Ancilla qubit error spreading to multiple data qubits?

Question

In this paper, the author considers a four qubit stabilizer circuit shown below. Note that the first four qubits from the top are data qubits and can be in any state (here we just put them in $\vert 0000\rangle$). The last qubit is the ancilla qubit used for the stabilizer measurement.

The claim is that an error on the ancilla qubit can "propagate to multiple data qubits". What does that exactly mean?

Concretely, suppose I have an $X$ error after the Hadamard on the ancilla qubit. Then, it appears to do nothing since the state is already in the $\vert +\rangle$ state. If I had a $Z$ error then, the ancilla gets flipped to $\vert -\rangle$ and then we apply the stabilizer measurment. How do I see this as an error on the data qubits?

Answer 1

Consider the following circuit.

Let the $|\psi\rangle$ state be some arbitrary state

$$|\psi\rangle = a|0\rangle + b|1\rangle$$

If there are no errors, then you can compute

$$|\psi\rangle |+\rangle \to |\psi\rangle |+\rangle$$ $$|\psi\rangle |-\rangle \to Z|\psi\rangle |-\rangle$$

Thus, when the target qubit is ∣+⟩, the control qubit is unchanged.

However, now suppose a phase-flip error happens to the target qubit right before the CNOT gate, when the target qubit is $∣+⟩$.

Since this flips the target qubit to $∣−⟩$, this means that the control qubit state becomes $Z|\psi⟩$, whereas it should have been $∣ψ⟩$:

$$|\psi\rangle |+\rangle \xrightarrow{Z_2 \text{error}} |\psi\rangle |-\rangle \xrightarrow{\text{CNOT}} Z|\psi\rangle |-\rangle \equiv (Z|\psi\rangle)(Z |+\rangle)$$

whereas, without the $Z_2$ error, state would've been just $|\psi\rangle |+\rangle\,.$ So you can see here that a single error on the target qubit became two errors.