# 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?”