Anderson, David

Morrison, John Ross

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VISUAL NOISE DUE TO QUANTUM INDETERMINACIESJOHN MORRISON AND DAVID ANDERSONAbstract. We establish that, due to certain quantum indeterminacies,there must be foundational colours that do not reliably cause any partic-ular experience. This report functions as an appendix to Morrison [5].1. IntroductionIn [5] Morrison argues that certain assumptions imply:noiseFor any observer and any context: If two foundational colours reliablyproduce different experiences, then there is a foundational colour be-tween them that does not reliably produce any particular experience(even if it will always produce some experience or another).Morrison’s argument focuses on the noise produced by internal, biologicalprocesses. We will establish that, given four assumptions, noise is trueeven for observers whose internal, biological processes do not produce anynoise. In particular, we will establish that certain quantum indeterminaciesby themselves suffice for noise.1The four assumptions:i. The relevant contexts are governed by the laws of quantum mechanics.ii. The visual systems of the relevant observers cannot take infinitelylong measurements.iii. The visual systems of the relevant observers produce at least threedifferent colour experiences.iv. Foundational colours are as fine-grained as spectral dispositions.Our argument focuses on an observer with an ideal visual system, whichis a visual system that does not have any computational limits, has per-fect information about the illuminant and has internal processes that do notcontribute any noise. If we can show that noise is true for all observerswith ideal visual systems then that will show that noise is true for all ob-servers. Consider that, if an observer with a non-ideal visual system were aDate: June 14, 2011.1We are using ‘quantum indeterminacies’ so that it includes any probability distributionon a set of measurements that is due to the laws of quantum mechanics.1counter-example to noise, then an observer with an idealization of that vi-sual system would also be a counter-example because better measurements,better mathematics and better information would make the visual systemeven more sensitive to small differences between foundational colours.Visual systems take measurements of the spectral power distribution oflight rays. These measurements involve measurements of both wavelengthsand intensities. There are many ways for a visual system to take thesemeasurements. For example, like the common spectrophotometer, a visualsystem might use a prism to separate light by wavelength and then measurethe intensity of each separated signal.Our strategy is to first consider an observer whose visual system onlymeasures the intensity of a light wave, and then consider an observer whosevisual system only measures the wavelength of a single photon. In each casewe will show that quantum indeterminacies by themselves suffice for noise.We will then generalize to an observer who measures both wavelengths andintensities.To simplify, we will focus on light after it has been reflected, emitted ortransmitted from a surface. Because reflection, emission and transmission in-troduce additional indeterminacies, we are effectively setting aside additionalreasons why noise is true.We are assuming that the foundational colours are as fine-grained as spec-tral dispositions (see (iv) above). Spectral dispositions are functions that,for a given content and location, specify the properties of the light reflected,emitted or transmitted by an object to that location. Therefore, given our as-sumption, if two objects produce light rays that are characterized by differentwave functions then those objects have different foundational colours.2. IntensityThe mean intensity of a light wave will be x ∈ <+. Any measurement ofthe intensity of that light wave will be a nonnegative integer m ∈ N+. It willbe a nonnegative integer because light is composed of indivisible photons.Due to quantum indeterminacies, for any measurement of finite length, theprobability of measuring m photons from a coherent light with mean intensityx is governed by the Poisson distribution:prob(m|x) = e−xxmm!(1)Incoherent light is noisier, so we’ll focus on coherent light. For more aboutthe physics behind the Poisson distribution see Fox [3, p.76-80] and Orszag[6, p.29-32].Let E1, E2 and E3 be three different colour experiences that the visualsystem produces. Let ΩE1 be the set of all measurements that lead that visualsystem to produce E1, ΩE2 be the set of all measurements that lead it toproduce E2, and ΩE3 be the set of all measurements that lead it to produceE3. How the ideal visual system decides which measurements should producewhich experience is an interesting question. No doubt it will rely on perfectinformation about the illuminant. But we don’t need to worry about thesedetails.Building on (1), the following function gives the probability that a lightwave with mean intensity x ∈ <+ will produce experience E1 in the relevantobserver:gE1(x) =∑m∈ΩE1e−xxmm!.(2)Likewise for gE2(x) and gE3(x).Because E1, E2 and E3 are different experiences, each measurement willproduce only one of them. That is:ΩE1 ∩ ΩE2 = ∅(3)ΩE1 ∩ ΩE3 = ∅(4)ΩE2 ∩ ΩE3 = ∅(5)As is evident from the fact that the Poisson distribution is a probabilitydistribution, for any value x ∈ <+:∞∑m=0e−xxmm!= 1(6)It follows from (2)-(6) that for any value x ∈ <+:gE1(x) + gE2(x) + gE3(x) ≤ 1.(7)Next, we’re assuming that the the relevant visual system produces each ex-perience (E1, E2 and E3), at least sometimes. It follows that:ΩE1,ΩE2,ΩE3 6= ∅(8)Building on (1), (8) and the fact that prob(m|x) > 0 for any x ∈ <+,m ∈ N+,it follows that for any value x:gE1(x), gE2(x), gE3(x) 6= 0.(9)Finally, to make things concrete, we will assume that the threshold forreliably production is at 0.5 so that, for instance, a light with mean intensityvalue x reliably causes experience E1 if and only if prob(E1|x) ≥ 0.5. Theproof straightforwardly generalizes to all thresholds above 0.5.Given these assumptions, let’s now prove that noise is true for an observerwho only measures the intensity of a light:Theorem 1. If for some value of x, y ∈ <+, gE1(x) ≥ 0.5 and gE2(x+ y) ≥0.5 then there is an i such that 0 < i < y and gE1(x+i), gE2(x+i), gE3(x+i) <0.5.Proof. The main point is that gE1, gE2, and gE3 are continuous for all x ∈<+. We will prove this for gE1, the proof for the other two functions beinganalogous.For the purposes of this proof, write:pm(x) =e−xxmm!for the Poisson distribution function. Also, letχm ={1 if m ∈ ΩE1;0 otherwisebe the indicator function for ΩE1. Using this notation, we can writegE1(x) =∞∑m=0χm pm(x).(10)By the uniform convergence theorem, it suffices to show the series on theright-hand side of (10) converges uniformly. For this, we will first prove thatanother, related function is continuous. Let N > 0. We claim that thefunctiong̃E1(x) =∞∑m=Nχm pm(x)(11)is continuous for all x in the interval [0, N ]. We will use the WeierstrassM -test. Here we need an easily verified property of the functions pm(x):For m ≥ N , the maximum value of pm(x) on the interval [0, N ] is pm(N) =e−NNmm!. Now apply the M -test with Mm = pm(N), and observe that theseries∞∑m=NMm =∞∑m=Ne−NNmm!converges, since it is dominated by∑∞m=0e−NNmm!= 1. Therefore the seriesin (11) converges uniformly for x in [0, N ], so g̃E1 is continuous on [0, N ].We will now prove that gE1 is continuous. First note that gE1 is the sumof g̃E1 and the (finitely many) continuous functions χm pm(x) for m < N .Therefore, gE1 is continuous on the interval [0, N ]. Because N is chosenarbitrarily, it follows that gE1 is continuous for all x ∈ <+.Similarly, gE2 and gE3 are continuous for all x ∈ <+Now suppose that for some values x, y ∈ <+, gE1(x) ≥ 0.5 and gE2(x+y) ≥0.5. By (7) and (9), it follows that gE1(x+y) < 0.5. Further, because gE1(x)is continuous for all x ∈ <+, it follows that there is a 0 ≤ k < y such thatgE1(x + k) = 0.5. By (7) and (9), it follows that gE2(x + k) < 0.5. Further,because gE2(x) is continuous for all x ∈ <+, there must be a value 0 ≤ δsuch that gE2(x+ k+ δ) = 0.5. Therefore, we can establish the theorem justby letting x+ k < i < x+ k + δ. At that point gE1(x), gE2(x), gE3(x) < 0.5.In fact, the consequent of the theorem will not just be true at a single point,it will be true for all points in the interval (x+ k, x+ k + δ). This proof establishes that, due to the quantum indeterminacies that aredescribed by (1), noise is true for an ideal observer who is measuring onlyintensity. In particular, if two foundational colours each reliably produce adifferent colour experience (E1, E2 or E3), and those foundational colourscorrespond to light with intensities x and x + y then there will be a foun-dational colour that corresponds to light with intensity x + i that does notreliably produce E1, does not reliably produce E2 and does not reliableproduce E3.3. WavelengthLet’s now consider an observer whose visual system is only measuringthe wavelength of a single photon. We will establish noise in two ways.The first way will appeal to the quantum indeterminacies that help producethermal Doppler shifting. The second way will appeal to indeterminacies inmeasurements of photons with certain wave functions.First way: thermal Doppler shifting. Let’s start with the simplest casefor the visual system: when the photon is monochromatic. This is the sim-plest case because, as we will see later, photons with mixed wavelengthscontribute additional indeterminacies. The wave function for a monochro-matic photon in free space is:ψ(x) = sin(2πωx)(12)where its wavelength, λ ∈ [400, 700], is the quotient of the speed of light, c,and its frequency, ω. That is, λ = cω.Due to quantum indeterminacies, when any photon passes through a gas itswavelength is thermally Doppler shifted. The probability of different shiftsdepends, in part, on the value of the self-diffusion constant of the gas, D,which depends on temperature and pressure. For details see chapter eight ofJeans [4]. To simplify, we treat D as a constant; perhaps the visual systemhas some way of determining the temperature and pressure of the gas. Dueto quantum indeterminacies, the probability that an ideal visual system willmeasure the photon’s wavelength within some interval of wavelengths can bedetermined by integrating the following function:p(α, λ) =2πDλ2(α− cλ)2 + (2πDλ2)2(13)For details, see Dickie [2].As before, for an arbitrary visual system that is ideal, let E1, E2 and E3be different experiences. Also, let ΩE1, ΩE2 and ΩE3 be measurable subsetssuch that ΩE1 is the set of all measurements that lead it to produce E1, ΩE2is the set of all measurements that lead it to produce E2, and ΩE3 is the setof all measurements that lead it to produce E3.WhereχE1(α) ={1 if α ∈ ΩE1;0 otherwisethe probability that a photon that is monochromatic with wacelength λ ∈[400, 700] will produce an E1 experience is:fE1(λ) =∫ 700400χE1(α)p(α, λ) dα(14)Likewise for fE2 and fE3.Without rehearsing all the details, in the same way that we established (7)and (9) in our discussion of intensity measurements, we can also establishthat for all λ ∈ [400, 700]:fE1(x) + fE2(x) + fE3(x) ≤ 1(15)fE1(x), fE2(x), fE3(x) 6= 0(16)Finally, let’s again assume that the relevant threshold is at 0.5.Given these assumptions, noise is true for an observer whose visual systemonly measures the wavelength of single monochromatic photons:Theorem 2. If for some λ and λ′, fE1(λ) ≥ 0.5 and fE2(λ′) ≥ 0.5 then thereis a λ′′ such that f1(λ′′), f2(λ′′), f3(λ′′) < 0.5.Proof. p(α, λ) is continuous for a fixed α. Just note that it is the quotientof two continuous functions and the denominator is never zero. Therefore,fE1(λ) is also continuous. Likewise for fE2 and fE3. Accordingly, the samereasoning employed at the end of the proof of Theorem 1 suffices for Theorem2. This proof establishes that, due to the quantum indeterminacies that producethermal Doppler shifting, noise is true for an idea observer who is measuringonly the wavelength of a single photon.Second way: mixed waves. One might try to avoid the indeterminaciesinherent in Doppler shifting by appealing to observers who not only know theaverage temperature and pressure of the gas but who also know the positionand momentum of every molecule. Such an observer would be equivalentto Maxwell’s demon and therefore might involve a violation of the laws ofthermodynamics (see Brillouin [1]). However, even if it were possible, thereare still quantum indeterminacies that suffice for noise.Let’s broaden our perspective to include mixed waves. For any assignmentof ai’s, the following describes a wave equation for a possible photon:ψ(x) =∑iai sin(2πωix).(17)The probability of measuring λi =cωiis proportional to a2i . Withoutaffecting wavelength measurements, we may assume ψ is normalized, so theprobability of measuring λi is equal to a2i . Therefore, for a given ψ, theprobability of measuring a wavelength in ΩE1 is equal to:fE1(ψ) =∑iχE1(ai)(λi)a2i(18)Similarly for ΩE2 and ΩE3, yielding fE2 and fE3.Because the visual system produces each of the experiences (E1, E2 andE3), at least sometimes:ΩE1,ΩE2,ΩE3 6= ∅,(19)We will assume that these are the only experiences produced by the rele-vant visual system, and that, in addition, every measurement produces someexperience, in which case:fE1(ψ) + fE2(ψ) + fE3(ψ) = 1(20)Our proof trivially generalizes to visual experiences that produce four, five,etc., different experiences.Finally, we will again assume that the threshold for reliably production isat 0.5.Given these assumptions, let’s now prove that noise is true for an observerwhose visual system is only measuring the wavelength of single photons:Theorem 3. If for some ψ and ψ′, fE1(ψ) ≥ 0.5 and fE2(ψ′) ≥ 0.5 thenthere is a ψ′′ such that fE1(ψ′′), fE2(ψ′′), fE3(ψ′′) < 0.5.Proof. Writeψ(x) =∑iai sin(2πωix)andψ′(x) =∑ibi sin(2πωix).Defineψ′′(x) = N1∑i:λi∈ΩE1bi sin(2πωi)+N2∑i:λi∈ΩE2ai sin(2πωi)+N3∑i:λi∈ΩE3(ai + bi2)sin(2πωi),whereN1 =1√3∑j:λj∈ΩE1 b2jis a normalization factor, and N2, N3 are defined similarly.Now we seefE1(ψ′′) =∑i:λi∈ΩE1N21 b2i =13,and similarly for fE2(ψ′′) and fE3(ψ′′). Note that by adjusting N1, N2 andN3 we can identify other wave fuctions that also establish the theorem. The preceding proves that, due to quantum indeterminacies, noise is truefor an ideal observer who is measuring only the wavelength of a single photonand who knows the position and momentum of each particle in the gas.4. Intensity and Wavelength: spectral power distributionsIt is straightforward generalize to an ideal observer who is measuring bothintensity and wavelength. Indeed, given pairs (x, ψ) and (x + y, ψ′), with xand x+y as in the hypothesis of Theorem 1 and ψ and ψ′ as in the hypothesisof Theorem 2 (or 3), the pair (x + i, ψ′′) chosen as in the conclusions ofTheorems 1 and 2 (or 3) will do. It also worth noting that when multiplephotons are involved, there might be entanglements, which will introduceadditional indeterminacies.We conclude that, given our assumptions, certain quantum indetermina-cies by themselves suffice for noise.John MorrisonDepartment of PhilosophyBarnard College, Columbia University3009 BroadwayNew York, NY 10027jmorrison@barnard.eduDavid AndersonDepartment of MathematicsUniversity of WashingtonC-138 Padelford HallSeattle, WA 98195dandersn@math.washington.eduReferences[1] Leon Brillouin. Maxwell’s demon cannot operate: information and entropy. Journal ofApplied Physics, 22:334–37, 1951.[2] Robert Dicke. The effect of collisions upon the doppler width of spectral lines. PhysicalReview, 89:472–73, 1953.[3] Mark Fox. Quantum Optics: an Introduction. Oxford University Press, New York, NY,2006.[4] James Jeans. An introduction to the kinetic theory of gases. Cambridge UniversityPress, 1967.[5] John Morrison. Colour in a physical world. Forthcoming in Mind.[6] Miguel Orszag. Quantum Optics: including noise reduction, trapped ions, quantumtrajectories, and decoherence. Harvard University Press, 1981.

Visual Noise Due to Quantum Indeterminacies

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Visual Noise Due to Quantum Indeterminacies

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