How come unpolarized light does not undergo destructive interference?

As far as I know, unpolarized light is defined as light which has no clear axis of polarization, but electric field vectors in all directions. However, I also know that electric fields can be superpositioned. So this leads me to the conclusion that if unpolarized light has equal $$\vec E$$ vectors in all directions, they should cancel.

Obviously, though, this is wrong (we've got light!). What's the explanation?

• Keep in mind that you're eyes are not sensitive to things that happen on time-scales similar to a single optical oscillation ($\approx 10^{-15}\,\mathrm{s}$), but only to averages over huge numbers of such oscillations. Jun 3, 2016 at 17:12
• A light source is made of larger number of small light emitters. In order to get the intensity you have to add the field amplitude from each one of them. Let us say you add field of two emitters you will get $I=I_1+I_2+2I_1I_2cos(\phi)$. As pointed out by several others if phase is not constant in time, the cosine term will average 0 and only intensity addition is seen. For destructive interference the two waves should be out of phase constantly. For two waves with polarization in opposite directions, if their phase relationship is not constant and there will be no destructive interference. Jun 4, 2016 at 17:49

One way of thinking about it is to consider light as made up of lots and lots of photons. You're not going to see destructive interference unless the polarizations of all of these photons are lined up and they all undergo destructive interference at the same time. If they're random, some of them will interfere destructively, some of them will interfere constructively, and on average the interference will cancel out and we'll see the average intensity of the light stay the same.

• But if they're random, wouldn't each photon find another photon that can interfere to it destructively? Jun 3, 2016 at 11:36
• @philip_0008: For that to work, wouldn't you need to be able to pair up the photons so that every single pair of them gives complete destructive interference? You can't do this if the photons are random. It's not like if two photons interfere destructively they cancel each other out and are never seen again. They probably aren't going in exactly the same direction, so when they stop overlapping they stop interfering. Jun 3, 2016 at 11:42
• It still confuses me. Especially when I look at the summation $\sum A\sin kx$ and randomizing from -A to A, the average of the randomization is zero. Jun 3, 2016 at 11:51
• @philip_0008: The average of the randomization is zero, but the average of the square of the randomization, which is the relevant quantity, is $n A^2$. Suppose you have $n$ random variables, each of them equally likely to a $+$1 and a $-$1. When you add them up, you get something whose absolute value is roughly $\sqrt{n}$. This is like the amplitude. Square it, to get the intensity, and you have something whose average is $n$, just the same as if you'd added $n$ values of $+1$ each. Jun 3, 2016 at 12:32
• @Alettix: If you flip a coin 1,000,000 times, the chances of getting exactly 500,000 heads and exactly 500,000 tails are around 0.08% (8 in 10,000) and not 100%. The difference between heads and tails will be on the order of 1,000, and when you square the difference, the expected value of the square will be 1,000,000. Jun 5, 2016 at 17:08

I tried to confirm the suggestion of @Peter Shor in the comments by creating a simple program that generates vectors by randomizing the angle $\phi$ from $0$ to $2\pi$ of a vector $\vec A_i = A\hat r_i$ that lies on a x-y plane. All the vectors have the same amplitude $A$. The program adds (by vector addition: summing all x-components and y-components) n = 10000 of the randomized vectors:
$\vec A_{tot} = \sum \vec A_i$.
After that, I obtain the square of the magnitude $A_{tot}^2$ of the resultant vector. Then repeating the procedure for 100000 times, and obtained the average.
Indeed, I got a result very near $(A_{tot}^2)_{ave} = nA^2 = 10000A^2$, which should make sense if we assume that $\vec A_i$ is the electric field of one photon with random axis of polarization.
Because the intensity of light is proportional to the square of the amplitude of electric field:
$I = \frac {1}{2}c\epsilon_0E_0^2$
then we would expect that by superimposing n photons with random polarization, the Intensity would increase to:
$I_{tot} = n\frac {1}{2}c\epsilon_0E_0^2 = \frac {1}{2}c\epsilon_0(nE_0^2) = \frac {1}{2}c\epsilon_0 E^2_{tot}$, and indeed, in the program we got $nA^2$.
The average energy per unit volume in one cycle is also proportional to the square of the amplitude of electric field: $\langle u \rangle = \frac {1}{2}\epsilon_0E_0^2$,
and following the same logic, this would mean no loss of energy after superimposing $n$ random photons.

• There's still something bothering me about this, idk. But I'm not completely convinced. Jun 4, 2016 at 14:48
• Another way to approach this with a visual aid is in my paper "single edge certainty "at the top of my page. Jun 5, 2016 at 19:42

Unpolarized light is more properly called light with random polarization. That makes it more clear what it means: the polarization state (circular, linear, elliptic) varies randomly over space, wavelength, and time.

Consider the scenario below, where a diffuse light source is converted to a collimated beam with a narrow range of wavelengths $\lambda\pm\Delta\lambda$. The beam is then split by a polarizer.

If you took an intensity profile of the two polarization components with a sufficiently short exposure time, you would get two light/dark patterns that would clearly show that some parts of the beam are clearly s-polarized (polarization vector perpendicular to the drawing plane), others clearly p-polarized, and yet others something in between or even dark. However, if you took another picture, the pattern would look completely different.

If the exposure time is too long, then you would get two images that are both a uniform 50% gray image. The maximum shutter time $\tau$ for which you can see a snapshot of the polarization would depend on the bandwidth $\Delta\lambda$ as $$\tau\approx \frac{\lambda^2}{2\pi c\Delta\lambda},$$ where $c=3\times10^8~\mathrm{m/s}$ is the speed of light. For example, if you do this with narrowband red light ($\lambda = 650\pm1$ nm), you would need an exposure time of 200 femtoseconds (2E-13 s). So, even though the light has a definite polarization at a particular point in time, you will not notice it in practice.

There is another effect that makes it difficult to notice the instantaneous polarization of randomly polarized light: the length scale of the spatial intensity fluctuations in the two sensor images will depend on the size of the pinhole. For a beam with 1 cm diameter and a lens with 10 cm focal length, the pinhole would need to be 20 $\mu$m or so to see the pattern clearly. If you increase the size of the pinhole, then the patterns will become finer and finer until you cannot resolve them anymore. And if the pinhole is so small and the exposure time is so short, you would need to start with a pretty bright light source in order to see anything at all.

• No you wouldn't. The intensity of each polarization separately is a totally random variable, and they are uncorrelated with each other. Jun 4, 2016 at 23:20
• @PeterShor - I stand corrected. I changed the sentence directly after the picture and made a change to the picture as well. Jun 5, 2016 at 6:42

I will try a classical light wave explanation and hope that someone smart will come up with a full quantum mechanical description to enlighten me as well.

Lets imagine we have a light source which is providing unpolarized light at a single frequency. When we measure the lights intensity, we are actually measuring the absolute value of the electric field vector. Lets assume the measured intensity is constant over time (we of cause measured long enough, to see the effective electrical field).

If we now put in a linear polarizer the intensity drops by a factor of 2, no matter how we put it in. From that we learn, that the polarization axis, i.e. the axis parallel to the oscillation direction of the electric field, is rotating perpendicular to the lights propagation direction. We can repeat this experiment with a circular polarizer and get the same result. So the rotation of the polarization axis cannot have a constant angular speed.

You can imagine that the polarization axis is turning around in a chaotic manner. This makes it impossible to find a polarizer which on a sufficient long time scale is able to separate the light in two different polarization types. If we want to construct this light field out of harmonic oscillations, this is not possible. Actually we need an infinitive number of wave packages which overlap in a way to get the chaotic polarization axis.

This figure helps to get an intuition of how light is made up of photons

Left and right handed circular polarization, and their associate angular momenta.

The purple sticks in the middle are the photons which build up the macroscopic light. The individual photons, as elementary particles are point like, and are characterized by their spin , +/-1, and energy E=h*nu, nu the frequency of the wave that will be built up by a multitude of photons.

Where does the electric field enter in the photon? In its complex wavefunction, whose complex conjugate squared gives the probability of a photon existing at (x,y,z,t) and building up the maxwell potentials and fields.

To get interference in building up the classical E and B fields, the phases in the probability amplitudes between zillions of photons should be in step, as is achieved in polarized light above, or laser beams. In randomly produced light the phases that control the probabilities of the photons location are random and any cancelations will be random and improbable.

In this blog essay one can see how in QED the classical fields emerge from the underlying quantum level.