Let's have a definition of massive particle as an irreucible representation of the Poincare group. Then, let's have a spinor field psi_(alpha alpha_1)...alpha_(n−1) beta brta_1...beta_(m-1)), which is equal to (m/2,n/2) representation of the Lorentz group.

Lokubovumn

Lokubovumn

Open question

2022-08-16

Let's have a definition of massive particle as an irreucible representation of the Poincare group. Then, let's have a spinor field ψ α α 1 . . . α n 1 β ˙ β ˙ 1 . . . β ˙ m 1 , which is equal to ( m 2 , n 2 ) representation of the Lorentz group. There is the hard provable theorem:
ψ α α 1 . . . α n 1 β ˙ β ˙ 1 . . . β ˙ m 1 realizes irreducible representation of the Poincare group, if
( 2 m 2 ) ψ α α 1 . . . α n 1 β ˙ β ˙ 1 . . . β ˙ m 1 = 0 ,
α β ˙ ψ α α 1 . . . α n 1 β ˙ β ˙ 1 . . . β ˙ m 1 = 0.
Can this theorem be interpreted as connection between fields and particles?

Answer & Explanation

Ashlynn Stephens

Ashlynn Stephens

Beginner2022-08-17Added 25 answers

The definition is that a particle in Minkowski space is a unitary irreducible representation of the Poincare group. So one needs to see how various P.D.E.s are related to the classification of unitary irreducible representations of i s o ( 3 , 1 ) or i s o ( d 1 , 1 ) in the case of d-dimensions instead of 4.
Note that these are all the Poincare-invariant constraints that can be imposed on the given field without trivializing the solution space (one could imposed ψ = 0 (gradient), which is Poincare-invariant but too strong as the field must be a constant).
The theorem is not hard to prove. One has to know how to construct irreducible representations of the Poincare group. Then one solves the equations by standard Fourier transform and shows that the solution space indeed equivalent to what is called a spin- m particle in Minkowski space.
There is nothing special about 4 d in defining spin- m field, so it is simpler to look at arbitrary dimension, where, say for bosons the above equations are equivalent to
( m 2 ) ϕ μ 1 . . . μ m = 0
ν ϕ ν μ 2 . . . μ m = 0
η ν ρ ϕ ν ρ μ 3 . . . μ m = 0
ϕ μ 1 . . . μ s is totally symmetric in all indices.
In 4 d one can use s o ( 3 , 1 ) s l ( 2 , C ) and the last algebraic constraint then trivializes - an irreducible spin-tensor is equivalent to an irreducible s o ( 3 , 1 )-tensor

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