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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.

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.

For binary linear codes $C\subseteq C^{\prime }$, the coset superpositions of the CSS code have $X$ checks labelled by $C$ and $Z$ checks by $(C^{\prime }{)}^{\perp }$. A bit-error pattern $a$ gives the classical syndrome for $C^{\prime }$, and a phase-error pattern $b$ gives the syndrome for the dual code $C^{\perp }$. Errors with $2t_X<{\min }_{a\in C^{\prime }\setminus C}\mathrm{wt}(a)$ and $2t_Z<{\min }_{b\in C^{\perp }\setminus (C^{\prime }{)}^{\perp }}\mathrm{wt}(b)$ are jointly correctable. Every pairwise Pauli-error product either anticommutes with a check, giving zero code compression, or its low-weight components belong to the stabilizer and give scalar compression. Thus the Knill--Laflamme condition covers coherent superpositions and quantum channels as well as definite bit and phase errors.

A nonselective measurement in quantum mechanics averages over outcomes that are not read or not communicated. Its quantum channel is $\rho \mapsto {\sum }_i{A}_i\rho A_i^{†}$, with ${\sum }_i{A}_i^{†}{A}_i=I$. The completeness condition makes the channel trace preserving. The average equals ${\sum }_i{p}_i{\rho }_i$, where ${\rho }_i$ are the conditional states from a selective quantum measurement. A nonselective projective measurement is the special case in which the operators are orthogonal projectors.

In the unsquared quantum fidelity convention, operation fidelity compares a purification of a density operator with its image under a quantum channel acting on the source subsystem:Its square is entanglement fidelity. Thus this convention is distinct from the squared quantity $F_e$; the two should not be installed as synonyms.

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.

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).

Use the Hilbert-Schmidt inner product normalized by the dimension: $(A,B)=\frac{1}{2}\mathrm{Tr}(A^{†}B)$. The identity and the three Pauli matrices are Hermitian, each has square $I$, and each nonidentity Pauli matrix has zero trace. For distinct nonidentity indices, the Pauli matrix multiplication law gives ${\sigma }_j{\sigma }_k=i{\sum }_l{\epsilon }_{jkl}{\sigma }_l$, which is traceless. ThereforeThere are four such orthonormal operators, and the complex vector space of two-by-two matrices has dimension four. Hence they form an orthonormal basis. Taking the inner product of any $E$ with each basis element proves, for arbitrary complex $E$ and not just Hermitian ones,For example, if $E=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, its four coefficients are $(a+d)/2$, $(b+c)/2$, $i(b-c)/2$, and $(a-d)/2$, respectively.

For the joint error, regard the two prescribed basis outputs as the columns of an environment-valued linear map. Its coefficient array in the usual output-row, input-column convention isThe source's other array is arranged as a column of the images of basis kets; it must not be mistaken for this coefficient array acting on an amplitude column. Applying the just-proved Pauli matrix expansion componentwise in any environment orthonormal basis gives the Pauli decomposition of a qubit-environment isometry:To check the potentially delicate sign of the Pauli Y gate coefficient, use $Y|0⟩=i|1⟩$ and $Y|1⟩=-i|0⟩$. The four terms then giveLinearity proves the expansion for every input pure state. The environment coefficient vectors need not be normalized or mutually orthogonal. Consequently this expansion includes coherent combinations of Pauli errors, rather than asserting that every quantum channel is a probabilistic Pauli channel.

For the recovery probability, use the usual independent-use interpretation of the given phase-flip channel: each of the three transmitted qubits has a Pauli Z gate error independently with probability $q$. The phase-flip repetition code uses logical codewords $|+++⟩$ and $|---⟩$. The decoding Hadamard gates convert physical phase flips to bit flips because $HZH=X$. We can then analyze the two decoding controlled-NOT gates and the final Toffoli gate directly, rather than assuming a recovery rule.

Let $b_1,b_2,b_3$ be the error bits after this basis conversion. For a logical basis input $t\in \{0,1\}$, the three data bits before inverse encoding are $t\oplus b_1,t\oplus b_2,t\oplus b_3$. The two controlled-NOT gates produceThe Toffoli gate flips the first wire when both of the last two bits are one. Its residual error isThe last two wires carry an error syndrome independent of $t$. Thus this transformation preserves the amplitudes of an arbitrary superposition: zero or one phase error returns the original unknown qubit, while two or three phase errors leave the logical Pauli X gate applied to it. The probability of universally successful recovery is thereforeEquivalently, the decoded quantum channel is $\rho \mapsto (1-p_L)\rho +p_{L}X\rho X$, with logical failure of the three-qubit phase-flip repetition code $p_L=3q^2-2q^3$. For $0<q<1/2$, the failure probability is below $q$, since $q-p_L=q(1-q)(1-2q)>0$. A special input that is a Pauli X eigenstate is unchanged even by the residual logical error; this does not make the code correct that error on an unknown input. Its squared quantum fidelity with a particular pure input is $1-p_L+p_L|⟨\chi |X|\chi ⟩{|}^2$. If the errors on different channel uses are correlated, the marginal probability $q$ alone is insufficient: the general universal success probability is the probability of at most one error.

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.

The Schumacher compression theorem concerns a memoryless quantum information source with average density matrix $\rho$ on a finite-dimensional Hilbert space. For $n$ uses the average state is ${\rho }_n={\rho }^{\otimes n}$. Compression consists of an encoding quantum channel into a space of dimension $D_n$ and a decoding channel back to the original block space, with asymptotic rate $n^{-1}{\log }_2{D}_n$ qubits per source use. The optimal reliable rate is $S(\rho )$: every larger rate is achievable, and no smaller rate is reliable. This holds for average pure-source fidelity; the direct construction also preserves a purification reference, and the converse for that stronger entanglement fidelity criterion holds with fidelity tending to zero below the threshold. The encoding and decoding are deterministic channels; no free classical transmission of the source label is allowed.

Diagonalize $\rho ={\sum }_j{\lambda }_j|j⟩⟨j|$ and omit zero-probability eigenvalues when sampling. Define the quantum typical subspace projector $P_{n,\delta }$ by retaining product eigenvectors withThe sampled random variable $-{\log }_2{\lambda }_j$ has expectation $S=S(\rho )$. The weak law of large numbers gives $w_n=\mathrm{Tr}({\rho }_n{P}_{n,\delta })\to 1$. Every retained eigenvalue is at least $2^{-n(S+\delta )}$, and the total probability is at most one, soThis proves the two typical subspace theorem estimates used in the direct part.

Write $P=P_{n,\delta }$. Encode its range isometrically into a compressed space and reserve one orthogonal failure flag. Decode the range back isometrically and the failure flag to a fixed pure state $|\phi ⟩$. This typical-subspace compression with a failure flag is a physical channel whose composite action isIndeed its Kraus operators are $P$ and $|\phi ⟩⟨e_a|$, where the $e_a$ form an orthonormal basis of $\ker P$, and their adjoint products sum to $I$. The Kraus formula for entanglement fidelity givesFor any pure-state ensemble of ${\rho }_n$, the average squared quantum fidelity is at least this number: apply convexity of $|z{|}^2$ separately to the amplitudes $⟨{\psi }_x|K_{\alpha }|{\psi }_x⟩$ with their source probabilities. Thus average fidelity also tends to one. The compressed dimension is at most $2^{n(S+\delta )}+1$, and its rate is at most $S+\delta +o(1)$. Choose $\delta <R-S$ for any prescribed $R>S$. One may also take ${\delta }_n=n^{-1/4}$: the finite spectral information variance and Chebyshev's inequality still give $w_n\to 1$, with rate tending to $S$. A pure source, $S=0$, can simply be re-prepared exactly without sending any qubits.

For the coherent converse, we prove the finite-dimensional quantum compression converse. If the compressed space has dimension $D$, the state of reference plus compressed system has a pure-state decomposition with Schmidt rank at most $D$. Applying decoder Kraus operators locally cannot increase the Schmidt rank of those vectors, so the final reference–output state has Schmidt number at most $D$. Let $|{\Psi }_{{\rho }_n}⟩$ purify the source, with squared Schmidt coefficients equal to the decreasing eigenvalues ${\mu }_1\ge {\mu }_2\ge \cdots$ of ${\rho }_n$. For any normalized vector $|v⟩$ of Schmidt rank at most $D$, its reference support lies in a rank-at-most-$D$ projector $Q$. HenceThe last bound follows by maximizing ${\sum }_j{\mu }_j⟨j|Q|j⟩$ subject to $0\le ⟨j|Q|j⟩\le 1$ and their sum at most $D$. Average over a Schmidt-rank-bounded decomposition of the final state to obtain $F_e\le {\sum }_{j=1}^D{\mu }_j$.

At rate $R<S$, choose $0<\delta <S-R$. The total source probability outside the quantum typical subspace tends to zero, while every eigenvalue inside it is at most $2^{-n(S-\delta )}$. Therefore for $D_n\le 2^{nR}$,This argument allows arbitrary encoding and decoding channels, not only subspace projection.

For completeness, average pure-source reliability also forces the same rate, even though it is a weaker criterion than entanglement fidelity. Here is a separate converse for that formulation. Let the emitted block pure states be ${\psi }_x$, with probabilities $p_x$, and let ${\sigma }_x$ be the decoded states. If their mean squared fidelity tends to one, then their mean trace distance ${\bar{\delta }}_n$ tends to zero, since $\frac{1}{2}\Vert {\sigma }_x-{\psi }_x{\Vert }_1\le \sqrt{1-⟨{\psi }_x|{\sigma }_x|{\psi }_x⟩}$. This inequality follows by choosing a purification with that overlap and using contraction of trace distance under a partial trace.

We need only an elementary entropy continuity from Jordan decomposition bound, derivable from Question 4. If $\delta =\frac{1}{2}\Vert \sigma -\tau {\Vert }_1$, write the positive and negative parts of $\sigma -\tau$ as $\delta {\omega }_{+},\delta {\omega }_{-}$. The common density matrixand the two mixture bounds giveHere $g$ is increasing and concave, $g(0)=0$. Apply this in block dimension $d^n$, both to the average states and to individual outputs, using convexity of trace distance and concavity of $g$. Since the original ensemble is pure, its Holevo quantity is $S({\rho }_n)=nS$. The decoded ensemble hasThe Holevo quantity under a quantum channel cannot increase. To specify the entropy ingredient, Strong subadditivity of Von Neumann entropy states $S(XB)+S(BE)\ge S(B)+S(XBE)$. Attach the classical source label $X$, dilate the decoder isometrically to $BE$, and trace out $E$. The displayed inequality says $I(X:B)\le I(X:BE)$; the isometry preserves the input mutual information. Since $I(X:B)$ of a labelled ensemble is its Holevo quantity, this proves the needed decoder monotonicity. Before decoding, the system has dimension $D_n$, so its Holevo quantity is at most ${\log }_2{D}_n$. Thus ${\log }_2{D}_n\ge nS-o(n)$ and the rate cannot be below $S$. Both operational fidelity formulations therefore have the claimed noiseless coding threshold.

The mixed-state expression in the PDF contains a mismatched bra in its second term. A probability mixture of the stated orthonormal eigenvectors must be interpreted as $\rho ={\sum }_k{p}_k|u_k⟩⟨u_k|$. Taken literally with a different bra, it need not be Hermitian or positive and is not generally a density operator.

Let $m={\sum }_k{p}_k{e}^{i{\varphi }_k}=\mathrm{Tr}(\rho U)$. A quantum channel and its measurement probabilities are linear in the density operator, so averaging the pure-state results givesTherefore the same frequency estimator yieldsMore generally, after the controlled operation and before the final Hadamard gate, the reduced control density operator isThis follows by a partial trace of the control blocks $\rho$, $\rho U^{†}$, $U\rho$ and $U\rho U^{†}$. It proves the mixed-state Hadamard-test quadratures for any auxiliary density operator, even one not diagonal in the eigenbasis. Conjugation by the final Hadamard gate maps $X$ to $Z$ and $Y$ to $-Y$, explaining the imaginary-part sign.

There is a necessary distinction between a failure bound averaged over the unknown state, a worst-case bound, and a separate bound for each candidate state. The last interpretation is false: a forger can discard the inputs and always prepare $|{\psi }_0{⟩}^{\otimes N}$. If the actual state is $|{\psi }_0⟩$, every authenticity test passes with probability one. The meaningful uniform statement concerns worst-case failure, or average failure under any fixed prior giving both states positive probability. The PDF does not specify that prior; the argument below gives an explicit bound for either interpretation.

Put $c=|⟨{\psi }_0|{\psi }_1⟩|\in (0,1)$. Any unconditional strategy producing $N$ output qubits, including ancillas, adaptive measurement in quantum measurements and correlated outputs, is a quantum channel $\mathcal{E}$. Its inputs and ideal outputs are the density operatorsThe projective measurements on the different banknotes commute, and their all-pass effect is exactly $P_i$. Therefore the probability of at least one rejection, conditional on input $i$, isThis remains true for entangled outputs; no independence of their individual test outcomes is assumed.

Use trace distance $D(\rho ,\sigma )=\frac{1}{2}\Vert \rho -\sigma {\Vert }_1$. The pure-state trace distance and fidelity identity givesFor completeness, two pure-state projectors differ by a matrix with nonzero eigenvalues $\pm \sqrt{1-|⟨u|v⟩{|}^2}$, which proves this identity. Also the pure-target upper bound on trace distance gives $D({\rho }_i,P_i)\le \sqrt{e_i}$. To see the mixed-state step, write ${\rho }_i={\sum }_j{q}_j|u_j⟩⟨u_j|$ and use convexity of the trace norm followed by the Cauchy-Schwarz inequality:Finally, trace-distance contraction under quantum channels gives $D({\rho }_0,{\rho }_1)\le D({\sigma }_0,{\sigma }_1)$. This follows from $D(\rho ,\sigma )={\max }_{0\le E\le I}\mathrm{Tr}[E(\rho -\sigma )]$: the adjoint channel is positive and unital, so its image of any effect $E$ is still an effect.

The triangle inequality now yieldsThus the cloning failure bound from trace-distance contraction isand in particular $\max (e_0,e_1)\ge {\Delta }^2/4$. For positive priors ${\pi }_0+{\pi }_1=1$, another Cauchy-Schwarz inequality givesso the average rejection probability satisfiesFor equal priors, one may take $p_0={\Delta }^2/8$; the same choice is a strict lower bound on worst-case rejection. This proves a quantitative version of the no-cloning theorem for the bank's all-pass test. It concerns an unconditional attempt: heralded success cannot be postselected while omitting failed attempts from the accounting.

There is a normalization error in the original PDF: the first term contains $I$ rather than $\rho$. For a normalized qubit density operator, that printed expression has trace $2(1-p)+p=2-p$, so it is not a quantum channel except at $p=1$. The intended depolarizing channel is the Pauli channel that leaves the input unchanged with probability $1-p$ and applies each nonidentity Pauli matrix with probability $p/3$, for $0\le p\le 1$. Its Kraus operators areand ${\sum }_{k=0}^3{K}_k^{†}{K}_k=I$, proving that it is trace-preserving and a completely positive map.

Write the input in its Bloch vector representation,Conjugation by a Pauli matrix keeps its own Bloch vector component and reverses the other two. More explicitly, ${\sigma }_k{\sigma }_j{\sigma }_k={\sigma }_j$ for $j=k$ and $-{\sigma }_j$ for $j\ne k$. HenceThe corrected quantum channel therefore givesThus its action isThe Bloch sphere is sent to a sphere of radius $|1-4p/3|$ about the origin of the Bloch ball. At $p=0$ the action is the identity. For $0<p<3/4$ the Bloch vector shrinks without changing direction. At $p=3/4$ every input becomes the maximally mixed density operator $I/2$. For $3/4<p\le 1$, the Bloch vector reverses direction and shrinks, reaching factor $-1/3$ at $p=1$. The reversal is compatible with complete positivity because this remains a probabilistic mixture of Pauli matrices. The parameter here is the total nonidentity-error probability, not the retention parameter used in some definitions of the quantum depolarizing channel.

For completeness, retain the PDF literally and call its output $F_p(\rho )$. The same calculation givesThis has no normalized-output Bloch ball interpretation as printed. If it is manually normalized, thenThat different transformation sends $p=0$ to $I/2$, rather than to the input, and agrees with the intended answer only at $p=1$. The literal and corrected conventions therefore cannot be silently identified.

Under the intended assumption that one use of the quantum channel transmits one qubit without noise, Alice can communicate two classical bits with certainty if she and Bob already share a Bell pair. She uses superdense coding. The prior distribution of the Bell pair must be a resource available before the permitted channel use; distributing it through that same channel would consume another use.

Take the shared Bell state to be $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$, with Alice holding the first qubit. For a two-bit message $(a,b)\in \{0,1{\}}^2$, Alice applies the local unitary operation $U_{ab}=X^a{Z}^b$ to her qubit. The four encoded pure states are, up to irrelevant global phases,They are orthogonal: for any matrix $V$ on the first qubit, $⟨{\Phi }^{+}|(V\otimes I)|{\Phi }^{+}⟩=\mathrm{Tr}V/2$, and the four Pauli operators are orthogonal in the trace inner product. Explicitly,Alice sends her qubit through the quantum channel once. Bob now holds both qubits and performs a Bell-basis measurement. Its four orthogonal projectors identify $(a,b)$ with certainty, givingBefore the transmission, Bob's reduced density matrix is $I/2$ for every message, so shared entanglement alone does not convey the message.

Without shared entanglement, a noiseless single-qubit transmission cannot perfectly distinguish four messages: perfect discrimination requires orthogonal state supports, and its two-dimensional Hilbert space has room for at most two nonzero mutually orthogonal supports. The Holevo bound also limits its accessible classical information to one bit. For the usual protocol starting with a shared two-qubit pure state and using local unitary encodings, exact two-bit transmission requires a maximally mixed Bob marginal. Indeed all encoded joint states are pure and have the same ${\rho }_B$, while Subadditivity of Von Neumann entropy gives their average entropy at most $1+S({\rho }_B)$. Four equally likely perfectly distinguishable messages have average Von Neumann entropy two, forcing $S({\rho }_B)=1$ and hence a maximally entangled state.

The specification “quantum channel” alone does not determine its dimension or noise, so the one-qubit, noiseless interpretation is needed for this particular conclusion. A noiseless four-dimensional channel could send four orthogonal states without any shared entanglement; a completely depolarizing qubit channel cannot transmit the message even with a shared Bell pair, since its output is independent of Alice's encoding. Shared entanglement is therefore the extra resource for the intended single-qubit channel, not a guarantee for every unspecified quantum channel.

Let $D={\mathcal{C}}_{\mathrm{ev}}$, and write ${\mathcal{C}}_{\mathrm{odd}}=t+D$ for any odd-weight word $t\in {\mathcal{C}}_H$. The parity functional on the four-dimensional binary linear code ${\mathcal{C}}_H$ is nonzero, so its kernel $D$ has dimension three and its other fibre is the coset $t+D$. For example $t=(1,\dots ,1)$ belongs to the standard length-seven Hamming code and has odd Hamming weight. Denote the two logical basis states by $|{\psi }_0⟩$ and $|{\psi }_1⟩$, corresponding to these two cosets. The eight computational orthonormal basis vectors in each sum are distinct, so both logical states have norm one; their disjoint supports make them orthogonal.

For a binary vector $w\in {\mathbb{F}}_2^7$, define the product of Pauli Z gateswhere the dot product is computed modulo two. The key diagonal overlap, with $t_0=0$ and $t_1=t$, isIf $w\in D^{\perp }$, every summand is one. If $w\notin D^{\perp }$, choose $x_0\in D$ with $w\cdot x_0=1$. Translation $x\mapsto x+x_0$ permutes $D$ and negates every summand. The sum must then be zero. ThusOff-diagonal logical overlaps vanish for every $w$:because a diagonal Pauli operator cannot move a computational basis word from one coset to the disjoint other coset.

The possible errors are $E_0=I$ and $E_j=Z(e_j)$ for $j=1,\dots ,7$. Set $e_0=0$. Each phase flip is its own adjoint and inverse, soIf $i\ne j$, the vector $e_i+e_j$ is nonzero and has Hamming weight one or two. By the given dual code identity $D^{\perp }={\mathcal{C}}_H$ and the minimum Hamming distance three of the Hamming code, it cannot belong to $D^{\perp }$. Hence all its diagonal logical overlaps vanish. If $i=j$, the product is the identity and its logical overlaps are ${\delta }_{ab}$. Combining both cases givesThis proves the Knill--Laflamme condition with an identity error-overlap matrix, not merely a possibly degenerate scalar matrix. In particular the eight two-dimensional error subspaces $E_i{\mathcal{X}}_{\mathrm{Steane}}$ are mutually orthogonal. This is precisely nondegenerate phase-flip correction in the Steane code.

To exhibit recovery rather than only the criterion, let $P$ be the orthogonal projector onto the code and $P_i=E_{i}P{E}_i^{†}$. The projective measurement consisting of these eight mutually orthogonal projectors, completed by $I-{\sum }_i{P}_i$, determines which error occurred. For an arbitrary encoded pure state $|\chi ⟩=\alpha |{\psi }_0⟩+\beta |{\psi }_1⟩$ affected by $E_j$, outcome $j$ occurs with certainty and leaves $E_j|\chi ⟩$ unchanged. Applying $E_j$ again recovers $|\chi ⟩$. The outcome gives the error syndrome and does not reveal $\alpha$ or $\beta$, so recovery preserves logical quantum superpositions.

An equivalent explicit error syndrome uses three commuting X-type stabilizer generators. Choose a parity-check matrix for ${\mathcal{C}}_H$ with the seven different nonzero binary columns:Its rows span ${\mathcal{C}}_H^{\perp }=D$. For a row $h_l$, the operator $X(h_l)={⨂}_j{X}_j^{(h_l{)}_j}$ permutes each of $D$ and $t+D$, and therefore has eigenvalue $+1$ on both logical states. A phase flip at position $j$ changes its eigenvalue to $(-1{)}^{H_{lj}}$, since $X(h_l)Z_j=(-1{)}^{H_{lj}}{Z}_{j}X(h_l)$. Measuring the three commuting stabilizer generators gives the $j$th column of $H$ as a binary syndrome; no error gives the zero column. All eight outcomes are different, and applying the identified $Z_j$ restores the code state. Measuring these stabilizer generators never distinguishes the two logical states.