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.

enter image description here

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?


1 Answer 1


Consider the following circuit.

enter image description here

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 $∣+⟩$.

enter image description here

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.

  • 1
    $\begingroup$ Note: OP previously mentioned (which he edited out now), "Can someone show an example (either of this circuit above or any other) where the error in the ancilla causes errors to propagate to the data qubits?" and hence this answer. $\endgroup$
    – FDGod
    Feb 28 at 0:28
  • $\begingroup$ Thanks, you were very quick with the answer, sorry I was still editing the question but your example helped a lot! Just one thing - In the circuit in the question, the ancilla is the control qubit and the data qubit is the target qubit which is the opposite of your example. I see that an error on the target spreads to the control in your circuit. Is it also the case that an error on the control qubit spreads to the target qubit (which I think the author of the paper claimed)? $\endgroup$ Feb 28 at 0:43
  • $\begingroup$ Yes, it would. To see this spread of errors, it is sometimes more convenient to think about errors evolving in the Heisenberg picture under the action of these gates, i.e., $\text{CNOT}$ and $CZ$ gates. $\endgroup$
    – FDGod
    Mar 1 at 5:57

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