Magic T state

The magic $T$ state is

$|T\rangle = T|+\rangle = \frac{1}{\sqrt2}\big(|0\rangle + e^{i\pi/4}|1\rangle\big), \qquad T = \begin{pmatrix}1 & 0\\ 0 & e^{i\pi/4}\end{pmatrix}.$

Why it is "magic." Every Clifford-stabilized single-qubit state has a Bloch vector pointing along $\pm\hat x, \pm\hat y,$ or $\pm\hat z$. $|T\rangle$'s Bloch vector is $(\cos\tfrac{\pi}{4}, \sin\tfrac{\pi}{4}, 0) = (1/\sqrt2, 1/\sqrt2, 0)$ — exactly bisecting the $X$ and $Y$ axes — so it is stabilized only by the non-Pauli, non-Clifford operator $\cos(\pi/4)X+\sin(\pi/4)Y$. States outside the stabilizer octahedron have positive stabilizer Rényi entropy (a standard magic monotone) and a Wigner representation that goes negative, which is precisely the resource Clifford circuits cannot generate from scratch and cannot amplify: the Gottesman–Knill theorem says Clifford circuits acting on stabilizer states can be simulated efficiently classically, so some non-stabilizer input is required for the circuit to be classically hard, i.e. potentially universal.

Gate teleportation with $|T\rangle$. To apply $T$ to an unknown state $|\psi\rangle$ fault-tolerantly: prepare $|T\rangle$ on an ancilla, apply CNOT with $|\psi\rangle$ as control, measure the ancilla in the $X$ basis, and apply a Clifford correction ($S$ or identity) depending on the outcome. This consumes one copy of $|T\rangle$ and one classical bit per $T$ gate, moving the hard part of fault tolerance from "implement a non-Clifford gate transversally" to "produce clean copies of one specific state."

Distillation. Because physical $|T\rangle$ preparation is noisy, Bravyi and Kitaev's 15-to-1 distillation protocol takes 15 noisy copies with error $\epsilon$ and outputs 1 copy with error $O(\epsilon^3)$, using only Clifford operations plus measurement — the polynomial suppression is what makes large fault-tolerant T-gate counts (needed for e.g. Shor's algorithm or quantum chemistry) practically achievable.