Stabilizer Groups

A stabilizer group is a special kind of subgroup of the Pauli group. It is the central object of this module: a stabilizer group can pin down a quantum state, define a quantum code, and be measured to detect errors. In this lesson we give the precise definition and the two conditions every stabilizer group must satisfy.

The definition

A subgroup $S \subseteq \mathcal{P}_n$ is a stabilizer group if it satisfies two conditions:

1. It is abelian — every pair of elements commutes.
2. It does not contain $-I$ (and therefore not $\pm iI$ either).

The second condition rules out a trivial obstruction: if $-I \in S$, then for any state $|\psi\rangle$ we would need both $g|\psi\rangle = |\psi\rangle$ and $-I|\psi\rangle = |\psi\rangle$, i.e. $-|\psi\rangle = |\psi\rangle$, forcing $|\psi\rangle = 0$. No nonzero state could be stabilized.

A state $|\psi\rangle$ is stabilized by $S$ when

i.e. $|\psi\rangle$ is a $+1$ eigenvector of every element of $S$. The set of all such states is the codespace of $S$.

Why abelian and why no $-I$

These two requirements are exactly what is needed for a nontrivial common $+1$ eigenspace to exist.

• Abelian. Operators can share a complete set of simultaneous eigenvectors only if they commute. If $g$ and $h$ anticommuted, then $gh = -hg$, and a common $+1$ eigenvector would give $|\psi\rangle = gh|\psi\rangle = -hg|\psi\rangle = -|\psi\rangle$, again forcing $|\psi\rangle = 0$.
• No $-I$. As shown above, $-I \in S$ collapses the codespace to zero.

When both hold, the simultaneous $+1$ eigenspace is guaranteed to be nonempty, and its dimension is controlled by the number of independent generators.

Generators and the size of a stabilizer group

A stabilizer group is finitely generated. If $S$ has $m$ independent generators $g_1, \dots, g_m$ (each of order $2$, so $g_i^2 = I$), then

because every element is a product $g_1^{a_1}\cdots g_m^{a_m}$ with each $a_i \in \{0,1\}$. The generators are independent if no product of a nonempty subset of them equals $I$.

On $n$ physical qubits, $m$ independent stabilizer generators cut the $2^n$-dimensional Hilbert space down to a codespace of dimension

Each generator imposes one binary ($\pm 1$) eigenvalue constraint and halves the dimension. When $m = n$ the codespace is one-dimensional: a single state, the subject of the next lesson.

Commutation as binary algebra

Checking the abelian condition does not require multiplying matrices. As we saw with the Pauli group, two strings commute or anticommute, and this is determined by a parity. Writing each Pauli (ignoring phase) as a binary vector $(x \mid z)$ — with $x_j = 1$ for an $X$ or $Y$ on qubit $j$ and $z_j = 1$ for a $Z$ or $Y$ — two operators $P$ and $Q$ commute exactly when their symplectic inner product

is zero. A generating set is abelian iff this vanishes for every pair of generators. We will build the full binary-matrix picture (the check matrix) two lessons from now.

Try it

Use the symplectic inner product to verify that the proposed generators of the three-qubit bit-flip code, $g_1 = Z_0 Z_1$ and $g_2 = Z_1 Z_2$, commute — so they generate a valid stabilizer group.

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