# Qubit State (theta,phi) Dependence on the Readout Error

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.

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 $$\sin^2\theta/2$$.
• Thanks for your answer! I don't completely understand yet where the $\sin^2 \theta/2 scaling comes from, could you elaborate a bit more on that? Mar 21 at 11:19 • sure, there's nothing smart about it - all I'm saying is that the error rate is$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\$.