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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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  • Many thanks! But is the wave packet you mention a real electromagnetic wave or a quantum mechanical probability amplitude (the two must be related, but I don't know how...)? Second, I agree about the group velocity. But taking the refractive index of normal glass and reasonable assumptions for the photon... is the group velocity then closer to c or to the phase velocity, meaning does it take more like one day, or one and a half in that experiment? – Charles Tucker 3 Jul 30 '22 at 22:37
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    @CharlesTucker3 since it's a single photon, it's just the wave function. You may be interested in this question. As for group velocity, this is given by group index, which for silica is a bit higher than the refractive index. So it takes more time to propagate than it would take in a vacuum. – Ruslan Jul 30 '22 at 23:03
  • According to the cited webpage the group index for say a 500 nm photon is even larger than the phase index. So if doing my experiment, the outcome is that on average I detect the photon even a bit later than 1.5 days, correct? – Charles Tucker 3 Jul 31 '22 at 06:53
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    @CharlesTucker3 well, given that you've chosen phase index as 1.5, then yes, most likely so. – Ruslan Jul 31 '22 at 07:58
  • Many thanks, question is answered! – Charles Tucker 3 Jul 31 '22 at 08:55
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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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  • So your response to the question is "neither 1. nor 2."? – garyp Jul 29 '22 at 11:11
  • @garyp I edited my answer to give more details. – Miyase Jul 29 '22 at 11:16
  • @Miyase The experiment is designed to treat the photon by its particle nature and since it's a single photon there is no group velocity. If one would treat the photon as a wave, the answer is phase velocity, I agree. The question is if a single photon is enough to treat it as a wave slowed down as interfering with secondary waves induced in the glass? – Charles Tucker 3 Jul 29 '22 at 11:49
  • In an experiment like this one, I'm not sure having a single photon like that, inside a non-vacuum medium, is possible. Also, I don't see anything in this thought-experiment that can prevent the photon from developing wave-like behavior. It's as if you wanted to design an experiment involving a single particle while ignoring quantum laws? – Miyase Jul 29 '22 at 11:53
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    @CharlesTucker3: "since it's a single photon there is no group velocity" - no. The statement assumes that a single photon implies a single frequency, which is not in general valid. A single photon can (and in general would) represent a spectrum of frequencies. – flippiefanus Jul 29 '22 at 12:30
  • @flippiefanus Could you explain a bit more about and maybe write an answer/recommend further reading? Think you point in the right direction... – Charles Tucker 3 Jul 29 '22 at 15:57
  • @Miyase I don't want to ignore quantum laws, actually the opposite is true. The point is, our common explanation of group velocity, phase velocity are all very classical... I'm happy for every quantum mechanical explanation! – Charles Tucker 3 Jul 29 '22 at 16:04
  • two things: 1) the glass has phase velocity: $v_g=\partial \omega / \partial k$ and you assumed that $n$ is constant, which is never true. So there is a group velocity. Secondly, such an experiment guarantees that the emission will always have a bandwidth and this bandwidth will travel at group velocity. For the emission to not have a bandwidth, i.e, a well defined frequency, then the lifetime of the state would have to be infinite, which is somewhat paradoxical as the photon would also need to be infinite in time. – José Andrade Jul 29 '22 at 19:04
  • I see @flippiefanus had now addressed the same point as me. By infinite in time, this means that there is no time truncation of its EM oscillation in time. There is no start and end time for its "existence". So there is always a bandwidth associated with finite events. – José Andrade Jul 29 '22 at 19:07
  • Lastly, wave and particle are well married together and all laws apply the same for the photon as a particle and an EM wave. @CharlesTucker3 you see, a single event does not rule out wave behaviour. If you ran statistics while doing single photon measurements you would find out that indeed the answer converges to the wave description of the system. – José Andrade Jul 29 '22 at 19:13
  • @Jose Andrade I do not doubt in the wave particle dualism... But still, what is the outcome of the thought experiment described? If I repeat it 1 mio times, on average the traveling time will be 1.5 days? – Charles Tucker 3 Jul 29 '22 at 21:59
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    If I could just think of a defined emission time and then a detection time, my bet would be on group velocity and not phase velocity. The problem that I am having is on how to define all terms in a more realistic thought experiment, because of the problem of lifetimes, linewidth of emission, etc. Nonetheless I almost always arrive at the conclusion that you would find a distribution that is consistent with group velocity. This would in principle be just a small correction to the 1.5 days (for green, silica, the group velocity is only 1.8% slower). – José Andrade Jul 30 '22 at 12:17