# What is the cross-section size of a photon?

How "wide" is a photon, if any, of its electromagnetic fields? Is there any physical length measurement of these two orthogonal fields, $$E$$ and $$M$$, from the axis of travel? When a photon hits a surface, and is absorbed by an electron orbital, this width comes into play, as there could have been more than one electron that could have absorbed the photon?

This is not my personal query, I found it while I was surfing the web, and found it interesting, so I posted it here.

• Travel axis is not something that you can always define. What is the travel axis of a spherical wave? Same story with photons and their size. What is the physical situation you are interested in? Plane wave hits surface? Gaussian beam? Spherical wave? Bessel beam? etc etc
– Cryo
Commented Jan 6, 2020 at 9:54
• The atom has a cross section, depending on the energy of the photon. For example graphene, where a single layer of carbon atoms absorbs a bit less than 1 % of visible light and IR (theory 1/137, the fine structure constant).
– user137289
Commented Jan 6, 2020 at 14:59

How "wide" is a photon,

The photon is a point particle in the standard model of particle physics. It has no extent to be described with "wide". Interactions of photons with charges or magnets can have a "width" in the sense of "measurable range"

if any, of its electromagnetic fields? Is there any physical length measurement of these two orthogonal fields, E and M, from the axis of travel ?

The photon has no electromagnetic fields . It has energy equal to $$h\nu$$ where $$\nu$$ is the frequency of the classical electromagnetic radiation, and $$h$$ Plancks constant. Its relations with classical electromagnetism come through its wavefunction, as it is a quantum mechanical entity. The complex conjugate squared of the photon's wavefunction gives the probability of finding the photon at $$(x,y,z,t)$$.

The classical wave can be shown mathematically to emerge from the superposition of very many photons of the given frequency $$\nu$$.

When a photon hits a surface, and is absorbed by an electron orbital,

It is absorbed by an atom (or molecule or lattice) by giving the energy (within the specific line width) to change the orbital of the electron to a higher energy level

this width comes into play,

No, it has nothing to do with the case as there is no such width identified with the photon.

as there could have been more than one electron that could have absorbed the photon?

No, the energy levels are what exist in the atom, and they are uniquely identified with quantum numbers also. The only width would be the size of the atom (molecule, lattice) and the quantum mechanical probability calculated with the given boundary conditions.

• But it would be somewhat peculiar to find a radio frequency photon whose wave function was localized in a space smaller than a cubic millimeter.
– TLDR
Commented Aug 12, 2023 at 13:53

Depending on how you define size, a photon can be considered either pointlike or an extended object. It's pointlike in the sense that, unlike a proton, for example, it has no internal structure (as far as we know) that would give it a basic size.

On the other hand, a photon can exist in the form of a wave-train that's arbitrarily large. (There are some niceties involved in worrying about how to say this in a technically correct way, since QFT says that there is no position-space wavefunction for the photon, but I don't think that's particularly relevant here.) For example, the light from a typical pen-pointer laser has coherence lengths on the order of millimeters or centimeters. This is, in some sense, the "size" of the photon. There is no upper limit on the size of the photon. One way to see this is that the only scale in quantum mechanics is provided by Planck's constant, which can't be converted to a length scale because it doesn't have units of length.

When a photon hits a surface, and is absorbed by an electron orbital, this width comes into play, as there could have been more than one electron that could have absorbed the photon?

In the Copenhagen interpretation, we describe this by saying that the photon's position was measured, and therefore the wavefunction collapsed. (Again, this is modulo the facts referred to above about QFT.)

For a back of the envelope calculation, it is often useful to associate with a particle it's Compton Wavelength. This is generally the most accurately you can know an object's position due to the uncertainty principle.

Other related length scales are usually more useful, factoring in features of the particular interaction in question. This can be expressed as a Mean Free Path, basically the flight time between interactions or a Scattering Cross-Section, measure of the rate of interactions.

• The Compton wavelength is $h/mc$, which is infinite for a photon. The mean free path is not a measure of the size of a particle. A cross-section such as the Compton scattering cross-section is not a measure of the size of the photon. Sometimes it happens that a cross-section is basically the same as the geometrical cross-section (e.g., for neutron absorption by a nucleus), but this is not normally the case, and certainly not the case here.
– user4552
Commented Jan 6, 2020 at 17:31
• I'd thought the Compton wave length of a photon is $hc/E$. Does that at least establish the smallest sensible region to which to isolate it? Aren't there conditions in which the scattering cross sections do correspond to photon wave lengths? For example, some wavelengths of photons will give you Compton Scattering from an electron, other wavelengths will give you Thomson Scattering. Commented Jan 6, 2020 at 17:46
• I'd thought the Compton wave length of a photon is hc/E. That's not the Compton wavelength, that's the de Broglie wavelength, normally just known as the wavelength.
– user4552
Commented Jan 6, 2020 at 19:19

This is a puzzle. I hear that you can slowly build up an interference pattern using an ultra low intensity beam. This suggests that the EM wave associated with each photon can interfere with other parts of itself to determine the probability that the photon will be absorbed at a particular point. When I was at MIT they had a grating spectrometer that could spread the spectrum over a width of 10 feet or more. That would require a good sized photon. On the other hand, the photons in a laser beam can only interact with things that fall within the width of the beam which can be focused down to the width of the track on a DVD (and the temporal length of the photon must be shorter than that of a bit on the track).

• see this sps.ch/en/articles/progresses/… . The footprint of each photon is a dot, consistent with the point definition of the standard model en.wikipedia.org/wiki/Standard_Model . Wave functions are not measurable quantities, and it is the wavefunctions that are waving, which , by the postulates of quantum mechanics, give interference patterns in probailities of interaction, not on the mass/energy spread as with classical waves Commented Jan 13, 2020 at 5:16

Photons are pointlike elementary particles, as defined in the Standard Model. They do not have any substructure, or spatial extent.

Now there is something called the cross section.

In physics, the cross section is a measure of probability that a specific process will take place in a collision of two particles. Cross section is typically denoted σ (sigma) and is expressed in terms of the transverse area that the incident particle must hit in order for the given process to occur. There is no simple relationship between the scattering cross section and the physical size of the particles, as the scattering cross section depends on the wavelength of radiation used. This can be seen when driving in foggy weather: the droplets of water (which form the fog) scatter red light less than they scatter the shorter wavelengths present in white light, and the red rear fog light can be distinguished more clearly than the white headlights of an approaching vehicle. That is to say that the scattering cross section of the water droplets is smaller for red light than for light of shorter wavelengths, even though the physical size of the particles is the same.

https://en.wikipedia.org/wiki/Cross_section_(physics)

So basically photons are pointlike particles, and even if you try to use the cross section to define any kind of size for the photon, you will end up running into problems:

1. the photon cross section varies depending upon what it is interacting with

2. it varies on things like photon energy and polarization

3. varies upon the type of interaction, absorption has different cross sections then scattering

If you want to go very basic, and want to know why microwave photons can't pass through the mesh of the microwave oven, then you can use wavelength as a basic indication. Contrary to popular belief, the wavelength of a microwave photon is larger then a baseball.

It’s a bit odd to think of a microwave photon as being larger than a baseball, but it is a simple way to explain how mesh reflectors work. The “size” of microwave and radio wave photons is simply too large to fit through the mesh, and so are reflected. Visible light is much smaller, so it easily passes through the mesh. It’s important not to take this model too literally, but it’s good enough for rough estimates.