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Background:

I originally asked this question Does a single photon propagate with phase velocity or front velocity through a dispersive material? about the speed of a single photon in a dispersive medium. I received no answer but some good comments which made me re-thinking and concentrate on the core question.)

Question:

Consider the following hypothetical experiment: We have a small volume, completely evacuated except for one single atom that we can excite. The volume is surrounded by a giant shell made of glass, refractive index $n=1.5$. It's expensive glass, so no absorption and imaginary part of the refractive index for our wavelength of interest. The glass shell is one light day thick and is surrounded by photodetectors.

We excite our atom in the middle at $t_0 = 0$. After some lifetime it will emit a photon in some direction, which needs to transmit the huge glass shell an is then detected. The question is, when.. The glass shell is so thick that we can neglect Heisenberg uncertainty for the lifetime (traveling time will be much larger). I think we can also neglect photon shot noise at the receiving part of the experiment - although I'm not sure at this point, maybe that is the missing link...

Do we detect the photon:

  1. after roughly one day because it's a photon travelling with $c$?

  2. after 1.5 days because the photon was traveling with the phase velocity $v=\frac{c}{1.5}$?

Update:

From comments and this related question Phase and group velocities in QFT / Quantum Optics I meanwhile learned that even a single photon is composed of a frequency distribution and therefore has a group velocity (although the cited post is about the quantum wave function and I don't know how this relates to the electromagnetic function). So phase velocity and group velocity (which I think is the true speed of propagation if asking about when the photon is expected to arrive in this thought experiment) can differ even for a single photon.

In addition, in an answer to this question What really causes light/photons to appear slower in media? (fig. 2) it is stated that photons inside glass still travel at $c$, but as the wavelength is shorter in glass, the phase velocity is smaller. This makes sense, even if the propagation speed is $c$, planes of same phase will propagate slower if the wavelength is shorter. However, this answer was downvoted, no idea if this means something...

So, for now I think the photon is expected after one day in my thought experiment above, although phase velocity in the glass was smaller than the vacuum speed of light $c$ (the group velocity was $c$). Could someone confirm or correct?

Charles Tucker 3
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2 Answers2

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Emission process is always finite in time. This means that the EM field associated with the photon will be something like a wave packet, with a reasonably* well defined beginning and end. The detection, since at the very least it must happen after the atom has been excited to subsequently emit the photon, will also happen after some time.

During the time between emission and detection the photon will propagate through the medium. How a wave packet travels through the medium is governed by medium's dispersion relations and wave packet's frequency spectrum (which is at the very least affected by natural broadening). The overall motion of the wave packet envelope is described by group velocity. By contrast, phase velocity will only make the ripples of the EM field move along the wave packet, but not affect the overall motion of the packet. In fact, it may even be backwards compared to wave packet motion.

Since when you detect a photon, detection probability is proportional to the square of the EM field, the peak of the wave packet arriving at the detector is the time when it's most probable for the detection event to happen. Thus, we can conclude that the time it takes between emission and detection is defined by group velocity.

*E.g. you could take the cut-off at the tail as 0.1% of peak amplitude, or whatever else that suits your needs.

Ruslan
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If you're talking about a real experiment, the answer is $c/v_g$, with $v_g$ the group velocity. In some cases $v_g$ may simply be equals to phase velocity, but it's still important to distinguish the two.

A single photon will always travel with velocity $c$, but inside matter light keeps being absorbed and reemitted, which creates an "effective" velocity inferior to $c$.

Edit: if the medium is a simple transparent medium, then you have the dispersion relation $k=n\frac{\omega}{c}=nk_0$ (with $k_0$ the wavenumber in vacuum and $n$ assumed constant), which implies that phase velocity and group velocity are equals.

Miyase
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