Transversal Gates

A transversal gate is the cleanest fault-tolerant construction: a logical operation built as a tensor product of single-qubit physical gates, one per qubit of the code block (or one per corresponding pair of qubits across two blocks). Because every physical gate touches only one qubit of any given block, a single faulty gate can corrupt at most one qubit of that block — exactly the propagation property fault tolerance demands.

Definition

Let a code use $n$ physical qubits per logical qubit. A single-block logical gate $\bar{U}$ is transversal if it can be implemented as

\bar{U} = u_0 \otimes u_1 \otimes \cdots \otimes u_{n-1},

where each $u_k$ acts on physical qubit $k$ alone. A two-block gate (such as a logical CNOT) is transversal if it is a tensor product of two-qubit gates, the $k$ -th one coupling qubit $k$ of block A to qubit $k$ of block B — never two qubits of the same block.

Contrast this with a non-transversal gate that, say, applies a Toffoli among three qubits of the same block. One faulty Toffoli could place errors on two qubits of that block at once, overwhelming a distance-3 code.

Transversal gates of the 3-qubit bit-flip code

The 3-qubit repetition (bit-flip) code encodes

|0_L\rangle = |000\rangle, \qquad |1_L\rangle = |111\rangle .

Two of its logical gates are transversal:

Logical $X$ is $X \otimes X \otimes X$ : it flips $|000\rangle \leftrightarrow |111\rangle$ , i.e. $|0_L\rangle \leftrightarrow |1_L\rangle$ .
Logical $Z$ is $Z \otimes Z \otimes Z$ : it acts as $+1$ on $|000\rangle$ and $(-1)^3 = -1$ on $|111\rangle$ , i.e. $\bar{Z}|0_L\rangle = |0_L\rangle$ , $\bar{Z}|1_L\rangle = -|1_L\rangle$ .

You can check that $\bar{Z}$ also equals $Z$ on any single qubit (since $Z \otimes Z \otimes Z$ and $Z \otimes I \otimes I$ agree on the code space), but the symmetric three-qubit form is the transversal one and is what fault tolerance prescribes.

The Eastin–Knill obstruction

Why not just make every gate transversal and be done? The Eastin–Knill theorem proves that no quantum code with the ability to detect arbitrary single-qubit errors admits a universal set of transversal logical gates. Some gate in any universal set must be implemented by a non-transversal (and more expensive) construction. This is the deep reason the rest of the module exists: the Clifford gates are often transversal, but the $T$ gate that completes a universal set is not, and must be supplied by magic-state injection.

Try it

The starter code prepares the logical $|+_L\rangle = (|000\rangle + |111\rangle)/\sqrt{2}$ . Apply the transversal logical $Z$ (a $Z$ on each of the three qubits). The grader checks the resulting state vector, which should be $|-_L\rangle = (|000\rangle - |111\rangle)/\sqrt{2}$ — amplitude $+1/\sqrt2$ on index 0 and $-1/\sqrt2$ on index 7.

Run your code to see the quantum state.