition, let $R_{\mu }$ be Kraus operators of a common recovery, so ${\sum }_{\mu }{R}_{\mu }^{†}{R}_{\mu }=I$. Since recovery after $E_a$ must return the pure input state, every vector $R_{\mu }{E}_a|\psi ⟩$ is parallel to $|\psi ⟩$. A linear operator on the code which sends every vector to a multiple of itself is scalar: apply it to two basis vectors and their sum to force the two multipliers to agree. Therefore $R_{\mu }{E}_{a}P=c_{\mu a}P$ with no dependence on $|\psi ⟩$. It follows thatThis proves necessity without assuming in advance that the physical errors have orthogonal syndromes.

For sufficiency, $C$ is positive semidefinite: for any normalized code state and coefficients $v_a$, $v^{†}Cv=\Vert {\sum }_a{v}_a{E}_a|\psi ⟩{\Vert }^2\ge 0$. Diagonalize $C$ by a unitary change of the error basis to get operators $F_j$ with $P{F}_j^{†}{F}_{l}P=c_j{\delta }_{jl}P$. If $c_j>0$, $V_j=F_{j}P/\sqrt{c_j}$ is a linear isometry from the code to an error-image subspace, and distinct such images are orthogonal. If $c_j=0$, then $F_{j}P=0$. A projective measurement of these image subspaces followed by $V_j^{†}$ recovers every logical state. Complete the recovery on the unused orthogonal complement in any trace-preserving manner. For a coherent error ${\sum }_j{v}_j{F}_j$, the measurement branch $j$ contains $v_j\sqrt{c_j}|\psi ⟩$; the branch probability is independent of the logical state, and discarding the syndrome leaves the original state. This is constructive recovery from the Knill-Laflamme condition.

For the usual family of distinct unitary Pauli errors modulo scalar phase, nondegenerate quantum code correction means that different physical errors have orthogonal image subspaces. The precise condition isFor nonunitary errors, replace the unit diagonal by positive norm factors. In an arbitrary error basis the Gram matrix need not initially be diagonal; its positive-eigenvalue modes give the independent syndrome spaces just constructed. A kernel of $C$ exhibits a linear combination of errors acting as zero on the code. In particular two distinct physical errors with the same action on the code cannot be distinguished and give degenerate correction, as in part (c).