# Difference between revisions of "Photon"

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{\displaystyle p\equiv \left|{\boldsymbol {p}}\right|=\hbar \,k={\frac {\,h\,\nu \,}{c}}={\frac {\,h\,}{\lambda }}~.}{\displaystyle p\equiv \left|{\boldsymbol {p}}\right|=\hbar \,k={\frac {\,h\,\nu \,}{c}}={\frac {\,h\,}{\lambda }}~.} | {\displaystyle p\equiv \left|{\boldsymbol {p}}\right|=\hbar \,k={\frac {\,h\,\nu \,}{c}}={\frac {\,h\,}{\lambda }}~.}{\displaystyle p\equiv \left|{\boldsymbol {p}}\right|=\hbar \,k={\frac {\,h\,\nu \,}{c}}={\frac {\,h\,}{\lambda }}~.} | ||

+ | ===Photons Travel through Space=== | ||

+ | Once a photon is emitted, it can be absorbed once, at the end of its “journey.” No interaction “less than an absorption” is possible. It’s either absorbed, ending its existence, or its not. If it’s not absorbed it doesn’t interact with anything, and there’s no detecting it. | ||

+ | |||

+ | Now, if you release photons one at a time, and there’s no other source of photons, and you detect one at the detector, you can infer that it passed from the source to the detector, but you can’t be 100% sure of that - it could have hit an air molecule along the way, gotten absorbed, and another photon could have gone off in some odd direction, only to be absorbed and replaced by a third photon that happens to go the detector. This would usually be unlikely, but it is a possible sequence of events. So there’s always that slim chance that no photon traversed a particular volume, even if it’s in the “obvious path.” | ||

===Photon Smearing=== | ===Photon Smearing=== |

## Revision as of 15:19, 14 January 2022

## Contents

## Full Title

A **Photon** is a quantum of energy typically emmited by one Electron and absorbed by another.

## Context

In empty space, the photon moves at c (the speed of light) and its energy and momentum are related by *E = p c* , where *p* is the magnitude of the momentum vector **p**. This derives from the following relativistic relation, with *m = 0* :

E2 = p2c2 + m2c4

need to enable displaystyle for this to work

{\displaystyle E^{2}=p^{2}c^{2}+m^{2}c^{4}~.}{\displaystyle E^{2}=p^{2}c^{2}+m^{2}c^{4}~.} The energy and momentum of a photon depend only on its frequency ({\displaystyle \nu }\nu ) or inversely, its wavelength (λ):

{\displaystyle E=\hbar \,\omega =h\,\nu ={\frac {\,h\,c\,}{\lambda }}}{\displaystyle E=\hbar \,\omega =h\,\nu ={\frac {\,h\,c\,}{\lambda }}} {\displaystyle {\boldsymbol {p}}=\hbar \,{\boldsymbol {k}}~,}{\displaystyle {\boldsymbol {p}}=\hbar \,{\boldsymbol {k}}~,} where k is the wave vector, where

k ≡ |k| = 2π / λ

is the wave number, and

ω ≡ 2 π ν is the angular frequency, and ħ ≡ h / 2π

is the reduced Planck constant.

Since p points in the direction of the photon's propagation, the magnitude of its momentum is

{\displaystyle p\equiv \left|{\boldsymbol {p}}\right|=\hbar \,k={\frac {\,h\,\nu \,}{c}}={\frac {\,h\,}{\lambda }}~.}{\displaystyle p\equiv \left|{\boldsymbol {p}}\right|=\hbar \,k={\frac {\,h\,\nu \,}{c}}={\frac {\,h\,}{\lambda }}~.}

### Photons Travel through Space

Once a photon is emitted, it can be absorbed once, at the end of its “journey.” No interaction “less than an absorption” is possible. It’s either absorbed, ending its existence, or its not. If it’s not absorbed it doesn’t interact with anything, and there’s no detecting it.

Now, if you release photons one at a time, and there’s no other source of photons, and you detect one at the detector, you can infer that it passed from the source to the detector, but you can’t be 100% sure of that - it could have hit an air molecule along the way, gotten absorbed, and another photon could have gone off in some odd direction, only to be absorbed and replaced by a third photon that happens to go the detector. This would usually be unlikely, but it is a possible sequence of events. So there’s always that slim chance that no photon traversed a particular volume, even if it’s in the “obvious path.”

### Photon Smearing

- Exclusive 𝜋0-production Cross Section Extraction Po-Ju Lin 2020-02-01

### How does absorption of virtual photons "smear" a particle?

The electron loses energy/momentum when "emitting the virtual photon" and therefore must be in a "lower orbit" until it reabsorbs said photon. As it is doing so all the time, its average position is nearer to the nucleus. I must say that this is just a pretty picture for experimentalists/laymen who do not understand QFT. Virtual particles are a mathematical concept which describes the effect of what is happening on the quantum level (up to some order), but not the thing itself. In the end there is no easy way to imagine what happens b/c during an interaction the particle picture (asymptotic states) breaks down and the, for us, unintuitive wave nature of the "particles" takes over. The electron field kind of interacts with itself and in doing so produces higher order effects like the one you are thinking about.

The self-energy is valid for any electron (or any particle w/ charge generally). But if the electron isn't bound, we can't observe it as a "lower orbit". So, it happens everywhere. I think it should be observable theoretically in any bound electron, but it is a very small correction (hyperfine structure). I would imagine that for most systems (eg bigger atoms, molecules and solid-state systems) our theoretical predictions are not precise enough (many body problem) for this correction to be measurable. –

Self-energy is only appearing second order in alpha in perturbation theory. At all energies we can achieve, alpha is very small (1/137). At very high energies the effect of the self-energy would become more important.

our calculations/models from which the not yet corrected values come are approximating more things, s.t. the uncertainty is much bigger than this correction. The hydrogen atom is a two-body problem and analytically solvable. Helium already needs some approximations to be made/numerical calculations to be solved.