How is light slowing down in a medium thought of in the photon picture? [duplicate]

Question

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?

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.