Why doesn't incoherent light cancel itself out?

What is the precise mathematical description of an incoherent single-frequency signal for any type of wave? The reason I'm asking is because of the following apparent paradox in which incoherent light cannot exist.

Consider sunlight, for example, which has passed through a polarizing filter and frequency filter, so that only waves with wave numbers very close to $$k_0$$ are allowed to pass through. Since sunlight is totally incoherent, it seems reasonable to model the signal as a sum of sine waves $$E_\alpha(x,t)=A_\alpha\sin(k_0x-\omega_0t+\phi_\alpha),$$ where $$E$$ is the electric field in the direction of the polarizing filter, $$\omega_0=ck_0$$, $$A_\alpha$$ is a random amplitude, and $$\phi_\alpha$$ is a random phase shift. If the light were coherent, then the $$\phi_\alpha$$'s would all be identical; so it seems reasonable that for "maximal incoherence" the $$\phi$$'s and $$A_\alpha$$'s would be different and uniformly distributed. But then for every component with phase shift $$\phi$$ and amplitude $$A$$, there exists a wave $$A_\alpha\sin(k_0x-\omega t-\phi-\pi)$$, which cancels the original. Hence all components cancel and there is no wave (spectrometer detects nothing).

So what's the flaw here? I'm guessing the model of incoherent light is where the problem lies, but maybe it's in the reasoning. I'm also curious whether or not the answer necessarily relies on quantum mechanics.

EDIT:

Since there are some votes to close based on the proposed duplicate, I'll just say that both questions get at the same idea, but I think mine (which could also have focused on polarization) is more specific, since I'm asking for a precise model and whether quantum physics is a necessary part of the explanation.

• Does this answer your question? How come unpolarized light does not undergo destructive interference? Jun 18, 2021 at 4:25
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– rob
Jun 19, 2021 at 14:51
• You might consider Feynman 's path integral mathematics instead of the classical superposition. Jul 3, 2021 at 1:34

Interference of classical electromagnetic waves is a complex phenomenon . Studied with lasers where a single frequency can be made to interfere with itself,one sees that in a completely destructive interference the energy of the beam goes back to the laser source! See this MIT video on this, "Destructive Interference — Where does the Light go?". (also on youtube)

So the question to ask is "if incoherent light could interfere destructively, where would the energy go?" Back to the sun? Each pulse of light coming from the sun is a mixture of zillions of wave coming from all the hot plasma surrounding the sun, all of them incoherent in time and space. The probability of a pure frequency pulse existing in sunlight and interfering with an exactly similar one in order to see the effect of the laser above, is very very small, if not improbable , since they must have come from the same delta(t) and delta(xy,z)

• I find that particular demonstration very interesting. The interference effect is not obtained with slits, but with the partial reflection of beamsplitters. With slit setup the interference effect occurs, presumably, as the electromagnetic waves undergo diffuse reflection on contact with the screen. With the beam splitter setup the interference effect occurs, by the looks of it, at the beam splitter surface. When no light reaches the screen there is constructive inference for light being reflected back to the source. Until now I had not quite appreciated the role of reflection. Jun 18, 2021 at 5:52
• More specifically on the role of reflection: of course mirror reflection will not give rise to interference effect. The types of reflection that do give rise to interference effect, by the looks of it, are forms of reflection that involve a form of randomness: the diffuse reflection of a screen, the partial reflection of a beam splitter, and partial reflections at the two surfaces of a thin layer. Jun 18, 2021 at 6:05
• I agree that probability that two randomly chosen sources will cancel is very small. But there are many sources uniformly distributed. By analogy, consider an ion surrounded by a uniformly charged spherical shell. If you choose two subsets of the shell at random, there is very low probability that their forces on the ion cancel. But since the charge is uniformly distributed, we are guaranteed that every subset of the shell does indeed have another subset such that the forces cancel. Jun 18, 2021 at 15:22
• see my answer above Jun 18, 2021 at 16:59

That phase is random does not mean that the waves of all the phases are present at any space point at any time. The averaging happens in the eye (or photodetector), which has the reaction time and the spatial resolution bigger than the coherence time and coherence length of the light. This is where the model described in the OP applies... except the eye/photodetector does not register the amplitude of the electromagnetic wave, but its intensity: $$I \propto \left[E(x,t)\right]^2$$ For this quantity the averaging gives a finite result.

Remarks

• If our eyes/detectors were measuring the wave amplitude rather than its intensity, then they would not be able to perceive even coherent light, due to averaging over times and lengths much greater than the period and wavelength of light.
• @uhoh has brought up a useful analogy in the comments: Why doesn't white noise cancel itself out? White noise actually cancels itself out in the same sense, as implied in the OP: it has zero (or constant) average. It is the intensity of the white noise that is not zero.

Supplementary: Modeling incoherent light
Incoherence may come from many sources:

• different atoms emit at different times, with different frequencies, different polarizations, and in different directions
• the light may come from different sources
• the observed light may be coming not directly from the source, but after multiple reflections

Thus, the light observed at point $$\mathbf{x}$$ is a sum of many waves: $$\mathbf{E}(\mathbf{x},t) = \sum_i \mathbf{E}_i(\mathbf{x},t)$$ Now, even if we assume that all these waves are plane waves with random amplitides and initial phases, we have $$\mathbf{E}(\mathbf{x},t) = \sum_i \mathbf{A}_i\cos(\mathbf{k}_i\mathbf{x} - \omega_i t +\phi_i)$$ We can now meaningfully consider this as a random wave field and characterize it by its correlation functions: $$K_{\alpha\beta}(\mathbf{x},t;\mathbf{x}',t') = \langle E_\alpha (\mathbf{x},t)E_\beta(\mathbf{x}',t')\rangle$$

Update
In more rigorous quantum optics terms, one uses correlation coefficient rather than the correlation function to characterize the coherence of light, see degree of first order coherence, and also the Loudon's The Quantum Theory of Light.

More references

• The key insight is that while the probability distribution is centered on zero (and has its mode there), the expected magnitude is not zero. Jun 18, 2021 at 13:28
• I think the issue of amplitude vs intensity is a bit confusing. I can see that the average amplitude (intensity) of a sine wave is zero (positive). However, if two overlapping sine waves cancel perfectly, then there really is no signal and so the average amplitude and intensity are both zero. Jun 18, 2021 at 18:49
• I am starting to understand the idea that the "imperfect" uniformity of the random distribution of components is what gives rise to a nonzero signal. But I still don't see what determines the correlation functions you describe, or how they play into the analysis. Jun 18, 2021 at 18:54
• @WillG one has to distinguish averaging and superposition of waves. Averaging over phase, i.e., integrating it from $0$ to $2\pi$ is done only after the amplitude is squared - because the detectors are quadratic. Summation of waves coming from different points occurs before squaring, but they do not necessarily have uniform continuum of phases in interval $[0, 2\pi]$ - we obly know that each phase is random and needs to be averaged afterwards. Jun 18, 2021 at 18:59
• I'm also still curious about the "quantum" part of my question: It seems like the final intensity will depend statistically on the number of photons. So could white light intensity have been used as an early way to determine the relation $E=\hbar\omega$? Jun 18, 2021 at 21:56

I'm not in a position to give a definite answer, but I would like to mention an example of a case where sunlight does give rise to interference effect. It may be interesting to compare the cases.

If there are puddles on the pavement, and some gasoline has been spilled, the gasoline forms a thin film. As we know, when the thickness is down to about the wavelength of light there are interference effects, arising as a consequence of some of the light being reflected back-and-forth internally. The result is that each thickness of the gasoline layer takes on a different hue, depending on which wavelengths of the constituting light have undergone destructive interference.

Temporal coherence
The time scale of the internal reflection is extremely short. On that short time scale the light can be treated as temporally coherent. Above a certain thickness of the layer the time scale of internal reflection is large enough that the fact that the source is not temporally coherent comes into play, and then there is no color effect to the interference effects.

Spatial coherence
The source does not have spatial coherence; the sunlight is entering the layer from all directions. When the layer is sufficiently thin there is no room to act independently. The source, sunlight, is not spatially coherent, but confinement of the reflections to the thin layer is in effect creating the condition of spatial coherence.

In order to get interference effect that is visible at macroscopic scale the requirements of both temporal coherence and spatial coherence must be met.

As you state, in the setup you describe there is no macroscopically visible interference effect. It must be the case that the requirements of temporal coherence and spatial coherence are not met. (Either of the two not met or both not met.)

I copy here comments I wrote to the answer by Physics SE contributor Anna V.

Anna V linked to a video with a demontration of obtaining interference effect with beam splitters.

I find that particular demonstration very interesting. The interference effect is not obtained with slits, but with the partial reflection of beamsplitters. With slit setup the interference effect occurs, presumably, as the electromagnetic waves undergo diffuse reflection on contact with the screen. With the beam splitter setup the interference effect occurs, by the looks of it, at the beam splitter surface. When no light reaches the screen there is constructive inference for light being reflected back to the source.

More specifically on the role of reflection: of course total mirror reflection will not give rise to interference effect. The types of reflection that do give rise to interference effect are, by the looks of it, forms of reflection that involve a form of randomness: the diffuse reflection of a screen, the partial reflection of a beam splitter, and partial reflections at the two surfaces of a thin layer.

The point is: the interference effect does not occur in transit. During transit the waves remain in lossless superposition. The recurring pattern is that interference effect occurs when propagating light enters into a process of interfacing with something else.

• see my answer above Jun 18, 2021 at 16:59

Light never cancels itself out, if it did that would be a violation of conservation of energy. Like waves in water or sound in air, the medium only transmits the wave it never destroys it. Water waves crash on the beach, sound gets absorbed in lossy materials etc .. the EM field never absorbs energy for light. Light is created by an excited electron and only absorbed by another electron.

What is amiss is that you are taking the word "interference" as you are taught in high school or first year physics as a superposition that nets a big zero! This math is not physics! Also the word "interference" was initially used in 1801 for Young's DSE ... as the image was similar to water wave interference. Essentially what you/we are still taught today is based on 1801! Quantum optics courses teach it correctly. It is note worthy that the mathematics of "interference" i.e. waves 180 degrees out of phase is very similar to the quantum math i.e. Feynman path integral.

Anna V mentions the MIT video above where the prof seems to say it's a mystery where the energy is going with his "destructive interference" setup ... he is in error .... when he sets the mirror for destructive interference his laser actually stops lasing, this could have been evidenced by noting the loss in power consumption on his laser power supply. External mirrors are no different than the internal mirrors of the laser .... upset the path length and the laser does not lase.

• I think it shows the energy of the interference beam going back to the source ,inot the laser stops lasing. you must not have watched the video. I added the youtube link Jun 18, 2021 at 18:52
• I think the OP doesn't claim that the net field vanishes - rather it follows from the model that they use. The energy argument is very general, but does little to resolve this problem. Since you have taken a formal course in quantum optics, perhaps you could throw in a few relevant equations for describing incoherent field, and solve this conclusively. Jun 18, 2021 at 19:12
• @RogerVadim if this is addressed to me , I am a retired experimental particle physicist and have taken no course in optics. I just try to ground physics questions to experimental facts, like conservation of energy as in my answer Jun 19, 2021 at 4:40
• @annav this is addressed to the answer above by PhysicsDave - the second paragraph makes claims that question is easily resolved with some math and basic quantum optics. I think the language is rather condescending, but I am trying to be constructive and call the author to justify their claims. Jun 19, 2021 at 5:06
• @RogerVadim "the energy argument is very general"? More correctly stated the energy argument is FUNDAMENTAL. The OP asks why light doesn't cancel itself out, in any ideal lossless medium superposition is transient, adding sine waves is fun when observing the medium directly (water, voltage, sound) but we can not observe the EM field directly. Drop stones in a lossless pond and the waves last forever, yes at some brief times the water can superimpose to zero but the waves reemerge, if you can get this concept then you have a basic understanding of QO, no formulas required. Jun 20, 2021 at 12:03