Stabilizer group

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

A stabilizer group excludes $-I$, so each element squares to $I$ and is Hermitian; an imaginary-phase Pauli element would square to $-I$. A nonidentity stabilizer contains at least one nonidentity local Pauli matrix and has zero trace, since the trace of a tensor product is the product of local traces. Its eigenvalues are therefore $\pm 1$ with equal multiplicities. This trace fact also gives $\dim {\mathcal{X}}_S=2^n/|S|$ from the stabilizer-projector formula.

A set of stabilizer generators is an independent commuting family of Hermitian Pauli operators whose products form a stabilizer group.

A stabilizer state on $n$ qubits is the unique simultaneous positive eigenstate of a stabilizer group with $n$ independent generators.

Stabilizer-state preparation starts from a computational-basis state and applies a Clifford circuit, producing a stabilizer state.

A stabilizer tableau represents each Pauli generator by a binary row $[x_1\cdots x_n\mid z_1\cdots z_n]$, with one additional bit when its sign is retained. Clifford gates update these rows by arithmetic over ${\mathbb{F}}_2$.

A graph state assigns one qubit to each vertex of an undirected graph. Begin with all qubits in $|+⟩$ and apply a Controlled-Z gate for every edge.

For an undirected graph $G=(V,E)$, the graph state is

The Controlled-Z gates commute, so their ordering does not change the result. On a computational-basis state $|x⟩$, each occupied edge contributes $(-1{)}^{x_a{x}_b}$; the amplitudes retain equal magnitude but acquire graph-dependent phases. An isolated vertex remains in the product state $|+⟩$, while edges can create entanglement. ^[ref]

Let $1$ be a leaf with neighbour $2$ and let the graph on the other vertices be generated by the Controlled-Z gate product $E_{\mathrm{rest}}$. If the leaf has Pauli Z gate frame $Z_1^r$, measuring it at angle $\alpha$ with result $s$ applies one-bit teleportation and leavesAny known Pauli Z gate frames on other vertices remain as additional factors before those vertices' input states. The identity $J(\alpha )Z^r=X^{r}J(\alpha )$ combines the leaf byproduct with the measurement result.

For a graph state $|G⟩={\prod }_{ij\in E}C{Z}_{ij}|+{⟩}^{\otimes |V|}$, measuring vertex $v$ in the computational basis with result $r$ gives the normalized stateEach Controlled-Z gate from $v$ to a neighbour acts as $Z_u^r$ after projecting $v$ onto $|r⟩$; all other edges remain unchanged. The projection contributes $1/\sqrt{2}$, so either result has probability $1/2$. For a four-cycle, deleting one vertex leaves a path and adds byproducts on its two endpoints.

The graph-state stabilizer generator at vertex $v$ is $S_v=X_v{\prod }_{u\in N(v)}{Z}_u$. Two such generators commute because adjacent vertices contribute two Pauli anticommutations and nonadjacent vertices contribute none.

Local complementation at a vertex $v$ toggles every edge between neighbors of $v$. It is implemented on graph states, up to global phase, by the local Clifford operator

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.

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