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The Bell basis diagonalizes the commuting Pauli operators $Z\otimes Z$ and $X\otimes X$. Two local anticommutations make the joint operators commute. Their signs encode parity and relative phase. The displayed projectors turn the signed eigenvalues into literal zero-or-one bit eigenvalues: even parity and plus phase have bit zero.

Store a Pauli group element with binary exponents $a_j,b_j$ and a phase exponent $s$ modulo four. This ordered $XZ$ convention uses $2n+2$ bits for $n$ qubits. Since $XZ=-iY$, phase factors are essential when converting to Hermitian Pauli operators and evaluating measurement probabilities.

The need for mutually exclusive experimental arrangements or incompatible observables to exhibit different aspects of a quantum system. For example, path discrimination and interference contrast obey a tradeoff, while sharp Pauli operator $X$ and $Z$ measurements have incompatible eigenbases. Which-path information is an operational record, rather than a requirement that a conscious observer reads it.

The guaranteed radius is the largest integer $t$ for which all unknown-location Pauli operators of weight at most $t$ can be corrected together. Pairwise products have weight at most $2t$, so $2t<d$ suffices by the Knill--Laflamme condition. Conversely, split a minimum-weight undetectable Pauli operator between two subsets of its support. If $d\le 2t$, this gives two errors of weight at most $t$ whose pairwise product violates that condition. Consequently $d$ is either $2t+1$ or $2t+2$; parity cannot be recovered from $t$ alone.

Split the support of a Pauli operator of weight at most $2t$ into two sets of size at most $t$. It is, up to a scalar phase, $E_a^{†}{E}_b$ for two errors in the correctable family. The Knill--Laflamme condition makes its compression to the code scalar, which is quantum error detection. Expansion in the Pauli basis extends the assertion to every operator supported on at most $2t$ qubits. The number $t$ counts affected qubits, not the cardinality of an arbitrary selected family of errors.

An $[[n,k,d]]$ quantum error correction code is a $2^k$-dimensional subspace of $n$ qubits. Its distance $d$ is the minimum weight of a Pauli operator $E$ for which $PEP$ is not a scalar multiple of the code projector $P$. Thus every operator supported on fewer than $d$ qubits has scalar compression to the code, by expansion in the Pauli basis. This definition includes degenerate codes; a low-weight operator acting as a scalar does not reduce the distance.

For the quantum depolarizing channel applying the identity with probability $1-p$ and each nonidentity Pauli operator with probability $p/3$, the entanglement-assisted classical capacity is the displayed number of bits per use. Here $H_4$ is Shannon entropy of the four listed probabilities. Its unassisted product-state capacity is $1-h_2(2p/3)$. Both vanish at $p=3/4$. If $q$ instead denotes replacement by $I/2$, then $p=3q/4$ and complete depolarization is at $q=1$. The original entanglement-assisted capacity paper gives the depolarizing-channel formula.

The five-qubit code encodes one logical qubit into five physical qubits and corrects an arbitrary error on one physical qubit. In a nondegenerate realization, the identity and fifteen one-qubit Pauli operators take the two-dimensional code into sixteen mutually orthogonal syndrome subspaces, filling the 32-dimensional physical Hilbert space.

// Target: quantum-error-correction.bigb

Charlie measures his qubit of a GHZ state in the $X$ basis. The two equally likely outcomes leave Alice and Bob in the Bell states ${\Phi }^{+}$ or ${\Phi }^{-}$, respectively. Charlie sends the outcome bit $c$ to Bob, who applies the Pauli operator $X{Z}^c$, giving ${\Psi }^{+}$ in either case. This is deterministic LOCC entanglement localization. If Charlie's record is ignored, Alice and Bob instead have the separable mixture $(|00⟩⟨00|+|11⟩⟨11|)/2$. The classical outcome is necessary to turn the conditional entanglement into a known usable Bell pair.

For a specified correctable set of distinct Pauli operators modulo phase, a nondegenerate code has mutually orthogonal error images $E\mathcal{X}$. Equivalently, the matrix in the Knill--Laflamme condition is diagonal for that Pauli error set. Each error therefore requires a separate syndrome subspace of dimension equal to the logical code dimension. Degenerate codes can have distinct errors with the same action on the code and do not satisfy this orthogonality counting argument.

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.

Here the number of errors means the number of affected physical qubits: assume that all errors supported on at most $t$ qubits are jointly correctable. It does not mean the cardinality of an arbitrary selected error list. Let $F$ be any Pauli operator of weight at most $2t$. Split its support into disjoint sets $A,B$, each of size at most $t$, and write $F=F_A{F}_B$. Each Pauli factor is Hermitian, so take $E_a=F_A$, $E_b=F_B$ to get $F=E_a^{†}{E}_b$, up to an irrelevant scalar phase if different Pauli representatives are used. The Knill--Laflamme condition givesThis is precisely quantum error detection: a projective measurement with outcomes $P,I-P$ either detects departure from the code or leaves the accepted logical state unchanged up to a scalar. Some errors can be harmless scalars on the code, particularly for degenerate correction; detection is not required to flag such an error with certainty.

Any linear operator supported on at most $2t$ qubits is a linear combination of Pauli operators on those qubits, so its code compression is scalar by linearity. ThusThis is correction implies detection at twice the weight. The premise about all locations and all error types is essential: merely correcting one selected operator gives no universal statement about every operator of twice its weight.

For an $n$-qubit quantum code with projector $P$, the quantum code distance isHere the weight counts nonidentity tensor factors. Thus every error supported on fewer than $d$ qubits is detectable, by the Pauli expansion of a quantum error. A scalar action on the code, including a nonidentity stabilizer, is harmless and does not reduce this distance. For a stabilizer code, the equivalent definition minimizes weight among Pauli operators which commute with every stabilizer generator but act nontrivially on the logical information. The Knill--Laflamme condition then guarantees correction of all errors of weight at most $t=\lfloor (d-1)/2\rfloor$, since each pairwise product has weight at most $2t<d$.

Suppose the code encodes $k$ logical qubits and is nondegenerate for all errors through weight $t$. There aredistinct Pauli errors modulo phase: choose the $j$ affected positions and choose $X,Y$ or $Z$ at each. Each error is unitary, so its image of the code has dimension $2^k$. Nondegeneracy makes these images mutually orthogonal, equivalently the combined vectors $E_a|i_L⟩$ are orthonormal. They all lie in the physical Hilbert space of dimension $2^n$. Counting dimensions proves the quantum Hamming boundThis dimension argument requires nondegeneracy. For a degenerate code, different errors can share an image, as in the Shor code, and this counting proof cannot be applied unchanged.

For a binary classical code of $M$ words, the Hamming bound is $M{\sum }_{j=0}^t\left(\genfrac{}{}{0ex}{}{n}{j}\right)\le 2^n$: disjoint Hamming balls around the words must fit into all binary strings. The quantum factor $3^j$ is needed because at each erroneous qubit there are three linearly independent nonidentity Pauli operators, not just a classical bit flip. A phase flip is invisible to a computational-basis classical record but changes a quantum superposition. A $Y$ error combines bit and phase effects, and the identity with $X,Y,Z$ spans every one-qubit operator. Quantum correction must protect amplitudes and coherences, and consequently count whole orthogonal logical subspaces rather than just classical words.

For parameters $[[5,1,3]]$, $t=1$, andThus a nondegenerate such code saturates the bound: the sixteen two-dimensional error images exhaust the physical space. It is a perfect quantum error-correcting code for single-qubit errors. It also has the smallest possible length for nondegenerate single-error protection of one logical qubit: for $1\le n\le 4$, the required inequality $1+3n\le 2^{n-1}$ fails.

Here is an explicit realization and its five-qubit stabilizer syndrome table. Write tensor products without multiplication symbols and useEach pair has an even number of anticommuting single-site factors, so the four stabilizer generators commute. Their binary Pauli labels are independent: no nonempty product has identity label. For example, the $X$-support at site three fixes the exponent of $g_3$ in such a product to zero, site five then fixes that of $g_2$, site one fixes that of $g_1$, and site two fixes that of $g_4$. Thus their group has sixteen elements and excludes $-I$. The common $+1$ eigenspace has projector $P={\prod }_{a=1}^4(I+g_a)/16$ and dimension $\mathrm{Tr}P=32/16=2$, since all nonidentity Pauli strings have zero trace.

Record a syndrome bit as one when an error anticommutes with the corresponding generator, in order $g_1,g_2,g_3,g_4$. Direct local commutation givesThese are all fifteen nonzero four-bit error syndromes; the identity has $0000$. Distinct errors in this list have a product with nonzero syndrome, so it anticommutes with some $g_a$. Inserting $g_{a}P=P$ on both sides gives $P{E}_b^{†}{E}_{c}P=-P{E}_b^{†}{E}_{c}P=0$. The diagonal product is the identity. Hence the Knill--Laflamme condition is exactly $C_{bc}={\delta }_{bc}$, proving nondegenerate correction. Measure the four commuting checks, apply the unique indicated Pauli error again, and every logical state is recovered. The Pauli expansion of a quantum error extends this recovery to an arbitrary one-qubit noise channel, not just the fifteen discrete errors.

The table also proves that every Pauli operator of weight one or two has nonzero syndrome: a weight-two operator is the product of distinct single-site errors, whose syndromes differ. Therefore the distance is at least three. The operators $\overline{X}=X^{\otimes 5}$ and $\overline{Z}=Z^{\otimes 5}$ commute with all four generators and anticommute with each other. They are nontrivial logical Pauli operators. Multiplication of $\overline{X}$ by $g_1$ gives $-IYYIX$, a weight-three operator with the same logical action; it cannot be scalar on the code because it anticommutes with the invertible logical operator $\overline{Z}$. This proves $d=3$ exactly.

The code consequently detects every two-qubit error and corrects two known-location erasures by the quantum erasure correction criterion, but it cannot correct all two-qubit errors of unknown location. A nontrivial weight-three logical operator can be split into a weight-one and a weight-two error, whose pairwise product violates the correction condition. It also saturates the quantum Singleton bound, since $n-k=4=2(d-1)$. These properties distinguish single-error perfectness from unrestricted protection against larger errors.

Fix the noise convention explicitly: $p$ is the total probability of a nonidentity Pauli error. The memoryless qubit quantum depolarizing channel iswith independent identical use on successive qubits, namely ${\mathcal{D}}_p^{\otimes n}$ on an $n$-qubit block. Its normalized Kraus operators are $\sqrt{1-p}I,\sqrt{p/3}X,\sqrt{p/3}Y,\sqrt{p/3}Z$.

Write the Bloch sphere representation of a density matrix as $\rho =(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, with $|\mathbf{r}|\le 1$. Its eigenvalues are $(1\pm |\mathbf{r}|)/2$, proving exactly this positivity condition. Conjugation by one Pauli operator preserves that coordinate of the Bloch vector and reverses the other two. Summing the three conjugations givessoThe Bloch ball contracts isotropically, collapses to the maximally mixed state at $p=3/4$, and contracts with reversal for $p>3/4$. That last interval still represents a physical Pauli mixture. If instead $q$ is the probability of replacing the state by $I/2$, then $\mathcal{D}(\rho )=(1-q)\rho +qI/2$, with $p=3q/4$ and $\eta =1-q$; the usual replacement-probability range $0\le q\le 1$ stops at complete depolarization.

Let $h_2(x)=-x{\log }_{2}x-(1-x){\log }_2(1-x)$, with endpoint limits. In bits per use, the unassisted product-state classical capacity and the entanglement-assisted capacity of a qubit depolarizing channel arewhere $H_4$ is Shannon entropy of the listed four probabilities. Equivalently $C_E=2+(1-p){\log }_2(1-p)+p{\log }_2(p/3)$. The first quantity is a classical capacity with product quantum inputs and potentially collective output measurements, not the quantum capacity. These are the requested capacity formulae; a proof of their coding theorems is not needed here. The maximally entangled input gives a Bell-diagonal joint state with those four probabilities, explaining the entropy appearing in $C_E$. The original entanglement-assisted capacity calculation gives the depolarizing-channel expression.

At $p=0$ the capacities are one and two. Both vanish at $p=3/4$. On the full Pauli-error interval they increase again beyond that point, reaching $1-h_2(2/3)$ and $2-{\log }_{2}3$ at $p=1$; on the replacement-probability plot both decrease to zero at $q=1$. The two panels make this convention distinction explicit.

For the high-noise capacity ratio for qubit depolarization, put $\eta =1-4p/3\to 0$. Taylor expansion of binary entropy around one half givesThe four probabilities for $C_E$ are $1/4+3\eta /4$ and three copies of $1/4-\eta /4$. For a perturbation ${\delta }_i$ of the uniform four-outcome distribution, with ${\sum }_i{\delta }_i=0$, entropy expansion givesHere ${\sum }_i{\delta }_i^2=3{\eta }^2/4$, hence $C_E=3{\eta }^2/(2\ln 2)+O({\eta }^3)$. ThereforeThe reciprocal ratio is $1/3$, and the same enhancement factor occurs as $q\to 1$ in the replacement convention.

By the Schmidt decomposition theorem, suitable local unitary operations put the normalized pure state into the formIt is entangled precisely when both Schmidt coefficients are nonzero. Put $s=2{\lambda }_0{\lambda }_1$, so $0<s\le 1$ for the case of interest. The Pauli operatorssatisfy $⟨Z\otimes Z⟩=1$ and $⟨X\otimes X⟩=s$ in this pure state: $Z\otimes Z$ fixes both basis terms, and $X\otimes X$ exchanges them.

Choose Alice's two observables as $A_0=Z$, $A_1=X$, and Bob's asThese are valid dichotomic projective measurements. Indeed, $XZ+ZX=0$ and $X^2=Z^2=I$, so $B_0^2=B_1^2=I$; each observable has outcomes $\pm 1$. For the CHSH inequality convention $|⟨A_0\otimes (B_0+B_1)+A_1\otimes (B_0-B_1)⟩|\le 2$, the quantum value isFor the original pure state, conjugate these four observables by the corresponding local unitary operators. Their outcomes and Pauli correlators are unchanged, and each party still measures only their own qubit. Thus every entangled pure two-qubit state violates CHSH for suitable local measurements, proving Gisin's theorem directly. The CHSH axes for an entangled pure two-qubit state approach a nonviolating configuration continuously as one Schmidt coefficient tends to zero; the strict violation therefore need not be large for weak entanglement.

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.

Finally, if a noise Kraus operator is a coherent linear combination $F={\sum }_j{c}_j{E}_j$, the same syndrome measurement separates its mutually orthogonal components $c_j{E}_j|\chi ⟩$. Conditional recovery returns $|\chi ⟩$ in every nonzero branch. Linearity then also corrects any quantum channel whose Kraus operators lie in this span. The argument therefore corrects arbitrary unknown single-site phase flips nondegenerately, including coherent phase-error amplitudes, rather than only a known error on a basis codeword.

Apply the Schmidt decomposition to the bipartite pure state. Local unitary operators put it in the form $|{\psi }_{\theta }⟩=\cos \theta |00⟩+\sin \theta |11⟩$, with $0<\theta \le \pi /4$ after interchanging basis vectors. Both Schmidt coefficients are nonzero because the state is entangled. Put $s=\sin 2\theta >0$.

For this state, direct application of the Pauli operators givesFor example ${\sigma }_x\otimes {\sigma }_x$ interchanges $|00⟩$ and $|11⟩$, giving $2\cos \theta \sin \theta$; the crossed products take these two kets into their orthogonal complement.

Choose the CHSH axes for an entangled pure two-qubit state:Their associated Bloch vectors have unit length, so each is a valid two-outcome projective measurement with projectors $(I\pm A_i)/2$ or $(I\pm B_j)/2$. The correlation combination isA local hidden-variable assignment with outcomes $\pm 1$ gives absolute value at most two: exactly one of $b_0+b_1$ and $b_0-b_1$ vanishes, while the other is $\pm 2$. Averaging these assignments preserves that bound, which is the CHSH inequality.

Therefore $\boxed{\mathcal{S}=2\sqrt{1+{\sin }^{2}2\theta }>2}$. For the original state $(U\otimes V)|{\psi }_{\theta }⟩$, conjugate Alice's observables by $U$ and Bob's by $V$. These conjugations rotate their measurement axes and preserve all the displayed correlations. Thus every entangled pure two-qubit state has the requested violating axes. The maximally entangled case gives $2\sqrt{2}$, while the unentangled limit merely saturates two.

Let $e_j$ be the bit string with one in position $j$ and zero elsewhere, and use the standard convention $Y=iXZ$. On the computational basis, the single-site Pauli operators act asThe identity leaves the basis state unchanged. A tensor-product Pauli operator applies these actions at each occupied site; operators at different sites commute. Their linear span is the full operator algebra, so correcting their span handles coherent combinations as well as random Pauli errors.

Let $P$ project onto the code $\mathcal{C}$, with orthonormal logical basis $|i_L⟩$. A set $\mathcal{E}=\{E_a\}$ is correctable exactly when the Knill--Laflamme condition holds:for a matrix $C$ independent of the logical indices. A set $\mathcal{D}$ is detectable whenThese conditions extend by linearity to the corresponding operator spans. Detection allows an error to act as a harmless scalar on the code; it need not assign a distinct syndrome to such an error.

For distinct Pauli representatives modulo phase, a nondegenerate quantum error-correcting code has orthogonal error-image subspaces, giving $C_{ab}={\delta }_{ab}$ because each Pauli is unitary. In the usual nondegenerate, or pure, detection convention, every nonidentity detectable Pauli has $PDP=0$, while $PIP=P$. More generally a nonsingular error-overlap matrix can be diagonalized and the resulting error operators rescaled to obtain orthogonal syndrome spaces; a singular overlap matrix expresses degeneracy of the specified independent error family.

For an $[[n,k,d]]$ code, the quantum code distance is the smallest Pauli weight for which scalar compression fails. Hence every Pauli of weight at most $d-1$ is detectable. If $E_a,E_b$ each have weight at most $t$, their product $E_a^{†}{E}_b$ has weight at most $2t$. When $2t<d$, its compression is scalar, so the Knill--Laflamme condition proves simultaneous correction of all such errors. ThereforeThe integer part is understood when the printed half-distance is not an integer. Arbitrary operators on such subsets are covered by their Pauli expansion.

For errors confined to a known set of $m$ positions, every pairwise product is still supported on that same set and has weight at most $m$, rather than $2m$. Thus $m<d$ supplies quantum erasure correction, with maximum guaranteed numberThis guarantee cannot hold for every set of $d$ positions: a minimum-weight undetectable Pauli has support in such a set, and the pair consisting of it and the identity violates the Knill--Laflamme condition. Particular larger location sets can sometimes be correctable; $d-1$ is the maximum uniform guarantee over all possible known sets.

For the Shor code, write $|G_{\pm }⟩=(|000⟩\pm |111⟩)/\sqrt{2}$. The logical codewords are $|G_{+}{⟩}^{\otimes 3}$ and $|G_{-}{⟩}^{\otimes 3}$. A phase flip $Z_j$ in one block interchanges $G_{+}$ and $G_{-}$ in that block, independently of its position within the block. Let $B_1=X_1{X}_2{X}_3$, $B_2=X_4{X}_5{X}_6$, $B_3=X_7{X}_8{X}_9$. Measure the two commuting stabilizer generators $S_1=B_1{B}_2$ and $S_2=B_2{B}_3$. Both are $+1$ on the entire logical code space. The Shor-code phase-flip syndrome isThese parity measurements reveal the erroneous block without revealing the logical amplitudes, so they preserve arbitrary superpositions of the two codewords. Applying the listed representative phase flip restores the state.

For any two physical positions $i,j$ in the same block, $Z_i{Z}_j$ acts as $+1$ on both $|000⟩$ and $|111⟩$, and hence on the code. Therefore $Z_{i}P=Z_{j}P$ and $P{Z}_i^{†}{Z}_{j}P=P$ even when $i\ne j$. Their error spaces coincide, and the error-overlap matrix has a nonzero off-diagonal entry. The correction is degenerate because different within-block phase errors have exactly the same action on every logical state.

Let $Q$ be Alice's unknown qubit, with pure state $|\varphi {⟩}_Q=\alpha |0⟩+\beta |1⟩$. Alice and Bob initially share the Bell state $|{\Phi }^{+}{⟩}_{AB}=(|00⟩+|11⟩)/\sqrt{2}$. Alice holds $Q,A$ and Bob holds $B$. Define the Bell basisExpanding in this orthonormal basis gives the Bell-basis teleportation identityFor example, the last branch on $B$ is $\alpha |1⟩-\beta |0⟩$, which is $XZ|\varphi ⟩$; the ordering of the Pauli operators therefore matters.

Alice performs a projective measurement in the Bell basis on her two local qubits. Each outcome has probability $1/4$, independent of the unknown amplitudes. She encodes the four possible outcomes as two classical bits and sends them to Bob. For outcomes ${\Phi }^{+},{\Phi }^{-},{\Psi }^{+},{\Psi }^{-}$, respectively, Bob applies the local unitary operationsThese are inverses of the corresponding branch Pauli operators, so Bob obtains exactly the unknown input qubit in every branch. Using $XZ$ instead of $ZX$ as the last correction also succeeds up to an irrelevant overall phase. Alice need not know $\alpha$ or $\beta$. Every quantum operation is local to one of the parties, and the only transmitted information is the two-bit classical outcome. The shared Bell pair is consumed.

Before the message arrives, Bob's density operator isso no unknown-state information is available without the classical communication. Alice's original qubit has been included in the destructive Bell-basis measurement; the protocol creates no extra copy. Since the corrected transformation is the identity on every input vector and is linear, it also preserves entanglement with an external reference and transfers mixed inputs. Thus this is deterministic quantum teleportation, using precisely one shared maximally entangled Bell pair, two classical bits, and local operations.