 Trapped-ion fluorescence count model

For an original illustrative readout model, take detected photon count $K$ in a fixed exposure to have a Poisson distribution, with stipulated means ${\lambda }_B=12$ for a bright state and ${\lambda }_D=0.2$ for a dark state. Classify $K\ge 3$ as bright. Direct summation givesWith equally likely preparations, the average misclassification probability is $0.0008354$. These numbers describe only the chosen count model. Actual fluorescence includes short-time photon correlations, detector effects and possible state changes; it need not follow a fixed-rate Poisson process. Longer exposure can improve separation of count distributions while increasing unwanted optical pumping, metastable decay or recoil.

If the prepared bright population is $p$, let $\alpha$ and $\beta$ denote the two conditional errors above. The observed bright-label probability is $q=(1-\alpha )p+\beta (1-p)$, hence $p=(q-\beta )/(1-\alpha -\beta )$ when the denominator is nonzero. Known-state controls must estimate $\alpha ,\beta$ in the actual apparatus; the formula cannot correct unknown drift. For $M$ independent preparations, the raw bright fraction has variance $q(1-q)/M$. Counting uncertainty, preparation error, pulse error and state-detection bias therefore enter different parts of the inference. Repeating a biased measurement reduces its sampling uncertainty without removing its bias.