I haven't yet gotten a good answer to this: If you have two rays of light of the same wavelength and polarization (just to make it simple for now, but it easily generalizes to any range and all polarizations) meet at a point such that they're 180 degrees out of phase (due to path length difference, or whatever), we all know they interfere destructively, and a detector at exactly that point wouldn't read anything.

So my question is, since such an insanely huge number of photons are coming out of the sun constantly, why isn't any photon hitting a detector matched up with another photon that happens to be exactly out of phase with it? If you have an enormous number of randomly produced photons traveling random distances (with respect to their wavelength, anyway), that seems like it would happen, similar to the way that the sum of a huge number of randomly selected 1's and -1's would never stray far from 0. Mathematically, it would be:

$$\int_0 ^{2\pi} e^{i \phi} d\phi = 0$$

Of course, the same would happen for a given polarization, and any given wavelength.

I'm pretty sure I see the sun though, so I suspect something with my assumption that there are effectively an infinite number of photons hitting a given spot is flawed... are they locally in phase or something?

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    $\begingroup$ You might be interested to read the article by @LubosMotl which discusses how classical fields emerge from a quantum theory of particles motls.blogspot.gr/2011/11/… $\endgroup$
    – anna v
    Jul 4, 2013 at 19:05
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    $\begingroup$ I've sometimes thought about similar questions, despite coherence and other details, about two people playing the same note with a trumpet. Should not it be silent ~1/2 of the attempts? :-D Music would be very different... $\endgroup$
    – Andrestand
    Sep 14, 2014 at 9:25
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    $\begingroup$ Re: "an insanely huge number of photons are coming out of the sun constantly" If I recall correctly, the sun turns 4 tons of its mass in to light every second, and the Earth intercepts 2 ounces worth. $\endgroup$
    – user95006
    Feb 11, 2016 at 21:38
  • $\begingroup$ If you are saying photons.. You already assumed light as a particle and particles don't interfere $\endgroup$
    – brainst
    Feb 14, 2016 at 13:49
  • $\begingroup$ I think @Andrestand makes a good point about cancelation. Two trumpets are unlikely to be tuned to an exact same frequency with all the variables - down to the type of metal they are made of, how much air escapes through the key holes, very finite things like exact cavity shape, air pressure, reed type etc... The variables are so infinitely vast that cancellation between 2 is impossible. Perhaps differences in EM frequency can be just as finite. Excellent question. $\endgroup$ Jun 18, 2020 at 19:40

7 Answers 7


First let's deal with a false assumption:

similar to the way that the sum of a huge number of randomly selected 1's and -1's would never stray far from 0.

Suppose we have a set of $N$ random variables $X_i$, each independent and with equal probability of being either $+1$ or $-1$. Define $$ S = \sum_{i=1}^N X_i. $$ Then, yes, the expectation of $S$ may be $0$, $$ \langle S \rangle = \sum_{i=1}^N \langle X_i \rangle = \sum_{i=1}^N \left(\frac{1}{2}(+1) + \frac{1}{2}(-1)\right) = 0, $$ but the fluctuations can be significant. Since we can write $$ S^2 = \sum_{i=1}^N X_i^2 + 2 \sum_{i=1}^N \sum_{j=i+1}^N X_i X_j, $$ then more manipulation of expectation values (remember, they always distribute over sums; also the expectation of a product is the product of the expectations if and only if the factors are independent, which is the case for us for $i \neq j$) yields $$ \langle S^2 \rangle = \sum_{i=1}^N \langle X_i^2 \rangle + 2 \sum_{i=1}^N \sum_{j=i+1}^N \langle X_i X_j \rangle = \sum_{i=1}^N \left(\frac{1}{2}(+1)^2 + \frac{1}{2}(-1)^2\right) + 2 \sum_{i=1}^N \sum_{j=i+1}^N (0) (0) = N. $$ The standard deviation will be $$ \sigma_S = \left(\langle S^2 \rangle - \langle S \rangle^2\right)^{1/2} = \sqrt{N}. $$ This can be arbitrarily large. Another way of looking at this is that the more coins you flip, the less likely you are to be within a fixed range of breaking even.

Now let's apply this to the slightly more advanced case of independent phases of photons. Suppose we have $N$ independent photons with phases $\phi_i$ uniformly distributed on $(0, 2\pi)$. For simplicity I will assume all the photons have the same amplitude, set to unity. Then the electric field will have strength $$ E = \sum_{i=1}^N \mathrm{e}^{\mathrm{i}\phi_i}. $$ Sure enough, the average electric field will be $0$: $$ \langle E \rangle = \sum_{i=1}^N \langle \mathrm{e}^{\mathrm{i}\phi_i} \rangle = \sum_{i=1}^N \frac{1}{2\pi} \int_0^{2\pi} \mathrm{e}^{\mathrm{i}\phi}\ \mathrm{d}\phi = \sum_{i=1}^N 0 = 0. $$ However, you see images not in electric field strength but in intensity, which is the square-magnitude of this: $$ I = \lvert E \rvert^2 = \sum_{i=1}^N \mathrm{e}^{\mathrm{i}\phi_i} \mathrm{e}^{-\mathrm{i}\phi_i} + \sum_{i=1}^N \sum_{j=i+1}^N \left(\mathrm{e}^{\mathrm{i}\phi_i} \mathrm{e}^{-\mathrm{i}\phi_j} + \mathrm{e}^{-\mathrm{i}\phi_i} \mathrm{e}^{\mathrm{i}\phi_j}\right) = N + 2 \sum_{i=1}^N \sum_{j=i+1}^N \cos(\phi_i-\phi_j). $$ Paralleling the computation above, we have $$ \langle I \rangle = \langle N \rangle + 2 \sum_{i=1}^N \sum_{j=i+1}^N \frac{1}{(2\pi)^2} \int_0^{2\pi}\!\!\int_0^{2\pi} \cos(\phi-\phi')\ \mathrm{d}\phi\ \mathrm{d}\phi' = N + 0 = N. $$ The more photons there are, the greater the intensity, even though there will be more cancellations.

So what does this mean physically? The Sun is an incoherent source, meaning the photons coming from its surface really are independent in phase, so the above calculations are appropriate. This is in contrast to a laser, where the phases have a very tight relation to one another (they are all the same).

Your eye (or rather each receptor in your eye) has an extended volume over which it is sensitive to light, and it integrates whatever fluctuations occur over an extended time (which you know to be longer than, say, $1/60$ of a second, given that most people don't notice faster refresh rates on monitors). In this volume over this time, there will be some average number of photons. Even if the volume is small enough such that all opposite-phase photons will cancel (obviously two spatially separated photons won't cancel no matter their phases), the intensity of the photon field is expected to be nonzero.

In fact, we can put some numbers to this. Take a typical cone in your eye to have a diameter of $2\ \mathrm{µm}$, as per Wikipedia. About $10\%$ of the Sun's $1400\ \mathrm{W/m^2}$ flux is in the $500\text{–}600\ \mathrm{nm}$ range, where the typical photon energy is $3.6\times10^{-19}\ \mathrm{J}$. Neglecting the effects of focusing among other things, the number of photons in play in a single receptor is something like $$ N \approx \frac{\pi (1\ \mathrm{µm})^2 (140\ \mathrm{W/m^2}) (0.02\ \mathrm{s})}{3.6\times10^{-19}\ \mathrm{J}} \approx 2\times10^7. $$ The fractional change in intensity from "frame to frame" or "pixel to pixel" in your vision would be something like $1/\sqrt{N} \approx 0.02\%$. Even give or take a few orders of magnitude, you can see that the Sun should shine steadily and uniformly.

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    $\begingroup$ Good answer. I like the explanation in terms of coin tosses and distance from exact cancellations. There is an air of Zeno's Paradox about this question - the OP's assumptions in the question could equally be used to show that objects do not radiate heat, the sea is perfectly flat, and that erupting volcanoes should be silent. $\endgroup$ Jul 4, 2013 at 9:20
  • $\begingroup$ Great answer, but I'm wondering about the meaning of "spatially separated photons" in your discussion. Any comment on that? $\endgroup$
    – baptiste
    Jul 4, 2013 at 14:03
  • $\begingroup$ Hi, thanks for the in depth response. I'm basically ready to accept it, but there seems to be some funky stuff in your math, to me: your line of $E = \sum _i e^{i\phi_i}$ seems to be the same as my original one, averaging the electric field at a point over a huge number of particles, but you use a sum rather than an integral. But then, later, when finding $\langle E \rangle$, you have an integral over the same variable, inside the sum..? I'm not sure why. $\endgroup$
    – user22862
    Jul 4, 2013 at 14:51
  • $\begingroup$ @ChrisWhite thanks for the clarification. I'm always wary of associating an intrinsic volume to the word photon, especially when the rest of the discussion would well apply to plane waves. $\endgroup$
    – baptiste
    Jul 4, 2013 at 14:56
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    $\begingroup$ @Chris White -- I believe there should be a "$j = i + 1$" in the sum index of equations (3) and (4) and possibly also below. For instance: $(\Sigma_{i = 1}^N X_i) (\Sigma_{j = 1}^N X_j) = \Sigma_{i = 1}^N X_i^2 + 2 (\Sigma_{i = 1}^N \Sigma_{j = i + 1}^N) X_i X_j$. $\endgroup$
    – user12262
    Jul 11, 2013 at 0:40

Chris White wonderfully addresses this with some Statistics, but there's a less math-y way of looking at it, too. Firstly, to dispel this notion:

So my question is, since such an insanely huge number of photons are coming out of the sun constantly, why isn't any photon hitting a detector matched up with another photon that happens to be exactly out of phase with it?

There's an equal chance that a photon will be matched up with another photon of the same phase as it will be with an opposite phase. The phase of each photon entering is an independent variable. If we're talking about two photons, then there's an equal chance of constructive interference a there is of destructive interference. This holds even if you scale up. (See last section if you're not convinced of this)

There are basically three things that you must note here:

  • The average value of a distribution is not always the most likely value. Indeed, it may not even be a possible value.
  • Our eyes measure intensity, not amplitude. We do not distinguish between positive and negative amplitude. Rhodopsin works by absorbing energy, which does not distinguish between the sign of the phase
  • Interference is local, not global. If one of your retinal rod cells receives positive-phase light and the other receives negative-phase light, there will be no cancelling out.

Energy conservation argument

Here's a very simple way of looking at it. Due to energy conservation, if there is destructive interference there must be constructive interference elsewhere. Otherwise one could cleverly place detectors and create/destroy energy at will.

Since the light from the sun is incoherent, at any given point in time, approximately half of the spots on a sphere drawn around it will have constructive interference, and half will have destructive (not necessarily full destructive, just that the net energy is less) interference. These spots will change at random -- if a spot had constructive interference at one moment, it could have destructive interference the next.

With this in mind, there will always be some significant fraction of your rod/cone cells (which take up a small sliver of this imaginary sphere) receiving constructively-interfered light. That's enough for you to be able to see.

Why it holds even when you scale up

I'm using + to mean positive phase, and - to mean negative phase. I'm neglecting the fact that phase is not just a binary value, as this involves calculations (see Chris White's answer). A number next to the sign is the new amplitude if it has changed.

The basic thing going on here is that the mean value is not always the most probable value. Take the case of three photons:

 1   2   3   Amplitude  Intensity
 +   +   +   +3         9
 +   +   -   +1         1
 +   -   +   +1         1
 +   -   -   -1         1 
 -   +   +   +1         1
 -   +   -   -1         1
 -   -   +   -1         1
 -   -   -   -3         9

(Avg intensity is 3)

Note the lack of a 0 in the output column. 0 is the mean output amplitude, but it never is observed as a value of the output phase. In the case of a continuous set of phases, a case of total destructive interference is possible, and it is the mean phase, however, there are man, many other final phase values that are more probable.

If you make this chart for any odd value, you'll always have no total destructive interference. If you make it for any even value, half of the time you get destructive interference, however the other half you get constructive interference, so total destructive interference does not occur. In all cases, the average intensity will always be equal to the number of incident photons. You can scale this up as much as you want, it won't change.


Your integral is a great representation for the sum of a collection of oscillators that are coherent in time and have the same amplitudes. But your critical mistake is in assuming that those oscillators have constant frequencies and amplitudes. That's just not true, because the sources for each of those oscillators is changing violently in time. (The "surface" of the sun is a violent place.) And that means that your integral isn't a good model for the sun.

In particular, all those different oscillators have different amplitudes. And your integral represents the limit of a sum of a really large number of oscillators, with all different amplitudes. So it should really be more like \begin{equation} \int_0^{2\pi} A(\phi)\, e^{i\phi} d\phi~, \end{equation} where $A(\phi)$ is the total amplitude of all the oscillators with phases between $\phi$ and $\phi+d\phi$ (loosely speaking). And this integral is not zero except for very special functions $A(\phi)$. And in random processes, "very special" things happen "almost never".

So the question becomes: what is $A(\phi)$? Well, it's time-dependent because it represents the state of the oscillators at that instant in time. But just thinking of one instant of time, it's the sum resulting from a pretty random distribution of oscillators. Now, you do have a really large total number of oscillators (because the sun is large), but it's still a finite number. And the integrand narrows that finite number down to an infinitesimal. So $A(\phi)$ won't actually be averaging over a large number of oscillators at all. Even if the average for $A(\phi)$ were zero, you would never really get zero; it would generally be some random nonzero number. It certainly won't be a constant function of $\phi$. And there's no reason for it to be periodic in $\phi$. Therefore, the integral will be nonzero in general.

In fact, the total value of the integral will essentially be a random number. So you can ask, what's the probability a random (real) number being exactly zero? And the answer is: zero. You will never see a perfect total cancellation of photons from the sun.

  • $\begingroup$ Hi, I don't think the "constant amplitudes and frequencies" thing is a problem, either. For the amplitude one, multiply the integral I gave above by an amplitude A, and integrate that from 0 to whatever you want. The inner (my original) integral is still 0. Likewise for the frequency (or wave number k, whatever). Instead of just having iphi in the exponent, have i(kr + phi), and integrate k from 0 to infinity. You can similarly pull out the phi part again, and the integral is still 0. $\endgroup$
    – user22862
    Jul 3, 2013 at 20:57
  • $\begingroup$ I ran out of space, but the point of all that is just that, with a huge amount of random waves, for any one, you can pair it up with an "opposite" one in every respect (phase, amplitude, wavelength, polarization, etc). $\endgroup$
    – user22862
    Jul 3, 2013 at 21:05
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    $\begingroup$ But your integral represents the limit of a sum of a really large number of oscillators. If you think of that sum, I'm saying that each oscillator would have a different amplitude. In the limit, that means you would have to multiply by an amplitude $A(\phi)$ corresponding to the total amplitude of all the oscillators with phases between $\phi$ and $\phi + d\phi$. That integral is not zero, unless $A(\phi)$ is constant. $\endgroup$
    – Mike
    Jul 3, 2013 at 21:07
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    $\begingroup$ You can't pull it out of the integral like that. Maybe you could make a statement about the expectation value of the integral, but not about its value -- and those will be different. $\langle A(\phi) \rangle \neq A(\phi)$. $\endgroup$
    – Mike
    Jul 3, 2013 at 22:09
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    $\begingroup$ As Chris White's excellent answer points out, the average value of the integral will be zero. But you will "almost never" see that value; you will see values in a range above and below zero. The typical size of that range will be something like the square root of the intensity, which is nonzero. $\endgroup$
    – Mike
    Jul 3, 2013 at 22:29

Even though you are talking of photons, you are not thinking of them as particles.

Particle means that given a screen depicted with an x and y axis , (or your retina), each individual photon will hit at a specific (x,y) point and be detected as a particle. The interference appears with a build up of many many individual hits on the screen, if there is the necessary phase coherence.

It is true that the classical wave framework of light blends smoothly with the photon particle framework but that does not mean that the individual photons are spread out all over the (x,y) plane. Each will hit one point. It might help if you contemplate the build up of the quantum mechanical probabilistic interference pattern one electron at a time in the two slit experiment which shows the interference pattern, a probability distribution. Photons are equally particles and quantum mechanical probability waves.

The zillions of photons from the sun are not coherent and their hits will appear randomly on the screen; or your retina creating an image of the sun, but be carefull to wear appropriate goggles so as not to burn it.

Edit in answer to comment:

The concept of light as waves works because there is consistency between the particle/probability_wave nature of the photon and the classical electromagnetic wave that creates interference patterns visible to the naked eye . When one mixes the two concepts, photon and classical wave, paradoxical situations seem to appear. Chris (photons) and Mike (classical waves) are giving you the mathematics of it. In your question you mix the two frameworks, classical wave and photons. When you say 1s and -1s will add up statistically close to zero you are using the particle concept, because the addition happens at a specific (x,y). When you are assigning the pluses and minuses you are using the classical concept , where the phase is kept over the whole x,y plane. This is not true for incoherent sources from the sun. It is true for lasers where the two frameworks overlap consistently and phases are kept over the x,y plane. The sun is not a laser. If it were a laser, depending on the position of the screen interference patterns would appear, and there would be regions with zero energy, the energy having gone to the bright regions. Energy is conserved in all physics frameworks.


Two photons of the same wavelength do not interfere destructively everywhere. Typically, you would obtain fringes. The total energy remains the same as two photons but distributed differently. For two other photons, you might obtain another pattern. If you add many many photons, all these patterns will merge so you won't see them (You can see interferences only when most of the photons are coherent). Overall, what you can see is a uniform irradiation.

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    $\begingroup$ I know they don't interfere destructively everywhere, but my point is that at a farther point, they'll again each be matched up with another "opposite" photon that will again interfere destructively. $\endgroup$
    – user22862
    Jul 3, 2013 at 20:10
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    $\begingroup$ I have the impression that you see photons as particles which get somehow destructed when they 'hit' each other. This is incorrect. They are, in some way, a field, or wave, that occupies a given volume. At the intersection of two of these volumes, the fields get scrambled. In some points it becomes 0, in others it becomes twice higher. The total energy is conserved, so when you launch two photons, you detect two photons. The destructive interference does not destroy the photons. They are simply located elsewhere. $\endgroup$
    – fffred
    Jul 3, 2013 at 20:16
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    $\begingroup$ But his point is that for every pair that interferes destructively, you have a massive number that don't in the same point in space, so you don't see the interfering photons but the millions of others that don't happen to be in phase at your point. $\endgroup$
    – tpg2114
    Jul 3, 2013 at 20:16
  • $\begingroup$ @fffred, I know they don't actually get destroyed, their E and B fields just cancel at that exact point. But my confusion is, at any given point, a nearly infinite number (which I'm assuming, maybe wrongly) of randomly out of phase photons should all interfere destructively, at that point, but also basically at every point. $\endgroup$
    – user22862
    Jul 3, 2013 at 20:30
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    $\begingroup$ @dmckee, correct me if I'm wrong, but you seem to be saying my assumption of a huge number of photons at every given point is right, but my following logic/math is wrong. Could you tell me how you'd change the above integral to match what you said? $\endgroup$
    – user22862
    Jul 3, 2013 at 20:38

I would prefer to deal with scattering waves instead of photons (it’s too hard for me to envision photons with frequency) but the answer is the same.

Naively, I first would say that light coming from the sun to the earth is an example of scattering in the forward direction and is in-phase. Why? Sunlight, coming from a far distance, scatters from the atmosphere and all the scattered wavelets add constructively (their light paths don’t change very much) with each other in the forward direction. Therefore, the waves all arrive at earth pretty much in-phase.

However, if we bring in some lateral scattering, then I think about it this way: sunlight arriving at earth’s atmosphere (made up of zillions of independent molecules randomly arrayed) will have secondary wavelets with phases that have no particular relationship to one another. That is, the wavelets arriving at some point P has a jumble of different phases and tend not to interfere in a sustained constructive or destructive fashion. So to answer your question: some photons do interference destructively but not in a sustain fashion.

This is best appreciated from a phasor viewpoint – as the wavelets arrive at some point P the phasors have randomly large phase angles differences with respect to each other. When adding the tips-to-tails, they sum up to zero, exactly like your integral shows.

enter image description here


Am I missing something or is the explanation much, much simpler than all the previous answers?

It's analogous to asking, "Aren't there so many waves in the ocean that they should all cancel?" - Waves only cancel at a point, then continue to pass through each other and this process does not destroy energy which is what our eyes actually see.

Photons from the Sun do not cancel often because it is nearly impossible to 2 photons to have been generated in the same space and time. If a ray travels from the Sun to your eye, it travels in a straight line (or reflects, refracts, etc). In order for another photon to be exactly 1/2 step (180 degrees) out of phase with it. Part of the actual wave-front would have to overlap with the first photon's wave-front and continue to do so all the way along that straight line. This geometrically gives rise to exactly 1 position that the photon could originate from (or travel through) at one exact quanta of time. If the H/He atoms in the Sun emitting the first photon also experience the second, cancelling photon emerging from behind it at that point in time, it is very likely to absorb and possibly re-emit it a short time later.

We see interference patterns in the two-slit experiment because the diffracted light rays are at an angle converging to each other, if they were parallel (or diverging) as they are in the Sun, one would expect absolutely no cancellation at long distances.

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    $\begingroup$ The statement in bold is true, but it's not an answer. Your understanding of interference is incorrect. Photons don't have to "have been generated in the same space and time" and converging/diverging doesn't eliminate interference. You might think of it more like this: the distances the photons have to have traveled must be different by half a wavelength. So if you want to cancel one photon, the possible origins of the second photon are anywhere on an infinite series of spheres -- each of which has infinitely many points. $\endgroup$
    – Mike
    Mar 4, 2015 at 13:37
  • $\begingroup$ Mike, thanks for the correction. I think I do understand interference but I'm having trouble with the geometric aspect of it. The set of spheres that you are describing would generate rays that can intersect only at a point in space, not along a line. Eyes absorb light in 3 dimensions, not at a point, so no single point cancellation persists. Would you mind helping me with this offline somehow? I don't want to dilute this question. $\endgroup$
    – WillMcLeod
    Mar 5, 2015 at 14:18
  • $\begingroup$ I think I understand the confusion. If this is correct, I'll post a new answer. The OP is modeling a photon hitting the eye like a pen marking a dot on a piece of paper (1D and at a point). It might more accurately be pictured as a meteorite crashing into a field of corn. The retna is not a one atom thick gold foil. It is a several (hundred trillion) atom deep cell array and if the waves of incoherent light cancel, they do so over a minute distance, and are absorbed once they pass through each other (like waves on the ocean) $\endgroup$
    – WillMcLeod
    Mar 5, 2015 at 14:52
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    $\begingroup$ But if the OP's argument were correct at any point, it would apply at every point in space. In fact, we can say that the electromagnetic field will be zero almost never -- that is, there is 0 probability of it being exactly 0 at any point. There's no intermediate-value theorem, or topological restriction, or anything like that; it's just infinitely unlikely to ever happen anywhere. $\endgroup$
    – Mike
    Mar 5, 2015 at 16:43