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So I was reading a document that stated that when traveling through a material (I'll use a liquid here, maybe water), a photon actually always traveled at 300,000 km/s, it was just that it occasionally 'hit', was absorbed and then re-emitted by water molecules along its path.

It was this absorption and re-emission that made the speed of a photon appear slower overall, but that "in between" hitting molecules it had the same 300,000 km/s speed.

My questions:

  • Is this a correct way of thinking about why a photon moves more slowly through water?

  • If it is, how would the photon "know" to exit the water molecule heading in the exact same direction that it was heading when absorbed?

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Yes, it's ok, but it's an explanation that has been stripped down to bare bones, and leaves out quite a bit. Here's a little more to help prop up the explanation.

First, it's important to realize that in a condensed phase like a solid or liquid the light is not interacting with molecules in isolation. Light is interacting with all of the molecules. This makes a big difference. Light absorbed and re-emitted by a single molecule can go off in (almost) any direction.

Next, remember that a photon is an excitation of a complete electromagnetic field. It is, unfortunately, often not helpful to think of a photon as a particle that exists at a particular place in space. Like all particles in quantum mechanics, there's a chance that it could exist anywhere. Individual interactions, on the other hand, can occur with a particular molecule at a particular location.

It's best to start thinking about your question in the realm of classical physics, and then modify it later to include quantum mechanics. Classically, when light interacts with a molecule, the electron is set into vibrational motion. The energy that the field gave to the molecule will stay with the molecule for a while while this oscillation occurs. The molecule, then, is a radiator and can generate its own light. That's the classical picture of the delay experienced by light during an interaction. After this re-emission, the light will travel at $c$.

How does it know to go straight? Here's where we need all the other molecules in the liquid. The incident field has a particular spatial pattern, say a sine pattern, and it excites in the molecules an oscillation pattern that exactly matches. The incident light was traveling in a particular direction with a well-defined phase front, and so too does the pattern of oscillation in the liquid. The light re-emitted from the molecules will also share that same pattern, although it will be delayed in time relative to the incident light. The re-emitted light adds to the portion of the incident light that passes unaffected. But the phase fronts, and hence the direction of travel, is parallel to that of the incident wave. It goes off in the same direction.

Photons: Almost the same story. Recall that a photon is an excitation of the light field. Instead of exciting the molecules into oscillation, the molecule temporarily absorbs a quantum of energy from the field (the "photon"), and the molecule is raised to an excited state. The molecule lives in this state for a short period of time, and then the energy is returned to the field ("photon emission"). But the business about phase fronts and directions still holds. The energy is added to a field propagating in the original direction, slightly delayed.

(Another detail we're leaving out. The absorption and re-emission I describe is a very fast process called "virtual transition". It is fast enough that the uncertainty principle $\Delta E \Delta t \leq \hbar/2$ allows energy conservation to be temporarily violated. So the frequency of the incident light does not have to match an absorption frequency of the molecule in this process, and the explanation works for transparent media.)

Note carefully that I refer to a photon as an excitation of the field, rather than as a particle. When the interaction between light and something else occurs at a particular place, as when it hits a particular pixel in a digital camera, it looks for all the world like a particle hit the pixel. But a more useful way of thinking about it is that the light field exists everywhere, but the interaction occurs at a particular place.

Hope that helps!

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  • $\begingroup$ from science.howstuffworks.com/question4041.htm "Glass, of course, falls into this last category. Photons pass through the material because they don't have sufficient energy to excite a glass electron to a higher energy level. Physicists sometimes talk about this in terms of band theory, which says energy levels exist together in regions known as energy bands. In between these bands are regions, known as band gaps, $\endgroup$
    – anna v
    Jun 25, 2014 at 18:52
  • $\begingroup$ where energy levels for electrons don't exist at all. Some materials have larger band gaps than others. Glass is one of those materials, which means its electrons require much more energy before they can skip from one energy band to another and back again. " So there is no individual molecules absorbing photons and re emitting them. It is the total latice. $\endgroup$
    – anna v
    Jun 25, 2014 at 18:53
  • $\begingroup$ One would have to think in terms of Feynman diagrams for individual photons and then build up the ensemble that makes the classical wave. see hyperphysics.phy-astr.gsu.edu/hbase/solids/band.html for glass, second paragraph $\endgroup$
    – anna v
    Jun 25, 2014 at 18:55
  • $\begingroup$ @annav The energy eigenstates in a solid are extended states, but they are constructed from linear combinations of localized molecular states. That's exactly what my description of the "oscillation pattern that exactly matches" is. $\endgroup$
    – garyp
    Jun 25, 2014 at 19:32
  • $\begingroup$ @annav Feynmann diagrams may help some people, but they are not necessary for the discussion. The books on QED on my shelf have exactly zero Feynmann diagrams in them. $\endgroup$
    – garyp
    Jun 25, 2014 at 19:35
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It was this absorption and re-emission that made the speed of a photon appear slower overall, but that "in between" hitting molecules it had the same 300,000 km/s speed.

It is correct to talk of light as the classical electromagnetic wave which refracts through a transparent medium ( and goes straight through only at normal incidence ) and the velocity of light enters the index of refraction as n=c/v , c the velocity of light and v the velocity in the medium.

Any light wave is built up by an ensemble of innumerable photons in a complex manner, and it is deceptive to think one can follow individual photons and their scattering. In transparent media the wavelengths are such that the photons cannot lose energy by transferring it to energy levels of the lattice atoms or molecules. At most they might interact with them with off mass shell photons, but two photon interaction is very small. It is the whole lattice that interacts with the ensemble of photons going through and the build up of the tiny interactions change the behavior of the wavefront to effectively give smaller velocities for the classical wave. It is not useful to be talking of individual photons which are quantum mechanical entities and can only be described by probability distributions.

My questions:

Is this a correct way of thinking about why a photon moves more slowly through water?

As I said, the individual photon is a member in a coherent ensemble of innumerable photons. One cannot talk of individual photon behavior per se, except probabilistically

If it is, how would the photon "know" to exit the water molecule heading in the exact same direction that it was heading when absorbed?

It is not absorbed, it might or might not interact with the whole lattice as a single photon. The coherent ensemble of photons interacts with the transparent solid so as to give the macroscopic behavior of refraction and thus the collective velocity. An analogy is a tiny bit of wood carried by a wave in water: how does it know to go straight? The behavior of light is collective behavior of an ensemble of photons.

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