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.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.
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?Formulas please
I think formulas in modern notation, i don't understand weinberg notation for sst , thanks !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?