Whenever I study the photoelectric effect and the Compton effect, I have always had a question about how a photon can possibly collide with an electron given their unmeasureably small size. Every textbook I've read says that the photo-electrons are emitted because the photons collided with them. But since the photons and electrons virtually have no size, how can they even collide? I have searched for the answer on the internet but I couldn't find any satisfying one.
Photoelectric Effect – How Can a Photon Collide with an Electron? Exploring Photons and Electrons
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Related Solutions
Here are real events relating to the last page of the pdf link you gave:
Fig.1 This bubble chamber picture shows some electromagnetic events such as pair creation or materialization of high energy photon into an electron-positron pair (green tracks), the Compton effect (red tracks), the emission of electromagnetic radiation by accelerating charges (violet tracks) (bremsstrahlung) and the knock-on electrons or delta ray (blue tracks)
Photons are invisible in bubble chambers as they interact only with direct collisions with electrons, called Compton scattering, or pair production in the field of a nucleus. Charge particles turn in the perpendicular to the plane magnetic field; this allow us to measure their momentum and charge and the ionisation of the tracks allows the identification of masses.
At the lower left of the picture, there is an electron( identified by its ionisation) which loses energy into a photon, and the photon pair produces some centimeters away, an electron positron pair.
in the middle right side, we see a positron that loses energy into a photon and the photon kicks an electron from the atoms of the chamber, this is a Compton scatter. This corresponds to the diagram in the last page of your link, except it has been reduced to one dimension. In reality there are two dimensions, because the photon gives part of its momentum/energy kicking at an angle . The following is the correct diagram kinematically:
It should not be surprising that classical scatters and particle scatters kinematically are the same, because momentum and energy conservation hold both classically and quantum mechanically. It is the probability of interaction that is different in the microcosm of elementary particles to the billiard ball particle scattering. In simple scattering experiments the kinematics are not different ( except that special relativity holds in the microcosm).
Edit after edits in question. I think the image , from a real experiment, answers whether a photon can hit an electron or not.
Now you ask:
What if I put the electrons behind a double slit apparatus, and treat individual photons as particles? Based on this "compton scattering," it's possible for the photon to be deflected any which way. I could claim that the diffraction pattern observed in the double-slit experiment is due to compton scattering, among other factors. Prove me wrong!
The difraction pattern of individual photons, even when sent one at a time, is a direct result of the quantum mechanical nature of the photon. The solution of the boundary conditions imposed by the two slits gives a probability distribution that displays an interference pattern. Even though electromagnetic interactions viewed as Feynman diagrams are similar, it is the boundary conditions that determine the probability of scatter, and two slits is different than two particle scatter, the fields are different and the solutions are different.
The original problem can be seen in terms of energy and momentum conservation.
Before scatter, there are two particles in the center of mass and the center of mass has an invariant mass larger than the mass of the electron. For total absorption of the photon there would be only the electron left. As the electron has a fixed mass and at the center of mass it should be at rest, the reaction cannot happen. It can only happen if a third particle is involved to conserve the overall energy and momentum , and this is what is happening with the photoelectric effect.
a)
The incoming photon interacts with an electron that is tied to the atom by a virtual photon . The whole system takes up the energy and momentum conservation.
The inverse problem happens with a gamma generating an e+e- pair. The gamma has zero invariant mass, the pair will have at least two electon masses at the center of mass, so a gamma cannot turn into an electron positron pair , a third particle has to be involved. The simplest is a virtual photon from some nucleus too. The Feynman diagram is the same as the one above with a different interpretation ( incoming e- is read as outgoing e+)
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Best Answer
This is an answer by a particle physicist that has been working with data for forty years:
Photons and electrons are quantum mechanical entities, and to really really understand their interactions, quantum mechanics has to be invoked.
When detected, the photon has a point particle footprint (as does the electron) consistent with the axiomatic particle table of the standard model.
The leftmost frame shows the collision of a countable single photons on a screen (in a double-slit experiment).
The accumulation of photons (light emerges in a calculable manner from many photons), shows the wave nature's interference effects. It is the probability of landing on the (x,y) of the screen that displays a wave behavior. Not the individual photons.
Here is another measurement of a photon
The original picture is here. That the single photon (gamma) electron interaction is at a point is evident.
Now let us see how what we call size for macroscopic particles in quantum mechanics appears. It is all dependent on probabilities of a particle being at an (x,y,z) to interact with another particle. Look what an electron around a hydrogen atom has as a probable location:
This is what defines the macroscopic charge distribution, and the probability of an incoming gamma ray to interact with the electron is a mathematical combination of this, and the coupling constants of the quantum mechanical interactions.
A free electron has a very small probability to be hit by a photon. That is why high density beams are used in high energy experiments. In general it will be the coupling constants which will give high probabilities the closer the two point particles are, and of course not to forget Heisenberg's uncertainty principle, which also will define a volume in space and momentum where interactions can happen.
The photoelectric effect involves electrons that are in orbitals and a large number of atoms and molecules, and the fact that it exists means that there is enough probability for an incoming photon to hit an electron in the orbitals distributions of the specific solid.