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.

## Best Answer

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"

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 $\nu$.

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

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