# Is the intrinsic parity necessarily $\pm 1$?

Intrinsic parities of various particles we know are $$\pm 1$$. My question is, can it be a more general phase?

It seems it's sometimes argued (like page 140 in "Introduction to elementary particles" by Griffiths) that because $$P^2=I$$, the eigenvalues has to be $$\pm 1$$. As far as I can go, by definition $$P^2$$ commutes with all Poincare transformations, so I can conclude for a given particle, i.e. in an irreducible rep, $$P^2$$ is proportional to the identity operator, and since there is always the freedom to choose the overall phase, we can set $$P^2=I$$ for that particle, but just one particle. I can't rule out the possibility that $$P^2$$ takes different phases for different particles (in different irreducible blocks of the representation).

In Griffiths, the argument is we will "of course" return to the same state by applying the parity operator twice, then you would have $$P^2=I$$ for all particles (at least in the same sector of super-selection rule). Is this an essential part of the definition of the parity? Can we have some operator that by conjugation it flips the spatial momentum while keeping the Hamiltonian and the angular momentum, it doesn't square to the identity and we still call it the parity operator?

It's also worth noting that in many QFT books (for example Weinberg's) the parity is given as a general phase. Are there any known models that intrinsic parities take value other than $$\pm 1$$? Or to be more precise, the phase difference of parities of two particles is not $$0$$ or $$\pi$$?

• from the definition here, en.wikipedia.org/wiki/Intrinsic_parity that uses the mainstream physics accepted model functions , it seems to me that the answer is "no" not in mainstream physics that is discussed on this site. It would need a new theory. Mar 5 at 5:33