Physics theoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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Ideal independent loss preserves normalized factorial correlation

Physical scope

A selected mode with a finite positive ideal mean is sent through an explicitly assumed independent, constant-efficiency counting channel with no background or dark counts. The theorem does not model detector dead time, mode dependence, correlated loss, or actual measured data.

Statement

Assume the ideal wavepacket-counting postulate for a selected mode $v \in H_{T}$ with $∥ v ∥ = 1$ , and let $ρ$ be the incoming density operator on $F_{s} (H_{T})$ (a positive self-adjoint trace-class operator with trace one). By that postulate, $P_{n} (v)$ are the mode-count projections and $p_{n} = tr (ρ P_{n} (v))$ are the count probabilities, so $p_{n} \geq 0$ and $\sum_{n = 0}^{\infty} p_{n} = 1$ . Define the incoming mean and second factorial moment by $μ : = \sum_{n = 0}^{\infty} n p_{n}, F_{2} : = \sum_{n = 0}^{\infty} n (n - 1) p_{n},$ and assume $0 < μ < \infty$ and $F_{2} < \infty$ . Define $g_{N}^{(2)} (0) : = F_{2} / μ^{2}$ ; for this input state $ρ$ , the postulate denotes the same statistic by $g_{v}^{(2)} (0)$ .

Let $0 < η \leq 1$ be a dimensionless fixed efficiency and assume no background or dark counts. To specify independent recording, put $b_{η} (0) = 1 - η$ , $b_{η} (1) = η$ , and $Ω : = {(n, x) : n \in N_{0}, x = (x_{1}, \dots, x_{n}) \in {0, 1}^{n}}, A : = P (Ω) .$ This is countable under the explicit encoding $(n, x) \mapsto (n, \sum_{j = 1}^{n} x_{j} 2^{j - 1})$ . Give the space the point masses $P ({(n, x)}) : = p_{n} \prod_{j = 1}^{n} b_{η} (x_{j}),$ where the empty product for $n = 0$ is $1$ . For each fixed $n$ , the product factors define the conditional product law $Q_{n}$ on ${0, 1}^{n}$ ; hence its coordinates are independent Bernoulli variables with success probability $η$ . Since $\sum_{x \in {0, 1}^{n}} Q_{n} ({x}) = (η + (1 - η))^{n} = 1$ , these point masses sum to one and define a probability measure $P$ on $(Ω, A)$ . Define $N (n, x) : = n$ and $K (n, x) : = \sum_{j = 1}^{n} x_{j}$ . For every nonnegative random variable $Y$ on this space, use the expectation $E Y : = \sum_{(n, x) \in Ω} Y (n, x) P ({(n, x)}) .$ The marginal law of $N$ is $(p_{n})$ , and the conditional count law of $K$ given $N = n$ is the binomial law induced by $Q_{n}$ . If $E K > 0$ , define $g_{K}^{(2)} (0) : = E [K (K - 1)] / (E K)^{2}$ . Then $E K = η μ, E [K (K - 1)] = η^{2} F_{2} .$ Consequently the normalized zero-delay factorial correlation is unchanged: $g_{K}^{(2)} (0) = \frac{E [K (K - 1)]}{(E K)^{2}} = \frac{F_{2}}{μ^{2}} = g_{N}^{(2)} (0) = g_{v}^{(2)} (0) .$ This is a model deduction conditional on independent constant-efficiency thinning; it is not a universal property of real detectors.

Statement references

Ideal one-mode intensity correlations separate model states

Tonelli's theorem for nonnegative measurable functions on a sigma-finite product

Steck, Quantum and Atom Optics

Glauber (1963)

Facts & Assumptions

Given: The count distribution $(p_{n})$ of the ideal mode model, its finite moments $μ > 0$ and $F_{2}$ , a fixed efficiency $0 < η \leq 1$ , and conditionally independent Bernoulli recording of each incoming photon.

[A1]

The conditional law on the $n$ th fiber of $Ω$ is $Q_{n} ({x}) = \prod_{j = 1}^{n} b_{η} (x_{j})$ , with $b_{η} (0) = 1 - η$ and $b_{η} (1) = η$ . For $j \leq n$ , the coordinate map $X_{n, j} (x) : = x_{j}$ is therefore Bernoulli with success probability $η$ , and the coordinate maps are mutually independent under $Q_{n}$ . The joint mass is $P ({(n, x)}) = p_{n} Q_{n} ({x})$ . This is the explicitly assumed thinning channel, not an empirical detector law.

[F1]

For this incoming law, the ideal mode theorem uses the wavepacket postulate's input statistic $g_{v}^{(2)} (0) = F_{2} / μ^{2}$ when $0 < μ < \infty$ and $F_{2} < \infty$ . Thus the Statement's $g_{N}^{(2)} (0)$ is exactly that postulate value for $ρ$ . Ideal one-mode intensity correlations separate model states

[F2]

Nonnegative countable sums may be averaged and regrouped by Tonelli's theorem. Tonelli's theorem for nonnegative measurable functions on a sigma-finite product

Proof

1.1A1algebra

Fix $n \in N_{0}$ and average on its finite fiber using $Q_{n}$ . For each $j \leq n$ , factorization gives $\sum_{x} Q_{n} ({x}) X_{n, j} (x) = η$ , and for distinct $i, j \leq n$ it gives $\sum_{x} Q_{n} ({x}) X_{n, i} (x) X_{n, j} (x) = η^{2}$ . Since $K (n, x) = \sum_{j = 1}^{n} X_{n, j} (x)$ and $K (n, x) (K (n, x) - 1) = \sum_{i \neq j} X_{n, i} (x) X_{n, j} (x)$ for zero-one coordinates, it follows that $\sum_{x} Q_{n} ({x}) K (n, x) = n η$ and $\sum_{x} Q_{n} ({x}) K (n, x) (K (n, x) - 1) = n (n - 1) η^{2}$ . For $n = 0$ or $n = 1$ the relevant sums are empty and the same formulas hold.

2.1A1F1F2step 1.1algebra∎

Using the Statement's encoding, identify $Ω$ with a subset of $N_{0} \times N_{0}$ . For either $Y = K$ or $Y = K (K - 1)$ , extend $f_{Y} (n, m) : = p_{n} Q_{n} ({x}) Y (n, x)$ by zero outside that subset. Counting measure on $N_{0}$ is sigma-finite, so [F2] applied to the product of the two counting-measure spaces permits the nonnegative sum to be regrouped over the finite fibers. Using step 1.1 gives $E K = \sum_{n} p_{n} n η = η μ$ and $E [K (K - 1)] = \sum_{n} p_{n} n (n - 1) η^{2} = η^{2} F_{2}$ . Both are finite by hypothesis. Since $η μ > 0$ , division gives $g_{K}^{(2)} (0) = η^{2} F_{2} / (η μ)^{2} = F_{2} / μ^{2} = g_{N}^{(2)} (0) = g_{v}^{(2)} (0)$ by [F1].

Ideal independent loss preserves normalized factorial correlation

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources