The speed of light in any medium besides vacuum is smaller than $c$. In a classical way, I just look at that as a wave that propagates less fast, the change in EM-field is passed on slower. How should I imagine that when thinking of light as being photons? Are they slowed down and accelerated again? What is the consequence of accelerating a massless particle?

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    $\begingroup$ possibe duplicate with a nice answer: physics.stackexchange.com/q/466 $\endgroup$
    – user46925
    Jul 3, 2015 at 22:50
  • $\begingroup$ The trivializing answer for the classical case is deceptive: it's not enough to say that the wave slows down! It also changes the state of the matter that it goes trough and part of the incident wave's energy is now in the polarization of the medium, so in a way you start with a wrong mental model already, if you think that "the classical wave slows down and that's it". $\endgroup$
    – CuriousOne
    Jul 3, 2015 at 22:57
  • $\begingroup$ It is a very nice answer, but it end with '... stay with this kind of picture, rather than talking in terms of photons', while I want talking in terms of photons $\endgroup$
    – Dries
    Jul 3, 2015 at 22:59
  • $\begingroup$ @CuriousOne I think, reading the answer igael suggested, I understand (more or less) what you mean $\endgroup$
    – Dries
    Jul 3, 2015 at 23:03
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    $\begingroup$ Photons are not little localized balls of energy. You have to think of light (and all other particles, actually) as a wave. The thing which slows down in the medium is the combination of the usual light wave and the responding polarization in the medium. $\endgroup$
    – DanielSank
    Jul 4, 2015 at 1:49

1 Answer 1


It turns out that one photon states of the electromagnetic field can be written in a way such that the state "propagates" fulfilling Maxwell's equations. This is an exact model as I discuss this in more detail in my answer here.

So we begin with a one-photon Fock state of the quantized electromagnetic field. Let's keep our discussion to one mode, so one quantum harmonic oscillator, so our pure state is $|1\rangle$.

Now, suppose we have a medium. Let's model the active medium "atoms" (they could be excited states of molecules as well) them as two-dimensional, one particle quantum states. Let's think particle number 1: it could fleetingly absorb the photon, so if it does so, the state of the whole system is state $|0\rangle\otimes |1,\,0,\,0,\,\cdots\rangle$, where the $|0\rangle$ on the LHS tells us that the EM field has dropped to its ground state, and $|1,\,0,\,0,\,\cdots\rangle$ representes atom 1 is in its first raised state, hanging onto the excitation. In a dielectric material the interaction conserves excitation number, so our total system base states are:

$$|1\rangle\otimes |0,\,0,\,\cdots\rangle;\quad |0\rangle\otimes |1,\,0,\,0,\,\cdots\rangle,\,|0\rangle\otimes |0,\,1,\,0,\,\cdots\rangle,\,|0\rangle\otimes |0,\,0,\,1,\,\cdots\rangle,\cdots$$

where the leftmost state is the one where the excitation is with the EM field and the notation $|0,\,\cdots,\,1,\,\cdots\rangle$ with one "1" in the $j^{th}$ position means that atom $j$ is holding onto the excitation.

The total excitation is now a quantum superposition of all the above base states. So we have a one photon state in superposition with excited matter states. This is the reason for the speed difference: the photon still has phase speed $c$ but the full superposition now has different states with different dispersion relationships from that of the freespace photon. You can also come up with a modified Maxwell equation description of the above, with a refractive index to model the changed dispersion relationships $\omega(k)$ of the modal eigenfrequecy as a function of wavenumber.

You can also think roughly, but not altogether correctly, of the medium repeatedly absorbing the photon then re-emitting it, thus accounting for the slower speed, as I discuss in my answer here. The quantum superposition idea is our best current description of these notions.

  • $\begingroup$ This is very technical ans,I could not follow. $\endgroup$
    – Paul
    Jul 4, 2015 at 17:19
  • $\begingroup$ @Paul Sorry. Try the answer I linked in my last paragraph on for size, then. $\endgroup$ Jul 5, 2015 at 2:45

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