Simulating MBQC Gates

This lesson turns the elementary gadget into something you can run. We simulate the one-bit teleportation circuit on a qubit simulator and watch a single-qubit gate appear on the ancilla as a consequence of measuring the input — the smallest complete MBQC computation.

The gadget as a circuit

The measurement-based Hadamard is a three-line circuit on two qubits:

Input $|\psi\rangle$ sits on qubit $0$ .
Ancilla qubit $1$ is prepared in $|+\rangle$ by a Hadamard on $|0\rangle$ .
Entangle the edge with a controlled- $Z$ , $CZ_{01}$ — this is precisely the two-qubit cluster state construction.
Measure qubit $0$ in the $X$ basis.

A state-vector simulator measures only in the computational basis, so we realise the $X$ -basis measurement the standard way: apply a Hadamard to qubit $0$ and then read it in the computational basis, since $H$ maps the $X$ basis $\{|+\rangle, |-\rangle\}$ onto $\{|0\rangle, |1\rangle\}$ .

Tracing the simulated state

Before any measurement, after the entangling step and the rotating Hadamard on qubit $0$ , the joint state is (from the gadget identity of the previous lessons, with the final $H_0$ applied):

\tfrac{1}{\sqrt2}\Big(|0\rangle_0 \otimes H|\psi\rangle \;+\; |1\rangle_0 \otimes XH|\psi\rangle\Big).

Reading qubit $0$ now gives outcome $s$ , and qubit $1$ collapses to $X^{s} H|\psi\rangle$ — the Hadamard, up to the byproduct $X^{s}$ .

Choosing a deterministic instance

To get a clean, outcome-independent check without post-selecting on a measurement, choose the input to be an eigenstate that the byproduct fixes. Take $|\psi\rangle = |0\rangle$ . Then

H|0\rangle = |+\rangle, \qquad X|+\rangle = |+\rangle,

so both branches leave qubit $1$ in $|+\rangle$ . The two outcomes are physically identical on the ancilla; tracing out the (still-entangled) measured qubit, the reduced state of qubit $1$ is the pure state $|+\rangle$ , with reduced Bloch vector

(x, y, z) = (1, 0, 0).

That is the signature of a successful Hadamard: the teleported logical qubit has been mapped from $|0\rangle$ to $|+\rangle$ .

Why this is the whole story in miniature

Nothing in this circuit ever applied a gate to qubit $1$ . The Hadamard on the logical information arose entirely from entangling and then measuring qubit $0$ . Replace the plain $X$ -basis measurement with a tilted one at angle $\xi$ and the same circuit realises $H R_z(\xi)$ instead; chain several edges and the gadgets compose into an arbitrary single-qubit unitary; add a cross-edge between two such chains and you get an entangling gate. Everything in the one-way model is this gadget, repeated and stitched together.

Try it

Build the Hadamard gadget for input $|0\rangle$ : prepare the ancilla qubit $1$ in $|+\rangle$ , entangle the edge with $CZ_{01}$ , and rotate qubit $0$ into the $X$ -measurement basis with a Hadamard. The grader reads the reduced Bloch vector of qubit $1$ , which must be $(1, 0, 0)$ — the teleported state $H|0\rangle = |+\rangle$ .

Run your code to see the quantum state.