= Sequential Bayesian photon counting
{title2=$p_k(n)=p_{k-1}(n)L_{s_k}(n;\phi_k)/Z_k$}
Consider an original balanced-readout model, rather than a fit to the experiment. Let $n\in\{0,\ldots,N_{\max}\}$, with a normalized <prior probability> $p_0(n)\geq0$ and $\sum_n p_0(n)=1$. A detected <chemical atom> has outcome $s\in\{e,g\}$. Define the <conditional probabilities>
$$
L_e(n;\phi)=P(e\mid n,\phi)=\frac{1+C\cos(\phi+n\chi)}2,\qquad
L_g(n;\phi)=1-L_e(n;\phi),\qquad 0\leq C\leq1.
$$
Here $\chi$ is the accumulated <quantum phase> per photon, $\phi$ the controlled analysis <quantum phase>, and $C$ the fringe contrast. The two <probabilities> sum to one for every $n$. Actual calibration may also require an offset and asymmetric detection errors; $C$ alone does not specify a <quantum instrument>.
Assume no <photon number> change during the record and fresh independently prepared atomic meters. For observed outcomes $s_1,\ldots,s_k$ at recorded settings $\phi_1,\ldots,\phi_k$, the <Born rule> and <Bayes' theorem> give
$$
Z_k=\sum_{m=0}^{N_{\max}}p_{k-1}(m)L_{s_k}(m;\phi_k),\qquad
p_k(n)=\frac{p_{k-1}(n)L_{s_k}(n;\phi_k)}{Z_k}
=\frac{p_0(n)\prod_{j=1}^k L_{s_j}(n;\phi_j)}{\sum_{m=0}^{N_{\max}}p_0(m)\prod_{j=1}^k L_{s_j}(m;\phi_j)}.
$$
The <posterior probability> is defined only for $Z_k>0$. Summing its numerator over $n$ proves normalization. This is also the update of diagonal elements of a field <density operator> under a <quantum measurement>; it does not assume that an initial <quantum superposition> was an unknown pre-existing integer. The observed settings are part of the record, especially when selected adaptively.
For an exact finite example take $p_0=(1,1,1,1)/4$, $C=4/5$, and $\chi=\pi/2$. An $e$ outcome at $\phi=0$ has <likelihood> $(9,5,1,5)/10$, giving $p_1=(9,5,1,5)/20$. Another $e$ at $\phi=-\pi/2$ gives $p_2=(9,9,1,1)/20$. A subsequent $g$ at $\phi=0$ gives
$$
p_3=\frac1{68}(9,45,9,5),\qquad P(s_1=e,s_2=e,s_3=g)=\frac{17}{200}.
$$
Thus three atomic records raise the <posterior probability> of $n=1$ from $1/4$ to $45/68$, while leaving finite uncertainty. No one of these outcomes identifies $n$ with certainty.
\Image[/sequential-photon-number-posterior.png]
{title=Synthetic atomic records progressively concentrate a photon-number posterior}
{description=Original simulation, not laboratory data: $n=0,\ldots,7$, uniform prior, $C=0.8$, $\chi=\pi/4$, four cyclic analysis phases, and a fixed simulated $n=3$. Cavity loss is omitted. Individual updates can broaden the distribution even though this record eventually concentrates.}
The periodic <likelihood function> imposes an identifiability limit. If $\chi=2\pi/q$ for an integer $q>0$, then $L_s(n+q;\phi)=L_s(n;\phi)$ for every setting: the <posterior odds> between $n$ and $n+q$ never change. Varying $\phi$ can resolve a cosine reflection ambiguity within a period, but cannot remove this exact period. Changing $\chi$ can supply additional information. A finite cutoff is therefore a physical assumption to justify, not evidence that excluded high <photon numbers> were absent. At $C=0$ no outcome changes the <prior probability>; at $0<C<1$ neither binary outcome has an exactly zero <likelihood>.
The no-loss assumption also has a time limit. In an original zero-temperature loss model, let $\kappa=1/T_c$. A <Fock state> $|n\rangle$ loses its first photon at rate $n\kappa$, so its survival <probability> is $\exp(-n\kappa t)$ and its mean lifetime is $T_c/n$ for $n>0$. Several records must be acquired within that lifetime to identify a plateau. When losses matter, first predict $p_k^-(n)=\sum_m T_{nm}(\Delta t)p_{k-1}(m)$ using a normalized transition <matrix>, then apply the atomic <Bayesian inference> update to $p_k^-$. Thermal excitation requires upward transitions too. A delayed or broadened inference after a jump need not indicate that an atom caused the jump.
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