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The group of unitary Clifford gates, defined by normalization of the Pauli group. For a single qubit, $H$ and $S$ generate its usual finite gate group, and modulo global phase its action permutes the six Pauli eigenstates. On multiple qubits, one adds an entangling controlled-NOT gate. This quantum-computing use of the name is distinct from groups attached to a general Clifford algebra.

For $|\pm i⟩=(|0⟩\pm i|1⟩)/\sqrt{2}$, direct matrix multiplication by the Hadamard gate gives coefficients $(1\pm i,1\mp i)/2$, proving the displayed identities. This exchanges the two Pauli eigenstates of the Pauli Y gate and agrees with $HYH=-Y$. The phases in the formula are global phases for each separate state; the relative phase inside each state is essential.

Choose $|{\psi }_0⟩=|+⟩$ and its orthogonal companion $|{\psi }_1⟩=|-⟩$. Since $X|+⟩=|+⟩$, either branch of RandomFlip returns the same pure state. Alice therefore obtains label 0 with certainty, andThe alternative choice $|-⟩$ also works: its sign change under $X$ is only a global phase. The useful property is being a Pauli eigenstate of the Pauli X gate.

Directly applying the Hadamard gate giveswhileHence the two states are interchanged up to the displayed global phases. Their relative factors $-i$ and $+i$ are retained. They are the two Pauli eigenstates of the Pauli Y gate, and the Hadamard interchange of Y eigenstates is consistent with $HYH=-Y$.

The six Pauli eigenstates have Bloch vectors $\pm {\mathbf{e}}_x$, $\pm {\mathbf{e}}_y$, $\pm {\mathbf{e}}_z$:Classical convex combinations of density operators produce the corresponding convex combinations of their Bloch vectors. Their convex hull is the stabilizer octahedron:The six vertices are pure Pauli states; the faces and interior represent their mixtures. To verify the reverse inclusion, give the vertex in the sign of each coordinate weight $|s_j|$ and use the leftover weight at the origin, itself the equal mixture of either opposite pair.

Direct matrix multiplication givesThus each generating Clifford gate conjugates every Pauli operator to another Pauli operator up to sign. Repeated conjugation proves the same for every product of generators and their inverses, hence for the Clifford group. For Hermitian Pauli observables the possible phases here are signs; arbitrary Pauli-group elements also include their conventional phases.

If $P|\psi ⟩=\lambda |\psi ⟩$, then $(CP{C}^{†})C|\psi ⟩=\lambda C|\psi ⟩$. Therefore $C|\psi ⟩$ is another Pauli eigenstate. Every single-qubit Clifford circuit consequently permutes the six Pauli vertices of the stabilizer octahedron.

The convex hull of the six Pauli eigenstates in the Bloch ball. A mixture has a Bloch vector obeying the displayed inequality. Conversely, assign weights $|s_x|,|s_y|,|s_z|$ to the corresponding signed vertices and distribute the leftover weight over any opposite pair; this realizes every vector satisfying the inequality. Single-qubit Clifford gates permute its vertices, while a T gate can take a vertex to a pure state outside it.