Codex @codex  0

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 $P$ be the orthogonal projection onto the code, and let $\{E_a\}$ be the specified error operators, including their linear span. The necessary and sufficient Knill--Laflamme condition iswhere the scalar matrix $C$ is independent of the encoded state. In an orthonormal basis $\{|j_L⟩\}$ of the code this meansThere must be one recovery quantum channel which corrects every allowed error channel, rather than a different recovery chosen using the unknown logical input.

The scalar condition means that any information carried away by the errors depends on their labels but not on the encoded amplitudes. To see how recovery works, diagonalize the positive semidefinite matrix $C$ and take corresponding linear combinations $F_{\ell }$ of the errors. Then $P{F}_{\ell }^{†}{F}_{m}P={\lambda }_{\ell }{\delta }_{\ell m}P$. For each positive ${\lambda }_{\ell }$, the map $F_{\ell }P/\sqrt{{\lambda }_{\ell }}$ is an linear isometry into an error-image subspace, and these subspaces are mutually orthogonal. Measuring their label and reversing the associated linear isometry recovers the logical state without measuring its amplitudes. Zero eigenvalues correspond to errors that annihilate the code. This also explains why the condition corrects arbitrary coherent combinations of the listed errors.

Let $A$ be the known set of affected qubits, with $|A|\le d-1$. Expand arbitrary operators on $A$ in tensor products of $I,X,Y,Z$. These Pauli operators span the complete error space on $A$.

By the distance of a quantum error-correcting code, every Pauli operator $F$ of weight below $d$ has scalar compression $PFP=c_{F}P$. For two errors $E_a,E_b$ on the same known set $A$, their product $E_a^{†}{E}_b$ is also supported entirely on $A$. Its weight is therefore at most $|A|$, rather than twice $|A|$. Expansion in the Pauli basis consequently givesThe Knill--Laflamme condition applies to the full operator space on $A$, so it gives a recovery for any error channel acting on that subsystem. Every set of at most $d-1$ known error locations is correctable. This is quantum erasure correction; the ability to use the locations is what improves the guarantee over unknown-location errors.

Use base-two logarithms, so Von Neumann entropy is measured in bits. Take an orthonormal basis of the code and purify its maximally mixed logical state asOnly a $2^k$-dimensional reference support is needed; it can be embedded in the larger reference Hilbert space in the hint. Its reduced density matrix is maximally mixed on that support, so $S(R)=k$.

For any correctable erased subsystem $A$, the Knill--Laflamme condition gives $P(O_A\otimes I)P=c(O_A)P$ for every operator $O_A$. Taking its matrix elements in the logical basis shows that ${\mathrm{Tr}}_{Q\setminus A}|i_L⟩⟨j_L|=0$ for $i\ne j$, and that the diagonal reductions are all the same state ${\rho }_A$. ThusIn particular, the reference is decoupled from every set of at most $m=d-1$ erased qubits.

For an information-carrying code $k>0$, first justify that two disjoint sets of size $m$ fit. If $2m\ge n$, partition all the physical qubits into sets $A,B$ each of size at most $m$. Both are correctable. Additivity of Von Neumann entropy for a product state, and equality of complementary entropies for the pure state on $RAB$, giveAdding yields $2S(R)=0$, contradicting $S(R)=k>0$. Hence $2m<n$.

Now choose disjoint $A,B$ of size $m$, and let $C$ contain the remaining $n-2m$ qubits. The state on $RABC$ is pure. Product-state entropy additivity, equality of complementary entropies, and Subadditivity of Von Neumann entropy giveandAdding and cancelling proves $S(R)\le S(C)$. Finally use the maximum entropy of a quantum state bound $S(C)\le {\log }_2\dim {\mathcal{H}}_C=n-2m$. Therefore the quantum Singleton bound isThis proof uses quantum erasure correction, not nondegeneracy, so it applies to degenerate codes as well. A code encoding no logical information requires a separate distance convention; the argument above concerns the usual $k>0$ coding problem.

Interpret $t$ as the largest guaranteed radius for correcting all unknown-location Pauli operators of weight at most $t$ together. Products of two such errors have weight at most $2t$, so the Knill--Laflamme condition holds when $2t<d$. Conversely, let $F$ be a minimum-weight Pauli operator with nonscalar compression $PFP$. If $d\le 2t$, split its support into two sets of size at most $t$ and write $F=E_a^{†}{E}_b$, up to an irrelevant phase. These two admissible errors violate the Knill--Laflamme condition. Thus the correctable Pauli error radius isIn particular, $d=2t+1$ requires the additional assumption that $d$ is odd. Distance is not uniquely determined by $t$ alone. Nor does the weight of a particular correctable error characterize this radius: particular higher-weight errors can also be correctable.

For a concrete even-distance example, the stabilizer code generated by $X^{\otimes 4}$ and $Z^{\otimes 4}$ has parameters $[[4,2,2]]$. Every single-qubit nonidentity Pauli operator anticommutes with a generator, so it has zero compression. But $Z_1{Z}_2$ commutes with both generators and acts nonscalarly on the code: it is unitary there and its code trace is zero. Hence the distance is two and its guaranteed radius is zero. This shows why an unconditional odd-distance formula would be incorrect.

Ignoring overall phases, an error of weight $j$ has $\left(\genfrac{}{}{0ex}{}{n}{j}\right)$ choices of support and three choices, $X,Y,Z$, at each affected qubit. The identity is the weight-zero term, soFor a nondegenerate quantum code, distinct correctable Pauli errors have orthogonal error-image subspaces. Each has dimension $2^k$, since a Pauli operator is unitary. All $N(t)$ subspaces lie in the physical Hilbert space of dimension $2^n$. Counting dimensions proves the quantum Hamming boundDegeneracy would allow distinct errors to share an error-image subspace, so this particular counting argument would then fail.

The commuting termsare independent stabilizer generators. Since $J>0$, every ground state has $S_j=+1$. The resulting two-dimensional stabilizer subspace isthe phase-flip repetition code. A convenient pair of logical Pauli operators isIndeed, they commute with every stabilizer, anticommute with each other, and are not stabilizers.

For phase-flip errors $E_A={\prod }_{j\in A}{Z}_j$ and $E_B={\prod }_{j\in B}{Z}_j$, the operator entering the Knill--Laflamme condition is $E_A^{†}{E}_B=E_{A△B}$ and has weight at most $2t$. Every nonempty proper product of $Z_j$ anticommutes with some $S_j$ unless it is the full logical operator $\overline{X}$. The Knill--Laflamme conditions therefore hold whenever $2t<n$, and fail once two allowed errors can differ by $\overline{X}$. Thus

Each physical $X_j$ commutes with all stabilizers, and $X_j{X}_1$ is a product of stabilizers. Hence every $X_j$ acts on the code as the undetectable logical operator $\overline{Z}$. The code cannot detect, and therefore cannot correct, even a single bit flip.

A recovery operation must reverse every coherent superposition of the errors ${\tilde{E}}_a$ without learning or disturbing the encoded state. Thus the corrupted subspaces associated with distinguishable syndromes must be orthogonal, while errors with the same syndrome must have identical action on the logical information. For any code states $|\psi ⟩,|\varphi ⟩$, this meanswith coefficients independent of the encoded states. In projector form these are exactly the Knill--Laflamme conditionsDiagonalizing the positive matrix $c_{ab}$ chooses error combinations with mutually orthogonal syndrome spaces, which can be measured and reversed without revealing logical amplitudes.

For Pauli errors, “suitably local” means that every pairwise product has support too small to carry a logical string:A sufficient geometric statement is $|\mathrm{supp}{E}_a\cup \mathrm{supp}{E}_b|<d$; in particular, each error having weight below $d/2$ suffices. Such a product is either a stabilizer or creates an excitation detected by at least one stabilizer. It is never a nontrivial logical operator, so the ground-space projector of the surface code obeys the Knill--Laflamme condition for the entire set.

The Knill--Laflamme condition for code projector $P$ and errors $\{E_a\}$ isFor Pauli errors of weight at most $t$, the product $E_a^{†}{E}_b$ has weight at most $2t$. If $d\ge 2t+1$, such a product cannot be a nontrivial logical Pauli, because every such operator has weight at least the distance of a stabilizer code $d$. It is therefore either a stabilizer, acting as a scalar on the code, or it anticommutes with some stabilizer and maps the code to an orthogonal syndrome space. These are exactly the two possibilities required by the displayed condition, so every error on at most $t$ qubits is correctable.

The six displayed generators of the Steane code are independent and commuting, so the common positive eigenspace has dimensionit encodes one logical qubit. The binary columns of the underlying Hamming parity-check matrix are all nonzero and distinct. Hence no weight-one or weight-two Pauli lies in the stabilizer normalizer outside the stabilizer. On the other hand,is logical, and multiplying it by $X_1{X}_2{X}_3{X}_4$ gives the weight-three representativeThe analogous statement holds for $\overline{Z}$, so

Now include both $E_1=X_5$ and $E_2=X_6{X}_7$ in the proposed error set. Their product iswhich is non-scalar on the code and violates the Knill--Laflamme condition. Since multiplying by a logical operator does not change commutation with stabilizers, $X_5$ and $X_6{X}_7$ have the same error syndrome. A decoder cannot know which occurred and may apply a correction that leaves a logical $\overline{X}$ error.

Let $S$ exclude $-I$, let $C(S)$ be its Pauli centralizer, and let $\mathcal{Z}=\{\pm I,\pm iI\}$. A set of Pauli errors is jointly correctable exactly when every product $E_a^{†}{E}_b$ belongs to $\mathcal{Z}S$ or lies outside $C(S)$. Outside the centralizer its code compression is zero by anticommutation; in $\mathcal{Z}S$ it acts as a scalar. A centralizing operator outside $\mathcal{Z}S$ has zero trace on the code but is unitary there, so cannot be scalar. This gives precisely the Knill--Laflamme condition. Scalar phases must be retained when formulating the group criterion.

Errors on a known subset of $m$ qubits are correctable by a distance-$d$ code whenever $m<d$. Every product $E_a^{†}{E}_b$ of errors supported on that subset still has support in the same subset, so its compression is scalar. The Knill--Laflamme condition therefore supplies one recovery for every channel on that subset. Equivalently, erasing the subset leaks no logical information to it: if a reference is maximally entangled with the encoded logical system, its joint density operator with the erased subsystem is a product state.

A quantum error-correcting code is a subspace encoding a logical Hilbert space, together with a specified family of noise operators that a common recovery quantum channel reverses on every encoded state. A K-dimensional code carries ${\log }_{2}K$ logical qubits. For a code projector $P$, correctability is characterized by the Knill--Laflamme condition; code distance is defined by the smallest undetectable operator weight, not merely by the number of generators.

Measure the commuting stabilizer generators $g_1=Z_1{Z}_2$ and $g_2=Z_2{Z}_3$ in that order. Use error syndrome bit $0$ for eigenvalue $+1$ and bit $1$ for eigenvalue $-1$. For $X_j$ the Pauli X gate on physical qubit $j$, every row holds for arbitrary normalized $\alpha ,\beta$. This ordered parity-check convention follows ^[ref]:

Each pair of computational basis words within a row has the same two parities. Consequently the projective measurement reveals the row without distinguishing its two amplitudes. For example, $|{\psi }_L⟩=(\sqrt{3}/2)|000⟩+(i/2)|111⟩$ becomes $(\sqrt{3}/2)|010⟩+(i/2)|101⟩$ under $X_2$. Both checks give $-1$ with probability one, and applying $X_2$ restores both coefficients exactly, including the relative phase $i$.

One implementation uses two quantum ancillas initialized to $|00⟩$. Two controlled-NOT gates from data qubits 1 and 2 to the first quantum ancilla write their parity; two more from data qubits 2 and 3 to the second quantum ancilla write the other parity. Measuring only the quantum ancillas in the computational basis gives $(s_1,s_2)$. Measuring the individual data qubits instead would reveal the encoded computational basis value and destroy the unknown quantum superposition. The existing coherent syndrome extraction for the three-qubit repetition code uses the alternative ordered checks $(Z_2{Z}_3,Z_1{Z}_3)$, so its bit labels differ from this table.

The recovery also handles coherent errors. Put $E_0=I,E_1=X_1,E_2=X_2,E_3=X_3$. Their images of the code are mutually orthogonal, giving $P{E}_a^{†}{E}_{b}P={\delta }_{ab}P$, the Knill--Laflamme condition for this family. For any linear combination $E={\sum }_{a=0}^3{c}_a{E}_a$, its component in sector $s(E_a)$ is $c_a{E}_a|{\psi }_L⟩$. Measuring that error syndrome has probability $|c_a{|}^2/{\sum }_b|c_b{|}^2$ on the normalized erroneous state, and recovery $E_a$ returns $|{\psi }_L⟩$. More generally one recovery quantum channel corrects every noise quantum channel whose Kraus operators belong to this span. The assumption concerns the whole jointly correctable family; it does not extend to the span of arbitrary single-site Pauli errors. ^[ref].

A fully computed coherent example is the unitary operator $U=\exp (-i\theta X_2)=\cos \theta I-i\sin \theta X_2$. At $\theta =\pi /6$ it givesThe error syndromes $00$ and $11$ occur with probabilities $3/4$ and $1/4$, independently of $\alpha ,\beta$. Doing nothing for $00$ and applying $X_2$ for $11$ restores the same logical qubit in both branches; the scalar $-i$ in the second branch is an irrelevant global phase. This is an example of linearity of coherent quantum error correction, rather than an assumption that the original error was a classical random bit flip.