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What is the physical meaning of the wavelength of light? This question has been asked before but I cannot find a satisfactory answer. Some respondents have said that the question is vague, I don't think so, but let me clarify. Suppose one plucks a guitar string and takes a photo of the vibrating string at a very high shutter speed so the image of the string is "frozen" in time. Further suppose that the string is vibrating with a single frequency. One can then use the image to define and measure the wavelength of the vibrating string. Now let us suppose that an advanced alien lands on earth and shows us how to "fire" a single photon from a laser gun, and, with advanced technology which is totally beyond our comprehension, is able to "freeze" the path of the single photon over a distance of several meters and can then produce a 3D hologram of the wave (or waves - electric and magnetic). Given that this 3D hologram can be rotated and expanded or contracted to suit our purposes, how would one measure the wavelength of the photon? Is there actually a real physical instantiation of the wavelength of light or is it just a useful abstract concept (like imaginary numbers) to be used in De Broglie and other equations without any real physical meaning?

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    $\begingroup$ You said it yourself, light is an electromagnetic wave; and the wavelength is the one of this wave. The electromagnetic field is a very real physical quantity. What don't you like about that? $\endgroup$ – sebhofer Aug 17 '15 at 14:52
  • $\begingroup$ Photons are a very tricky thing to think about. While a single photon can have a well defined wavelength, it is not really equivalent to a classical wave with that wavelength. $\endgroup$ – Javier Aug 17 '15 at 15:21
  • $\begingroup$ Yes, there is a "real physical instantiation of the wavelength of light". But you need to look into four-potential. See the wiki electromagnetic radiation article where you can read that "the curl operator on one side of these equations results in first-order spatial derivatives of the wave solution, while the time-derivative on the other side of the equations, which gives the other field, is first order in time". One's the spatial derivative, the other's the time derivative. If it was a water wave and you were in a canoe, the tilt of your canoe is E and the rate of change of tilt is B. $\endgroup$ – John Duffield Aug 17 '15 at 19:26
  • $\begingroup$ The wavelength is a length of space after which an oscillating field is similar to itself (self-similar), whether it is its phase and amplitude (optionally) or its phase only $-$ strictly speaking, the wavelength is defined for a specific frequency only and may be easily recognized in simple propagating waveforms (plane, spherical...). It is mostly different for each frequency (from $\lambda = c_{material}(f) / f$ for EM), so the final field formed as superposition of these components of each frequency may have a complex final shape where a specific wavelength might not be easily noticed. $\endgroup$ – gox Aug 17 '15 at 19:53
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Well the imaginative experiment you have put forth is a fairly satisfactory method I would say. The wavelength of a guitar string is simply the distance between two consecutive peaks or troughs of the oscillating string (for simplicity let the frequency be constant). The wavelength of the electromagnetic field, or light, is just the corresponding value for an oscillating electric field. I can think of two possible objections to that definition -

Q. Guitar strings are real, are fields real objects or just mathematical constructs?

This would be a justifiable question, exacerbated by how fields are taught in school. But rest assured fields are about as much a mathematical construct as objects are. Quantum field theory, which is a very successful description of the world, treats objects as fields and so in a way every guitar strings are field excitations. So yes, fields are as real as you can get, not a convenient construct.

Q. The second objection would be, so can we take a snapshot of the photon as you mention?

Think about how you take the snapshot of the guitar string. Billions of photons emitted by some light source strike the string at every instant and with a camera you could capture those photons. If your camera is fast enough you could see the difference between the photons that hit the string at say every nanosecond of its oscillation and that would give you a series of snapshots locating the string at every nanosecond. So taking a snapshot of the electromagnetic waves boils down to 1. finding the analog of light used to see the guitar string. If you had some particle that was small enough that even though they would be deflected by electromagnetic field, they would not actually change the path of the field. Moreover they must be much smaller than the distances of electromagnetic wavelength (micrometer - nanometer) to resolve those distance in this futuristic camera. 2. A proper camera. It has to be sensitive enough to detect this minuscule particle whose energy/mass is so small that it does not even deflect light. Secondly it has to be fast enough to resolve the time period of light which is (10^-15 seconds).

Neither of those requirements inherently violates any laws of physics. Such aliens might well exist, or we might be them in the distant future.

So, the wavelength of light is just like the wavelength of any other field, including guitar strings. If EM waves are a convenient mathematical construct, then so is everything else.

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