Many paper about Randomized benchmarking will be related to state preparation and measurement errors (SPAM), I am reading the paper "Efficient learning of quantum noise", https://arxiv.org/pdf/1907.13022.pdf, it says state preparation and measurement errors can bring systematic bias, so the state preparation and measurement errors refer to what?


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State preparations errors arise because there's some non-ideality of your quantum computer such that your qubits don't always start in the $|0\rangle$ state, and measurement errors mean that when you go to read out your qubit, there's some chance you'll get the wrong result (e.g. measuring $|1\rangle$ when the qubit is really $|0\rangle$).

SPAM errors are usually neglected in randomized benchmarking because their effect is constant vs circuit depth. For example, say the platform has a 95% gate fidelity and an 90% SPAM error rate. Then if I want to do benchmarking with circuit depth $N$, I'll have a total error rate of $0.9 \times (0.95)^N$, where the constant 0.9 at the front is the systemic bias from the SPAM errors.

  • $\begingroup$ The effect of SPAM errors is constant vs circuit depth, but what is the reason they can be neglected in randomized benchmarking? Can you make an addition? Thanks a lot! $\endgroup$
    – Mengr
    Aug 11, 2022 at 11:29
  • $\begingroup$ You get rid of SPAM by varying N. If you take data at multiple values of N, this gives you enough equations that you can solve for the unknown SPAM (.9) and gate errors (.95) separately. $\endgroup$ Aug 11, 2022 at 16:03
  • $\begingroup$ Yeah, just to clarify Jahan's response, the gate fidelity that comes from randomized benchmarking is a better metric for understanding how well your platform will scale with more complicated algorithms. If you say that neglecting SPAM errors, your 100 step algorithm has 90% fidelity, then you know a more complicated 200 step algorithm will have 81% and a 300 step one will have 73%. Of course, you'll actually see all of these hit by the SPAM error rate, but that's just a given from using that system and not because of the depth of your algorithm. Does that answer your question? $\endgroup$
    – Chris E
    Aug 11, 2022 at 20:47

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