Clifford group

Remarks

The Clifford group contains all scalar matrices $e^{i\theta} I$, so it is infinite; one often works modulo these phases. Up to phases, it is generated by the Hadamard gate, the phase gate and the controlled-NOT gate: $$H = \frac{1}{\sqrt2}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, \qquad S = \begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix}, \qquad \mathrm{CNOT} = |0\rangle\langle 0| \otimes I + |1\rangle\langle 1| \otimes X.$$ Here $H$ and $S$ act on any one qubit, and CNOT acts on any ordered pair of qubits. For example, $H$ is a Clifford element: $HXH = Z$, $HZH = X$ and $HYH = -Y$, where $X, Y, Z$ are the Pauli matrices. Similarly $SXS^* = Y$ and $SZS^* = Z$.

The $T$ gate $\mathrm{diag}(1, e^{i\pi/4})$ is not a Clifford element, since $TXT^* = (X + Y)/\sqrt2$ is not in $\mathcal{P}_1$.

Conjugation by elements of $\mathcal{C}_n$ induces symplectic maps on the quotient $\mathcal{P}_n / \set{\pm I, \pm iI} \cong \F_2^{2n}$, which carries the symplectic form of the Pauli group. Let $N$ be the subgroup of $\mathcal{C}_n$ generated by $\mathcal{P}_n$ and the phases. This gives an isomorphism of the quotient $\mathcal{C}_n / N$ onto the symplectic group $Sp(2n, \F_2)$. Here $\F_2$ is the field with two elements.

Circuits built from Clifford gates acting on computational basis states can be simulated efficiently on a classical computer. Adding any non-Clifford gate, such as $T$, gives a universal gate set.