Magic-State Injection

We now cash in the central promise of the module: given one copy of the magic state $|T\rangle = T|+\rangle$ and only Clifford operations, apply a $T$ gate to an arbitrary data qubit. The technique is gate teleportation, and the specific instance is $T$ -gate injection.[^gottesman-chuang]

The injection circuit

Let the data qubit be $|\psi\rangle = a|0\rangle + b|1\rangle$ on wire $0$ , and prepare the ancilla on wire $1$ in the magic state

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

Apply a single $\mathrm{CNOT}$ with the data as control and the ancilla as target, then measure the ancilla in the computational ( $Z$ ) basis. Writing the joint state and applying $\mathrm{CNOT}_{0\to1}$ ,

|\psi\rangle|A\rangle \xrightarrow{\ \mathrm{CNOT}_{0\to1}\ } \tfrac{1}{\sqrt2}\Bigl[\bigl(a|0\rangle + b\,e^{i\pi/4}|1\rangle\bigr)|0\rangle_a + \bigl(a\,e^{i\pi/4}|0\rangle + b|1\rangle\bigr)|1\rangle_a\Bigr].

Reading off the two ancilla branches:

Ancilla measures $0$ (probability $\tfrac12$ ): the data collapses to $a|0\rangle + b\,e^{i\pi/4}|1\rangle = T|\psi\rangle$ . Done — no correction.
Ancilla measures $1$ (probability $\tfrac12$ ): the data collapses to $a\,e^{i\pi/4}|0\rangle + b|1\rangle \;\propto\; a|0\rangle + b\,e^{-i\pi/4}|1\rangle = T^\dagger|\psi\rangle$ .

The Clifford correction

In the outcome- $1$ branch we have $T^\dagger|\psi\rangle$ instead of $T|\psi\rangle$ . We repair it with a single $S$ gate, which is Clifford:

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

So the recipe is: if the ancilla read $1$ , apply $S$ to the data. Either way the data ends up in exactly $T|\psi\rangle$ , and every operation we performed — $\mathrm{CNOT}$ , $Z$ -measurement, and a classically conditioned $S$ — is Clifford. The lone non-Clifford ingredient was the state $|A\rangle$ , prepared offline.

Why the correction must be classically conditioned

The byproduct $S$ is not optional decoration — without it, the outcome- $1$ branch applies $T^\dagger$ , the inverse of what we wanted, and the computation is corrupted. The measurement result must be fed forward to decide whether to apply $S$ . This feed-forward of measurement outcomes is the operational heart of fault-tolerant non-Clifford gates, and it is why a fault-tolerant machine needs fast classical control.

Try it

Verify the protocol from first principles. For a real test input $|\psi\rangle = \cos(0.4)|0\rangle + \sin(0.4)|1\rangle$ , compute the corrected data state in both measurement branches (apply $S$ in the outcome- $1$ branch), and return the worst-case fidelity $F = |\langle T\psi | \text{corrected}\rangle|^2$ . A correct protocol gives $F = 1$ on both branches.

[^zhou-leung-chuang]: Zhou, Leung, Chuang, Methodology for quantum logic gate construction, arXiv:quant-ph/0002039.

Run your code to see the quantum state.