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I How does number density relate to probabilty density

  1. Mar 18, 2016 #1
    I was just wondering how number density relates to probability density in general, within particle physics?

    Thank you!!!
  2. jcsd
  3. Mar 18, 2016 #2


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    This question is way too vague.
  4. Mar 18, 2016 #3
    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.
  5. Mar 18, 2016 #4


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    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 [itex]N = \int n \diff^3x = \int \frac{\diff^3 p}{(2 \pi)^3} f(p) [/itex]
    1. you already wrote how they are related....
  6. Mar 24, 2016 #5
    p = <v|v> where v represents a wavefunction, can that also be a distribution function?
  7. Mar 24, 2016 #6


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    Distribution of what?
  8. Mar 26, 2016 #7


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    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.
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