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How is the following classical optics phenomenon explained in quantum electrodynamics?

  • Color

According to Schroedinger's model of the atom, only particular colors are emitted depending on the type of atom and the state of its electrons. Then, how it is possible that the same type of matter changes color in relation to temperature? For example, water is transparent with a light blue tint, but snow is white (which means, pretty much, that photons have no particular color in the visible spectrum).

Split by request: See the other part of this question here.

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up vote 4 down vote accepted

It's important to distinguish different phases of the material. While it's true that water and snow consists of the same building blocks of $H_2 O$, those individual blocks are actually not as important as the resulting material. You can build a storage room and pyramids just out of bricks (at least in principle).


Let's start with the simplest case. Simplest in the sense that there is not much going on, just molecules flying around. Let us just assume ideal gas where molecules don't interact with each other. Then to explain everything it suffices to look at just one molecule.

The color of the molecule of course arises because of its absorption spectrum. This in turn depends on the molecule's energy levels. For atoms these are pretty nice discrete levels. For more complicated molecules you also have rotational and vibrational degrees of freedom to take into account and besides discrete levels you'll also see continuous strips that consist of very fine energy levels corresponding to that. For illustration, see the wikipedia article on $H_2$ hydrogen.

In any case, if you are somehow able to obtain that energy spectrum you can then investigate the macroscopic properties by the means of the usual Boltzmann statistics

$$\hat W = Z^{-1} \exp(-{\beta \hat H}) = Z^{-1} \sum_n\exp(- \beta E_n) {\hat P}_n $$

with $Z = {\rm Tr} \exp(-{\beta \hat H})$ being the partition function, $\beta$ the inverse temperature and $P_n$ projector on $n$-th energy level.

Using this you can see that as you increase temperature the energy level distribution will change to occupy higher levels and this in turn will change the probability of individual absorption processes between certain levels.

Now, the only place where you need to talk about QED is the connection between absorption spectrum and energy levels. Recall that in quantum mechanics energy levels are stable. They don't change, so there would be neither emission nor absorption. To resolve this apparent paradox we have to recall that we forgot to quantize the electromagnetic field. If you take this into account then atom's excited energy levels are no longer stable because of the fluctuations of electromagnetic field. Or in particle terms: because the number of particles is not conserved and it's easy to create photon out of nothing. Of course, energy is conserved so only photon's corresponding to the energy difference are possible. And the same can be said for absorption.

Liquids and solids

For now I won't talk about these phases because the answer is both becoming long and because optical properties of solids would constitute a book of its own. As for liquids, I have a feeling that the same analysis as for gases should hold except it's of course no longer sufficient to talk about individual molecules because of non-negligible interactions. But I guess starting from ideal gas's statistics and just switching temperature should give a reasonable first approximation.

Remark: Looking back on the answer, I didn't really explain your question completely. But I guess it's because it's quite hard to encompass everything that is going on. Maybe another split and specialization (e.g. to color of solids) of the question would be in order? :-)

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