Like we all probably know, today’s NISQ computers are, as their name implies - very noisy. Hence, if we desire to obtain valuable results then we should come up with circuit designs that minimizes the expected errors to an accepted level we can handle with.

It seems to me that it’s possible to compute an expected error rate for a specific circuit design given some characterization data (T1, T2, gate error rates, measurement error rates..). But I haven't saw any practical example of this so far.

The question is - are there any systematic tools to asses an expected overall error rate when running a specific circuit on a specific quantum computer?


  • $\begingroup$ To a first order isn’t the overall fidelity of a circuit roughly the fidelity of each non-Clifford gate to the power of the depth of non-Clifford gates? $\endgroup$ Commented Oct 30, 2022 at 15:45
  • $\begingroup$ @Mark S Do you have any source for that? Anyway, what I imagine that with respect to a specific quantum hardware specification / noise model, we will have a method to know what is the meximum circuit depth and CNOT count for some error rate (chose depth and CNOTs as benchmarks for noise while other factors are relatively negligible to my understanding, though I might be wrong). $\endgroup$
    – Ohad
    Commented Oct 30, 2022 at 21:25
  • $\begingroup$ I’m not super sure but Google’s landmark Sycamore paper made much of the “straight-line fidelity” showing that as depth increases, fidelity decreases loglinearly, which they took to mean that Sycamore wasn’t picking up any strange correlations at larger depth… $\endgroup$ Commented Oct 30, 2022 at 21:53
  • $\begingroup$ @Mark S - I was digging into the Sycamore supremacy experiment lately, and it seems like indeed Google have estimated the fidelty of a circuit by multiplying all gate-fidelties of the gates used in the circuit. But I was wondering - isn't it too sloppy? For example - how does it consider relaxtion of idle qubits? $\endgroup$
    – Ohad
    Commented Feb 14, 2023 at 9:24
  • $\begingroup$ I don’t know! Sorry. But didn’t Google’s circuit activate every qubit (or every other qubit) at each step? That is - I don’t think there was a lot of idling going on, or if there was, it was split evenly amongst the qubits. Or maybe it only used a small portion of the fidelity budget… not sure! $\endgroup$ Commented Feb 14, 2023 at 12:59


Your Answer

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