Do the readout errors on the publicly available IBM quantum computers have any dependency on the state being measured? That is, if we are measuring a qubit in the state

$\cos({\theta/2}) |0> + e^{i\phi} \sin({\theta/2}) |1>$ ,

does the error rate have any $\theta$ or $\phi$ dependence? If so, what does the dependence look like?

Thanks in advance for any replies.


1 Answer 1


the answer is yes - in general $p(m=0|s=1)> p(m=1|s=0)$, where $s$ is the state at the beginning of the measurement, and $m$ is the measurement outcome.

This is because the qubit has a probability to decay from 1 to zero during the measurement which is much higher than the probability of a false excitation from 0 to 1, by the Boltzmann factor $\exp({\hbar\omega_{01}/k_BT})$. So the dependence is on $\theta$, and it scales approximately as $\sin^2\theta/2$.

This is disucssed explicitly, for example, in fig. 2 here.

  • $\begingroup$ Thanks for your answer! I don't completely understand yet where the $\sin^2 \theta/2 scaling comes from, could you elaborate a bit more on that? $\endgroup$
    – user206444
    Mar 21, 2022 at 11:19
  • 1
    $\begingroup$ sure, there's nothing smart about it - all I'm saying is that the error rate is $p_e = p(0|1)p(1) + p(1|0)p(0)$, and since the $p(0|1)$ and $p(1|0)$ are system parameters, and $p(0|1)$ is dominant, $p_e \simeq p(0|1)p(1) = p(0|1)\sin^2\theta/2$. $\endgroup$
    – Lior
    Mar 21, 2022 at 13:04
  • $\begingroup$ Ah, I see thank you. $\endgroup$
    – user206444
    Mar 21, 2022 at 13:57

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