The T Gate and Magic

The $T$ gate, also called the $\pi/8$ gate, is the diagonal single-qubit unitary

T = \begin{pmatrix} 1 & 0 \\ 0 & e^{i\pi/4} \end{pmatrix} = e^{i\pi/8}\begin{pmatrix} e^{-i\pi/8} & 0 \\ 0 & e^{i\pi/8} \end{pmatrix}.

The second form, factoring out the global phase $e^{i\pi/8}$ , explains the historical name: $T$ is a rotation by $\pi/4$ about the $z$ -axis of the Bloch sphere, i.e. $T = e^{i\pi/8} R_z(\pi/4)$ where $R_z(\theta) = e^{-i\theta Z/2}$ . Note that $T^2 = S = \operatorname{diag}(1, i)$ , the phase gate, and $S^2 = Z$ . So $T$ is a "square root of $S$ " and a "fourth root of $Z$ ."

Where T sits in the Clifford hierarchy

The Clifford hierarchy organizes gates by how they conjugate Paulis. Level $\mathcal{C}^{(1)}$ is the Pauli group itself; level $\mathcal{C}^{(k)}$ consists of unitaries $U$ such that $U P U^\dagger \in \mathcal{C}^{(k-1)}$ for every Pauli $P$ . The Clifford group is $\mathcal{C}^{(2)}$ . The $T$ gate lives at level three:

T X T^\dagger = \frac{1}{\sqrt2}\,(X + Y), \qquad T Z T^\dagger = Z .

Conjugating $X$ by $T$ produces a Clifford (here $\tfrac{1}{\sqrt2}(X+Y)$ , which is in $\mathcal{C}^{(2)}$ ), not a Pauli — so $T \in \mathcal{C}^{(3)} \setminus \mathcal{C}^{(2)}$ . This is the formal sense in which $T$ is the first "non-Clifford" gate, and why $\{H, S, \mathrm{CNOT}, T\}$ achieves universality that $\{H, S, \mathrm{CNOT}\}$ cannot.

The associated magic state

To each gate in $\mathcal{C}^{(3)}$ we can associate a resource state by applying it to a stabilizer state. For $T$ the natural choice is

|T\rangle \;\equiv\; T|+\rangle \;=\; \frac{1}{\sqrt2}\bigl(|0\rangle + e^{i\pi/4}|1\rangle\bigr).

This is the $T$ -type magic state. On the Bloch sphere it points along the direction

\hat{n} = \bigl(\cos\tfrac{\pi}{4},\, \sin\tfrac{\pi}{4},\, 0\bigr) = \Bigl(\tfrac{1}{\sqrt2},\, \tfrac{1}{\sqrt2},\, 0\Bigr),

i.e. it lies in the equatorial $xy$ -plane, bisecting the $+X$ and $+Y$ axes. It is a pure state on the Bloch sphere — yet, as the next lessons show, no Clifford circuit can prepare it from $|0\rangle$ . That is exactly its value.

Why this matters for fault tolerance

We saw in the previous lesson that $T$ cannot generally be made transversal. But preparing the fixed state $|T\rangle$ is a state-preparation task that can be done offline and then cleaned up by distillation. Gate teleportation (next lessons) lets us spend one $|T\rangle$ to enact one logical $T$ . So the entire non-Clifford content of a universal fault-tolerant computation is bottled up in copies of this one single-qubit state.[^gottesman-chuang]

Try it

Prepare the magic state $|T\rangle = T|+\rangle$ on a single qubit, starting from $|0\rangle$ . The grader checks your full state vector against the reference, so getting the relative phase $e^{i\pi/4}$ right is essential.

[^gottesman-chuang]: Gottesman, Chuang, Quantum Teleportation is a Universal Computational Primitive, arXiv:quant-ph/9908010.

Run your code to see the quantum state.