Undergrad Understanding the Spin-Statistics Theorem: A Modern Notation Approach

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The discussion centers on the equation u(a,b) = Cab/sqrt(2m), where Cab represents Clebsch-Gordan coefficients and m is the mass of a particle. The reasoning provided indicates that for massive particles, transformations on u(ab)(0,σ) do not alter momentum through boosts, only involving rotations. Participants express a desire for formulas in modern notation, as they find Weinberg's notation for the spin-statistics theorem challenging to understand. The conversation highlights the need for clarity in the application of these concepts to facilitate better comprehension. Overall, the focus remains on bridging the gap between classical and modern representations of these theoretical constructs.
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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 !
 

Thread 'Two equivalent statements of time reversal symmetric Hamiltonian'

Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?
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