Is The Wavelength The “Size” of The Photon Particle?

The wavelength of a photon particle is perfectly well-defined but only when it’s in a momentum eigenstate (i.e., when it has perfectly well-defined momentum–and energy). However, this never happens since a photon is always in a superposition of momentum eigenstates:

\lvert\alpha\rangle=\int{d^3p}\ f(\mathbf{p}) \lvert \mathbf{p}\rangle

Perhaps, the quantity that we might want to consider as the “size” of a photon is the width of the Fourier transform of the momentum-space wave function, \lvert, which is to say, the uncertainty in the photon’s position. This uncertainty could be anything that is between zero and infinity. Let’s ignore the effects of the Planck-scale for a moment since it can be arbitrarily close to zero it’ll make sense to call the photon a “point particle”.

Now, assuming that the uncertainty, in momentum, is proportional to the magnitude of the momentum–possibly the only thing we’re able to assume if we know not a damn thing about the state–the uncertainty in position would be proportional to Planck’s constant divided by {p} (which is the magnitude of the momentum). Since {p} is [inversely] proportional to the wavelength then the uncertainty in position is proportional to the wavelength.

In theory, it would make sense to consider the possibility of the wavelength to be the “size” of the photon–or at least proportional to the “size”. Though seemingly strange, it is somewhat consistent with the fact that microwaves (that have wavelengths of order of one centimeter) will not go through a metal net that has millimeter-sized holes such as the net that covers the window of your microwave oven. But they will go through a net with considerably larger holes.

So, what’s the effect of the Planck-scale? Well, a Planck-energy photon will have a wavelength [read: “size”] near that of the Planck length, meaning that it implies that the energy density of the region where the photon is located will be just large enough to form what most deem as a “black hole”.

Understand that the state of the photon is in general a superposition of many wavelengths and the same state can be described as a superposition of many different positions. See, if the position space wave function has its peak it would then make sense to say that the width of this peak is the “size” of the photon. Also, you could say that the location of the peak of the momentum space wave function is the “momentum” of the photon.

None of the concepts, “size”, “momentum” or “wavelength” are well-defined here. But with a little effort things can be found in the quantum mechanical formalism that corresponds to all of these classical concepts.

I do not find this at all to be strange for quantum mechanics, I just find it strange for someone wanting to find the size of something that’s based on a pre-defined understanding of what size is. Take the de Brogile wavelength as an example. It is a wave function that has absolutely nothing to do with size. The position operator is just that–a position. It is not a “width” of an object. If you look at the definition of what a photon is you’ll notice the lack of a “size” as being a part of that definition. However, there is a characteristic length that is associated with photons and that is the wavelength of the light that we measure, although, you shouldn’t misconstrue that as being the photon’s [physical] size. When you ask for a physical size of a photon [and a photon was never defined as having a physical size] the question begins to venture into undefined territories.

You’d be better off asking, “What does the color blue smell like?”

-Desmond (DTO™)

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Onanel

Desmond J. Watson is a Co-Founder and the Chief Executive Officer of Hexagon Lavish®, a scientific R&D start-up that specializes in informational interpretation software development and unique IR techniques. He also manages the blog, Arheliean. You can follow him on Twitter, @TheRealOnanel. If you're rich and want to invest in Hexagon Lavish®, then do so.

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  • OK, I have a follow-up question. A specific wavelength of red light comes about when an electron in a hydrogen atom moves from one state to another. I might be wrong but I imagine the distance between the two energy levels is equal to half of this red wavelength. Would it emit one photon because the electron’s moved with an electric field from a crest to a trough?

    Also if you consider a radio-transmitter aerial. Are there formulae to calculate how many radiowave photons are emitted for every cycle of its frequency, depending on the signal’s amplitude (Let’s say it’s tuned to exactly 100MHz, so the period for each cycle is one hundred-millionth of a second).

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    • sorry just to clarify, i meant “Would the electron emit one photon because it has an electric charge and moves from a crest to a trough defining a waveform?”

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    • Mike Corballis,

      My assumption of your observation would be that an unbound or “free” electron is one that has enough energy to get far away from the nucleus of a hydrogen atom. This is not the same thing as saying that an electron “moves” around freely in an atom, it doesn’t. Another thing to bear in mind is that, in quantum mechanics, there are no paths nor orbits, so, for stationary states per se, there is a time-independent Schrodinger equation that determines the wave function that can be interpreted as probability density amplitudes, which in turn, aid in the locality of an unbound electron in a [hydrogen] atom.

      However, in regards to an electron’s “position” in an atom, electrons do not “move” from one place to another. The very instant anything is done to identify the electron’s position, it appears in a position/displacement.

      Again, rolling on an assumption here, pertaining to your last inquiry regarding an electron emitting one photon….I believe so since it’s presumable that a static electric field would easily perform work on charges. So, what about the other [electromagnetic] fields generated by the electron and/or the photon? Keep in mind, that an atom is nearly impossible to isolate since there are always electromagnetic fields present. Also, you need to take into account the momentum of the photon

      \frac{h}{ \lambda}

      Conservation of momentum dictates that this momentum must be in the opposite direction of the photon. Perhaps, the answer to your inquiry lies within this particular observation. Also, when it comes to a hydrogen atom, the difference between an electron and a photon is that an electron can give to an atom part of its 12.5 eV energy; a photon has to give all of its 12.5 eV energy or nothing at all. And, for a photon, everything is dependent upon the state of the energy as well.

      Falling back on excitation in a hydrogen atom, think of heat in place of “red wavelength”. When a hydrogen atom [H2] or a hydrocarbon are broken by heat, oxygen initiates a decay of the excited electron state, releasing a photon. The [hydrogen] atoms emit their characteristic spectrum lines associated with their energy eigenstates and molecular/atomic orbitals. As an example, I can add NaCl salt to an alcohol flame and the action will generate a brilliant yellow light. It turns out to be monochromatic and can be used to do Newton rings and other wavelength/coherence sensitive optical experiments–and that’s the way it was done before the invention of the laser.

      On a spectrogram, light from an evenly heated solid object will have continuous spectra whereas atomic transitions from vaporized atoms or diatomic molecules tend to have distinct spectral lines. The sun is peculiar light source when you factor-in that intense pressure appears to favor blackbody emission characteristics of continuous spectra and yet, you’ll come across darker points in this spectrum that are seemingly discontinuous points such as hydrogen’s Ha frequency ~650nm–Balmer and other emission lines.These [emitting] points, more than likely, are where the energy being absorbed is enhanced. Since the thermal agitation from the sun is so high, the likelihood of an electron falling to the ground state of energy rather than a thermal disturbance which will cause random wavelength light to be emitted. Electrons, in an excited state, will fall a very short distance when they are “trapped” near an impurity. You can say that an electron remains “excited” for an extended period of time and the original waveform of when the [excited] photon was absorbed has long since diffused into obscurity.

      Summarily, it’s not surprising different colors will be emitted from an atom or molecule than the original wavelength and that is when the original wavelength has an energy output equivalent to a sum of more than one excited state.

      Desmond

      Like

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