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?
 A: 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.  
A: 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.
A: 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.
A: 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.
A: 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.
