# Photon interaction during refraction

1. Feb 21, 2012

### harrylin

This is a follow-up of the following thread on refraction:

In a nutshell, an optical photon that traverses a solid dielectric extends over many atoms. While the photon traverses the dielectric, it continuously transmits some of its energy into a forced oscillation of charges which return the energy by emitting secondary waves (perhaps better to say wavelets) with a slight, progressive phase delay. The result is a delayed photon.

However, in post #19 the following question came up:
Indeed, it is no conceptual problem for those of us who regard a photon as a kind of wave packet, and the Planck energy as the characteristic energy of emission and detection.
However, it may be a problem for those who regard a photon as an indivisible particle, and/or the Planck energy as a true atom of energy, independent of observation. With such an interpretation, how does one explain refraction?

Last edited: Feb 21, 2012
2. Feb 21, 2012

### Cthugha

At first it should be stated that bare photons are never eigenstates of the light field in matter.

Now, the question is how one treats the difference. In the weak coupling regime one can still use perturbation theory up to second order which gives a refractive index like
$$n\approx 1 +\left|\sum_z \frac{\langle i|H_D|z\rangle \langle z|H_D|i\rangle}{E_z-E_i-\hbar \omega}\right|^2$$.

It should be emphasized that the absorption and emission processes involved lead to virtual states, not real ones. That treatment of course must fail on resonance, that means when the denominator approaches zero.

In that case, the light field and the material excitation behave light coupled harmonical oscillators in the simplest case which can exchange energy effectively, so that spontaneous emission can become reversible and you now have real absorption and emission instead of virtual ones. In that case perturbation theory must be replaced by exact bookkeeping of these interactions. What one typically does is a diagonalization of the interaction Hamiltonian via a Bogoliubov transform which leads to new eigenstates. These are now partially photon-like and partially excitation-like and typically called polaritons.

The energy is in both cases taken from the photonic mode in the time-average only which is no problem for any model, I suppose.

3. Feb 22, 2012

### harrylin

Thanks, I guess that one could see the photon as a particle that exchanges energy with the medium. But I'm afraid that this doesn't really address the question of Antiphon, who holds that 'the photons are absorbed and re-emitted by the phonons'.

Also in your description, at any time only some of the photon's energy is exchanged, and with a great number of electrons - thus I think that also as your picture it, less than a quantum of energy is exchanged at any time, and the maximum additional kinetic energy of each oscillating electron is again much smaller than that. Is that correct?

4. Feb 22, 2012

### Cthugha

The point of view that photons are generally being absorbed and re-emitted by photons is plain wrong and leads to inconsistencies - unless of course absorption to virtual states is implied. The FAQ of these forums already has an entry on that: https://www.physicsforums.com/showpost.php?p=899393&postcount=1 where it is written that "So the lattice does not absorb this photon and it is re-emitted but with a very slight delay." Although I must admit that the term re-emitted is a bit off as it is explicitly stated that absorption does not happen.

Real absorption and reemission processes happen way too slow to explain refraction and should also randomize the direction in which light is traveling.

On very short timescales as implied by uncertainty the "total" photon energy can go to other modes. However, the energy of the photon is of course not too well defined on these timescales and as transitions to these modes are not allowed due to energy mismatch the energy will stay in the photon mode in the long run. Averaging over timescales longer than the typical timescales on which uncertainty is important then gives you an effective portion of energy that goes into these transitions on average, but that does not mean that at some instant the photon loses only some percentage of its energy.