As in this question, let ψp,σ be a single-particle 4-momentum eigenstate, with σ being a discrete label of other degrees of freedom.
Weinberg discusses the effect of a homogenous Lorentz transformation U(Λ,0) or U(Λ) on these states, and concludes that U(Λ)ψp,σ is a linear combination of ψΛp,σ′.
U(Λ)ψp,σ=∑σ′Cσ′σ(Λ,p)ψΛp,σ′
Again, is there any physical information that we can extract from this? (I realize that ψΛp,σ′ represent physical states after Lorentz transformation).
Answer
Yes, indeed, first of all it tells you that a particle of momentum p in one frame looks like a particle of momentum Λp in another frame which is related to the first by Lorentz transformation Λ. That means p is a Lorentz vector and hence gives you more certainty that it indeed captures momentum of the particle.
If you read a few more pages in the book, you will see that σ would correspond to spin for a massive particle and helicity for a massless particle. So, the above equation is telling us that not only do you find a different momentum for the particle but also that you might find the particle carrying a different spin (or more concretely the probabilities to find the particle in different spin states would be changed) when you do a Lorentz transformation.
Compare this with a usual case one encounters in QM. We have a massive motionless particle as our physical system. And we now want to study how the physical state transforms under a rotation U(R)ψp=0,σ=Cσ′,σ(R,0)ψp=0,σ′ Here σ labels spin of the particle and U(R) forms a representation of rotation group which is generated by the angular momenta J1,J2,J3.
So, this equation is telling you that transformation of single particle states in relativistic quantum mechanics is a straightforward union of the transformations that we expect from our earlier study of relativity and quantum mechanics.
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