The book I'm reading about optics says at some point that "each color (wavelength) contained in the white light interferes only with itself". But why is this so?
Edit: I moved the rest of the question elsewhere.
Optics – Why Don’t Waves with Different Wavelengths Interfere in White Light?
interferenceoptics
Related Solutions
The thickness of the AR coating is chosen such that the reflections from the two interfaces cancel out (at the wavelength for which the AR coating was designed):
See Anti-reflective coating in Wikipedia.
As endolith points out in the comments, to explain how the transmission is enhanced, you have to draw a few more rays in the diagram. Here's another illustration, from the Wikipedia article for Fabry–Pérot interferometer, which shows a few higher-order reflections:
For the anti-reflective coating, you choose the thickness such that R1 and R2 cancel while T1 and T2 constructively interfere. Note that this is dependent on the wavelength, the angle of incidence, and the index of refraction of whatever is being coated. With other thicknesses, you can make a high-reflectivity coating, or a coating of whatever reflectivity you want.
Your assertion that
Usually, "white light" is described or defined as an uniform mixture of waves
is pretty much completely incorrect: this is not how the term "white light" is treated in the literature. The meaning of the term is relatively well captured by this glossary at Plastic Optics:
light, white. Radiation having a spectral energy distribution that produces the same color sensation to the average human eye as average noon sunlight.
However, the term is not normally taken to have a strict technical meaning, a fact which is well reflected by the observation that in the first page of a search for "optics glossary" only a single resource has an entry for "white light".
The meaning of the term is even more complicated because it depends on who is using it:
If it is a spectroscopist that needs a white-light source to obtain a reflectivity or absorptivity spectrum, they will usually require the light to have a broad bandwidth, with support over the entire visible-light range, to be called "white".
However, if it's a manufacturer of light bulbs, they will only require that the light be perceived as white, even if it is produced e.g. by three-colour LEDs with narrow-band spectra like this one, and their use of the term will be completely justified.
In terms of its use within the physics literature, it is much more usual to require a broadband source, with a large continuum of wavelengths contributing significantly to the spectrum. However, there isn't a requirement that all the frequencies contribute equally (partly because, as you note, that doesn't even begin to make sense).
Thus, a flat wavelength spectrum (over a broad enough range) will normally be called "white", but so will a flat frequency spectrum over an equivalent range. Moreover, many of the standard models of white light do not have a flat spectrum in either representation, with the most famous model being, of course, blackbody radiation. This has frequency and wavelength spectral distributions of the form $$ P_\nu(\nu,T) = \frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/k_BT}-1} \quad\text{and}\quad P_\lambda(\lambda,T) = \frac{2hc^2}{\lambda^5} \frac{1}{e^{hc/\lambda k_B T}-1} \quad \text{resp.}, $$ and at high enough temperature (i.e. $T\approx 5500\:\mathrm K$) it models white sunlight. At lower temperatures, such as those in incandescent light bulbs, it produces a rather different spectrum, which is still called white light in the literature.
Best Answer
At any instant in time, light of different wavelengths can be said to interfere. However, because of the extremely high frequencies of visible light, any cross interference will get time-averaged away very quickly unless the two waves are very close in frequency.