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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

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  • $\begingroup$ 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 '20 at 19:37
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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).

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