advanced32 min studyLesson 473 in Physics

Connect field moments to detector counts

Use normal ordering, finite efficiency, and factorial moments to distinguish a source state from its recorded photon-count distribution.

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What you’ll understand

  • Explain why ideal absorptive detection uses normally ordered field correlations.
  • Compute binomial thinning of a fixed photon occupation.
  • State detector assumptions under which normalized factorial correlations survive loss.

A photon detector is a physical measurement device, not a direct readout of every electromagnetic fluctuation. An ideal absorbing detector responds to the positive-frequency field that removes excitation, leading to normally ordered correlations. Finite efficiency then changes the recorded count distribution even when the incident state is known exactly. We will use a single selected temporal mode and an explicit independent-loss model, so the distinction between incident occupation, detected counts, and quadrature vacuum noise remains visible.

Absorption selects an operator ordering

The positive-frequency field is proportional to an annihilation operator for a selected mode. The excitation rate of an ideal weak absorptive detector is therefore proportional to the expectation of the negative-frequency field followed by the positive-frequency field. This ordering gives mean occupation rather than occupation plus one half. Vacuum has no photons for an ideal passive detector to absorb, despite its nonzero quadrature variance. A detector with dark counts, an active pump, or another excitation mechanism requires additional terms.

Two absorptions involve a normally ordered four-field product. At zero delay in one mode this becomes the expectation of number times number minus one, rather than the second number moment. The subtraction has an operational meaning: the same single photon cannot supply two separate absorption events. Normal ordering is consequently tied to the specified detection process. It does not imply that all possible quantum measurements ignore vacuum fluctuations; phase-sensitive quadrature measurements have a different observable and calibration.

R1∝⟨E^(−)E^(+)⟩∝⟨n^⟩,⟨a^†2a^2⟩=⟨n^(n^−1)⟩R_1\propto\langle\hat E^{(-)}\hat E^{(+)}\rangle\propto\langle\hat n\rangle,\qquad \langle\hat a^{\dagger2}\hat a^2\rangle=\langle\hat n(\hat n-1)\rangle
R₁ is a count rate in s⁻¹ after including detector coupling and mode normalization. The relations describe weak absorptive photodetection of a selected mode, with normal ordering and no background counts.↔ Scroll wide equations horizontally.

An inefficient detector thins the incident occupation

Assume each incident photon is independently recorded with probability eta, independent of photon number and time within the chosen detection window. Conditional on exactly n incident photons, the recorded count is binomial. Its mean is eta times n and its variance includes the randomness of missed photons. For a general incident number distribution, average over n after performing this conditional calculation. The output variance then contains both source fluctuations and detection-loss fluctuations.

A coherent input remains Poissonian after this ideal loss, with reduced mean, while a number-state input becomes binomial and loses its perfect number certainty. The binomial instrument matches the conditional fixed-number case exactly. It does not model the entire coherent-state distribution, a random arrival-time process, detector dead time, or afterpulsing. Source and detector assumptions must be stated before comparing a measured variance with a shot-noise reference.

P(m∣n)=(nm)ηm(1−η)n−m,⟨m⟩=η⟨n⟩,Var⁡(m)=η2Var⁡(n)+η(1−η)⟨n⟩P(m|n)=\binom nm\eta^m(1-\eta)^{n-m},\quad \langle m\rangle=\eta\langle n\rangle,\quad \operatorname{Var}(m)=\eta^2\operatorname{Var}(n)+\eta(1-\eta)\langle n\rangle
η is a dimensionless constant detection efficiency, and n and m are incident and detected counts in the same selected window. The formula assumes independent loss with no dark counts, saturation, or number-dependent response.↔ Scroll wide equations horizontally.

Factorial moments expose source correlations

The detected second factorial moment acquires eta squared because two distinct incident photons must both be detected. Dividing by the detected mean squared cancels this efficiency factor. Thus an ideal normalized second-order correlation is unchanged by uniform independent linear loss when its mean is nonzero. This useful invariance does not mean loss leaves the count histogram unchanged, nor does it guarantee that a poorly resolved experimental correlation estimate is unbiased.

Background counts add uncorrelated events and usually pull measured correlations toward one. Finite detector timing convolves a time-dependent correlation, and dead time can imitate a deficit of nearby events on a single detector. A split-beam two-detector arrangement helps measure coincidences without assigning two counts to the same detector event, but still needs calibrated backgrounds and timing. The operational content of a correlation is therefore incomplete without its window, mode selection, and detector response.

⟨m(m−1)⟩=η2⟨n(n−1)⟩,g(2)(0)=⟨n(n−1)⟩⟨n⟩2=⟨m(m−1)⟩⟨m⟩2\langle m(m-1)\rangle=\eta^2\langle n(n-1)\rangle,\qquad g^{(2)}(0)=\frac{\langle n(n-1)\rangle}{\langle n\rangle^2}=\frac{\langle m(m-1)\rangle}{\langle m\rangle^2}
The final equality requires η>0, nonzero incident mean, independent uniform loss, and no background. This single-mode factorial correlation is dimensionless; resolving a continuous-time g²(τ) requires a temporal detector model.↔ Scroll wide equations horizontally.

Thin a three-photon input

A number state with exactly three incident photons enters a detector of efficiency η=1/2, with no background. Find the probabilities for zero through three counts, mean and variance, and the normalized second factorial correlation.

  1. Binomial thinning gives probabilities 1/8, 3/8, 3/8, and 1/8 for m=0,1,2,3. They sum to one, and counts above three are impossible in this model.

  2. The mean is 3/2 and variance is 3×(1/2)(1/2)=3/4. Missed photons introduce count randomness even though the incident occupation was exact.

  3. The incident factorial moment is 3×2=6, so the detected value is 6/4=1.5. Dividing by the detected mean squared, 2.25, gives g²(0)=2/3, the same as for the incident number state.

The result

The count histogram is binomial with mean 1.5 and variance 0.75; g²(0)=2/3 survives this ideal loss although the histogram changes.

Explore the model

Interactive lab

Interactive lab · probability

Test fixed-number detection loss

For fresh runs Reset and set BOTH p=0.5, trials n=3; then Reset and set p=0.25, n=3. Interpret p as independent detection efficiency for a three-photon input. Predict means 1.5 and 0.75 and variances 0.75 and 0.5625. This exact binomial law contains no arrival times, background, or coherent-state Poisson process.

Test fixed-number detection lossThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forDetector loss creates conditional count variance without changing the ideal normalized factorial correlation.

Check your understanding

Why does ideal absorptive direct detection register no counts from a vacuum input?

Try it yourself

A coherent incident mode has mean photon number four and is detected with efficiency one quarter and no background. Find detected mean, variance, and zero-count probability.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

An ideal linear-loss channel preserves coherent-state Poisson statistics while reducing its mean.

Compare with the solution

The detected mean is one and its variance is one. The zero-count probability is e⁻¹≈0.3679. Substituting incident variance four into the thinning formula gives (1/4)²×4+(1/4)(3/4)×4=1, confirming the result. This Poisson law is not a finite-trial binomial distribution.

Take it with you

  • Normally ordered moments describe ideal absorptive photodetection.
  • Independent efficiency loss gives binomial counts conditional on fixed incident number.
  • Normalized factorial correlations survive uniform loss only under stated detector assumptions.

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