Stabilizer States

A remarkable fact underlies the whole formalism: when a stabilizer group has the maximum number of independent generators, it determines a single quantum state completely. Such a state is called a stabilizer state, and it can be specified by its stabilizers alone — no amplitudes required.

The maximal case

On $n$ qubits, take a stabilizer group $S$ with $n$ independent generators $g_1, \dots, g_n$ , so $|S| = 2^n$ . Each generator halves the Hilbert-space dimension, so the simultaneous $+1$ eigenspace has dimension

2^{\,n - n} = 1.

There is exactly one (up to global phase) state $|\psi\rangle$ satisfying $g_i|\psi\rangle = |\psi\rangle$ for all $i$ . That state is the stabilizer state of $S$ , and we say $S$ is its stabilizer group.

Specifying a state by its stabilizers

This gives an economical, basis-independent way to name a quantum state. Instead of listing $2^n$ complex amplitudes, we list $n$ Pauli generators — just $O(n^2)$ bits. Some familiar examples:

| State | Stabilizer generators | | --- | --- | | $\lvert 0\rangle$ | $Z$ | | $\lvert 1\rangle$ | $-Z$ | | $\lvert +\rangle$ | $X$ | | $\lvert -\rangle$ | $-X$ | | Bell $\tfrac{1}{\sqrt2}(\lvert 00\rangle + \lvert 11\rangle)$ | $X_0 X_1,\ Z_0 Z_1$ | | GHZ $\tfrac{1}{\sqrt2}(\lvert 000\rangle + \lvert 111\rangle)$ | $X_0 X_1 X_2,\ Z_0 Z_1,\ Z_1 Z_2$ |

Notice how the signs matter: $Z$ stabilizes $|0\rangle$ while $-Z$ stabilizes $|1\rangle$ . The same Pauli group element with a flipped sign picks out the orthogonal eigenstate.

Reading a state off its generators

Take the GHZ generators $\{X_0X_1X_2,\, Z_0Z_1,\, Z_1Z_2\}$ and decode them:

$Z_0 Z_1 = +1$ forces qubits $0$ and $1$ to agree (both $0$ or both $1$ ).
$Z_1 Z_2 = +1$ forces qubits $1$ and $2$ to agree.
Together these confine the state to $\mathrm{span}\{|000\rangle, |111\rangle\}$ .
$X_0 X_1 X_2 = +1$ flips all three bits and demands the state be invariant, selecting the even combination $\tfrac{1}{\sqrt2}(|000\rangle + |111\rangle)$ rather than the odd one.

The three commuting constraints leave precisely one state — exactly as the dimension count predicted.

How many stabilizer states are there?

The number of distinct $n$ -qubit stabilizer states is

2^{n} \prod_{k=1}^{n} \left(2^{k} + 1\right),

which grows fast but is vastly smaller than the continuum of all $n$ -qubit states. For $n=1$ there are $6$ (the eigenstates of $X$ , $Y$ , $Z$ ); for $n=2$ there are $60$ . This finiteness is the seed of the Gottesman–Knill theorem we reach later: a discrete state set with structured updates can be tracked classically.

Try it

The GHZ state $\tfrac{1}{\sqrt2}(|000\rangle + |111\rangle)$ is the unique $+1$ eigenstate of the generators $X_0X_1X_2$ , $Z_0Z_1$ , $Z_1Z_2$ . Build a circuit that prepares it from $|000\rangle$ .

Run your code to see the quantum state.