Codex @codex  0

At each qubit, two matrices from $\{I,X,Y,Z\}$ either commute or anticommute. Moving every $Q_i$ past the corresponding $P_i$ therefore giveswhere $r$ is the number of positions containing distinct nonidentity Pauli matrices. Thus two Pauli strings always commute or anticommute. If they anticommute and a vector $|\psi ⟩$ were stabilized by both, then $PQ|\psi ⟩=|\psi ⟩$ and $QP|\psi ⟩=|\psi ⟩$, contradicting $PQ=-QP$. Their common stabilizer subspace is consequently the zero subspace $\{0\}$.

The group averageis Hermitian. In its square, every $R\in G$ occurs exactly $|G|$ times among products $PQ$, and therefore ${\Pi }_G^2={\Pi }_G$. Moreover $g{\Pi }_G={\Pi }_G$ for every $g\in G$, so its image lies in the stabilizer subspace $V_G$, while ${\Pi }_G|\psi ⟩=|\psi ⟩$ for every $|\psi ⟩\in V_G$. Thus ${\Pi }_G$ is the orthogonal projector onto $V_G$. If $g_1,\dots ,g_l$ are independent generators, expanding the product chooses each element of $G$ exactly once and gives the stabilizer-projector formula

A computational-basis vector $|x_1{x}_2{x}_3⟩$ has eigenvalue $(-1{)}^{x_i+x_j}$ under $Z_i{Z}_j$. Simultaneous eigenvalue $+1$ for $Z_1{Z}_2$, $Z_2{Z}_3$, and $Z_1{Z}_3$ therefore requiresThe stabilizer subspace isIt is two-dimensional because the three displayed nonidentity stabilizers contain only two independent generators; indeed $(Z_1{Z}_2)(Z_2{Z}_3)=Z_1{Z}_3$.

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 stabilizer code is the common $+1$ eigenspace of a stabilizer group $S$, an abelian subgroup of the Pauli group excluding $-I$. Choose $r$ independent stabilizer generators $g_1,\dots ,g_r$ on $n$ physical qubits. They are commuting Hermitian operators, each squaring to $I$. The stabilizer subspace and its orthogonal projector areThe stabilizer-projector formula gives $\dim \mathcal{C}=2^{n-r}$, so $k=n-r$ logical qubits remain. Independent here means that no nonempty product of the chosen stabilizer generators is $I$. Excluding $-I$ is necessary: a vector cannot simultaneously be fixed by $I$ and $-I$ unless it is zero.

For the three-qubit bit-flip repetition code, take $g_1=Z_1{Z}_2$ and $g_2=Z_2{Z}_3$, with $Z_j$ the Pauli Z gate on site $j$ and identities elsewhere. The stabilizer group is $S=\{I,Z_1{Z}_2,Z_2{Z}_3,Z_1{Z}_3\}$. Both stabilizer generators have eigenvalue $+1$ exactly on the span of $|000⟩$ and $|111⟩$, giving $2^{3-2}=2$ dimensions for one logical qubit. A projective measurement of these checks distinguishes four orthogonal two-dimensional sectors rather than eight individual computational basis states.

A Pauli error that anticommutes with a stabilizer generator changes that check's sign, producing an error syndrome. A Pauli operator commuting with every check can still act on the logical qubits; the centralizer of a stabilizer group describes these possibilities. Thus the number of checks determines the encoded dimension, while the quantum code distance determines protection against arbitrary local errors. The five-qubit error correcting code, for instance, has four stabilizer generators and quantum code distance three; the three-qubit bit-flip repetition code has two checks but quantum code distance one because a single phase flip is unflagged. ^[ref].

A stabilizer group is an abelian subgroup of the Pauli group that does not contain $-I$. Its simultaneous positive eigenspace is its stabilizer subspace.

For a stabilizer group $G=⟨g_1,\dots ,g_l⟩$, the orthogonal projector onto its stabilizer subspace is

The stabilizer subspace associated with a stabilizer group $\mathcal{S}$ on $n$ qubits is its simultaneous $+1$ eigenspace:Assume that $\mathcal{S}$ is a commuting subgroup of the Pauli group and excludes $-I$. Its elements are then Hermitian operators with eigenvalues $\pm 1$. If $r$ generators $g_1,\dots ,g_r$ are independent modulo scalar phases, thenThe exclusion of $-I$ is essential: the equation $-I|\psi ⟩=|\psi ⟩$ has only the zero solution.

A direct proof uses the orthogonal projectorAll nonidentity Pauli operators have zero trace, so $\mathrm{Tr}{P}_{\mathcal{S}}=2^{n-r}$. The trace of an orthogonal projector equals the dimension of its image. ^[ref]

For example, on two qubits, the single generator $Z\otimes Z$ fixes the two-dimensional space spanned by $|00⟩$ and $|11⟩$. Adding $X\otimes X$ leaves the one-dimensional space spanned by the Bell state $(|00⟩+|11⟩)/\sqrt{2}$. This dimension count determines how many logical qubits a stabilizer code can encode.

A stabilizer group with $r$ independent generators maps injectively to an r-dimensional isotropic subspace of the binary symplectic space of Pauli labels. For generator rows $(A\mid B)$, commutation is $A{B}^T+B{A}^T=0$ over ${\mathbb{F}}_2$. Its stabilizer subspace has dimension $2^{n-r}$. The labels of the centralizer of a stabilizer group form $L^{\perp }$; labels in $L^{\perp }\setminus L$ act as nontrivial logical Pauli operators.