# Noise model correspondence to qiskit implementation for $CNOT$ gate

My question especially relates to the CNOT because I'd like to carefully understand how Qiskit simulate a CNOT model.

Specifically, from my understanding, two independent qubits are affected by a noise well modeled by i.i.d. pauli operators. This can be implemented with the pauli_error model to each qubit.

When it comes to the $$CNOT$$ I don't know if I can stil model it by tensoring the two wires, e.g.:

error = pauli_error([('I', 1 - 3*p), ('X', p), ('Y', p), ('Z', p)])
error = error.tensor(error)


Or it is more appropriate to model it with the depolarizing_error method as follows:

error = depolarizing_error(p,2)


If they are equivalent, the first one is better because the simulation runs much faster.

EDIT: To me, a noisy $$CNOT$$ gate can be modeled as a perfect $$CNOT$$ which mixes up i.i.d. operators through the wires -- according to propagation rules --. So at the end of the gate, the noise is mixed, but, it can still be expressed with i.i.d. Pauli operators.

Reading the documentation for depolarizing_error, they are not the same. Under two-qubit depolarizing noise the $$X \otimes X$$ error case is as likely as the $$X \otimes I$$ case, whereas if you instead apply two single-qubit depolarizing noise channels then $$X \otimes X$$ will be much less likely because it requires two simultaneous failures.
You can easily check this for yourself by applying depolarizing noise to the state $$|00\rangle$$ then measuring it and seeing how often you see the result $$11$$. It should be much higher for depolarizing_error(p, 2) than for pauli_error([('I', 1 - 3*p), ('X', p), ('Y', p), ('Z', p)]).tensor(pauli_error([('I', 1 - 3*p), ('X', p), ('Y', p), ('Z', p)])) especially as p becomes small (e.g. less than 1%).
• Perhaps the point is that, due to propagation rules, an $X\otimes X$ is indeed as likely as $X\otimes I$ (with first qubit being the target). Jun 8 at 21:10
• No yes, not all. I meant that specific case. Assuming $X \sim p$, i.i.d. on the two wires. At the end of a $CNOT$ gate, the probability of $X\otimes X$ and $X\otimes I$ are equally likely ($p$), while $I\otimes X$ should be $p^2$. But maybe you meant that depolarizing_error just put all equally likely.. Jun 8 at 21:23
• May I equivalently model the i.i.d by adding an identity noisy gate $I$ before each $CNOT$? Instead of tensoring the error for the $CNOT$ I mean. Jun 8 at 21:33