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?
[Physics] Physical meaning of wavelength of an EM wave
electromagnetic-radiationphotonsquantum mechanicswavelengthwaves
Related Solutions
Fields
First you need to understand what a field is. There is a very good answer by dmckee on what a field really is which you can (and should read), but I'll try my own version. Mathematically, a field is something that has a value at every point of space and time. A typical example is temperature. The air in your room has a different temperature at every point and this temperature may change with time, so to each point in space and time we associate a number $T$. We might write $T(x, y, z, t)$, indicating that the temperature is a function of $x, y, z$ (space) and $t$ (time).
Temperature is a scalar field because at each point it is a scalar (i.e., a number). But we can have different kinds of fields. For example, the air in your room might be moving around, and so at each point it will have some velocity $\mathbf{v}(x,y,z,t)$. This velocity is a vector field, because at each point it has a magnitude and a direction (if you don't know what a vector is, picture it as a small arrow; the direction tells you which way the air is moving at that particular point, and the length of the arrow tells you how fast it is moving).
Waves
Air can carry waves, which we call sound. Sound is nothing more than a bunch of air molecules oscillating together in such a way that they carry energy from one place to another, in the same way that we see waves in water. With our fancy fields we can describe a wave by saying that at any given point the velocity is oscillating back and forth, and the phase of this oscillation changes as we move from place to place.
Temperature and velocity are fields that, in a sense, don't physically exist by themselves: they describe some property of a fluid, but it is the fluid that has physical reality, not its properties. But there are fields that are not a property of anything else, and the electromagnetic field is the most important among them.
Electromagnetic field
The electromagnetic field is described by two vector fields $\mathbf{E}$ and $\mathbf{B}$, called the electric and magnetic field respectively. For the purposes of light we can forget about $\mathbf{B}$ and just talk about the electric field. Just like the velocity of a fluid, this field can be represented by an arrow at every point in spacetime. Its physical intepretation is that if you place a charge somewhere, there is a force felt by the charge that points in the direction of $\mathbf{E}$ and is proportional to its magnitude. (Also there are magnetic effects but we're ignoring those). This is simply a more sophisticated view of the idea that like charges repel and opposite charges attract; instead of thinking of a force between the charges, we say that one charge creates an electric field near it, which is in turn felt by the other charge.
An electromagnetic wave is simply an oscillation of the electric and magnetic fields. At each point, the field's magnitude is increasing and decreasing with time. Wikpiedia has some nice gifs showing this process in time and space. The wavelength is a physical distance: it's the distance between two maxima or two minima of the field. The amplitude is not a distance, however: it measures how strong the field is, and so it is measured in units of field (Newton per Coulomb or Volt per meter for the electric field in SI units).
You can see in the usual pictures that an EM wave is a transverse wave; that is, the direction of the fields is perpendicular to the direction of propagation of the light. This is in contrast to a sound wave, which is longitudinal: that is, the molecules oscillate back and forth, and the move in the same line that the wave travels.
So, let's answer your questions:
a) The peaks and troughs are the points where the magnitude of the field is maximum in one direction or the other. As such, it doesn't make much sense to distinguish between peaks and troughs, because if you look from the other side they switch places.
b,c,d) A wave doesn't really take up space. There might be fields over a region of space, but the arrows you see in the animations don't have a physical length. They represent the magnitude of the fields, but they don't occupy physical space. Remember that there are two arrows (because of $\mathbf{E}$ and $\mathbf{B}$) at every point in space. As I've said before and has been said in the comments, wavelengths are lengths because they are the distance between two maxima, but amplitudes are not lengths.
The mental picture you describe in your question is, if you forgive me, a mess. You're mixing this description of EM waves with the quantum mechanical point of view, which is almost sure to lead to errors. QM usually deals in terms of particles, so the basic idea is that now light is thought of as a bunch of particles (photons), with a certain probability at each point in space to find a photon. The thing with quantum mechanics is that it's extremely weird and even the very best physicists have trouble forming an intuitive mental image of how it works. So please just forget about photons until you really understand the classical waves I've described in this post.
We could spend forever playing whac-a-mole with all of the confusing/confused statements that continue popping up on this subject, on PhysicsForums and elsewhere. Instead of doing that, I'll offer a general perspective that, for me at least, has been refreshingly clarifying.
I'll start by reviewing a general no-go result, which applies to all relativistic QFTs, not just to photons. Then I'll explain how the analogous question for electrons would be answered, and finally I'll extend the answer to photons. The reason for doing this in that order will probably be clear in hindsight.
A general no-go result
First, here's a review of the fundamental no-go result for relativistic QFT in flat spacetime:
In QFT, observables are associated with regions of spacetime (or just space, in the Schrödinger picture). This association is part of the definition of any given QFT.
In relativistic QFT, the Reeh-Schlieder theorem implies that an observable localized in a bounded region of spacetime cannot annihilate the vacuum state. Intuitively, this is because the vacuum state is entangled with respect to location.
Particles are defined relative to the vacuum state. By definition, the vacuum state has zero particles, so the Reeh-Schlieder theorem implies that an observable representing the number of particles in a given bounded region of spacetime cannot exist: if an observable is localized in a bounded region of spacetime, then it can't always register zero particles in the vacuum state.
That's the no-go result, and it's very general. It's not restricted to massless particles or to particles of helicity $\geq 1$. For example, it also applies to electrons. The no-go result says that we can't satisfy both requirements: in relativistic QFT, we can't have a detector that is both
perfectly reliable,
localized in a strictly bounded region.
But here's the important question: how close can we get to satisfying both of these requirements?
Warm-up: electrons
First consider the QFT of non-interacting electrons, with Lagrangian $L\sim \overline\psi(i\gamma\partial+m)\psi$. The question is about photons, and I'll get to that, but let's start with electrons because then we can use the electron mass $m$ to define a length scale $\hbar/mc$ to which other quantities can be compared.
To construct observables that count electrons, we can use the creation/annihilation operators. We know from QFT $101$ how to construct creation/annihilation operators from the Dirac field operators $\psi(x)$, and we know that this relationship is non-local (and non-localizable) because of the function $\omega(\vec p) = (\vec p^2+m^2)^{1/2}$ in the integrand, as promised by Reeh-Schlieder.
However, for electrons with sufficiently low momentum, this function might as well be $\omega\approx m$. If we replace $\omega\to m$ in the integrand, then the relationship between the creation/annihilation operators becomes local. Making this replacement changes the model from relativistic to non-relativistic, so the Reeh-Schlieder theorem no longer applies. That's why we can have electron-counting observables that satisfy both of the above requirements in the non-relativistic approximation.
Said another way: Observables associated with mutually spacelike regions are required to commute with each other (the microcausality requirement). The length scale $\hbar/mc$ is the scale over which commutators of our quasi-local detector-observables fall off with increasing spacelike separation. Since the non-zero tails of those commutators fall off exponentially with characteristic length $\hbar/mc$, we won't notice them in experiments that have low energy/low resolution compared to $\hbar/mc$.
Instead of compromising strict localization, we can compromise strict reliability instead: we can construct observables that are localized in a strictly bounded region and that almost annihilate the vacuum state. Such an observable represents a detector that is slightly noisy. The noise is again negligible for low-resolution detectors — that is, for detector-observables whose localization region is much larger than the scale $\hbar/mc$.
This is why non-relativistic few-particle quantum mechanics works — for electrons.
Photons
Now consider the QFT of the elelctromagnetic field by itself, which I'll call QEM. All of the observables in this model can be expressed in terms of the electric and magnetic field operators, and again we know from QFT $101$ how to construct creation/annihilation operators that define what "photon" means in this model: they are the positive/negative frequency parts of the field operators. This relationship is manifestly non-local. We can see this from the explicit expression, but we can also anticipate it more generally: the definition of positive/negative frequency involves the infinite past/future, and thanks to the time-slice principle, this implies access to arbitrarily large spacelike regions.
In QEM, there is no characteristic scale analogous to $\hbar/mc$, because $m=0$. The ideas used above for electrons still work, except that the deviations from localization and/or reliability don't fall off exponentially with any characteristic scale. They fall of like a power of the distance instead.
As far as this question is concerned, that's really the only difference between the electron case and the photon case. That's enough of a difference to prevent us from constructing a model for photons that is analogous to non-relativistic quantum mechanics for electrons, but it's not enough of a difference to prevent photon-detection observables from being both localized and reliable for most practical purposes. The larger we allow its localization region to be, the more reliable (less noisy) a photon detector can be. Our definition of how-good-is-good-enough needs to be based on something else besides QEM itself, because QEM doesn't have any characteristic length-scale of its own. That's not an obstacle to having relatively well-localized photon-observables in practice, because there's more to the real world than QEM.
Position operators
What is a position operator? Nothing that I said above refers to such a thing. Instead, everything I said above was expressed in terms of observables that represent particle detectors (or counters). I did that because the starting point was relativistic QFT, and QFT is expressed in terms of observables that are localized in bounded regions.
Actually, non-relativistic QM can also be expressed that way. Start with the traditional formulation in terms of the position operator $X$. (I'll consider only one dimension for simplicity.) This single operator $X$ is really just a convenient way of packaging-and-labeling a bunch of mutually-commuting projection operators, namely the operators $P(R)$ that project a wavefunction $\Psi(x)$ onto the part with $x\in R$, cutting off the parts with $x\notin R$. In fancy language, the commutative von Neumann algebra generated by $X$ is the same as the commutative von Neumann algebra generated by all of the $P(R)$s, so aside from how things are labeled with "eigenvalues," they both represent the same observable as far as Born's rule is concerned. If we look at how non-relativistic QM is derived from its relativistic roots, we see that the $P(R)$s are localized within the region $R$ by QFT's definition of "localized" — at least insofar as the non-relativistic approximation is valid. In this sense, non-relativistic single-particle QM is, like QFT, expressed in terms of observables associated with bounded regions of space. The traditional formulation of single-particle QM obscures this.
Here's the point: when we talk about a position operator for an electron in a non-relativistic model, we're implicitly talking about the projection operators $P(R)$, which are associated with bounded regions of space. The position operator $X$ is a neat way of packaging all of those projection operators and labeling them with a convenient spatial coordinate, so that we can use concise statistics like means and standard deviations, but you can't have $X$ without also having the projection operators $P(R)$, because the existence of the former implies the existence of the latter (through the spectral theorem or, through the von-Neumann-algebra fanciness that I mentioned above).
So... can a photon have a position operator? If by position operator we mean something like the projection operators $P(R)$, which are both (1) localized in a strictly bounded region and (2) strictly reliable as "detectors" of things in that region, then the answer is no. A photon can't have a position operator for the same reason that a photon can't have a non-relativistic approximation: for a photon, there is no characteristic length scale analogous to $\hbar/mc$ to which the size of a localization region can be compared, without referring to something other than the electromagnetic field itself. What we can do is use the usual photon creation/annihilation operators to construct photon-detecting/counting observables that are not strictly localized in any bounded region but whose "tails" are negligible compared to anything else that we care about (outside of QEM), if the quasi-localization region is large enough.
What is a physical consequence?
What is a physical consequence of the non-existence of a strict position operator? Real localized detectors are necessarily noisy. The more localized they are, the noisier they must be. Reeh-Schlieder guarantees this, both for electrons and for photons, the main difference being that for electrons, the effect decreases exponentially as the size of the localization region is increased. For photons, it decreases only like a power of the size.
Best Answer
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.