Checkpoint: Detect an Error

This is a cumulative checkpoint for Module 14. You will assemble the complete three-qubit bit-flip detection circuit — encoding, error injection, and syndrome measurement — in a single exercise.

What you need to build

The three-qubit bit-flip code stores a logical qubit across three physical qubits $q_0, q_1, q_2$ . Logical $|0\rangle$ encodes as $|000\rangle$ ; logical $|1\rangle$ encodes as $|111\rangle$ . Two extra syndrome ancilla qubits, $q_3$ and $q_4$ , will hold the parity check results.

Step 1 — Encoding

With $q_0 = |0\rangle$ and $q_1 = q_2 = |0\rangle$ , the encoding circuit is two CNOTs:

|0\rangle|00\rangle \xrightarrow{\text{CNOT}_{0\to1}} |0\rangle|00\rangle \xrightarrow{\text{CNOT}_{0\to2}} |000\rangle.

Because the data qubit starts in $|0\rangle$ both ancillas remain $|0\rangle$ . (Had the data been $|1\rangle$ , the CNOTs would have spread it to $|111\rangle$ .)

Step 2 — Injected error

An adversarial bit flip $X$ strikes $q_1$ , turning the data from $|000\rangle$ to $|010\rangle$ . The circuit stub already includes this line — do not remove it.

Step 3 — Syndrome measurement

Two parity checks identify the faulty qubit without revealing the logical value:

| Syndrome bit | Parity checked | Circuit | |---|---|---| | $s_0$ into $q_3$ | $q_0 \oplus q_1$ | CNOT $q_0 \to q_3$ , then CNOT $q_1 \to q_3$ | | $s_1$ into $q_4$ | $q_1 \oplus q_2$ | CNOT $q_1 \to q_4$ , then CNOT $q_2 \to q_4$ |

Each target qubit accumulates the XOR of its control qubits because each CNOT flips the target once iff the control is $|1\rangle$ . After two CNOTs, target $= c_1 \oplus c_2$ — the standard fanout trick for parity extraction.

With the injected error ( $q_1 = 1$ , $q_0 = q_2 = 0$ ):

s_0 = q_0 \oplus q_1 = 0 \oplus 1 = 1, \qquad s_1 = q_1 \oplus q_2 = 1 \oplus 0 = 1.

Syndrome $11$ is the unique signature of a bit flip on $q_1$ . A flip on $q_0$ would give $10$ ; a flip on $q_2$ would give $01$ ; no error gives $00$ .

Expected final state

After the correct circuit the five-qubit statevector has a single non-zero amplitude: the computational-basis state

|q_0\, q_1\, q_2\, q_3\, q_4\rangle = |0\;1\;0\;1\;1\rangle.

Data qubits hold $|010\rangle$ (the corrupted codeword), syndrome qubits hold $|11\rangle$ (flagging qubit 1 as the error site). The grader checks the full statevector.

Try it

Complete the three missing sections — encoding, syndrome for $s_0$ , and syndrome for $s_1$ . The error injection on line c.x(1) is already in place.

Run your code to see the quantum state.

Once your circuit passes, inspect the Statevector panel: only the $|01011\rangle$ basis state should carry amplitude, confirming that the syndrome bits correctly flag the injected error.