How can one argue that the $S$-gate is Clifford while $T$-gate is not?

Question

How can one argue that $S$-gate is Clifford while $T$-gate is not?

Answer 1

By definition, conjugation by a Clifford gate preserves the Pauli group $G = \langle X, Y, Z\rangle$. It is easy to check that $SXS^\dagger = Y, SYS^\dagger = -X$ and $SZS^\dagger = Z$. Since $G = \langle -X, Y, Z\rangle$ we see that $S$ is Clifford. On the other hand, $TXT^\dagger = \begin{pmatrix} & e^{-i\pi/4}\\ e^{i\pi/4}&\end{pmatrix} \notin G$, so $T$ is not.

Answer 2

Geometrically: For a single qubit, we have the Bloch sphere and the stabiliser states span an octahedron inside it. Unitaries act in the adjoint representation as $SO(3)$, i.e. they induce rotations of the Bloch sphere.

It is easy to see, that only rotations about the $X,Y,Z$ axis with angles $\theta=\pi/2,\pi,3\pi/2,2\pi$ preserve the octahedron. This exactly correspond to unitaries of the form $$ U = \exp(i \theta/2 P), \quad P=X,Y,Z $$ Now, $S$ is a $\theta=\pi/2$ rotation, while $T=\sqrt{S}$ is a $\pi/4$ rotation.

EDIT: Of course, this makes use of the fact that Cliffords are exactly those unitaries which map stabiliser states to stabiliser states!