 Photon correlation loss and background model

Independent loss alone leaves normalized factorial moments invariant: if every input photon is retained with probability $\eta$, binomial thinning conditional on a total gives $⟨K⟩=\eta \mu$ and $⟨K(K-1)⟩={\eta }^2⟨N(N-1)⟩$. Both powers cancel from $g^{(2)}(0)$. Fewer coincidences still mean poorer statistical precision.

Now let each channel contain a signal rate $s_i$ and an independent Poisson process background rate $b_i$, independent also of the signal. Put ${\rho }_i=s_i/(s_i+b_i)$. The pair density is $s_1{s}_2{g}_{\mathrm{sig}}^{(2)}+s_1{b}_2+b_1{s}_2+b_1{b}_2$. Dividing by $(s_1+b_1)(s_2+b_2)$ proves the displayed formula, for the same bin and timing response. For chosen equal signal fractions ${\rho }_1={\rho }_2=4/5$, ideal signal values $0,1,2$ become $9/25,1,41/25$. Background pulls both dips and peaks toward one.

This calculation assumes additive, mutually independent background counts. Scattered coherent light that interferes with the signal, correlated electronic noise, or recovery effects require a different model. Correcting a dip without independently estimating the backgrounds cannot establish photon antibunching.