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A dark bin of positive length has positive coherent probability

Example

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Work on $D = R / Z$ identified with $[0, 1)$ and Lebesgue measure, and take the two normalized amplitudes $a (x) : = 1, b (x) : = e^{2 π i x} (x \in [0, 1)),$ so that $∥ a ∥ = ∥ b ∥ = 1$ in $L^{2} (D)$ and $∣ a (x) + b (x) ∣^{2} = 2 + 2 \cos (2 π x)$ (The complex exponential by its power series, The addition formulas for sine and cosine). Then:

(a) the coherent law of Coherent and incoherent two-path bin laws and the marker overlap is $p_{c o h} (B) = \int_{B} (1 + \cos (2 π x)) d x$ for every measurable $B \subseteq [0, 1)$ , and $p_{m i x} (B) = ∣ B ∣$ for the balanced incoherent law $w = \frac{1}{2}$ , where $∣ B ∣$ is Lebesgue measure; in particular $p_{c o h} (B) \leq p_{m i x} (B)$ on bins where the cosine is negative on average;

(b) the density $1 + \cos (2 π x)$ vanishes at the single point $x_{0} = \frac{1}{2}$ , the ideal dark point; nevertheless for every $ε \in (0, \frac{1}{2})$ the symmetric dark bin $B_{ε} = (\frac{1}{2} - ε, \frac{1}{2} + ε)$ has $p_{c o h} (B_{ε}) = 2 ε - \frac{\sin (2 π ε)}{π} = \frac{4 π^{2}}{3} ε^{3} + O (ε^{5}) > 0,$ so a bin of positive length containing the dark point is not forbidden;

(c) for the same bin $p_{m i x} (B_{ε}) = 2 ε$ , so $p_{c o h} (B_{ε}) < p_{m i x} (B_{ε})$ and the two specified probability measures assign different probabilities to this bin. This is a comparison of bin probabilities only; no claim about individual observations is made.

The example is a computation in the specified two-amplitude model; it asserts no detector law, no measured counts and no empirical conclusion.

Verification

Given: the amplitudes $a = 1$ , $b (x) = e^{2 π i x}$ on $[0, 1)$ and bins $B_{ε}$ as above.

Proof technique: direct evaluation of the two bin laws and of the elementary integral $\int \cos (2 π x) d x$ , together with the pointwise vanishing of the density at $x_{0} = \frac{1}{2}$ .

[F1] $∥ a + b ∥^{2} = \int_{0}^{1} ∣ 1 + e^{2 π i x} ∣^{2} d x$ and, for every measurable bin $B$ , $p_{c o h} (B) = ∥ a + b ∥^{- 2} \int_{B} ∣ a + b ∣^{2}$ ; for unit $a, b$ the balanced incoherent law is $p_{m i x} (B) = \frac{1}{2} \int_{B} ∣ a ∣^{2} + \frac{1}{2} \int_{B} ∣ b ∣^{2}$ (Coherent and incoherent two-path bin laws and the marker overlap clauses (a), (b)).

[F2] Euler's formula gives $e^{2 π i x} = \cos (2 π x) + i \sin (2 π x)$ , hence $1 + \cos (2 π x) = \frac{1}{2} ∣ 1 + e^{2 π i x} ∣^{2} \geq 0$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ). The fibre theorem says $e^{2 π i x} = - 1 = e^{i π}$ exactly when $x = \frac{1}{2} + n$ for $n \in Z$ , so the density vanishes on $[0, 1)$ only at $x_{0} = \frac{1}{2}$ ( $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ). Also $\int_{0}^{1} \cos (2 π x) d x = 0$ , and $\frac{d}{d x} \sin (2 π x) = 2 π \cos (2 π x)$ with $\sin (π \pm θ) = \mp \sin θ$ for the second fundamental theorem of calculus applied to the continuous primitive $\frac{\sin (2 π x)}{2 π}$ (The derivatives of sine and cosine are cosine and minus sine, The addition formulas for sine and cosine, Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ ).

1.1givenF1F2algebra

[The two laws] By [F2], $∥ a + b ∥^{2} = \int_{0}^{1} (2 + 2 \cos (2 π x)) d x = 2$ , so $p_{c o h} (B) = \frac{1}{2} \int_{B} (2 + 2 \cos (2 π x)) d x = \int_{B} (1 + \cos (2 π x)) d x$ by [F1]. Since $∣ a ∣ = ∣ b ∣ = 1$ , $p_{m i x} (B) = \frac{1}{2} ∣ B ∣ + \frac{1}{2} ∣ B ∣ = ∣ B ∣$ . This is (a).

2.1givenF2step 1.1algebra

[The dark point and the dark bin] The density $1 + \cos (2 π x)$ equals $1 + \cos π = 0$ at $x_{0} = \frac{1}{2}$ , a single point of Lebesgue measure zero. For $0 < ε < \frac{1}{2}$ , [F2] gives $\int_{1 / 2 - ε}^{1 / 2 + ε} \cos (2 π x) d x = \frac{\sin (2 π x)}{2 π} ∣_{1 / 2 - ε}^{1 / 2 + ε} = \frac{\sin (π + 2 π ε) - \sin (π - 2 π ε)}{2 π} = \frac{- \sin (2 π ε) - \sin (2 π ε)}{2 π} = - \frac{\sin (2 π ε)}{π}$ , so $p_{c o h} (B_{ε}) = 2 ε - \frac{\sin (2 π ε)}{π}$ . By [F2], $1 + \cos (2 π x)$ is nonnegative and vanishes on $[0, 1)$ only at $x = \frac{1}{2}$ . The point $x_{*} : = \frac{1}{2} + \frac{ε}{2}$ lies inside $B_{ε}$ and the density is positive there; continuity (A function differentiable at $c$ is continuous at $c$ ) gives a subinterval of positive length on which it is bounded below by a positive constant, so the bin probability is strictly positive; the Taylor expansion of $\sin$ gives $2 ε - \frac{\sin (2 π ε)}{π} = \frac{(2 π ε)^{3}}{6 π} + O (ε^{5}) = \frac{4 π^{2}}{3} ε^{3} + O (ε^{5})$ . This is (b).

3.1givenstep 1.1step 2.1algebra∎

[Comparison and scope] By step 1.1, $p_{m i x} (B_{ε}) = 2 ε$ , and $0 < 2 π ε < π$ implies $\sin (2 π ε) > 0$ by Pi is the first positive zero of sine, so the exact formula of step 2.1 gives $p_{c o h} (B_{ε}) < p_{m i x} (B_{ε})$ ; this is (c). The statements are properties of the two computed bin laws on the specified amplitudes, and no observed frequency is claimed.

A dark bin of positive length has positive coherent probability

Example

Verification

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