Stabilizer States vs Magic

A single-qubit pure state is a point on the Bloch sphere. The previous lesson placed the magic state $|T\rangle$ on the equator at $45^\circ$ between $+X$ and $+Y$ . To say precisely why that point is "magic," we contrast it with the stabilizer states — the states the Clifford machine can reach for free.

Stabilizer states

An $n$ -qubit pure state $|\psi\rangle$ is a stabilizer state if it is the unique simultaneous $+1$ eigenstate of a maximal abelian subgroup (a stabilizer group) of the Pauli group $\mathcal{P}_n$ . Equivalently, $|\psi\rangle$ is a stabilizer state iff it can be produced from $|0\rangle^{\otimes n}$ by a Clifford circuit. For one qubit there are exactly six of them — the eigenstates of the three Pauli operators:

|0\rangle,\ |1\rangle\ (\pm Z);\qquad |+\rangle,\ |-\rangle\ (\pm X);\qquad |{+}i\rangle,\ |{-}i\rangle\ (\pm Y).

These six points are the vertices of an octahedron inscribed in the Bloch sphere. They sit on the three coordinate axes; the magic state, sitting on the equator at $45^\circ$ , sits maximally far from all of them.

The stabilizer polytope and mixed states

Allowing classical mixtures (and ancillas, measurements, and Clifford unitaries — the full set of "stabilizer operations") fills in the convex hull of the pure stabilizer states. For one qubit this hull is the solid octahedron $\mathrm{STAB}_1$ with the six axis points as vertices. Every state inside the octahedron can be prepared and manipulated entirely within stabilizer operations and is therefore classically simulable.[^gottesman-knill] States outside the octahedron — but still inside the Bloch ball — are exactly the magic states: they have positive distance to $\mathrm{STAB}_1$ and cannot be reached by stabilizer operations alone.

Why the distinction is the whole game

The Gottesman–Knill theorem says a circuit composed of (i) stabilizer-state inputs, (ii) Clifford gates, and (iii) Pauli measurements is efficiently classically simulable in the Heisenberg picture by tracking how the stabilizer generators evolve.[^gottesman-knill] No such circuit can be more powerful than a classical computer. The only way to break out is to inject a non-stabilizer — i.e. magic — resource. This is why the entire architecture of fault-tolerant universality reduces to producing good magic states:

Stabilizer states: free (Clifford-preparable), classically simulable, inside $\mathrm{STAB}_n$ .
Magic states: not free, the source of all quantum computational advantage, outside $\mathrm{STAB}_n$ .

A resource theory

This dichotomy is formalized as a resource theory of stabilizer computation: stabilizer operations are the "free operations," stabilizer states are the "free states," and magic is the resource that free operations cannot create or increase.[^veitch-resource] A function $M(\rho)$ that is zero on the octahedron and non-increasing under stabilizer operations is a magic monotone — the subject of a later lesson. For now the key conceptual point is the cleanest possible one:

\text{magic} \;=\; \text{the part of a state that lives outside the stabilizer polytope.}

Distillation, which the next lessons build, is nothing but the act of taking many slightly-outside-the-octahedron copies and concentrating their magic into a few copies that sit much further out, much closer to the ideal $|T\rangle$ vertex of magic — all without ever using a non-stabilizer operation.

[^gottesman-knill]: Gottesman, The Heisenberg Representation of Quantum Computers, arXiv:quant-ph/9807006. [^veitch-resource]: Veitch et al., The resource theory of stabilizer quantum computation, arXiv:1307.7171.