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I am writing a note about the Poincare group and I am trying to explain that argument that one-particle state transforms under irreducible unitary representations of the Poincare group. However, there are some ambiguities which confuse me.

  1. Is a composite particle transformation under an irreducible unitary representation of the Poincare group? For example, a proton consists of some quarks. It seems they can't be separated by just rotating or translating or boosting, therefore it does form an irreducible unitary representation of the Poincare group, even though it's not a fundamental particle.
  2. The irreducible unitary representation of the Poincare group is characterised by mass and spin, which are given by two Casimir operators ($P^2$ and $W^2$). However, the electron and positron have the same mass and spin, but definitely different particles. So it seems the original argument has a loophole and we should somehow incorporate CPT transformation to modify the original argument. If I'm right, how to incorporate CPT in the original argument?

Edit: I just found that both questions are explained in Winberg Chapter Two, but I would still appreciate it if someone would like to explain it again.

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  1. You likely have already dealt with composite particles in non-relativistic quantum mechanics, under the guise of "addition of angular momentum." The upshot is that you can always choose to represent a composite state (a member of a tensor product of irreducible representations) as a member of a direct sum of irreducible representations.

  2. The statement is that a particle state is defined by the value of the Casimir operators, plus additional internal quantum numbers that are invariant under Poincaire transformations. In the case of the electron and positron, the internal quantum number is the electric charge.

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  • $\begingroup$ Thank you for your answer, I still have a question about the second question. As mass and spin can be obtained from Casimir operators, what corresponds to these internal quantum numbers from a Group-theoretical perspective? $\endgroup$
    – David Shaw
    Commented Nov 12, 2023 at 9:59
  • $\begingroup$ @DavidShaw Electric charge as an operator commutes with all the generators of the Poincaire group. $\endgroup$
    – Andrew
    Commented Nov 12, 2023 at 17:08
  • $\begingroup$ So you mean it is also a Casimir element in this case? $\endgroup$
    – David Shaw
    Commented Nov 13, 2023 at 3:07
  • $\begingroup$ @DavidShaw In a trivial sense I suppose it is. The electric charge would just be proportional to the identity as far as the Poincaire algebra is concerned. The actual group would be $ISO(3,1) \otimes U(1)$, where $ISO(3, 1)$ is the 3+1-dimensional Poincaire group and $U(1)$ is the group of electric charge. The tensor product means these two groups don't "interact", all their generators commute. You could also replace $U(1)$ with a more complicated non-Abelian gauge group, which happens in the standard model. But those symmetries are purely internal and do not mix with the spacetime symmetries. $\endgroup$
    – Andrew
    Commented Nov 13, 2023 at 3:15
  • $\begingroup$ Thank you very much! $\endgroup$
    – David Shaw
    Commented Nov 13, 2023 at 3:40

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