# Circuit of quantum error correction code

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

In the paper, there is a circuit. Note that .

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 $$IIZXI$$?

• I think IZ acts on the two qubits affected by the gate. Then you just need to do the propagation. Aug 20 '20 at 19:37

## 1 Answer

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).