How does a photon of light reflect?

What happens when a photon hits a mirror?

Visited again after some negative votes.

The photon is an elementary particle in the Standard Model of particle physics. This means that it is a quantum mechanical "particle" that is described by wave functions that indicate the probability of the interaction of a particular photon for each interaction. In the case of a mirror, ray optics describe the most likely path of a photon before and after an interaction.

If a particle meets matter in a solid, it can elastic scatter with the collective electric field of the medium it encounters. In order to have a mirror, all photons must scatter elastically from the solid-state lattice that is the mirror.

Elastic means that the photon that leaves an interaction only changes the direction in the center of mass. The center of gravity of a photon and a mirror is effectively the laboratory framework as the mirror has a mass of ~ 10 ^ 23 molecules. Thus, the elastically scattered photon does not lose energy and the colors of the images it builds up do not change. How classical states emerge from the underlying quantum field theoretical state is described here.

A photon is absorbed when its energy is given by E. = hν matches a certain energy level of the atoms (molecules, system) it hits, and then a re-emitted photon can refer to both the direction and the energy change the original photon, that is, when the reflected photon changes frequency due to re-emission, and loses the phase which it cannot contribute to a faithful image. The photon goes naturally with the velocity c (like all photons) regardless of their direction (elastic scattering only means change of direction and no Energy).

The diagrams describing the photon scattering are in the first order similar to the following.

Where the electrons are virtual, the interactions with the mirror grating and the outgoing photons have the same frequency / energy.

In elementary particles, "equal" can only have the meaning of certain variables in certain interactions. In elastic scattering, the photon entering the interaction and the exiting photon have the same frequency (energy) and each photon has a probability of being scattered at an angle. The classical wave, which is built up by the millions of photons in the superposition of their wave functions, has to maintain the phases so that the macroscopic images can keep their color and dimension, ie are "reflected".


Your first paragraph seems to affirm that the speed of the photon can assume all values ​​on [-c, c], since it scatters elastically. This is misleading at best. A photon is not a classic object with a "primitive dies", it is an oscillation in a field. There is no point in talking about a photon like it is slowing down and changing direction. The group speed changes.

anna v

@ user1247 !!! Elastic scattering only means change of direction, not momentum value, also in classical physics. Elastic scattering cross-sections exist for all scattering of elementary particles including photons. If there is a slowdown, of course in the case of particles with mass, one speaks of inelasticity.


You gave the billiard ball as an example. Let's think of it as a one-dimensional spread. Its speed changes continuously because the acceleration is not infinite. Infinite acceleration is not only not physical, but also wrong and, in the case of a photon, misleading. Group speed can do things like that, but a single photon description breaks down and doesn't correspond to any physical reality.

anna v

@ user1247 what do you mean by "dispersion in one dimension". The photon is not one-dimensional, it is four-dimensional.


You gave the billiard ball as an example. When a billiard ball collides with another billiard ball and scatters elastically, its speed changes during the collision. It is a simple fact. The number of dimensions doesn't matter, but of course it's easiest to consider a 1d collision.