2
$\begingroup$

I read an interesting paper of QEC with only two extra qubits.

In the paper, there is a circuit. Note that enter image description here.

Here are some contexts of the paper:

"The stablizer is $XZZXI$. If the second gate fails and introduces an $IZ$ fault, then this fault will propagate through the subsequent gates to become an $IIZXI$ error on the data."

Since the stablizer on the second gate is $Z$, how could it introduce the $IZ$ fault? $IZ$ is equal to $Z$, isn't it? Why the subsequent gate becomes $IIZXI$?

enter image description here

Improve this question
$\endgroup$
1
  • $\begingroup$ I think IZ acts on the two qubits affected by the gate. Then you just need to do the propagation. $\endgroup$
    – M. Stern
    Aug 20, 2020 at 19:37

1 Answer 1

Reset to default
1
$\begingroup$

enter image description here

As you can see in th picture, the (I,Z) error occurs on 2 qubits going into a CNOT. I happens on the control and Z error happens on the target ancilla qubit, and then propagates upwards through the next CNOT, then continues from the ancilla to the following CNOT and becomes an X error on the 4th data qubit because of the Hadamard gate, and finally the 5 data qubits have an error pattern (I on the first qubit, I for the second qubit, Z on the third one, X for the fourth one, and I on the 5th one).

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.