Adding the T Gate

The Gottesman–Knill theorem told us that Clifford gates alone yield no speedup; we must add a gate from outside the Clifford group. The standard choice is the $T$ gate, and adding it to $\{H, S, \mathrm{CNOT}\}$ produces a finite set that is universal: it can approximate any unitary to arbitrary precision.

The T gate

$T$ is the diagonal single-qubit gate

T = \begin{pmatrix} 1 & 0 \\ 0 & e^{i\pi/4} \end{pmatrix} = \mathrm{diag}\!\left(1,\, e^{i\pi/4}\right) .

It applies a $\pi/4$ phase to the $|1\rangle$ amplitude (sometimes written $\pi/8$ because of an overall-phase convention, hence the nickname " $\pi/8$ gate"). Its relatives in the same diagonal family are

Z = \mathrm{diag}(1, e^{i\pi}), \qquad S = \mathrm{diag}(1, e^{i\pi/2}), \qquad T = \mathrm{diag}(1, e^{i\pi/4}).

So $T$ is the square root of $S$ and the fourth root of $Z$ :

T^2 = S, \qquad T^4 = Z, \qquad T^8 = I .

The exercise verifies the first of these directly.

Why T is not Clifford

Conjugate $X$ by $T$ :

T X T^\dagger = \frac{X + Y}{\sqrt{2}} ,

which is a $45^\circ$ rotation of $X$ in the $X$ – $Y$ plane — not a Pauli. Because $T$ does not normalize the Pauli group, it escapes the stabilizer formalism, and the Gottesman–Knill simulation no longer applies. This is exactly the property we needed.

Universality of Clifford + T

The set $\{H, S, \mathrm{CNOT}, T\}$ is universal for quantum computation: any $n$ -qubit unitary can be approximated to any precision $\varepsilon$ by a finite circuit drawn from this set. Two ingredients make this work:

$H$ and $T$ generate a single-qubit rotation about an axis whose angle is an irrational multiple of $\pi$ . Repeatedly composing such a rotation densely fills a great circle of the Bloch sphere; combining two such non-commuting rotations densely fills all of $SU(2)$ .
Adding CNOT lifts single-qubit density to density on all of $SU(2^n)$ .

Density alone shows approximability; the Solovay–Kitaev theorem (a later lesson) makes it efficient, guaranteeing a short word in the generators for any target precision.

The price of T

Because the Clifford gates are typically transversal (cheap), but the $T$ gate is not transversal for the standard codes, $T$ is the expensive ingredient. It is supplied fault-tolerantly through magic-state distillation and injection (next lessons), a process that dominates the qubit and time budget of realistic fault-tolerant algorithms. Keep this asymmetry in mind: in Clifford+ $T$ circuits, counting $T$ gates is counting cost.

Try it

Show that two $T$ gates compose into the Clifford phase gate $S$ . Build $T \cdot T$ on one qubit; the grader checks that the resulting unitary equals $S = \mathrm{diag}(1, i)$ .

Run your code to see the quantum state.