Feb 21, 2015 · In all electromagnetic interactions a photon is involved, either on shell ( mass zero) or virtual within a Feynman diagram. There exists a quantum mechanical equation which gives a wave function for the photon. It is a form of Maxwell's equation the differentials treated as operators for example here.

Dec 18, 2014 · The photon appears as a mediator between charged particles. At the same time while a moving charged particle has its self electric field, and magnetic field that depend on the particle velocity, the photon, the carrier of the electromagnetic energy is void of these self-fields because it has no charge and mass.

Oct 27, 2016 · A photon is typically described as a point particle, with no size whatsoever. This makes sense when considering that there is no limit to the number of photons that can exist in a fixed volume of space, the fact that individual photons can interact with very tiny objects, and the fact that photons are absorbed in their entirety and all at once.

Nov 19, 2020 · So this assumes photon-photon scattering process or a pair-production reaction : $$ \gamma_1 + \gamma_2 \to e^+ e^- $$ or as a Feynman diagram : Pitty that this process involves highly-energetic photons, with order of couple dozen $\text{GeV}$ energy. That much energy can be only produced in CERN or similar particle accelerator laboratories.

Sep 16, 2022 · That's wrong. A single photon can be linearly or elliptically polarized. I’ve also had someone tell me that a photon doesn’t have an electric field and magnetic field. Photons are the electromagnetic field. They don't have a field in the way that, say, electrons do. Field lines don't end on photons. The photons just are the field.

Dec 22, 2014 · This equation, if you can follow the link, is a quantized version of Maxwell's equations in their potential form, acting on the photon wave function. The state function of each photon is described by a complex number, there exists an amplitude whose square gives the probability of finding the photon at (x,y,z) at time t, and a given phase .

No, a photon does not need to have mass to be able to interact with matter. In fact it is its energy which is important in interactions. For the photoelectric effect the incoming photon must contain enough energy to displace the electrons on the metal of the photodiode.

Dec 25, 2017 · Having read that a photon has both wave and a light characteristics it is natural for a novice like myself to wonder "what is the amplitude? " Naturally I have looked on wiki and other sources and there are a number of answers some of which appear not so much to contradict one another of varying opinions of which is the most meaningful in physics.

The fact that the photon's rest mass is zero means that there isn't any "rest frame of a photon". Such a rest frame would have to move by the speed of light relatively to any proper inertial system and with such a fast motion, all the factors such as $\gamma$ would be singular and ill-defined.

Each quark couples to the photon through its (different) charge. So the photon couples to both isoscalars (I=0) and isovectors (I=1), and electromagnetism violates strong isospin. However, if you imagine this coupling term to preserve strong isospin, then you must represent the photon as a linear combination of (strong) isoscalar and isovector.