Understanding the Spin-Statistics Theorem: A Modern Notation Approach

Click For Summary

Discussion Overview

The discussion revolves around the Spin-Statistics Theorem and its representation within the context of the Lorentz Group, specifically focusing on the coefficient functions of fields as described in Weinberg's work. Participants are exploring the mathematical formulations and reasoning behind the Clebsch-Gordan coefficients and their application to massive particles.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant questions the expression for the coefficient function, u(a,b) = Cab/sqrt(2m), where Cab represents the Clebsch-Gordan coefficients and m is the mass of the particle.
  • Another participant suggests that the transformations on u_{ab}(0,\sigma) for a massive particle do not alter momentum through boosts, indicating that these transformations are limited to rotations, thus relating to Clebsch-Gordan coefficients.
  • There is a request for formulas, indicating a desire for mathematical clarity.
  • A participant notes that Weinberg's text contains the necessary formulas but expresses uncertainty about the notation used in the context of the Spin-Statistics Theorem.

Areas of Agreement / Disagreement

Participants appear to have differing levels of understanding regarding Weinberg's notation and the specific formulas related to the Spin-Statistics Theorem. There is no consensus on the clarity of Weinberg's text or the need for additional formulas.

Contextual Notes

Participants express uncertainty regarding the notation used by Weinberg and the specific mathematical formulations, indicating potential limitations in understanding the material presented in the text.

Physics news on Phys.org
filip97 said:
i read that question and Weinberg book (A,B)-Representation of Lorentz Group: Coefficient functions of fields

why u(a,b)=Cab/sqrt(2m) ?

where Cab is Clebsch-Gordan coefficients, and m is mass of particle
Basically because transformations on ##u_{ab}(0,\sigma)## for a massive particle are transformations that don't alter momentum by boosting to another frame, i.e. occur in one frame and don't involve translation. The only transformations then are rotations, thus they are Clebsch-Gordan coefficients.
 
DarMM said:
Basically because transformations on ##u_{ab}(0,\sigma)## for a massive particle are transformations that don't alter momentum by boosting to another frame, i.e. occur in one frame and don't involve translation. The only transformations then are rotations, thus they are Clebsch-Gordan coefficients.

Formulas please
 
filip97 said:
Formulas please
Weinberg has all the formulae, I'd just be repeating him. I thought you wanted the reasoning as such, what gap does Weinberg's text have for you?
 
DarMM said:
Weinberg has all the formulae, I'd just be repeating him. I thought you wanted the reasoning as such, what gap does Weinberg's text have for you?
I think formulas in modern notation, i don't understand weinberg notation for sst , thanks !
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 87 ·
3
Replies
87
Views
8K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 16 ·
Replies
16
Views
7K