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I apologise for the very non-technical nature of this question. I am new to QED and perhaps am interpreting things in the wrong way, but I'll ask anyway, and hopefully someone can provide a non-technical response.

There are lots of questions on here about virtual particles travelling faster than the standard speed of light such as this one. However, in Feynman's book QED, The Srange Theory of Light and Matter, it seems that Feynman is not saying that virtual photons can travel faster than light (which is what these questions are asking about), but that there is a probabilty that (real) photons will travel faster (or slower) than $c$ but that these probabilities cancel out over longer distances. (I have added quotes at the bottom to support this).

Is this, like the virtual photons, just a mathematical construction and not to be taken as reality? From reading the rest of the book I would guess not since Feynman uses words like appear frequently when describing what light appears to do.

As a secondary question, Feynman also seems to suggest that photons do not only travel in a straight line. Instead, they can take all paths, but the probabilities of these are very low and once again cancel out.

Is Feynman describing this in a different way from usual? Or am I misinterpreting what he is trying to say? Or is it really true that over short distances photons can travel faster than light (and seemingly violate relativity)?

Edit:

Here is a quote from Feynman's book (p89):

"...there is also an amplitude for light to go fsster (or slower) than the conventional speed of light. You found out in the last lecture that light doesn't only go in straight lines; now, you find out that it doesn't only go at the speed of light!"

Later he goes on to say:

"The amplitudes for these possibilities are very small compared to the contribution from speed c; in fact, they cancel out when light travels over long distances."

Community
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bnosnehpets
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  • Feynman uses the words "the arrows cancel out". The lenght of the arrows squares gives the probability, although I don't know their mathematical defintion. – bnosnehpets Apr 28 '15 at 22:44
  • @Chris I have added some quotes at the bottom. I hope that helps. – bnosnehpets Apr 28 '15 at 22:56
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    The arrows are representing the wave function, or rather infinitesimal pieces of the wave function of the photons. What Feynman is describing in layman's terms is the path integral in quantum mechanics. – Raskolnikov Apr 28 '15 at 22:56
  • @Raskolnikov So is not correct to say that the protons really can travel above the speed of light? – bnosnehpets Apr 28 '15 at 22:59
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    I'd say no. But someone else might disagree. What is certain is that it's the computational procedure that requires adding arrows for all processes. Even if they involve apriori unphysical things. – Raskolnikov Apr 28 '15 at 23:02
  • @Raskolnikov Ok, is this to do with virtual particles or is that something different? – bnosnehpets Apr 28 '15 at 23:05
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    Feynman also warned to take the path integral picture for a representation of actual physical reality. It's a method of calculating results in quantum field theory by "integrating" kernels over very hard to define infinite dimensional spaces. That doesn't mean that nature solves reality as a path integral by moving virtual and real photons around all the time. Indeed, nothing in nature moves around on all possible paths, in reality it's a field that permeates the vacuum which has quantized solutions. The path integral is just one way of calculating its dynamics. – CuriousOne Apr 28 '15 at 23:13
  • I find it strange that Feynman does not warn of this at any point in the lectures. But I guess he is describing such a complicated theory in layman's terms, so perhaps it is not surprising – bnosnehpets Apr 28 '15 at 23:17
  • @bnosnehpets: I agree, one has to consult multiple sources to get the right picture. Personally I am not a big fan of the Feynman lectures. I do like his papers and other writings of his, though, like QED. I think the concepts are clearer presented in there. Maybe it's just a matter of taste. – CuriousOne Apr 29 '15 at 00:34
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    I am old enough to have heard Feynman lecture on QCD when QCD was brand new and he had just accepted its existence. He had his own POV and method, very confusing to those struggling to understand the new concepts introduced. It confused me so that I do not remember his personal way of looking at it. He had a special way of seeing nature and the mathematics, out of the box, and that is how his contributions are brilliant, but could also be confusing to others. – anna v Apr 29 '15 at 04:27
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    Is he suggesting that there is a higher probability of photons getting faster than light over short distances? – lurscher Jan 17 '18 at 05:16
  • @Raskolnikov Your explanation can be confusing. Nobody knows and will never be able to directly measure how nature behaves between the initial state and the measurement. But in order to obtain correct answers related to measurement, the calculations must also take into account the contribution of the components corresponding to the velocity v> c and v <c. Therefore, it is wrong to say that photons never move at a different speed. – Arman Armenpress Feb 08 '21 at 07:59

4 Answers4

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Yes, parts of a wave function can travel faster than light, but from my understanding, much of it has to do with the uncertainty of the position of the particle the wavefunction represented in the first place.

For example, there is active research into how to interpret the results of quantum tunneling experiments that indicate "superluminal tunneling." This recent article from Quanta Magazine‘ explains that research area well. There are several competing definitions on the tunneling time, because time duration is not an quantum observable.

It is thought, but as far as I know not proven, that attempting to use this vanishingly small superluminal part of the wavefunction to send information will always be less efficient than sending the light directly, because for a large barrier, nearly all of the wavefunction is reflected.

(I don't know about how to reason about the case Feynman described, because not enough context about the quote is given.)

Jonathan Jeffrey
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I just saw a related question in another forum, and a commenter there noted that the non-classical paths "aren't actually taken." But Feynman addressed this too, in his QED lectures that are available on YouTube. In particular, he described an experiment with a mirror, and showed that the answer was dominated by the classically reflecting bit of the mirror, and the contributions from the rest of the mirror cancelled out. However, he then proceeded to turn the mirror into a diffraction grating by removing the pieces well away from the classical path that contributed "negative phase." And in that case the mirror does reflect at an odd angle.

However, if you actuall did this experiment with a diffraction grating designed to make the entire surface reflect (i.e., the grating gradually became a normal mirror in the classically reflecting region), and used a continuous light source, then I'd think the photons that travelled longer path would leave earlier. So your total response at the sensor would be composed of components that were from different cycles of the source.

Kip Ingram
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  • In fact, the answer is simpler. In the photon propagator, it is necessary to take into account the contribution from all admissible diagrams. That's all. – Arman Armenpress Feb 08 '21 at 07:29
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It is important to see a direct quote of where and how Feynman uses the word "light" and where and how "photon".

The quote you give speaks of light.

Light emerges in a complicated quantum mechanical superposition from zillions of photons. Photons are not light, though they are marked by the frequency that light built up by them will display, by E=h.nu. Photons are zero mass particles with spin + or - 1 to their direction of motion. Light in the superposition of the photon wave functions displays all the wave properties of classical Maxwell equations.

I have found that this plot gives an intuition of how this happens:

enter image description here

Even though the photons follow straight paths with just + or -spin, light displays polarization, a complicated function in space, here seen with the electric field vector of the classical electromagnetic wave. The connection to the quantum comes in the spin orientation of the photon.

How the quantum formalism of QFT handles this needs mathematics, and it is outlined here .

My guess, as I do not have the book, is that when Feynman is talking of cancellations he is talking of the classical light field functions built up by the photons.Collective wave behavior with group velocity and phase velocity create complications in light propagation not relevant for photon behavior, which , in my books, always travels at c.

For light:

In vacuum, the phase velocity is c = 299 792 458 m/s, independent of the optical frequency, and equals the group velocity. In a medium, the phase velocity is typically smaller by a factor n, called the refractive index, which is frequency-dependent (→ chromatic dispersion). In the visible spectral region, typical transparent crystals and glasses have refractive indices between 1.4 and 2.8. Semiconductors normally have higher values.

I would be interested in a direct Feynman quote where it says photons may travel faster than c.

anna v
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  • Feynman uses the word "light" as a photon, there is no need to add unnecessary entities here. In free theory, a photon is an elementary excitation of a quantum electromagnetic field. But if you include interaction, you will have to take into account the effects of higher orders, including "virtual" processes involving virtual massive pairs and virtual photons with infinite speed. This is the perturbation theory. – Arman Armenpress Feb 08 '21 at 04:24
  • Even though the photons follow straight paths with just + or -spin wrong. – Arman Armenpress Feb 08 '21 at 08:48
  • @ArmanArmenpress This is your personal theory. Photons have to have to have the spn aligned with their direction of motion because there is no 0,It has to be +pr - 1 – anna v Feb 08 '21 at 08:59
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Technically no, because if something went faster than c it would be classified as a Tachyon. If they did however, they would only be a Tachyon by definition, physically they would still be Photons.

DragonSlayer3
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