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

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Visibility bound in a specified two-beam model

Physical scope

Conditional result for an ideal scalar two-beam classical phasor model with a common ensemble, finite second moments, a controllable full relative-phase sweep, and mean-square amplitude as the detector-response model. The $Γ_{12}$ here is the cross-moment between two beam channels; it is not the transverse-polarization cross-component of one beam. The visibility is the model contrast, not a finite-sample estimate. No detector calibration, noise model, empirical observation, or unconditional claim about measured fringe visibility is made.

Statement

Let $A_{1}, A_{2} : Ξ \to C$ be jointly measurable classical phasor amplitudes at one fixed observation point for a single selected scalar field component, on one probability space, each in V m $^{- 1}$ , with $E (∣ A_{1} ∣^{2} + ∣ A_{2} ∣^{2}) < \infty$ . Define $I_{j} : = E [∣ A_{j} ∣^{2}]$ and $Γ_{12} : = E [A_{1} \overline{A_{2}}]$ . Adopt the ideal scalar two-beam model with a phase multiplier ranging over the full unit circle $T : = {ζ \in C : ∣ ζ ∣ = 1}$ . For each $ζ \in T$ , set $A_{ζ} : = A_{2} + ζ A_{1}$ and $I (ζ) : = E [∣ A_{ζ} ∣^{2}]$ . The quantities $I_{j}$ and $I (ζ)$ are mean-square field coordinates in (V m $^{- 1}$ ) $^{2}$ , not calibrated irradiances. Set $I_{\max} : = \max_{ζ \in T} I (ζ)$ and $I_{\min} : = \min_{ζ \in T} I (ζ)$ . When $I_{1} + I_{2} > 0$ , define the model visibility by $V : = \frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}} .$ Then $I (ζ) = I_{1} + I_{2} + 2 Re (ζ Γ_{12}), V = \frac{2 ∣ Γ_{12} ∣}{I_{1} + I_{2}} \leq 1.$

Statement references

Classical two-component coherency matrix

Stokes cone and deterministic Jones transformations

Facts & Assumptions

Given: The common probability space, square-integrable beam amplitudes $A_{1}, A_{2}$ , a full controllable phase sweep, and the ideal mean-square detector model stated above.

[F1]

Classical two-component coherency matrix: applying its second-moment construction to $W = (A_{1}, A_{2})^{T}$ gives $Γ_{11} = I_{1}$ , $Γ_{22} = I_{2}$ , and $Γ_{12} = E [A_{1} \overline{A_{2}}]$ . Here the coordinates are distinct beam channels, not polarization components.

[F2]

Stokes cone and deterministic Jones transformations: every such second-moment matrix is positive semidefinite, so its entries obey $I_{1} I_{2} \geq ∣ Γ_{12} ∣^{2}$ .

[F3]

Linearity, monotonicity, and the modulus bound for expectation: expectation of the integrable cross terms is linear and commutes with complex conjugation under the componentwise complex-integral convention.

[F4]

Real and imaginary parts, complex conjugation, and modulus: write a complex number as $a + i b$ , take its conjugate as $a - i b$ , and define its modulus by $\sqrt{a^{2} + b^{2}}$ .

[A1]

The ideal detector output is the ensemble mean of the squared modulus of the coherently summed phasor; no finite-sample or noisy detector response is assumed.

Proof

1.1givenF1F3A1algebra

Expanding the squared modulus in [A1] gives $∣ A_{ζ} ∣^{2} = ∣ A_{2} ∣^{2} + ∣ A_{1} ∣^{2} + ζ A_{1} \overline{A_{2}} + \overline{ζ} A_{2} \overline{A_{1}}$ . Taking expectations, using [F1] and [F3], yields $I (ζ) = I_{1} + I_{2} + 2 Re (ζ Γ_{12})$ .

2.1step 1.1givenF4algebra

Put $S = I_{1} + I_{2}$ and $c = Γ_{12}$ . Write $ζ = a + i b$ and $c = u + i v$ . Since $ζ \in T$ , $a^{2} + b^{2} = 1$ . By [F4], $∣ c ∣^{2} = u^{2} + v^{2}$ and $Re (ζ c) = a u - b v$ . Therefore $∣ c ∣^{2} - (Re (ζ c))^{2} = (a^{2} + b^{2}) (u^{2} + v^{2}) - (a u - b v)^{2} = (a v + b u)^{2} \geq 0,$ so each $Re (ζ c)$ lies in $[- ∣ c ∣, ∣ c ∣]$ . If $c \neq 0$ , set $r : = ∣ c ∣ = \sqrt{u^{2} + v^{2}} > 0$ and $ζ_{0} : = (u - i v) / r$ . Then $∣ ζ_{0} ∣^{2} = (u^{2} + v^{2}) / r^{2} = 1$ , so $ζ_{0} \in T$ , and direct multiplication gives $ζ_{0} c = r$ while $(- ζ_{0}) c = - r$ . Both endpoints are therefore attained. If $c = 0$ , the cross term is identically zero. Hence $I_{\max} = S + 2 ∣ c ∣$ and $I_{\min} = S - 2 ∣ c ∣$ .

3.1givenstep 2.1F2algebra∎

By [F2], $S^{2} - 4 ∣ c ∣^{2} = (I_{1} - I_{2})^{2} + 4 (I_{1} I_{2} - ∣ c ∣^{2}) \geq 0$ ; since $S \geq 0$ , this gives $2 ∣ c ∣ \leq S$ . Under the given $S > 0$ condition, $I_{\max} + I_{\min} = 2 S > 0$ , so substitution of step 2.1 into the visibility definition gives $V = 2 ∣ c ∣ / S \leq 1$ .

Remarks

The displayed $V$ is a phase-extremized ensemble-model contrast. A finite set of detector readings need not attain it; estimating visibility from data requires separate sampling, calibration, uncertainty, and detector-response assumptions. This theorem supplies none of those empirical inputs.

Visibility bound in a specified two-beam model

Statement

Statement references

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources