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Given the fact that the photons arrive at different places on the screen, does the speed of each photon is different ?

Abc2000ro
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  • Plane waves of light always travel at the speed of light, but with photons it's a bit more tricky. Most importantly, "a photon" is a one time measurement, after we find one, it's gone, so we can't measure it twice, to test its velocity. I could try to explain it, but Feynman has done it so much better in his book "QED: The strange Theory of Light and Matter", that I rather point to that than try my hand on a short explanation. – CuriousOne Jul 26 '16 at 07:31
  • Interesting question since physics stated that all paths are possible and a photon is resambled at the observation screen. So the path length is different but the time from the source to the screen is equal for all paths. Distance divided by time is velocity and that is not a constant number. – HolgerFiedler Jul 26 '16 at 08:21
  • @HolgerFiedler: that's wrong; the time from the source to the screen is different for the two paths, except in the center of the screen. The *difference* in length between the two paths is an integer multiple of the wavelength, and the difference in time between the paths is just the difference in length times the speed of light. – Peter Shor Jul 26 '16 at 11:51
  • @Peter Shor. Did anyone actually measured the speed of the photons for various points on the screen ? Or is just wishful thinking to say that the time from the source to the screen differs depending on the position on the screen where the photon hits ? – Abc2000ro Jul 27 '16 at 08:28
  • If you don't assume that the travel time differs, it becomes very hard to explain the interference patterns you see on the screen. But I don't think anybody has measured it. – Peter Shor Jul 27 '16 at 10:41

1 Answers1

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No, the light rays all travel at the speed of light $c$ but because they travel different distances they arrive at slightly different times. And indeed it is the difference in the light travel time that causes the interference pattern.

If we take the phase of the light at the slits as our reference, then the phase change once the light reaches the screen is given by:

$$ \phi = 2\pi\frac{\ell}{\lambda} $$

where $\lambda$ is the wavelength of the light and $\ell$ is the distance travelled by the light from the slits to the screen. The distance $\ell$ is related to the travel time $t$ by:

$$ \ell = ct $$

So we can rewrite our phase equation as:

$$\begin{align} \phi &= 2\pi\frac{ct}{\lambda} \\ &= 2\pi \nu t \end{align}$$

where $\nu$ is the frequency of the light and we're using the fact that $\nu=c/\lambda$.

So if you have two light rays (one from each slit) with travel times $t_1$ and $t_2$ then the difference between their phases will be:

$$ \Delta\phi_{12} = 2\pi \nu (t_1 - t_2) $$

When $\Delta\phi_{12}$ is a multiple of $2\pi$ we get constructive interference and a bright spot, and when $\Delta\phi_{12}$ is a multiple of $2\pi$ plus an extra $\pi$ we get a dark spot.

So the light travel times are different and it's precisely this difference that causes interference.

Response to comment:

Have a read through my answer to What is the relation between electromagnetic wave and photon?. It is rarely useful to describe interference by considering photons because interference is a wave phenomenon. It can be done but you would need to resort to a full quantum field theory calculation. If you really want to do this see for example Young's Double Slit Experiment in Quantum Field Theory by Masakatsu Kenmoku and Kenji Kume.

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John Rennie
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  • +1, because it answers the question in the headline, but that doesn't answer the question about photons in the body of the question, or shall we not try to answer that, since it's tricky and, honestly, could be confusing? Feynman ended up writing a small book about it, after all. – CuriousOne Jul 26 '16 at 07:29
  • What wavelength ? I'm talking about individual photons. The interference is caused by its quantum nature. – Abc2000ro Jul 26 '16 at 07:46
  • @Abc2000ro: I've added a footnote to my answer to respond to your comment. – John Rennie Jul 26 '16 at 08:03
  • @Abc2000ro: Interference is not caused by quantum physics. We can find perfectly classical interference phenomena on classical waves, e.g. surface waves on water or acoustic waves in air. Interference alone is not enough to define quantum behavior and one can produce systems without interference with quantum mechanics, as well. Permanent magnetism is a perfect example of a system that requires quantum mechanics but shows no interference, at all. – CuriousOne Jul 26 '16 at 08:10
  • "So if you have two light rays (one from each slit) with travel ... then the difference between their phases will be: ..." Once again, why it is supposed that all phases have the same amount (are max or are min) at the slit? Not supposing this, there could not be any intensity distribution on the screen. – HolgerFiedler Jul 26 '16 at 11:23
  • @HolgerFiedler: This answer is supposing that the distance from the source to the slit is the same for both slits, which means that if the phase for both light rays is the same at the source, it will be the same at the two slits. – Peter Shor Jul 26 '16 at 11:54
  • @Peter Shor The shortest and nice formulated argumentation I ever heard. Stays the question why the intensity is higher at the centre of the screen and falls to the right and to the left. Why at all the radiation is bend around a rigid body? Note that the argumentation with circular waves from every point of a wave could explain the decrease only if one suppose dissipation processes, but this could not be because it would be change the wavelength of the EM radiation. – HolgerFiedler Jul 26 '16 at 12:11
  • Dissipation exists, and it doesn't change the wavelength. It changes the probability of observing the photon. – Peter Shor Jul 26 '16 at 12:26