How does number density relate to probabilty density

[thegirl]

I was just wondering how number density relates to probability density in general, within particle physics?

Thank you!

 

[Khashishi]

This question is way too vague.

 

[thegirl]

This question is way too vague.

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.

 

[ChrisVer]

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...

 

[thegirl]

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...

p = <v|v> where v represents a wavefunction, can that also be a distribution function?

 

[mfb]

Distribution of what?

 

[vanhees71]

A wave function (applicable in non-relativistic quantum theory or at very low-energy relativistic cases, where the non-relativistic approximation is not too bad) is always of the structure $\langle a|\psi \rangle$, where $|a \rangle$ denotes a orthonormal system of (generalized) common eigenvectors of a complete set of compatible observables and $|\psi \rangle$ is a representing normalized vector of a pure state. An expression like $\langle v|v \rangle$ is just a scalar product of a vector with itself, i.e., its norm squared and cannot be a wave function, because it's just a single number.