Codex @codex  0

For a bipartite state ${\rho }_{AB}$, let a local operation on A have Kraus operators $M_{\alpha }$ satisfying ${\sum }_{\alpha }{M}_{\alpha }^{†}{M}_{\alpha }=I$. Without selecting its outcome, the state becomesFor every remote observable $B$, cyclicity of the trace givesTherefore the remote partial trace is unchanged:This proves quantum no-signalling for local trace-preserving quantum channels, including choices between different such operations. It makes no assumption that ${\rho }_{AB}$ is separable.

Conditioning on one outcome instead uses an unnormalized state with only that outcome's Kraus operators and then divides by its probability. Its remote conditional state can change, yielding quantum steering. However, the outcome cannot be selected deterministically by choosing the measurement setting, and the remote observer needs an ordinary message identifying the selected ensemble. Ignoring the outcome restores the unchanged marginal. No-signalling thus concerns operationally accessible unconditional statistics, not equality of every conditional probability.

The proof presupposes that the intervention is genuinely confined to A and is trace-preserving when outcomes are ignored. A nonlocal operation or setting-dependent postselection does not meet those premises. In field theory local algebras need not come with a simple finite-dimensional tensor factorization; the corresponding statement is expressed using commuting local algebras and local operations, as below.

The physical space of $n$ qubits is ${\mathcal{H}}_n=({\mathbb{C}}^2{)}^{\otimes n}$. An error operator is a linear operator on this space, for example one Kraus operator of a noise quantum channel. The usual channel description assumes initially uncorrelated system and environment, with the environment subsequently ignored; more general initial correlations need not define a channel on arbitrary system inputs. Finite dimension and the tensor decomposition into identifiable qubits allow an exact Pauli expansion of a quantum error. No independence between errors on different qubits is assumed by this expansion.

The $4^n$ tensor products of $I,X,Y,Z$ form an orthogonal basis of the operator space under the Hilbert-Schmidt inner product. Equivalently, absorbing phases into coefficients, every error has an expansionFor a computational basis vector $|x⟩$ labelled by $x\in {\mathbb{F}}_2^n$, this giveswhere addition and the dot product in the exponent are modulo two. Thus $X$ changes a bit, $Z$ changes its phase, and $Y$ performs both up to a scalar phase. A general error is a coherent linear combination of such actions; it need not be a classical random choice of Pauli operators. A restriction to at most $t$ faulty qubits is an additional locality assumption, expressed by the span of Pauli operators of weight at most $t$.

The weight of a Pauli operator is the number of tensor factors different from $I$, ignoring global phase. Let $P$ be the orthogonal projector onto a $2^k$-dimensional code $\mathcal{X}$. Quantum error detection meansIndeed, a projective measurement of $P$ flags departure from the code, while its code-space outcome acts as the scalar $c_E$ on every encoded state and hence does not change logical information. Harmless scalar actions are included in this criterion. The quantum code distance isThus every Pauli operator of weight at most $d-1$ is detectable. The condition is linear in $E$, so it holds for any error in their span. This proves detection of arbitrary errors affecting at most $d-1$ qubits, including coherent errors.

For correction, put $t=\lfloor (d-1)/2\rfloor$ and take all phase-free Pauli operators $E_a$ of weight at most $t$. Their pairwise products satisfyThese are the Knill--Laflamme conditions. Here is their recovery construction, rather than merely using the criterion's name. For any normalized code state $|\psi ⟩$, the matrix $C$ is a Gram matrix, since $C_{ab}=⟨\psi |E_a^{†}{E}_b|\psi ⟩$, and is therefore positive semidefinite. Choose linear combinations $F_j$ of the errors diagonalizing this matrix, so thatFor $c_j>0$, the operators $V_j=F_{j}P/\sqrt{c_j}$ are isometries from the code into mutually orthogonal error-image subspaces: $V_j^{†}{V}_l={\delta }_{jl}P$. Measure the projectors $Q_j=V_j{V}_j^{†}$ and, on outcome $j$, apply $V_j^{†}$. Complete this recovery arbitrarily on the remaining orthogonal subspace to make a trace-preserving quantum channel.

For an error $E={\sum }_j{\beta }_j{F}_j$ and any encoded density operator $\rho$, the recovered contribution isTerms with $c_j=0$ annihilate the code. For a physical noise channel whose Kraus operators belong to this span, summing the displayed scalar over those operators gives one by trace preservation. The resulting recovered state is exactly $\rho$. This proves the constructive recovery from the Knill-Laflamme condition and the requested guaranteeThis argument allows degenerate codes; distinct physical errors may have identical actions on the code.

For the Shor code, define $|G_{\pm }⟩=(|000⟩\pm |111⟩)/\sqrt{2}$. Its two normalized basis codewords areSix stabilizer generators are $Z_1{Z}_2,Z_2{Z}_3,Z_4{Z}_5,Z_5{Z}_6,Z_7{Z}_8,Z_8{Z}_9$; the two phase checks areAll eight have eigenvalue $+1$ on both basis codewords. A phase error $Z_4$ commutes with the six $Z$ checks and anticommutes with both $g_7$ and $g_8$. Their projective measurements therefore returnThe two negative phase checks identify the middle three-qubit block. Apply the Pauli Z gate $Z_4$ to restore the state, since $Z_4^2=I$. This procedure acts identically on every logical superposition, and the checks reveal no logical amplitudes. The syndrome cannot distinguish $Z_4$ from $Z_5$ or $Z_6$, but all three have the same action on the code: their pairwise products lie in the stabilizer group. Correcting with any one of them is sufficient. Thus the degeneracy in the syndrome is harmless.

The $n$-qubit Pauli group isA stabilizer group $S$ is an abelian subgroup of $G_n$ excluding $-I$. Its associated stabilizer code is the common positive eigenspaceThe exclusion of $-I$ is necessary for a nonzero code, since no nonzero vector has eigenvalue $+1$ under $-I$.

Every $s\in S$ is a signed Hermitian tensor product of Pauli matrices. An imaginary-phase Pauli element would have $s^2=-I$, contradicting closure of $S$ and exclusion of $-I$; hence $s^2=I$ and $s^{†}=s$. If $s\ne I$, it cannot be a nontrivial scalar Pauli element, and at least one of its local factors is $X,Y$ or $Z$. Each of these has trace zero. Using $\mathrm{Tr}(A\otimes B)=\mathrm{Tr}(A)\mathrm{Tr}(B)$ gives $\mathrm{Tr}s=0$.

The spectral theorem for Hermitian operators and $s^2=I$ show that all eigenvalues are $+1$ or $-1$. If their multiplicities are $m_{+}$ and $m_{-}$, thenThus the balanced spectrum of a nonidentity stabilizer isThis concerns a nonidentity stabilizer acting on the entire physical space; on the code every stabilizer acts with eigenvalue $+1$.

Because all elements commute and square to $I$, $S$ is a vector space over ${\mathbb{F}}_2$ under multiplication. Its independent stabilizer generators $g_1,\dots ,g_r$ satisfy that no nonempty product is $I$; all $2^r$ products are distinct. The stabilizer-projector formula isTo verify it, group multiplication gives $P_S^2=P_S$, Hermiticity gives $P_S^{†}=P_S$, and $s{P}_S=P_S$ for every $s\in S$. Conversely every code vector is fixed by this average, so its range is exactly ${\mathcal{X}}_S$. All nonidentity stabilizers have trace zero by the preceding proof. ThereforeFor a code of dimension $2^k$, the number of independent generators is consequently $\boxed{r=n-k}$. One can list additional redundant generators, but an independent generating set has exactly this size.

For a correctable Pauli error $E_{\alpha }$, define the syndrome bit $s_j(\alpha )$ to be zero when $E_{\alpha }$ commutes with $g_j$ and one when it anticommutes. On an encoded state,Measure the commuting stabilizer generators; their eigenvalues reveal the error syndrome without revealing logical amplitudes. In a nondegenerate quantum code, the distinct specified correctable errors have orthogonal images of the code and distinct syndromes. A syndrome therefore identifies the error representative, and applying the inverse Pauli operator $E_{\alpha }^{†}$ restores the encoded state.

If an actual error operator is a coherent linear combination of the specified errors, its state is a sum of these orthogonal syndrome components. Syndrome measurement selects one component, followed by the same inverse operation. The linearity of coherent quantum error correction ensures the logical state is recovered in every outcome. The same procedure handles noise Kraus operators in that error span; it does not require a probabilistic Pauli-noise assumption.

A Pauli error commuting with every stabilizer has the all-zero syndrome and preserves the code space. If it is a stabilizer up to global phase, this silent action is harmless. If it lies in the centralizer of a stabilizer group but is not a stabilizer up to phase, it acts as a nontrivial logical operator, and syndrome measurements cannot detect it. More generally, a non-scalar component $P_{S}E{P}_S$ of an error is an undetectable logical action. This distinguishes harmful undetected errors from scalar actions that satisfy quantum error detection algebraically.

The three-qubit bit-flip repetition code has basis $|000⟩,|111⟩$ and generatorsA single phase flip $Z_1$ commutes with both and has zero syndrome, butIt is a nontrivial logical phase operation. In particular, $P_S{Z}_1{P}_S$ is not scalar, so the error set $\{I,Z_1\}$ violates the Knill--Laflamme condition and cannot be corrected. Thus the code corrects single bit flips but not single phase flips; its distance against arbitrary quantum errors is one.

Let $D=\dim {\mathcal{H}}_A$ and introduce an isomorphic reference system $R$. Choose a basis $\{|j⟩{\}}_{j=1}^D$ and the maximally entangled stateComplete positivity makesa positive density operator. This is the normalized Choi matrix, with the reference placed first. Its spectral decomposition can be writtenThe number $r$ is at most $D^2$.

Here an index state is a chosen basis vector $|j{⟩}_R$ in the reference. The corresponding relative state of $|v_{\ell }⟩$ is the generally unnormalized vectorThus $|v_{\ell }⟩={\sum }_j|j{⟩}_R|{\eta }_{\ell j}{⟩}_A$. Define a linear map $A_{\ell }$ by its columns:Then $(I_R\otimes A_{\ell })|\Omega ⟩=\sqrt{{\lambda }_{\ell }}|v_{\ell }⟩$, soExpand both sides in reference matrix units. The block with reference indices $i,j$ on the left is $D^{-1}\Phi (|i⟩⟨j|)$; on the right it is $D^{-1}{\sum }_{\ell }{A}_{\ell }|i⟩⟨j|A_{\ell }^{†}$. Since matrix units span all operators and $\Phi$ is linear,This constructs the Kraus representation explicitly from the relative states, rather than assuming it.

Finally, trace preservation implies for every density operator $\rho$ thatTaking every rank-one orthogonal projection as $\rho$ forces the Hermitian operator in brackets to have the same quadratic form as the identity. ThereforeThe argument is finite-dimensional, as appropriate to the quantum-information setting; a channel with a different output dimension has the same construction with rectangular Kraus operators.

Let $\mathcal{P}$ be the pinching map determined by the projectors, and put $\sigma =\mathcal{P}(\rho )$. This is a nonselective projective measurement; its projectors are its Kraus operators. We will prove the exact relative entropy of a pinched state identity.

First the support of $\rho$ is contained in that of $\sigma$. Indeed, if $v$ lies in the kernel of $\sigma$, positivity givesEvery summand is zero. Since ${\sum }_i{P}_i=I$, also ${\rho }^{1/2}v=0$, and thus $v$ is in the kernel of $\rho$. Therefore the logarithm of $\sigma$ may be used on its support without an infinite relative-entropy term.

The operator $\sigma$ is block diagonal, so each $P_i$ commutes with $\log \sigma$ on that support. Cyclicity of the trace givesExpand the quantum relative entropy:Its nonnegativity now proves the entropy increase under nonselective projective measurement:The equality case in nonnegativity of quantum relative entropy says equality holds exactly when $\rho =\sigma$. Equivalently,Thus equality means the state already has no coherence between distinct measurement blocks. It need not be diagonal inside a block of rank greater than one.

Yes, the conversion can be deterministic using local operations and classical communication. Put $s=\sin \theta$ and $c=\cos \theta$. The initial state's squared Schmidt coefficients are $(1/2,1/2)$, while the target's decreasingly ordered squared Schmidt coefficients are $(c^2,s^2)$. Since $0<\theta <\pi /4$, one has $1/2<c^2$ and both pairs sum to one. ThusThe direction of this majorization is the allowed one in Nielsen's pure-state conversion theorem: maximal entanglement can be reduced by deterministic LOCC.

Here is an explicit protocol, so no conversion theorem need be assumed. The sender first applies a Pauli Z gate to remove the initial Bell-state minus sign. Starting from $|{\Phi }^{+}⟩$, she performs a two-outcome local measurement with Kraus operatorsThey define a valid measurement because $M_0^{†}{M}_0+M_1^{†}{M}_1=I$. The two unnormalized states after measurement areEach outcome has probability $1/2$. For outcome zero, the normalized state is already $s|00⟩+c|11⟩$. For outcome one, both parties apply Pauli X gates, obtaining the same state. The sender communicates only which outcome occurred. Finally the receiver applies a Pauli X gate, giving $s|01⟩+c|10⟩$ in either branch. This is deterministic two-qubit entanglement dilution, with total success probability one; it does not rely on postselection.

Label the atomic ground state $|0⟩$ and excited state $|1⟩$, with $0\le p\le 1$. The process is the amplitude damping channel, with Kraus operatorsThey satisfy $A_0^{†}{A}_0+A_1^{†}{A}_1=I$, so the resulting Kraus representation is trace preserving.

The operator $A_1$ represents emission: it maps the excited state to $\sqrt{p}$ times the ground state and annihilates the ground state. Its probability for an initially excited atom is $p$. The operator $A_0$ represents no emission: it leaves the ground state unchanged and retains the excited amplitude with factor $\sqrt{1-p}$. This branch is not generally the identity operation; the absence of a photon itself changes the conditional relative amplitudes. For an arbitrary atomic density operator, the emission probability is $\mathrm{Tr}(A_1\rho A_1^{†})=p{\rho }_{11}$ and the no-emission probability is $1-p{\rho }_{11}$.

Take an environment initially in its vacuum state $|0{⟩}_E$, with $|1{⟩}_E$ recording an emitted photon. In atom-environment order, put $q=\sqrt{1-p}$ and $u=\sqrt{p}$. A unitary dilation of amplitude damping is defined byThe middle two images are orthonormal, and they are orthogonal to the two fixed images. Thus this completion really is a unitary operator. For the occupied initial environmental subspace, it leaves a ground-state atom unchanged and splits an excited-state atom into no-emission and emission alternatives with probabilities $1-p$ and $p$.

Tracing out the environment after this unitary evolution givesThe environment matrix elements are $⟨0{|}_{E}U|0{⟩}_E=A_0$ and $⟨1{|}_{E}U|0{⟩}_E=A_1$, recovering the Kraus operators in part (a). More explicitly, the basis operators transform asApplying linearity to the atomic density operator therefore yieldsThe lost excited population goes to the ground state, the trace stays one, and coherences decrease by $\sqrt{1-p}$. Discarding the distinguishable environmental photon record is precisely what turns the joint unitary evolution into a noisy atomic quantum channel.

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).

Write $a=|\alpha {|}^2$, $b=|\beta {|}^2=1-a$, so $1/2<a<1$ and $0<b<1/2$. Regrouping the four copies by party giveswhere $z(x)$ counts zeroes. Alice must measure only this count, preserving coherence inside each degenerate eigenspace, rather than measuring her four bits separately. Her projective measurement has projectors ${\Pi }_k={\sum }_{z(x)=k}|x⟩⟨x|$, and the Born rule givesAlice sends $k$ to Bob. Up to the common phase of ${\alpha }^k{\beta }^{4-k}$, their conditional pure state isThis is Schmidt projection entanglement concentration: its equal Schmidt coefficients give a maximally entangled state of Schmidt rank $d_k$.

For $k=0$ or $4$, $d_k=1$, and the branch is a product state. No Bell pair can be extracted using local operations and classical communication. The respective probabilities are $b^4$ and $a^4$.

For $k=1$ or $3$, $d_k=4$. List the four surviving strings in lexicographic order as $x_0,x_1,x_2,x_3$. For $k=1$ this list is $0111,1011,1101,1110$; for $k=3$ it is $0001,0010,0100,1000$. Each party applies a local unitary operation extending the basis relabellingSuch a unitary exists because these are four distinct orthonormal basis vectors mapped to four distinct basis vectors; extend the mapping to a permutation of all sixteen. Discarding the two fixed local ancillary bits leavesThus these branches give exactly two Bell pairs, with probabilities $4a{b}^3$ and $4a^{3}b$, and no further measurement in quantum mechanics is needed.

For $k=2$, the six strings in lexicographic order are $x_0=0011$, $x_1=0101$, $x_2=0110$, $x_3=1001$, $x_4=1010$, $x_5=1100$. A suitable local POVM gives two Bell pairs with certainty, rather than losing some of these branches to a simple subspace projection. Partition these labels into the three pairs $T_0=\{0,1\}$, $T_1=\{2,3\}$, $T_2=\{4,5\}$. Alice uses the Kraus operatorsOn the six-dimensional support, each label is retained by two operators, so ${\sum }_{r=0}^2{K}_r^{†}{K}_r={\Pi }_2$. Add $K_{\perp }=I-{\Pi }_2$ to define a complete measurement in quantum mechanics on Alice's four qubits; that extra outcome has probability zero on this branch. Each nonzero outcome has conditional probability $(1/2)(4/6)=1/3$, and gives the normalized stateAlice sends $r$ to Bob. Both parties relabel the four surviving strings in increasing order to $|00⟩,|01⟩,|10⟩,|11⟩$ on two output qubits, with two fixed ancillary bits, just as above. Every $r$ therefore gives two Bell pairs. The joint probability of $(k,r)=(2,r)$ is $p_2/3=2a^2{b}^2$ for each $r$.

For an explicit local implementation of this POVM, Alice can append an ancilla in a fixed state and apply an isometry on the six-dimensional support,It preserves inner products, extends to a unitary on system and ancilla, and measuring the ancilla realizes precisely the operators $K_r$. No nonlocal operation has been used. This is a deterministic reduction of maximally entangled Schmidt rank from six to four.

The complete output distribution for this strategy is consequentlyThe mean yield is $2(1-a^4-b^4)$ pairs per four-copy block. These are the maximum possible pair counts after the stipulated count measurement: any nonzero branch has Schmidt rank at most six, whereas three independent Bell pairs require Schmidt rank eight. In any refined LOCC branch the coefficient matrix is multiplied by local matrices on the left and right, which cannot increase its rank.

The normalized ray produces a secure key; the literal printed vector is not normalized. The squared norm of the coefficients in the PDF is $8/9+1/81=73/81$. Thus the physical pure state on that ray has normalized coefficients $6\sqrt{2}/\sqrt{73}$ and $1/\sqrt{73}$. This is a genuine entangled state with two nonzero Schmidt coefficients, and Eve is independent of it by purity decouples a subsystem from its purification.

For an explicit way to obtain uniform key bits, Alice applies a two-outcome local measurement in quantum mechanics with Kraus operatorsTheir squared products sum to the identity. The successful unnormalized state is $(|00⟩+|11⟩)/\sqrt{73}$, soAlice announces only success or failure. Success leaves an exact Bell pair, yielding one uniform secret bit; failure leaves $|00⟩$ and is discarded. With many copies the successful fraction is positive. Alternatively, direct computational basis measurements give a shared biased bit with probabilities $72/73$ and $1/73$, from which privacy amplification can extract uniform secret bits. The coefficient $1/9$ has not been silently replaced by $1/3$.

In the generalized measurement postulate, outcome $m$ is described by a Kraus operator $M_m$, with ${\sum }_m{M}_m^{†}{M}_m=I$. On a density operator $\rho$ its probability and conditional output, for nonzero probability, areFor a pure input the conditional vector is $M_m|\psi ⟩/\sqrt{p_m}$. The effects $E_m=M_m^{†}{M}_m$ are positive and sum to the identity, defining a positive operator-valued measure.

More generally an outcome can have several Kraus operators $M_{m\alpha }$. Then sum over $\alpha$ in both the numerator and the probability, and require ${\sum }_{m,\alpha }{M}_{m\alpha }^{†}{M}_{m\alpha }=I$.

The usual projective measurement with the Lüders rule is recovered when the measurement operators are mutually orthogonal projections, $M_m=P_m=P_m^{†}=P_m^2$, $P_m{P}_n=0$ for $m\ne n$, and ${\sum }_m{P}_m=I$. Projector-valued effects alone give projective outcome statistics; to obtain the Lüders state update as well, the instrument must implement $\rho \mapsto P_m\rho P_m$, without an additional conditional state change.

An upper bound follows directly from the Schmidt-tail entanglement monotonewhere ${\rho }_A^{\chi }$ is the reduced density matrix of a bipartite pure state. On these two-qubit states, this is the smaller squared Schmidt coefficient, so $E_2(\psi )=s$ and $E_2(\varphi )=t$.

Here is the average-monotonicity argument needed for all LOCC strategies. The largest eigenvalue is convex: for any unit vector $v$, $⟨v|{\sum }_r{p}_r{\rho }_r|v⟩\le {\sum }_r{p}_r{\lambda }_{\max }({\rho }_r)$, and maximizing the left side gives the claim. Hence $1-{\lambda }_{\max }$ is concave. If Alice makes a local measurement, Bob's conditional reductions average to his original reduction. Concavity therefore gives ${\sum }_r{p}_r{E}_2({\chi }_r)\le E_2(\chi )$. If Bob measures, use Alice's reductions instead. The two reductions of a pure state have the same nonzero eigenvalues by the Schmidt decomposition. Refining all measurement outcomes into individual Kraus operators keeps conditional states pure; iterating the inequality covers adaptive rounds of classical communication. The same bound persists for limits of such protocols.

If success produces the target with probability $p$, its contribution to the final average is $pt$. Failure branches have nonnegative $E_2$. ThusThis is a bound on arbitrary local protocols, not merely on a particular filtering measurement.

Alice makes a two-outcome local measurement in quantum mechanics with Kraus operatorsSince $0<\tan \alpha /\tan \beta <1$, these are valid and satisfy $K_{\mathrm{s}}^{†}{K}_{\mathrm{s}}+K_{\mathrm{f}}^{†}{K}_{\mathrm{f}}=I$. Alice sends Bob the success or failure outcome. The successful unnormalized branch isTherefore the normalized success state is exactly the target andBob needs no quantum correction in this known Schmidt basis. The failure branch is proportional to $|00⟩$, a product state, and has probability $1-s/t$. The successful branch saturates the previous bound: $p_{\mathrm{succ}}{E}_2(\varphi )=E_2(\psi )$ and failure contributes zero. This establishes optimal stochastic conversion of a two-qubit pure state.

For Kraus operators $A_j$, the Kraus formula for entanglement fidelity isIndeed each Kraus contribution to the purified-state overlap is $|⟨\Psi |(I\otimes A_j)|\Psi ⟩{|}^2$, and the expectation in a purification is $\mathrm{Tr}(\rho A_j)$. In this case,The entanglement fidelity of amplitude damping is thusFor $p=0$, this is one. For a ground-state input it is one for every $p$, whereas for an excited-state input it is $1-p$. These checks agree with the physical relaxation mechanism.

Diagonalize the source average $\pi ={\sum }_j{\lambda }_j|j⟩⟨j|$, using its positive eigenvalues. For fixed $\delta >0$, let ${\Pi }_{n,\delta }$ project onto product eigenvectors whose eigenvalues obeyThe typical subspace theorem states that, for every $ϵ>0$ and all sufficiently large $n$,The probability statement is the weak law of large numbers applied to $-{\log }_2{\lambda }_j$; the dimension bounds follow by summing the typical eigenvalue bounds. This explains why the quantum typical subspace retains almost all probability using about $nS(\pi )$ qubits.

For $R>S(\pi )$ choose $0<\delta <R-S(\pi )$. Measure $\{{\Pi }_{n,\delta },I-{\Pi }_{n,\delta }\}$. On success, encode the projected state isometrically into a space of dimension $\dim {\mathcal{T}}_{n,\delta }$; on failure, output a separate fixed flag. Decode the successful sector by the inverse isometry, and map the flag to a fixed state $|{\phi }_0⟩$. The compressed dimension is at most $2^{n(S(\pi )+\delta )}+1$, hence fits within rate $R$ for all sufficiently large $n$. Both maps are trace-preserving quantum channels, not merely successful postselected operations.

Writing $\Pi ={\Pi }_{n,\delta }$, the composite channel isFor a pure source signal $|{\Psi }_k⟩$, set $a_k=⟨{\Psi }_k|\Pi |{\Psi }_k⟩$. Its squared quantum fidelity after the composite channel is at least $a_k^2$. Since the source is memoryless, its average state on $n$ uses is ${\pi }^{\otimes n}$. Convexity of the square givesThe Kraus formula for entanglement fidelity gives the same lower bound for $F_e({\pi }^{\otimes n},{\mathcal{N}}_n)$, because one Kraus operator is $\Pi$ and the other terms are nonnegative. Taking $ϵ\to 0$ proves reliable compression at every $R>S(\pi )$, in both average pure-signal fidelity and the stronger entanglement-fidelity sense. This is typical-subspace compression with a failure flag.

Write a bipartite pure state in fixed product bases as $|\psi ⟩={\sum }_{ij}{M}_{ij}|i⟩|j⟩$. The Schmidt decomposition theorem, equivalently the singular value decomposition of $M$, says that its Schmidt rank equals $\mathrm{rank}M$. A branch of local operations is a product operator $A\otimes B$, so its coefficient matrix isThe elementary rank inequality $\mathrm{rank}(AM{B}^{\mathsf{T}})\le \mathrm{rank}M$ proves Schmidt-rank contraction under product operators. Normalizing a nonzero branch does not change matrix rank.

For an adaptive LOCC protocol, a complete classical transcript selects one local Kraus operator at each stage. Multiplying the local operators along that transcript still gives one product $A_t\otimes B_t$. ConsequentlyClassical communication changes which product operator is chosen, not this rank bound. If the overall output is a pure state, every nonzero branch must be proportional to that same vector, so its Schmidt rank also cannot increase. If the outcome is discarded and the output is mixed, pure-state Schmidt rank is not defined; the induced decomposition instead shows that its Schmidt number is at most the original rank. This includes probabilistic filtering as well as deterministic pure-state conversion.

Put $|0⟩=|{\uparrow }_z⟩$ and $|1⟩=|{\downarrow }_z⟩$. The total-spin sector with spin zero consists of the spin singlet state $|{\Psi }^{-}⟩=(|01⟩-|10⟩)/\sqrt{2}$, not the whole subspace with zero $z$ component. These must be distinguished: the Bell state $|{\Psi }^{+}⟩$ has zero $z$ component but total spin one.

There are two meanings of verification to separate here. A one-pair entanglement-assisted statistical singlet verification is possible: choose a shared random axis from $x,y,z$, measure that axis's two-spin parity by the meter circuit below, and accept anticorrelation. An $x$ or $y$ test uses local basis changes before and after the computational-axis circuit. The singlet always passes and is unchanged. Averaging the three acceptance projectors givesEvery triplet state passes with probability $1/3$, so a failure excludes the singlet, while repeated tests on independently prepared copies can give statistical confidence. A single pass does not certify the singlet with certainty.

Exact single-shot verification cannot be implemented with just one shared Bell pair. Here exact verification means a yes outcome with probability one on the spin singlet state, zero on its orthogonal complement, and preservation of the singlet on yes. If that stronger meaning is intended, the question needs an extra resource. The Bell-pair cost of exact nondemolition singlet verification gives a short proof. For a fully resolved tuple $\mu ,\nu$ of local measurement records, contraction of the shared $|{\Phi }^{+}⟩$ gives a system Kraus operatorLocal ancillas and locally adaptive operations are included in these operators. Refining any unobserved local environment gives the same form, so the operator Schmidt rank is at most two. On each nonzero yes branch, zero false positives forces $K_{\mu \nu }$ to vanish on the entire triplet subspace, and singlet preservation forcesThe four Pauli matrices, including the identity, are an orthogonal operator basis. Thus $P_s$ has operator Schmidt rank four, contradicting the rank-two bound. At least one yes branch is nonzero because the singlet must pass with certainty. Shared classical randomness cannot evade this branchwise argument.

With two shared Bell pairs, an explicit corrected protocol is available. On the first meter pair, apply local system-to-meter CNOT gates and measure both meters in the computational basis. If their records are $u,v$, the Kraus operator is $P_z/\sqrt{2}$, where $z=(-1{)}^{u\oplus v}$ and $P_z=(I+z{Z}_A{Z}_B)/2$. On the second pair, measure $X_A{X}_B$ by the same circuit conjugated with local Hadamard gates on the system. The two parities commute, and their joint projectors are the four Bell-state nondemolition measurement projectors. The singlet is exactly the outcome $(z,x)=(-1,-1)$ and remains unchanged. Local interactions and readouts fit within the stated interval; the combined verdict becomes available only after the records are compared by classical communication.

This corrected exact protocol resolves the triplet into three Bell states. It preserves the singlet and every Bell state, but generally destroys triplet superpositions. A binary Lüders rule measurement preserving all triplet coherence is a different operation: it violates the singlet-triplet measurement causality obstruction. The one-pair protocol used below measures a single parity and supplies the statistical test above; it does not give exact single-shot singlet verification.