thegirl
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I was just wondering how number density relates to probability density in general, within particle physics?
Thank you!
Thank you!
The discussion revolves around the relationship between number density and probability density in the context of particle physics. Participants explore theoretical aspects, definitions, and implications of these concepts, particularly in relation to Lorentz invariant phase space and wave functions.
Participants express differing views on the definitions and roles of probability density and distribution functions, indicating that there is no consensus on these concepts. The discussion remains unresolved with multiple competing interpretations.
Limitations include potential misunderstandings of terminology, the need for clearer definitions, and the complexity of relating different mathematical representations in particle physics.
I'm currently studying particle physics and when talking about the lorentz invariant phase space factor in the notes it starts off with the probability density of a free relativistic particle being p=2E|N|^2 and then goes on to say the lorentz invariant number density becomes dn=d^3r/(2pi)^3p where r is the momentum of the particle and p the probability density. I was just wondering why the probability density is included within the number density formula and how it related to the number density. The number density formula was derived from a single particle in a quantised box.Khashishi said:This question is way too vague.
p = <v|v> where v represents a wavefunction, can that also be a distribution function?ChrisVer said:are you sure p is probability density? I thought it was a distribution function ... giving the occupation within a volume in the 6D phase space (3 momenta and 3 positions). Such that the integral you have written N = \int n \diff^3x = \int \frac{\diff^3 p}{(2 \pi)^3} f(p)
1. you already wrote how they are related...